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Transcript of Signals and Systems Lecture Notes
ECE 370 SIGNALS AND SYSTEMS - FALL 2012
Text: These lecture notes and board notes.
Optional Texts: H.P. Hsu, Signals & Systems. Shaum's Outline: McGraw{Hill, 1995.
This is essentially a good source of worked examples, all in the standard EE notation that
we use in ECE 370. Reasonably priced!
C.L. Phillips, J.M. Parr, and E.A. Riskin, Signals, Systems, & Transforms. 4th Edition,
Prentice-Hall, 2008. Only get this if you are inclined to read additional books!
Prerequisites: ECE 225 Electric Circuits, MA 238 Di®erential Equations.
Course Goals: To provide electrical engineers with physical understanding and compe-
tence in mathematical signal analysis and systems theory.
Instructor: Robert W. Scharstein, Houser 209
phone 205-348-1761, e-mail [email protected]
O±ce hours: to be announced and by appointment.
Lecture: 12:00 noon { 12:50 pm, Mon, Wed, Fri; SERC 1013.
Problem Session: Wednesday 7:00-8:00 pm, Houser 301, optional. Bring your questions!
Daily Homework: Homework problems and textbook reading will be assigned for each
class.
FFF !!!THESE ARE YOUR PRIMARY RESPONSIBILITY!!! FFFTests will consist of homework problems, modi¯ed homework problems, new problems,
and concepts and derivations from the text and lecture. Details and discussion of the
MATLAB homework problems will appear on tests.
Tentative Grades: All quizzes and tests are closed-book.
Eleven Friday quizzes : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 1/3
Test 1, Friday September 21 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :1/6
Test 2, Friday November 2 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 1/6
Final Exam, Monday December 10 11:30 am { 2:00 pm : : : : : : : : : : : : : : : : : : : : : : : : : : 1/3
Instructor may determine ¯nal course grades using any combination of the above. However,
the ¯nal exam will not be weighted less than 1/3.
Friday quiz dates: 1: Aug 24, 2: Aug 31, 3: Sept 7, 4: Sept 14, 5: Sept 28, 6: Oct 12
7: Oct 19, 8: Oct 26, 9: Nov 9, 10: Nov 16 11: Nov 30
\Our goal is not to develop all the applications, but to prepare for them
- and that preparation can only come by understanding the theory."
Gilbert Strang Linear Algebra and Its Applications, Academic Press, 1976 page x
Typeset by AMS-TEX
0
ECE 370 SIGNALS AND SYSTEMS - ROUGH LECTURE NOTES
Robert W. Scharstein
23 July 2012
CONTENTS
BACKGROUND AND TIME-DOMAIN SYSTEMS
Complex Numbers : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 1.
A Couple of Singularity Functions : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 7.
Symmetry : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :10.
A Note on Mathematical Functions and Units/Physical Dimensions : : : : : : : : : : : : : : : : : 13.
Linear System : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 15.
Time-Invariant System : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 16.
Input/Output Relationship for a LTI System : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :17.
Convolution : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 18.
Another Convolution Example: The Convolution of Two Gaussian Pulses : : : : : : : : : : : 21.
Yet Another Convolution Example : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :22.
Causal System : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 28.
Bounded Input/Bounded Output Stability : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 28.
Singular Functions and Integrals : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 29.
Homework Problems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :35.
Relationship Between Impulse Response and Step Response : : : : : : : : : : : : : : : : : : : : : : : : : 37.
Time-Domain RC Circuit Analysis : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :38.
Step Response of an RC Circuit : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 41.
Step Response of Another RC Circuit : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :43.
Homework Problems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :45.
Step and Impulse Response of Series RLC Circuit : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 46.
Homework : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :53.
FOURIER INTEGRAL
Evaluation of an Important Integral via the Laplace Transform : : : : : : : : : : : : : : : : : : : : : :54.
Spectral Form of the Dirac-Delta Function : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 57.
Fourier Transform : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 58.
Fourier Transform Exercises : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 59.
Tables of Integral Transforms : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 60.
Fourier Transform of the Gaussian : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 63.
Fourier Transform of the Heaviside Unit Step Function : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :65.
Duality Property of the Fourier Transform : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 68.
Half-Power Bandwidth or Pulse width : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 70.
Homework : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :73.
Uncertainty Principle for the Fourier Integral : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 74.
i
*Riemann-Lebesgue Lemma (two serious treatments) : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 76.
Riemann-Lebesgue Lemma (one reasonable treatment) : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :78.
Convergence of the Fourier Integral Representation : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :79.
System Transfer Function : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :81.
Superposition Interpretation of the System Transfer Function : : : : : : : : : : : : : : : : : : : : : : : 83.
Transfer Function for Linear Electric Circuits : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 84.
Parseval's Theorem for the Fourier Integral : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 87.
Other Conventions for the Fourier Transform Pair : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :88.
Cascade of Two LTI Systems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 90.
Linear Dispersionless Filter : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 90.
Rectangular Pulse Response of Ideal Low Pass Filter : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :91.
Ideal Band-Pass Filter : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :95.
Complex Phasors for Time-Harmonic Circuit Analysis : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :96.
Homework Problem on Uncertainty Principle : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 98.
*Approximate Analysis of Dispersion : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 99.
Fourier Integral Solution of Potential Problem : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 103.
FOURIER SERIES
Fourier Transform of the comb : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :106.
Fourier Transform of a Periodic Signal : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 107.
Periodic Signals and Fourier Series : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :110.
Orthogonality : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 110.
Kronecker delta : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 111.
Orthogonality of the Fourier basis : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 111.
Derivation of Fourier Coe±cients by Minimizing the Mean Square Error : : : : : : : : : : : :112.
Fourier Coe±cients by a Direct Inner Product : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 114.
Trigonometric Form of the Fourier Series : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 114.
Two Standard Notations for the Trig Fourier Series : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 116.
Riemann-Lebesgue Lemma: Fourier Coe±cients : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :117.
Gibbs' Phenomenon: Example Square Wave : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 118.
Pointwise Convergence of the Fourier Series : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :119.
On the Convergence and Summation of Some Series : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :126.
Kummer Transform : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 129.
Homework Problem: Convergence of Fourier Series : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :131.
Fourier Series of Full-Wave Recti¯ed Sine Wave : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :132.
Parseval's Theorem for the Fourier Series : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 134.
Power Supply Performance : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 135.
Periodic Excitation of a Simple Parallel GLC Circuit : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :139.
Homework Problems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 140.
Simple Low-Pass and Hi-Pass Filter Response to a Periodic Input : : : : : : : : : : : : : : : : : : 141.
The Vibrating String : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 147.
Homework Problems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 150.
Poisson Sum Formula : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 151.
ii
TRANSITION TO DISCRETE-TIME SYSTEMS
Sampling and Reconstruction : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :153.
Discrete-Time Signals, Systems, and Transforms : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 157.
Exercises : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 159.
Discrete-Time Systems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 159.
Linear Time-Invariant Discrete-Time Systems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :160.
Input/Output Relationship : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 160.
Example of a LTI Discrete-Time System : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 161.
The Z-Transform : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 163.
Fibonacci Sequence - Di®erence Equation and Z-Transform : : : : : : : : : : : : : : : : : : : : : : : : 165.
Convolution Theorem of the Z-Transform : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 166.
Z-Transform of a Delayed Sequence : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 166.
System Transfer Function : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 166.
Discrete-Time Fourier Transform : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 167.
Discrete Fourier Transform : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 167.
Exercises : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 169.
DFT Approximation of the Continuous-Time Fourier Transform : : : : : : : : : : : : : : : : : : : :170.
Numerical Example of Using the FFT to Evaluate the Fourier Integral : : : : : : : : : : : : : 171.
Example: DFT Approximation of a Fourier Sine Integral : : : : : : : : : : : : : : : : : : : : : : : : : : :174.
Discrete-Time Approximation for the Step Response of Series RLC Circuit : : : : : : : : : 177.
* Sections marked by the asterisk are A+ level, so put these last in your prioritized list of
study topics.
\... mathematics is learned by doing it, not by watching other people do it ... "
M. Reed and B. Simon, Functional Analysis, Academic Press, 1980, page ix.
iii
Complex Numbers
j = +p¡1
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x
y
²
.....................................................................................................................................................................................
Á
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..............................r
z
complex z = x+ jy plane
If z 2 C and x; y; r; Á 2 R, then we write the complex number z in rectangular and polarform as
z = x+ jy = rejÁ:
Refzg = x and Imfzg = yThe modulus or magnitude of z is
jzj = rand the argument or angle of z is
argfzg = Á:The complex conjugate of z = x+ jy = rejÁ is
z¤ = x¡ jy = re¡jÁ:
z + z¤ = 2x and z ¡ z¤ = 2jyzz¤ = jzj2 = r2
Euler's identity:
ejÁ = cosÁ+ j sinÁ
Taylor series for the exponential function, written as a function of the complex variable z
is
ez =
1Xn=0
zn
n!:
1
Let z = jy and separate the real and imaginary parts via the even and odd powers in
ejy =
1Xn=0
(jy)n
n!=
1Xm=0
(jy)2m
(2m)!+
1Xm=0
(jy)2m+1
(2m+ 1)!
Note that j2m = (¡1)m and j2m+1 = j(¡1)m so that
ejy =
1Xm=0
(¡1)my2m(2m)!
+ j
1Xm=0
(¡1)my2m+1(2m+ 1)!
= cos y + j sin y
There was nothing special about the real number y, and we also write
ejÁ = cosÁ+ j sinÁ:
The sum of ejÁ and its conjugate gives
ejÁ + e¡jÁ = 2cosÁ
and the di®erence gives
ejÁ ¡ e¡jÁ = 2j sinÁwhich are the important representations
cosÁ =ejÁ + e¡jÁ
2
sinÁ =ejÁ ¡ e¡jÁ
2j
Many (all?) of the trig identities are easily seen in complex form: Consider
1 = jejÁj2 =³ejÁ´³e¡jÁ
´=³cosÁ+ j sinÁ
´³cosÁ¡ j sinÁ
´= cos2 Á+ sin2 Á:
Now observe
z = x+ jy = rejÁ = r£cosÁ+ j sinÁ
¤Equating real and imaginary parts separately gives
x = r cosÁ and y = r sinÁ:
jzj2 = zz¤ = r2 = (x+ jy)(x¡ jy) = x2 + y2r = +
px2 + y2
Á = tan¡1³yx
´Note that the two-argument arctangent function (that maintains full information about
quadrant) is required. For example, observe that
z1 = ¡1 + j =p2ej3¼=4 where tan¡1
μ1
¡1¶=3¼
4
is certainly di®erent from
z2 = 1¡ j =p2e¡j¼=4 where tan¡1
μ¡11
¶= ¡¼
4
Your calculator has the two-argument arctangent function as part of its rectangular-to-
polar conversion. In MATLAB, it's called atan2(y,x) just like Fortran and probably
C§.
2
Consider z1 = x1 + jy1 and z2 = x2 + jy2. Then our four basic operations are:
addition z1 + z2 = x1 + jy1 + x2 + jy2 = (x1 + x2) + j(y1 + y2)
multiplication by a real constant c: cz1 = c(x1 + jy1) = cx1 + jcy1
multiplication z1z2 = (x1 + jy1)(x2 + jy2) = (x1x2 ¡ y1y2) + j(x1y2 + x2y1)and z1z2 = (r1e
jÁ1)(r2ejÁ2) = r1r2e
j(Á1+Á2)
divisionz1
z2=x1 + jy1
x2 + jy2=(x1 + jy1)(x2 ¡ jy2)(x2 + jy2)(x2 ¡ jy2) =
(x1x2 + y1y2) + j(x2y1 ¡ x1y2)x22 + y
22
andz1
z2=r1e
jÁ1
r2ejÁ2=r1
r2ej(Á1¡Á2)
Generally, since the real number line is a subset of the complex plane, all of our ordinary
functions f(x) of a real variable x have extensions (or continuations) to a function f(z) of
a complex variable z. One notable di®erence between the real number line and the more
general complex plane is the lack of ordering on z: That is, there is no inequality on the
complex plane. Inequality is only de¯ned on the reals, i.e.
x1 > x2; y1 · y2; 7 < 13; etc
but something like z1 < z2 is meaningless.
Triangle inequalities.
jz1 + z2j · jz1j+ jz2jjz1 ¡ z2j ¸
¯̄jz1j ¡ jz2j¯̄
3
Proof of the ¯rst triangle inequality:
jzj2 = x2 + y2
jzj =px2 + y2 ¸ jxj
jzj ¸ jRe(z)jjzj ¸ Re(z)
jz1z¤2 j ¸ Re(z1z¤2) (1)
jz1 + z2j2 = (z1 + z2)(z¤1 + z¤2) = z1z¤1 + z2z¤2 + z1z¤2 + z¤1z2= jz1j2 + jz2j2 + 2Re(z1z¤2) (2)
(jz1j+ jz2j)2 = jz1j2 + jz2j2 + 2jz1jjz2j = jz1j2 + jz2j2 + 2jz1z¤2 j (3)
compare (2) and (3) in view of (1)
jz1 + z2j2 · (jz1j+ jz2j)2jz1 + z2j · jz1j+ jz2j
..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
z1
z2
z1 + z2
Extensions ¯̄̄̄¯NXn=0
fn(z)
¯̄̄̄¯ ·
NXn=0
jfn(z)j
¯̄̄̄¯̄bZa
f(t) dt
¯̄̄̄¯̄ ·
bZa
jf(t)j dt
(Sometimes called the monotonicity property of the integral.)
F Prove the second triangle inequality.
Hint: let z1 = ³1 ¡ ³2 and z2 = ³2 in the ¯rst triangle inequality.
4
Multivalued functions.
The nth roots of unity
1 = ej2k¼ (k = 0;§1;§2; : : : )11=n = ej2k¼=n (k = 0; 1; 2; : : : ; n¡ 1) distinct roots
Examples.
11=2 =
½ej0 = 1 (k = 0)
ej¼ = ¡1 (k = 1)...................................................................................
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²²
11=3 =
8>>>><>>>>:ej0 = 1 (k = 0)
ej2¼=3 =¡1 + jp3
2(k = 1)
ej4¼=3 =¡1¡ jp3
2(k = 2)
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²
²
²
logarithm
ln(z) = ln£rejÁ
¤= ln
£rej(Á+2k¼)
¤= ln(r) + j(Á+ 2k¼) (k = 0;§1;§2; : : : )
The value for k = 0 is called the principal branch of the logarithm.
5
powers: Let z = x + jy and w = u + jv be two complex numbers, that is z;w 2 C andx; y; u; v 2 R.
zw = exp£ln(zw)
¤= exp
£w ln z
¤and we know how to interpret/evaluate
ez = ex+jy = ex£cos y + j sin y
¤
Example.
ln(j) = j(¼=2 + 2k¼)
jj = ej¢j(¼=2+2k¼) = e¡(¼=2+2k¼)
principal value is jj = exp(¡¼=2) = 0:2079 ¢ ¢ ¢
Problem: Show that
μHint: use the ¯nite geometric series
n¡1Pk=0
ak on page 164
¶n¡1Xk=0
ej2¼k=n = 0
Fundamental theorem of algebra. The polynomial
Pn(z) = anzn + an¡1zn¡1 + ¢ ¢ ¢+ a2z2 + a1z + a0
with an 6= 0 is said to be of degree n. It has exactly n roots (zeros) counting multiplicity.
Proposition: If the coe±cients ak (k = 0; 1; 2; : : : ; n) of the polynomial Pn(z) are all real,
then any complex roots must occur in complex conjugate pairs. That is, if Pn(z) = 0 then
Pn(z¤) = 0.
proof:
anzn + an¡1zn¡1 + ¢ ¢ ¢+ a2z2 + a1z + a0 = 0
take the complex conjugate
an(z¤)n + an¡1(z¤)n¡1 + ¢ ¢ ¢+ a2(z¤)2 + a1z¤ + a0 = 0
6
A Couple of Singularity Functions
(or \Distributions" or \Generalized Functions")
..................................................................................................................................................................................................................................................................................................................................................................
0 t
u(t) 1
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....N
0 t
±(t) (1)
Heaviside unit-step function
u(t) =
½0; t < 0
1; t > 0=) u(t¡ T ) =
½0; t < T
1; t > T:
Dirac delta-function (de¯nition)
±(t) = 0 if t6= 0 AND
1Z¡1
±(t) dt = 1
From this strange de¯nition, the sampling or sifting property follows:
bZa
f(t)±(t¡ t0) dt =½f(t0); a < t0 < b
0; o.w.
if f(t) is continuous at t = t0. This is e®ectively an operational de¯nition of the Dirac-delta.
It is under the operation of integration that the Dirac delta-function enjoys meaning, and
thus it is sometimes said that \every Dirac-delta function is begging to be integrated."
Relationship:tZ
¡1±(¿) d¿ = u(t) () ±(t) =
du(t)
dt
Dirac-Delta Sequences: There are lots (1!) of ordinary functions that satisfy, in somelimit, the two-statement de¯nition of the Dirac-delta function. Here are several:
lim²!0
1
²¦
μt
²
¶= ±(t) where ¦
μt
T
¶, u(t+ T=2)¡ u(t¡ T=2) =
½1 jtj < T=20 jtj > T=2
lim²!0
²=¼
t2 + ²2= ±(t) Cauchy or Lorenz distribution
lim®!1
r®
¼e¡®t
2
= ±(t) Gaussian distribution
lim−!1
sin−t
¼t= lim−!1
1
2¼
−Z¡−
e§j!td! = ±(t) Dirichlet kernel
7
Study the time behavior of the above \Dirac-delta sequences" to see how they approach
the ideal delta function. Also examine the Heaviside unit-step sequence
lim®!1
12 [1 + tanh®t] = u(t)
and its time-derivative.
Note: we don't ever try to evaluate ±(0). You might want to, and some books might say
that \±(0) =1 ", but I would prefer we avoid this point. We don't ever have to attach a
value to the Dirac delta function when its argument is zero. But, in the words of Lennon
& McCartney, it's \nothin' to get hung about" from Strawberry Fields Forever.
Another note: (along the same line) If a Dirac-delta function appears in the integrand
of some integral, the integral is evaluated from the sampling property.
Warning: the integration limits cannot \split" a Dirac-delta function \down the middle,"
for example1Z0
±(t) dt
should be either 1Z0¡
±(t) dt = 1 or
1Z0+
±(t) dt = 0:
You either capture all of the delta-function or you don't get any of it. It cannot be divided.
Or you will have to agree on some scheme to cut it up while minimizing bloodshed.
Proof of the SAMPLING PROPERTY: No harm is done in taking t0 = 0, so that
we want to showbZa
f(t)±(t) dt =
½f(0); a < 0 < b
0; o.w.
for f(t) continuous in a neighborhood of t = 0. For de¯niteness, use the limit of the
rectangular pulse to represent
±(t) = lim²!0
1
²¦
μt
²
¶and represent the continuous f(t) by its Taylor series about t = 0:
f(t) = f(0) + tf 0(0) +t2
2!f 00(0) +
t3
3!f 000(0) +O(t4):
Then the integral of interest is, for a < 0 < b,
bZa
f(t)±(t) dt = lim²!0
1
²
²=2Z¡²=2
·f(0) + tf 0(0) +
t2
2!f 00(0) + : : :
¸dt:
8
Note²=2Z
¡²=2
dt = ²;
²=2Z¡²=2
t dt = 0;
²=2Z¡²=2
t2 dt =²3
12
and sobZa
f(t)±(t) dt = lim²!0
1
²
·²f(0) +
²3
24f 00(0) +O(²5)
¸= f(0):
If, for a < b, either 0 < a or b < 0, then the integral is zero since ±(t) = 0 for t6= 0. ¥
F Provide an alternate proof of the sampling property by appealing to the mean value
theorem of the integral.
\Heavy formalism is not required to get across fundamental ideas. If it can't be said simply,
it's not worth being said at all."
Bernard H. Lavenda, Statistical Physics, 1991, p. viii.
\The pursuit of excellence is gratifying and healthy. The pursuit of perfection is frustrating,
neurotic, and a terrible waste of time."
Edwin Bliss
9
Symmetry1
symmetric or even fe(¡x) = fe(x)
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x
cos(x)
...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
....................................................................................................................................................................................................................................................................................................................................................................................................................................................................
......................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...........
asymmetric or odd fo(¡x) = ¡fo(x)
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x
sin(x)
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x
x2 ¡ 25.............................................................................................................................................................................................................................................................................................................................................................................................
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x
x
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x
18(35x
4 ¡ 30x2 + 3)....................................................................................................................................................................................................................
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x
12 (5x
3 ¡ 3x)
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............................................................................................................................................................................................
1\Symmetry is immunity to a possible change." J. Rosen, Symmetry Rules, Springer, 2008, p. 4
10
If a function f(x) is continuous to all orders (the function f(x) and all of its nth derivatives
f (1)(x); f (2)(x); : : : are continuous) then f(x) can be represented by a Taylor series about
the origin
f(x) =
1Xn=0
f (n)(0)
n!xn:
The Taylor series of an even function will have only even powers of x
fe(x) =
1Xn=0
f (2n)(0)
(2n)!x2n
and therefore all of its odd derivatives (at the origin) vanish
f (2n+1)e (0) =d2n+1f(x)
dx2n+1
¯̄̄̄x=0
= 0:
In particular, note that f 0e(0) = 0 (if fe(x) is continuous and di®erentiable at the origin).
Similarly, the Taylor series of an odd function will have only odd powers of x
fo(x) =
1Xn=0
f (2n+1)(0)
(2n+ 1)!x2n+1
and therefore all of its even derivatives (at the origin) vanish
f (2n)o (0) =d2nf(x)
dx2n
¯̄̄̄x=0
= 0:
In particular, note that fo(0) = 0 (if fo(x) is continuous at the origin).
Even-Order Derivatives Preserve Symmetry and Odd-Order Derivatives Re-
verse Symmetry
f 0(t) = lim¢t!0
f(t+¢t=2)¡ f(t¡¢t=2)¢t
f 0(¡t) = lim¢t!0
f(¡t+¢t=2)¡ f(¡t¡¢t=2)¢t
Even function fe(t) = fe(¡t)
f 0e(t) = lim¢t!0
fe(t+¢t=2)¡ fe(t¡¢t=2)¢t
f 0e(¡t) = lim¢t!0
fe(¡t+¢t=2)¡ fe(¡t¡¢t=2)¢t
= lim¢t!0
fe(t¡¢t=2)¡ fe(t+¢t=2)¢t
= ¡ lim¢t!0
fe(t+¢t=2)¡ fe(t¡¢t=2)¢t
= ¡f 0e(t) =)d
dtfe(t) is odd.
Similarly,d
dtfo(t) is even where fo(¡t) = ¡fo(t) is an odd function.
This is also seen by appealing to the Taylor series for even and odd functions.
11
Consider the integral of f(t) between limits that are symmetrically disposed about the
origin, that is look atTZ
¡Tf(t) dt
where T is some ¯xed, positive constant.
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t
fo(t)
...........................................................................................................................................
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.............................................................................................................................
.........................................................................................
¡T.........................................................................................
T
TZ¡T
fo(t) dt = 0
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......
t
fe(t)
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....
¡T.......................................
T
TZ¡T
fe(t) dt = 2
TZ0
fe(t) dt
Any function f(t) (neither symmetric nor antisymmetric) can always be decomposed into
a linear combination of a symmetric plus an antisymmetric part since
f(t) =f(t) + f(¡t)
2+f(t)¡ f(¡t)
2= fe(t) + fo(t):
Example.
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t0 T
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¡T T
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..
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..
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.. ............................................
¡TT
= +
t
t
F Homework. Decompose the Heaviside unit step function u(t) into the sum of a
symmetric plus antisymmetric signal.
12
A Note on Mathematical Functions and Units/Physical Dimensions
Consider almost any \ordinary" or \regular" function2 of an independent variable x, such
as
f(x) = cos(x) or sin2(x) or 3x2 ¡ 7x or ex:
The argument or independent variable x is always dimensionless. To see this, examine the
Taylor series expansion, such as
f(x) = cos(x) = 1¡ x2
2!+x4
4!¡ : : : :
If x has units such as meters (m), then the sum would be nonsense since the ¯rst term is
dimensionless, the second term has units (m)2, the third term has units (m)4, etc. Also
note that the function f(x) itself is dimensionless, too.
This may seem strange and somewhat contrary to your previous experience. For example,
you have dealt with a time-domain voltage signal such as
v(t) = 12 cos(t+ 45±) or 12 cos(t+ ¼=4):
If the time variable t is measured in (s), then there really is a transparent (or hidden)
frequency ! = 1 (s¡1) multiplying the t so that
v(t) = V0 cos(!t+ Á):
Now the argument !t+ Á of the cosine is clearly dimensionless. The constant V0 = 12 (V)
carries the dimensions of the voltage signal v(t). Also note that we often say ! has units
of (rad/s), but a radian is not dimensioned: Its use in (rad/s) is simply a reminder that
we are using the angular frequency ! and not the common experimentalist's frequency
f = !=2¼ (cycle/s=Hz).
The operators di®erentiation and integration impart units: Consider the time-
domain current signal
i(t) = I0 cos(!t):
In our standard mksC (SI) units, the current amplitude I0 is in (A), the frequency ! is in
(s¡1), and the time t is in (s). Di®erentiation with respect to the dimensioned variable timparts the units of 1=t or (s¡1) since
di(t)
dt= ¡!I0 sin(!t):
We can also see that the di®erential operator has to have units of inverse time by appealing
to the de¯nition of the derivative
df(t)
dt= lim
¢t!0
f(t+¢t)¡ f(t)¢t
:
The (s¡1) comes from the denominator ¢t that is in (s).
2Not ±(x) or u(x) or jxj, to name a few exceptions.
13
Integration with respect to the dimensioned variable t imparts the units of t or (s) since
tZt0
i(¿) d¿ =
tZt0
I0 cos(!¿) d¿ =I0
!
£sin(!t)¡ sin(!t0)
¤:
Note the good practice of using a dummy time variable for the integration, to avoid con-
fusion with the ¯nal time that is the upper integration limit. We can also see that the
integral operator has to have units of time by appealing to the de¯nition of the integral in
terms of a limit on the Riemann sum
bZa
f(t) dt = limN!1
NXn=1
f(tn)¢tn:
The prime notation for derivatives. When dealing with a generic function such as
f(t) or g(x), we commonly denote derivatives using the prime
df(t)
dt= f 0(t) and
dg(x)
dx= g0(x):
The prime denotes di®erentiation to the entire argument of the function, as in
f 0(») =df(»)
d»or f 0(~) = df(~)
d~ :
Note thatd
dtf(») =
df(»)
d»
d»
dt= f 0(»)
d»
dt:
14
Linear System
x(t) y(t)T
............................................................................................................................................................................................................................................................................................................................................................................
........................................................... ............. ........................................................... .............
Assume the zero-state response y(t) of a system to the input x(t) is described by an
operator T , such that
y(t) = Tfx(t)g:Another name for T is system transformation rule. Examples might be
T1fx(t)g = Ax(t); T2fx(t)g = x(t¡ t0); T3fx(t)g = d
dtx(t)
T4fx(t)g =μc2d2
dt2+ c1
d
dt+ c0
¶x(t); T5fx(t)g =
tZ¡1
x(¿) d¿
T6fx(t)g = x2(t); T7fx(t)g = Ax(t) +B:
Linear System: If the input/output relationship or system operator or system transfor-
mation rule satis¯es the superposition principle
Tfc1x1(t) + c2x2(t)g = c1Tfx1(t)g+ c2Tfx2(t)g;
then the system is said to be linear.
F HW 1 Test the examples T1; T2; : : : ; T7 above for linearity.
15
Time-Invariant System:x(t) y(t)
T
............................................................................................................................................................................................................................................................................................................................................................................
........................................................... ............. ........................................................... .............
x(t¡ t0) y(t¡ t0)T
............................................................................................................................................................................................................................................................................................................................................................................
........................................................... ............. ........................................................... .............
Let the response of a system to the input signal x(t) be the output signal y(t). If the
response to the delayed input x(t¡ t0) is the delayed response y(t¡ t0), the characteristicsof the system did not change in time, and the system is said to be time-invariant.
Note: In our linear RLC circuit analysis, the equation-of-motion is typically a linear
integrodi®erential equation, with constant coe±cients, such as
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..........................................................................................................................................................
+¡vg(t)
R L..........................................................................
..........................................................................
.................................
.................................
C.....................................................
...................................
..........................................................................................................................................................................................................................................................................................................................................................................................................................
.............................................................................................
......................... ................
i(t) ........................................................................................................................................................................................................................................................................................................
......................................................... ................ ......................................................... ................vg(t) i(t)
Ldi(t)
dt+Ri(t) +
1
C
tZ¡1
i(¿) d¿ = vg(t)
which is easily converted to an ordinary di®erential equation by di®erentiating w.r.t. t so
that
Ld2i(t)
dt2+R
di(t)
dt+1
Ci(t) = v0g(t):
Again, note the constant coe±cients L, R, and 1=C. If the equation-of-motion of a system
is a di®erential equation with non-constant (i.e. time varying) coe±cients, then the system
is a time-variant system.
F HW 2 Give an example of a linear, time-varying system.
16
Input/Output Relationship for a LTI System
(zero-state or driven response)
x(t) y(t) = Tfx(t)gT
............................................................................................................................................................................................................................................................................................................................................................................
........................................................... ............. ........................................................... .............
This fundamental concept is easy to see and derive, starting with a little \trick" by way
of the notation. First, invoke the sampling property of the Dirac-delta function to write
x(t) =
1Z¡1
x(¿)±(t¡ ¿) d¿:
Now exploit the linearity of the operator T and note that Tf¢g operates only on functionsof time t. In particular, T treats x(¿) as a constant, and acts only on the time-domain
function ±(t¡ ¿)
y(t) = Tfx(t)g = T8<:
1Z¡1
x(¿)±(t¡ ¿) d¿9=; =
1Z¡1
x(¿)Tf±(t¡ ¿)g d¿:
The response of our system to the Dirac-delta function or impulse function is called the
system impulse response and it is denoted by the signal h(t). It is a fundamental
characterization of any linear, time-invariant system.
±(t) h(t)T
............................................................................................................................................................................................................................................................................................................................................................................
........................................................... ............. ........................................................... .............
±(t¡ ¿) h(t¡ ¿)T
............................................................................................................................................................................................................................................................................................................................................................................
........................................................... ............. ........................................................... .............
Since the system is also time-invariant,
Tf±(t¡ ¿)g = h(t¡ ¿)and the result above is
y(t) =
1Z¡1
x(¿)h(t¡ ¿) d¿:
This integral operator that takes two functions of time x(t) and h(t), and produces a third
function of time y(t), is called the convolution integral; we write
y(t) = x(t)~ h(t):
17
Convolution
The convolution operation is not restricted to our above LTI system application. It can
operate on any two (reasonably) arbitrary signals, say f(t) and g(t):
f(t)~ g(t) =1Z
¡1f(¿)g(t¡ ¿) d¿:
Example. Evaluate s(t) = f(t) ~ g(t) where f and g are causal, decaying exponentialsignals
f(t) = e¡atu(t) and g(t) = e¡btu(t) with 0 < a < b:
case: t < 0
................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
¿0
f(¿) = e¡a¿u(¿)
...................................................................................................................................................................................................................................................................................................................................................................................................................
................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
¿0
g(t¡ ¿) = e¡b(t¡¿)u(t¡ ¿)
..............................................................................................
..............................................................................................................................................................................................................................................................
t
................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
¿0
f(¿)g(t¡ ¿)
...........
t
s(t) =
1Z¡1
f(¿)g(t¡ ¿) d¿ =1Z
¡10 d¿ = 0
18
case: t > 0
................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
¿0
f(¿) = e¡a¿u(¿)
...................................................................................................................................................................................................................................................................................................................................................................................................................
................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
¿0
g(t¡ ¿) = e¡b(t¡¿)u(t¡ ¿)
.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.......................................................................................................................................................................................................................................................................
t
................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
¿0
f(¿)g(t¡ ¿)
.............................................................................................
................................................................................................................................................................................
t
s(t) =
1Z¡1
f(¿)g(t¡ ¿) d¿
=
0Z¡1
0 d¿ +
tZ0
e¡a¿e¡b(t¡¿)d¿ +
1Zt
0 d¿
= e¡bttZ0
e(b¡a)¿d¿ =e¡at ¡ e¡btb¡ a
19
Now combine both expressions (¯rst one for t < 0 and second one for t > 0) into a single
equation valid for all t
s(t) =e¡at ¡ e¡btb¡ a u(t):
.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
t0
s(t)
............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
tm
Let's look for the maximum of s(t) analytically: The derivative
s0(t) =be¡bt ¡ ae¡at
b¡ a u(t) +e¡at ¡ e¡btb¡ a ±(t)
=be¡bt ¡ ae¡at
b¡ a u(t)
sincee¡at ¡ e¡btb¡ a
¯̄̄̄t=0
= 0:
Recall that Ã(t)±(t) = Ã(0)±(t). The derivative vanishes at time t = tm, so write s0(tm) = 0
so that
ae¡atm = be¡btm
e(b¡a)tm =b
a
tm =ln(b=a)
b¡ a
20
Another Convolution Example: The Convolution of Two Gaussian Pulses
Evaluate z(t) = f(t)~ g(t) with f(t) = exp(¡at2) and g(t) = exp(¡bt2).
Both f(t) and g(t) are \on" for all time, so we don't have to worry about sketching them
to see when they turn on or o®.
z(t) =
1Z¡1
f(¿)g(t¡ ¿) d¿ =1Z
¡1e¡a¿
2
e¡b(t¡¿)2
d¿
= e¡bt2
1Z¡1
e¡[(a+b)¿2¡2bt¿ ]d¿
complete the square
(a+ b)¿2 ¡ 2bt¿ =μp
a+ b¿ ¡ btpa+ b
¶2¡ (bt)2
a+ b
z(t) = exp
·μb2
a+ b¡ b¶t2¸ 1Z¡1
exp
"¡μp
a+ b¿ ¡ btpa+ b
¶2#d¿
let
x =pa+ b¿ ¡ btp
a+ b
dx =pa+ b d¿
¿ = §1 ¡! x = §1recall or use (\and it is a trick!"see page 63)
1Z¡1
e¡x2
dx =p¼
z(t) = exp
·μb2
a+ b¡ b¶t2¸ 1Z¡1
e¡x2 dxpa+ b
=
r¼
a+ bexp
·¡μb¡ b2
a+ b
¶t2¸
=
r¼
a+ bexp
·¡ ab
a+ bt2¸
The convolution of two Gaussians is a third Gaussian.
21
Yet Another Convolution Example
Evaluate s(t) = f(t)~ g(t) where f(t) and g(t) are the given rectangular pulses:
f(t) = u(t)¡ u(t¡ a) =½1; 0 < t < a
0; t < 0 or t > a
g(t) = u(t)¡ u(t¡ b) =½1; 0 < t < b
0; t < 0 or t > b
Here the pulse width a of f(t) has been arbitrarily selected to be greater than the pulse
width b of g(t).
..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
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............................................................................................................................................................................................................................................................................................................................................................... t0 a
1f(t)
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..................................................................................................................................................................................................................................................................... t0 b
1g(t)
s(t) = f(t)~ g(t) =1Z
¡1f(¿)g(t¡ ¿) d¿
Since the arguments of our individual signals are ¿ (and a shifted, rotated version of ¿)
in the convolution integral, let's redraw our functions f and g as functions of the dummy
variable ¿ .
..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
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............................................................................................................................................................................................................................................................................................................................................................... ¿0 a
1f(¿)
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..................................................................................................................................................................................................................................................................... ¿0 b
1g(¿)
22
Let's leave f(¿) alone (as in the version of the convolution integral above), and shift and
rotate g(¿) (as it appears in the version of the convolution integral above).
..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
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..................................................................................................................................................................................................................................................................... ¿0 t t+b
g(¿ ¡ t)
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..................................................................................................................................................................................................................................................................... ¿0 tt¡b
g(t¡ ¿)
Now let's vary the parameter t (it survives the ¿ integration), compute (graph!) the
product of f(¿) and g(t ¡ ¿), and evaluate the in¯nite range integral from ¿ = ¡1 to
¿ = +1. That is, we must calculate the total area under the curve of f(¿)g(t¡ ¿).
case: t < 0
..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
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............................................................................................................................................................................................................................................................................................................................................................... ¿0 a
......
.....
b
f(¿)
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..................................................................................................................................................................................................................................................................... ¿0t¡ b t
g(t¡ ¿)
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¿0
f(¿)g(t¡ ¿)
s(t) = f(t)~ g(t) =1Z
¡1f(¿)g(t¡ ¿) d¿ =
1Z¡1
0 d¿ = 0
23
case: 0 < t < b
..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
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............................................................................................................................................................................................................................................................................................................................................................... ¿0 a
......
.....
b
f(¿)
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..................................................................................................................................................................................................................................................................... ¿0t¡ b t
g(t¡ ¿)
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............................................................................................................................................................................................................... ¿0 a
......
.....
b...........
t
f(¿)g(t¡ ¿)
s(t) = f(t)~ g(t) =1Z
¡1f(¿)g(t¡ ¿) d¿ =
tZ0
d¿ = t
case: b < t < a
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............................................................................................................................................................................................................................................................................................................................................................... ¿0 a
......
.....
b
f(¿)
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..................................................................................................................................................................................................................................................................... ¿0 t¡ b t
g(t¡ ¿)
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..................................................................................................................................................................................................................................................................... ¿0 a
......
.....
b...........
tt¡ b
f(¿)g(t¡ ¿)
s(t) = f(t)~ g(t) =1Z
¡1f(¿)g(t¡ ¿) d¿ =
tZt¡b
d¿ = b
24
case: a < t < a+ b
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............................................................................................................................................................................................................................................................................................................................................................... ¿0 a
......
.....
b...........
a+ b
f(¿)
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..................................................................................................................................................................................................................................................................... ¿0 a
......
................
a+ btt¡b
g(t¡ ¿)
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........................................................................................................................................................................... ¿0 a
......
................
a+ bt...........
t¡b
f(¿)g(t¡ ¿)
s(t) = f(t)~ g(t) =1Z
¡1f(¿)g(t¡ ¿) d¿ =
aZt¡b
d¿ = a+ b¡ t
case: t > a+ b
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............................................................................................................................................................................................................................................................................................................................................................... ¿0 a
......
.....
b...........
a+ b
f(¿)
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..................................................................................................................................................................................................................................................................... ¿0 a
......
................
a+ b tt¡b
g(t¡ ¿)
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.....
¿0
f(¿)g(t¡ ¿)
s(t) = f(t)~ g(t) =1Z
¡1f(¿)g(t¡ ¿) d¿ =
1Z¡1
0 d¿ = 0
25
All together now.
s(t) = f(t)~ g(t) =
8>>>>><>>>>>:
0; t < 0
t; 0 < t < b
b; b < t < a
a+ b¡ t; a < t < a+ b
0; t > a+ b
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....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... t
0 a...........
......
.....
b...........
a+ b
s(t)
...........b
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26
Homework Problems
1. In the above example on page 18, ¯nd r(t) = f(t)~ f(t), that is let b = a. Show thatyour result can also be obtained from the limit of the above as b! a
r(t) = limb!a
s(t):
Use l'Hopital's rule to disarm the 0=0 form.
2. Show that f(t)~ ±(t¡ t0) = f(t¡ t0).
3. Show that
f(t)~ u(t) =tZ
¡1f(¿) d¿:
4. Evaluate the convolution of the standard rectangular pulse with itself ¦(t=T )~¦(t=T ).Express the result in terms of the standard triangular pulse ¤(t=T ).
..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
¡T=2 T=20
1
¦(t=T )
t.............................................................................................................
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......................................................................................................................................................................................................................................................................................................................................................................................................................................
.........................................................................................................................................................................................................................................................................................................................................................................................................................
.......................... 1
¡T 0 Tt
¤(t=T )
¦(t=T ) =
½1; (jtj < T=2)0; (jtj > T=2)
¤(t=T ) =
½1¡ jtj=T; (jtj < T )0; (jtj > T )
5. Find an expression for, and graph, each of the derivatives
d
dt¤(t=T ) and
d
dt¦(t=T ):
27
Causal system: If the input x(t) = 0 for t < t0, then the zero-state response y(t) = 0
for t < t0. The impulse response of a causal system is therefore a causal signal.
If f(t) and g(t) are both causal signals, then so is their convolution f(t)~ g(t).
Bounded input/bounded output stability: If the input signal is bounded, such that
jx(t)j · A <1 for all time t, then the zero-state output y(t) is also bounded if the impulse
response is absolutely integrable:
jy(t)j =¯̄̄̄¯̄1Z
¡1h(¿)x(t¡ ¿) d¿
¯̄̄̄¯̄ ·
1Z¡1
¯̄h(¿)x(t¡ ¿)¯̄ d¿ = 1Z
¡1
¯̄h(¿)
¯̄¯̄x(t¡ ¿)¯̄ d¿
· A1Z
¡1
¯̄h(¿)
¯̄d¿ <1
if 1Z¡1
¯̄h(t)
¯̄dt <1:
28
Singular Functions and Integrals
Consider a real-valued function f of a real variable x.
With A some positive and ¯nite constant, if jf(x)j · A < 1 for all x, then we say the
function f(x) is bounded.
If jf(x0)j = 1 for some particular x = x0, then we say f is singular at x0, and x0 is a
singular point of f .
Examples.
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x
1=x
1
xis singular at x = 0
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..
x
sin(x)=x
sin(x)
xis bounded at x = 0.....................................
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..
x
cos(x)=x
cos(x)
xis singular at x = 0
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29
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......
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......
...
0 7
x
1=(x¡ 7)2
1
(x¡ 7)2 is singular at x = 7..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... .........................................................................................................................................................................................................................................................................................................................................................
..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
0
x
1=px
1pxis singular at x = 0
.....................................................................................................................................................................................................................................................................................................................................................................................................
Consider now the de¯nite integral of f(x) between the limits x = a and x = b. These
limits are arbitrary, but with the restriction that a · b.
If
¯̄̄̄¯bZa
f(x) dx
¯̄̄̄¯ · A; then the integral converges, and it exists.
If
¯̄̄̄¯bZa
f(x) dx
¯̄̄̄¯ =1; then the integral diverges, and it does not exist.
From the triangle inequality
¯̄̄̄¯bZa
f(x) dx
¯̄̄̄¯ ·
bZa
jf(x)j dx:
30
If f(x) = jf(x)j (f is non-negative, that is f(x) ¸ 0), and if b > a, thenbZa
f(x) dx ¸ 0:
Therefore, for simplicity, let's temporarily restrict our attention to non-negative f(x).
Basically, there are two ways to get a divergent integral:
(1) A ¯nite range integral (¡1 < a < b <1) having an integrand with a non-integrable(too big!) singularity.
(2) An in¯nite range integral (a = ¡1 and/or b = 1) having a bounded integrandthat doesn't decay fast enough at 1.
Examples.
1Z0
dx
x2diverges because the integrand has a non-integrable singularity at the origin.
With ² > 0, if we avoid the second-order pole at the origin, the integral evaluates to
1Z²
dx
x2=1
x
¯̄̄̄¯²
x=1
=1
²¡ 1:
Clearly this integral is unbounded (singular) if we let ²! 0.
1Z0
dx
x= ln(x)
¯̄̄1x=0
= 0¡ ln(0) =1 is divergent.
1Z0
dxpx= 2x1=2
¯̄̄1x=0
= 2 is convergent.
1Z1
dx
x2=1
x
¯̄̄̄¯1
x=1= 1¡ 0 = 1 is convergent.
31
1Z1
dx
x= ln(x)
¯̄̄̄¯1
x=1
=1¡ 0 =1 is divergent.
Cauchy Principal Value
Sometimes we relax our concept of the integral to obtain a ¯nite result and \dance around"
singularities in the integrand that are, strictly speaking, non-integrable. Here is one exam-
ple that demonstrates what is meant by the Cauchy principal value which is one possible
interpretation of an otherwise divergent integral.
If f(x) is bounded everywhere and if a < c < b, then we can always break the integration
domain from a to b at the point c and write
bZa
f(x) dx =
cZa
f(x) dx+
bZc
f(x) dx:
We have already seen that the function f(x) = 1=x has a non-integrable singularity at the
origin, since1Z0
dx
xis divergent.
Also observe that0Z
¡1
dx
xis divergent.
If we want to integrate 1=x from x = ¡1 to x = 1, we should not break it as1Z
¡1
dx
x6=
0Z¡1
dx
x+
1Z0
dx
x:
But under the Cauchy principal value interpretation, if we exploit the odd symmetry of
the integrand, ignoring the singularity at the origin, then we can say
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......
x
1=x
PV
1Z¡1
dx
x=
1Z¡1|
dx
x= lim²!0
24 ¡²Z¡1
dx
x+
1Z²
dx
x
35 = 0:..............................................................................................................................................................................................................................................
.......
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........
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...................................................................
..................
¡11
32
SHOW THAT
1Z0
¯̄̄̄sinx
x
¯̄̄̄dx IS DIVERGENT
x
k¼ (k + 1=2)¼ (k + 1)¼
.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
....................................................................................................................................................
(k+1)¼Zk¼
¯̄̄̄sinx
x
¯̄̄̄dx >
¼=2
(k + 1=2)¼=
1
2k + 1
1Z0
¯̄̄̄sinx
x
¯̄̄̄dx =
¼Z0
¯̄̄̄sinx
x
¯̄̄̄dx+
1Xk=1
(k+1)¼Zk¼
¯̄̄̄sinx
x
¯̄̄̄dx >
¼Z0
¯̄̄̄sinx
x
¯̄̄̄dx
| {z }¯nite
+
1Xk=1
1
2k + 1| {z }divergent
=1
\... better to say something new and stimulating to the interested students than to show
routine steps to the uninterested ones."
E.G. Peter Rowe Geometrical Physics in Minkowski Spacetime, Springer, 2001, p. v
33
SHOW THAT
1Z0
sinx
xdx IS CONVERGENT
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..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... x
.......
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..
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..
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......
..
2k¼ (2k + 1)¼
(2k + 2)¼
(2k+2)¼Z2k¼
sinx
xdx =
(2k+1)¼Z2k¼
sinx
xdx +
(2k+2)¼Z(2k+1)¼
sinx
xdx
| {z }let t=x¡¼
=
(2k+1)¼Z2k¼
sinx
xdx ¡
(2k+1)¼Z2k¼
sin t
t+ ¼dt
=
(2k+1)¼Z2k¼
sinx
μ1
x¡ 1
x+ ¼
¶dx
= ¼
(2k+1)¼Z2k¼
sinx
x2 + ¼xdx < ¼
(2k+1)¼Z2k¼
dx
x2=
1
4k2 + 2k
1Z0
sinx
xdx =
2N¼Z0
sinx
xdx+
1Xk=N
(2k+2)¼Z2k¼
sinx
xdx <
2N¼Z0
sinx
xdx
| {z }¯nite
+
1Xk=N
1
4k2 + 2k| {z }¯nite
34
Homework Problems
1. Express each of these four functions (signals) in terms of shifted, modulated, unit steps.
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¡TT
+A
¡A
f(t)
0
................... 1
..........................................................................................................................................................................................................................................................................................................................................................................................................................................
t¡T T
¤
μt
T
¶
0
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¡TT
B
¡B
g(t)
0
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...................
...................
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t
T1 T2
.......
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.....
x(t)
0
...................A
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2. Discuss the existence and evaluate (if appropriate!) the following integrals. First sketch
the integrand (except in an extreme case!).
1Z¡1
u(t) dt;
1Z¡1
u(t)e¡¾t dt (¾ > 0);
1Z¡1
u(t)e¡¾te¡j!t dt
1Z¡1
sin(!0¿)±(¿) d¿;
1Z¡1
cos(!0¿)±(¿) d¿;
¼Z0
sin(x2)
ln(x=¼)±(x¡ 2¼) dx
1Z0
cosx
xdx;
1Z¡1
cosx
xdx;
1Z1
dx
x2;
1Z0
dx
x2;
1Z0
dxpx;
1Z0
dxpx
2¼Z¡p¼
·exp[j sin2(!0t)] ±(t+ ¼)
t3 tan(7t) + ln(sin t)¡ ¼2 sin(3t=2) ±(t¡ ¼)t2 + 2t¡ 4 sin¡1(t=¼)
¸dt
35
3. Prove that convolution is commutative:
f(t)~ g(t) = g(t)~ f(t):
4. Evaluate u(t)~ u(t).
5. Show that ±(t) = ±(¡t) is an even function.
6. If f(t) is continuous at t = t0, show that f(t)±(t¡ t0) = f(t0)±(t¡ t0).
36
Relationship Between Impulse Response and Step Response
±(t) h(t)
............................................................................................................................................................................................................................................................................................................................................................................
........................................................... ............. ........................................................... .............
u(t) s(t)
............................................................................................................................................................................................................................................................................................................................................................................
........................................................... ............. ........................................................... .............
±(t) =d
dtu(t) h(t) =
d
dts(t)
u(t) =
tZ¡1
±(¿) d¿ s(t) =
tZ¡1
h(¿) d¿
To see this, consider the general linear system input/output relationship.
x(t) y(t)h(t)
............................................................................................................................................................................................................................................................................................................................................................................
........................................................... ............. ........................................................... .............
y(t) = x(t)~ h(t) =1Z
¡1x(¿)h(t¡ ¿) d¿ =
1Z¡1
h(¿)x(t¡ ¿) d¿
d
dty(t) =
1Z¡1
x(¿)d
dth(t¡ ¿) d¿ =
1Z¡1
h(¿)d
dtx(t¡ ¿) d¿
= x(t)~ h0(t) = x0(t)~ h(t)
The output of these two systems is identical, since the order of di®erentiation and passing
through the linear system having a given impulse response h(t) does not matter. More
concisely, the operators h(t)~ and d=dt commute.
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ex(t) ey(t)h(t)d
dt........................................................................ ..........
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ex(t) ey(t)h(t)d
dt........................................................................ ..........
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....
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37
Time-Domain RC Circuit Analysis
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t = 0
R
C
+
¡
v(t)+
¡V0 cos(!t)u(t) = vg(t)
The capacitor is initially charged, such that v(t) = Vc0 prior to the closing of the switch
at time t = 0. Find an expression for the signal v(t) that is valid for all time t ¸ 0.The constants R, C, !, V0, and Vc0 are given and thus known.
The initial condition on the voltage v(t) that applies to our circuit for time t ¸ 0 is v(0+)and v(0+) = v(0¡) = Vc0 since the voltage across a capacitor must be a continuous functionof time. That is, if the capacitor current is bounded, because the voltage/current terminal
relationship of a capacitor is
iC(t) = Cd
dtvC(t) (1)
where iC(t) and vC(t) are related by the passive sign convention. Inverting this, that is,
writing the voltage as a function of the current, we have
vC(t) =1
C
tZ¡1
iC(¿) d¿ =1
C
0Z¡1
iC(¿) d¿ +1
C
tZ0
iC(¿) d¿ = vC(0) +1
C
tZ0
iC(¿) d¿: (2)
The Kirchho® current law applied to the (single unknown) node of potential v(t), for time
t ¸ 0 isCdv(t)
dt+v(t)¡ vg(t)
R= 0 (3)
which is rearranged asdv(t)
dt+
1
RCv(t) =
V0
RCcos(!t)u(t): (4)
Here we are already working in the time zone t ¸ 0, so with the stipulation that t ¸ 0, theHeaviside unit step is probably unnecessary so we can write
dv(t)
dt+
1
RCv(t) =
V0
RCcos(!t): (5)
The homogeneous solution is the solution to
dvh(t)
dt+
1
RCvh(t) = 0 (6)
38
and it is
vh(t) = Ke¡t=RC (7)
where the constant K cannot be determined until we construct the complete solution. The
particular solution is some speci¯c or particular function vp(t) that yields the right-hand
side (forcing term) when operated upon by the di®erential operatorμd
dt+
1
RC
¶vp(t) =
V0
RCcos(!t): (8)
There are no unknown constants (unknown by the di®erential operator) in the particular
solution.3 Since the forcing function is cos(!t) and since cos(!t) is not a homogeneous
solution, then we see that the particular solution must be a linear combination of cos(!t)
and sin(!t), as in
vp(t) = A cos(!t) +B sin(!t): (9)
The constants A and B come directly from the di®erential equation (8) , which readily
surrenders them (to us): All we have to do is insert the form (9) into the di®erential
equation (8). To do this, ¯rst write the required derivative
dvp(t)
dt= !B cos(!t)¡ !A sin(!t): (10)
Insertion of (9) and (10) into (8) yieldsμA
RC+ !B
¶cos(!t) +
μB
RC¡ !A
¶sin(!t) =
V0
RCcos(!t): (11)
The only way this can be true for all t ¸ 0 is if the coe±cients of the cos(!t) and the
sin(!t) terms are separately equal on both sides of the equation, so that
A
RC+ !B =
V0
RC(12)
B
RC¡ !A = 0: (13)
The solution to this two-by-two system of simultaneous, linear equations is
A =V0
1 + (!RC)2and B =
!RCV0
1 + (!RC)2; (14)
and so our particular solution (9) is
vp(t) =V0
1 + (!RC)2
£cos(!t) + !RC sin(!t)
¤: (15)
3In other words, the particular solution comes only from the di®erential equation, that is, the particular
solution is completely independent and ignorant of any initial conditions.
39
Again, note that vp(t) comes completely from the di®erential equation alone. We now can
assemble the complete solution v(t) = vh(t) + vp(t) or
v(t) = Ke¡t=RC +V0
1 + (!RC)2
£cos(!t) + !RC sin(!t)
¤: (16)
Application of the one initial condition v(0) = v(0+) = Vc0 (note that it is the plus side of
zero that applies to our di®erential equation, because the di®erential equation is valid for
time t ¸ 0) to (16) gives
Vc0 = K +V0
1 + (!RC)2or K = Vc0 ¡ V0
1 + (!RC)2: (17)
Let's write out the complete solution in two forms. Firstly, observe that for t ¸ 0
v(t) =
·Vc0 ¡ V0
1 + (!RC)2
¸e¡t=RC| {z }
homogeneous solution
+V0
1 + (!RC)2
£cos(!t) + !RC sin(!t)
¤| {z }
particular solution
(18)
is exactly in the form in which we constructed the solution, that is, the sum of the homo-
geneous plus particular solutions. If we instead group terms proportional to Vc0 and V0,
we have for t ¸ 0
v(t) = Vc0e¡t=RC| {z }
zero-input response
+V0
1 + (!RC)2
hcos(!t) + !RC sin(!t)¡ e¡t=RC
i| {z }
zero-state response
: (19)
If there is no input signal (or generator voltage vg(t) or forcing term), then V0 = 0 and
the output signal v(t) is caused solely by the initial energy (or non-zero initial state of
the system) stored in the capacitor, and is proportional to the non-zero initial condition
voltage Vc0. On the other hand, if the initial state of the system is zero, so that Vc0 = 0,
then the response due to the input vg(t) that is proportional to V0 is called the zero-state
response.
40
Step Response of an RC Circuit
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R
C
+
¡
v(t)+
¡V0u(t) = vg(t)
The step response is, by de¯nition, a zero-state response so the capacitor C is initially
uncharged prior to the turning on of the input step function at time t = 0. Therefore, by
the continuity of capacitor voltage we have the initial condition
v(0+) = v(0¡) = 0: (20)
Ultimately, the step response s(t) = y(t) should be the response due to a non dimensioned
input signal u(t) = x(t), but in doing circuit analysis I ¯nd it better to explicitly express
all physical signals such as voltages and currents in appropriate mksC (SI) units. At the
end we can normalize by V0, for example. In systems engineering (\black-box") notation,
we have the general form of a zero-state response y(t) to some arbitrary input x(t) like
this:
................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.......................................................................... ................ .......................................................................... ................x(t) y(t)
For the step response, we have:
................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.......................................................................... ................ .......................................................................... ................u(t) s(t)
Our ensuing circuit analysis regards the input as x(t) = vg(t) = V0u(t) and our output
signal is y(t) = v(t). The Kirchho® current law applied to the node of potential v(t) in
the circuit is for time t ¸ 0
Cdv(t)
dt+v(t)¡ vg(t)
R= 0 (21)
which is rearranged as
dv(t)
dt+
1
RCv(t) =
V0
RCu(t): (22)
41
Here we are already working in the time zone t ¸ 0, so with the stipulation that t ¸ 0, theHeaviside unit step is probably unnecessary so we can write
dv(t)
dt+
1
RCv(t) =
V0
RC: (23)
The homogeneous solution is (still) of the form
vh(t) = Ke¡t=RC (24)
where K is some (new) unknown constant. But now the particular solution, due to a
constant forcing term, is itself a constant, call it
vp(t) = D: (25)
This constant D is, again, completely determined (that is known by) the di®erential equa-
tion (24). Insertion of (25) into (23) gives the rather trivial
D
RC=V0
RCor D = V0: (26)
The complete solution is now v(t) = vh(t) + vp(t) or
v(t) = Ke¡t=RC + V0; (27)
whereupon the initial condition (20) gives K = ¡V0 so thatv(t) = V0
£1¡ e¡t=RC¤ (t ¸ 0): (28)
A better (systems engineering style!) way to write this is
v(t) = V0£1¡ e¡t=RC¤u(t): (29)
The unit step not only eliminates the need for the disclaimer t ¸ 0, but its inclusion is
invaluable in emphasizing the causal nature of the response, and it is critically important
if (when!) we di®erentiate the response.
Now we can normalize by V0, as advertised, and obtain the true, non dimensional step
response
s(t) =£1¡ e¡t=RC¤u(t): (30)
The impulse response is the derivative of the step response
h(t) =d
dts(t) =
1
RCe¡t=RCu(t) +
£1¡ e¡t=RC¤±(t)
=1
RCe¡t=RCu(t): (31)
Note that the units of h(t) are [(−)(F)]¡1 = (s)¡1. In this particular instance, the chain-rule di®erentiation that also di®erentiated the u(t) to get a ±(t) might appear super°uous,
since the contribution£1¡ e¡t=RC¤±(t) = £
1¡ e¡t=RC¤¯̄̄t=0
±(t) = 0±(t) = 0 (32)
is zero. But that will not happen in general.
42
Step Response of Another RC Circuit
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..
R
C
+
¡
v(t)+
¡V0u(t) = vg(t)
+ ¡vC(t)
In this circuit, we take our output signal to be the voltage v(t) across the resistor, but the
capacitor voltage vC(t) must also be speci¯cally addressed because of the importance of
continuity of capacitor voltage for determining initial conditions. The zero-state response
demands that there is no initial energy storage in the capacitor so vC(0¡) = 0 prior to the
unit step input turning on at time t = 0. By the continuity of capacitor voltage
vC(0+) = vC(0
¡) = 0: (33)
The application of the Kirchho® current law to the node of potential v(t) gives
v(t)
R+ C
d
dt
£v(t)¡ vg(t)
¤= 0 (34)
ordv(t)
dt+
1
RCv(t) =
dvg(t)
dt: (35)
The forcing term is now the derivative of the voltage generator, which puts us in the
unfamiliar realm of solving a di®erential equation with a wild Dirac-delta forcing term
dv(t)
dt+
1
RCv(t) = V0±(t): (36)
If we stay away from t = 0, then the delta function is zero and we have (note the time
restriction)dv(t)
dt+
1
RCv(t) = 0 (t > 0): (37)
It looks as though we have lost our excitation: The di®erential equation is now homoge-
neous. Information about the excitation, embodied by the source strength V0, is included
in the initial condition. Observe in the circuit that
vg(t) = vC(t) + v(t) (38)
43
and at time t = 0+ this is
vg(0+) = vC(0
+) + v(0+): (39)
The initial condition (33) on the capacitor voltage together with vg(0+) = V0 reveals that
the non-zero initial condition on our output signal v(t) is v(0+) = V0. No particular
solution is required for the complete solution to our homogeneous di®erential equation
(37), which is readily seen to be
v(t) = V0e¡t=RC (t > 0): (40)
Inclusion of the Heaviside unit step function gives
v(t) = V0e¡t=RCu(t) (41)
so that the time zone restriction t > 0 can be removed. The non dimensional step response
is therefore
s(t) = e¡t=RCu(t) (42)
and the corresponding impulse response is
h(t) =d
dts(t) = e¡t=RC±(t)¡ 1
RCe¡t=RCu(t)
= ±(t)¡ 1
RCe¡t=RCu(t) (43)
since
e¡t=RC¯̄̄̄¯t=0
= 1: (44)
Observe that the proper use of chain-rule di®erentiation and explicitly including the u(t)
in the step response s(t) is necessary to get the true impulse response h(t), which most
de¯nitely contains a ±(t) term.
44
Relationship Between the Two RC Circuits
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..
.......
.......
.......
.......
.......
.......
.......
.......
..
............................................
............................................
²
²R
C
+
¡
±(t)
+
¡
ha(t)
............................
............................
............................
............................
..................................................................................................................................................................................................................................................................................................
......................................................................................................
.......
.......
.......
.......
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..
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..
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.......
......
.......
.......
........................................................................
........................................................................
..........................................................
......
.......
......
.......
..
²
²C
R
+
¡
±(t)
+
¡
hb(t)
In the lab, simply connect R and C in series and hang the series leg across the output
terminals of a (ideal?) voltage source. Then we see that the output of circuit (a) is the
voltage across C and the output of circuit (b) is the voltage across R. Kirchho®'s voltage
law clearly shows the simple relationship that is displayed by comparing (30) and (42) and
by comparing (31) and (43).
.......
...............................................................................................................................................................................
.........................................................................................................................................................................................................................................................................................................................................................................................
............................................................................................................................................................................................................................
........................................................................
..........................................................
........
......
.......
.......................................................................................................................................................................................................................................................................
+
¡vg(t)
R
+
¡vb(t)
C
+
¡va(t)
vg(t) = va(t) + vb(t)
±(t) = ha(t) + hb(t)
u(t) = sa(t) + sb(t)
Homework Problems
1. Observe that the circuit on page 41 is a special case of the circuit on page 38, if Vc0 = 0
and ! = 0. Therefore, show for yourself that the response (29) is the corresponding special
case of the response (19).
2. Use convolution to ¯nd the zero-state (or driven) response of the RC circuit on page 41
to a rectangular input pulse vg(t) = V0[u(t)¡u(t¡T )]. Graph this response using MATLABfor several numerical values of the important (non dimensional) parameter T=RC.
3. Go backwards, and obtain the two step responses (30) and (42) by integrating the
corresponding impulse responses (31) and (43), since
h(t) =d
dts(t) () s(t) =
tZ¡1
h(¿) d¿:
45
Step and Impulse Response of Series RLC Circuit
...................................................................................................................................................................
..............................................
..............................................
................................................................................................................
.......
.......
.......
.......
.......
.......
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.......
..
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.......
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.......
.......
..
................................................................................................................................................................................................................................................
..................................... ........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......................................................................
........................................................................
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.......
.......
.......
.......
.......
.......
.......
.......
.......
...
²
²
.......
.....................................
........................................................................................................................................................................................
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
................................................................
.............................................................................................................................................
+
¡vg(t)
+ vC(t) ¡L
C
R vR(t)
+
¡
.......................................................................................................................................................................................................................................................................
......................................................................................................................................................................................................
.............................
..............................................................
..........................................................................................................................................................................................................
.................. ................
i(t)
................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
.....
Fig. 1. Circuit layout.
..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
............................................................................................................................. ................ ............................................................................................................................. ................vg(t) = x(t) y(t) = Ri(t)
Fig. 2. System block diagram.
The system step response is a \zero-state" response: No energy is initially stored in the
system prior to the onset of the exciting Heaviside step function that turns on at t = 0.
For de¯niteness and to keep track of the physical units, let
vg(t) = V0u(t) (1)
46
with the (real) constant V0 being the amplitude of the exciting step function. Since the
instantaneous energy stored in the inductor and capacitor at time t are
wL(t) =12Li
2(t) and wC(t) =12Cv
2C(t); (2)
respectively, zero initial (at time t = 0¡) energy storage requires
i(0¡) = 0 and vC(0¡) = 0: (3)
If the inductor voltage
vL(t) = Ldi(t)
dt
is to be bounded, then its current must be a continuous function of time, and hence
i(0+) = i(0¡): (4)
Similarly, if the capacitor current
iC(t) = CdvC(t)
dt
is to be bounded, then its voltage must be a continuous function as in
vC(0+) = vC(0
¡): (5)
The equation-of-motion for the series circuit of Fig. 1 is Kirchho®'s voltage law
Ldi(t)
dt+Ri(t) +
1
C
tZ¡1
i(¿) d¿ = vg(t); (6)
an integrodi®erential equation in the unknown series current i(t). With the forcing function
vg(t) known, the unique solution of equation (6), whose highest derivative term is of degree
one, requires the speci¯cation of one initial condition. The physics of the circuit dictates
i(0+) = 0; (7)
from (3) and (4). If we di®erentiate (6) once to eliminate the integral operator, the resulting
second-order di®erential equation
Ld2i(t)
dt2+R
di(t)
dt+1
Ci(t) = v0g(t) (8)
now requires two initial conditions for a complete (unique) solution. This second initial
condition was already contained in (6), before we di®erentiated. Evaluation of (6) at time
t = 0+ is
Li0(0+) +Ri(0+) + vC(0+) = vg(0+): (9)
The capacitor voltage, from (3) and (5), starts o® at zero
vC(0+) = 0; (10)
so that together with (1) and (7), statement (9) gives
i0(0+) =V0
L: (11)
Note that the dimensions are consistent in that (A/s)= (V/H). X
47
The proper formulation of the above initial value problem, especially stating the correct
number (two) and type of initial conditions (on the current and its derivative), required
our detailed understanding of the electrophysics embodied in the circuit. In one view
the ensuing construction of the solution to the di®erential equation may seem like a pure
exercise in mathematics, but that is not the view we will take. Instead, we want to let each
step in our solution be guided by the physics that the mathematics represents, so that our
mathematics will be e±cient and keep us in touch with the important physics that is our
quest. If we do this right, then our symbols and formulas and solution will be in a form
that not only allows, but even begs for physical interpretation.
Let's restate our initial value problem:
Ld2i(t)
dt2+R
di(t)
dt+1
Ci(t) = v0g(t) (12)
i(0+) = 0 (13)
i0(0+) =V0
L: (14)
Since our voltage generator is vg(t) = V0u(t), the forcing function on the right-hand side
of (12) is a Dirac-delta function, i.e.
Ld2i(t)
dt2+R
di(t)
dt+1
Ci(t) = V0±(t): (15)
That looks troublesome, I believe, but if we stay away from t = 0, then the delta function
is zero and our equation is homogeneous
Ld2i(t)
dt2+R
di(t)
dt+1
Ci(t) = 0 (t > 0): (16)
It looks as though the homogeneous di®erential equation (16) has lost the source, but note
that initial condition (14) has information about the original source in the circuit. If there
is no initial energy storage to drive the \dead" circuit, the excitation of a nonzero response
has to come from an explicit input. And our input (1) is proportional to the constant V0that explicitly appears in the nonzero (\nonhomogeneous") initial condition (14).
Homogeneous constant-coe±cient linear ordinary di®erential equations (LODE's) are par-
ticularly easy to solve, since the solutions are exponentials, say of the form i(t) = exp(rt)
where two such linearly independent forms (i.e. two di®erent r values), are required in our
case of a second-order equation. Before proceeding, it is expedient to condense the notation
and in fact, combine the element values R, L, and C by in a manner that anticipates the
solution. Of course, no one could truly do this in advance if they had not already solved
the problem (or read someone else's solution). But that's not the point: We choose the
best path because it draws attention to what is important and because it is e±cient and
48
because it works. Firstly, normalize (16) so that the coe±cient of the highest derivative is
unityd2i(t)
dt2+R
L
di(t)
dt+
1
LCi(t) = 0
and then introduce constants
® =R
2Land !20 =
1
LC(17)
to writed2i(t)
dt2+ 2®
di(t)
dt+ !20i(t) = 0: (18)
Insertion of i(t) = exp(rt) into (18) gives£r2 + 2®r + !20
¤ert = 0;
and since the exponential cannot equal zero, we arrive at the characteristic polynomial for
the two roots
r2 + 2®r + !20 = 0: (19)
The quadratic formula provides the two roots immediately
r1;2 = ¡®§q®2 ¡ !20 : (20)
Two distinct possibilities exist: The case of an over-damped circuit having ® > !0 and an
under-damped circuit when ® < !0. The case when ® = !0 is called critically-damped, and
can be obtained from either of the other two cases in the limit as ® ! !0. Or, it can be
constructed directly.
Under-damped circuit. If ® < !0, introduce yet another parameter (constant)
¯ = +
q!20 ¡ ®2 (21)
so that the two roots (20) are a complex-conjugate pair
r1;2 = ¡®§ j¯: (22)
The solution to the homogeneous di®erential equation (18) is a linear combination
i(t) = K1er1t +K2e
r2t = K1e¡(®¡j¯)t +K2e
¡(®+j¯)t = e¡®t£K1e
¡j¯t +K2e+j¯t
¤= e¡®t [A cos(¯t) +B sin(¯t)] : (23)
It is critically important to recognize that a linear combination of exp(§jx) is equivalentto a (di®erent) linear combination of cosx and sinx. Although either form is ¯ne, the
49
trig form is better here because it facilitates the application of initial conditions. Initial
condition (13) gives
i(0) = A = 0 (24)
immediately! The derivative of the remaining form
i(t) = Be¡®t sin(¯t) (25)
is
i0(t) = Be¡®t£¯ cos(¯t)¡ ® sin(¯t)¤ (26)
which has initial value
i0(0) = ¯B: (27)
Initial condition (14) is now
¯B =V0
L(28)
and the complete solution to the initial value problem is
i(t) =V0
¯Le¡®t sin(¯t) (t > 0)
or, with attention drawn to its \turned-on" nature,
i(t) =V0
¯Le¡®t sin(¯t)u(t): (29)
Note that the dimensions of ® and ¯ are both (s¡1) and that the dimensions of the product¯L are (H/s= −), so that the current is in (A). The signal that is regarded as the output
in the linear system diagram of Fig. 2 is the voltage vR(t) = Ri(t), so that when the input
vg(t) is regarded as
x(t) = V0u(t); (30)
the output is
y(t) =RV0
¯Le¡®t sin(¯t)u(t): (31)
If we normalize both signals by the amplitude V0, then the nondimensional input/output
signals are
x(t) = u(t) and y(t) = s(t) =R
¯Le¡®t sin(¯t)u(t): (32)
The response to the Heaviside unit step function is called the step response, and is denoted
here as s(t).
The derivative of the convolution
z(t) = f(t)~ g(t) =1Z
¡1f(¿)g(t¡ ¿) d¿ =
1Z¡1
g(¿)f(t¡ ¿) d¿ (33)
50
is
z0(t) = f 0(t)~ g(t) = f(t)~ g0(t): (34)
Therefore, the system impulse response is the derivative of its step response
h(t) = s0(t): (35)
The impulse response of the RLC circuit of Fig. 1 is therefore
h(t) =R
Le¡®t
·cos(¯t)¡ ®
¯sin(¯t)
¸: (36)
Note that the units of h(t) are (s¡1), as are the units of ±(t).
0 2 4 6 8 10 12 14 16 18 20¡0:75
¡0:50
¡0:25
0.00
0.25
0.50
0.75
1.00
¯t
¯L
Rs(t)
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..........................................................................................................................................................................................................................................................................................................................................
Fig 2. Normalized step response of underdamped RLC circuit.
Sold curve: ®=¯ = 0:1; Dashed curve: ®=¯ = 0:5
51
Over-damped circuit. If ® > !0, then both roots of the characteristic quadratic poly-
nomial (19) are real, so with
r1 = ¡®1 with ®1 = ®¡q®2 ¡ !20 (37)
and
r2 = ¡®2 with ®2 = ®+
q®2 ¡ !20; (38)
we see that 0 < ®1 < ®2 and the solution to the homogeneous di®erential equation (16)
can be written as the linear combination
i(t) = k1e¡®1t + k2e¡®2t (t > 0): (39)
Initial condition (13) is i(0+) = 0 so that k2 = ¡k1 and therefore
i(t) = k1£e¡®1t ¡ e¡®2t¤: (40)
The required derivative is
i0(t) = k1£®2e
¡®2t ¡ ®1e¡®1t¤
(41)
and at time t = 0+ this must be
i0(0+) = k1£®2 ¡ ®1
¤=V0
L(42)
and therefore
k1 =V0
(®2 ¡ ®1)L: (43)
The solution for the current is now
i(t) =V0
(®2 ¡ ®1)L£e¡®1t ¡ e¡®2t¤u(t) (44)
and the nondimensional step response is
s(t) =R
(®2 ¡ ®1)L£e¡®1t ¡ e¡®2t¤u(t): (45)
52
Homework. Given that a (scaled) rectangular pulse
x(t) =1
T¦
μt
T
¶behaves as the Dirac-delta function in the limit as T ! 0, how small does T have to be so
that the response y(t) of a system to the pulse x(t) is a good approximation to the true
system impulse response h(t) ? Consider the underdamped RLC circuit with parameters ®
and ¯. (Note that specifying ® and ¯ is equivalent to, and in fact more illuminating than,
specifying the circuit element values R, L, and C.) Use MATLAB to compare graphs of
h(t) with y(t) for several values of the pulse-width T . The important parameter(s) of the
problem should be nondimensional.
Homework Exercises. Find the impulse response of these four circuits.
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....................................................................................................................................................................................................
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.......
......
.......
......
............................................................................................................................................................................
²
²R
L
+
¡
±(t)
+
¡
hc(t)
............................
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..
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.......
.......
.......
.......
.......
..
............................................
............................................
²
²R
C
+
¡
±(t)
+
¡
ha(t)
............................
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.......
......
.......
......
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.......
.......
......
.......
...
²
²L
R
+
¡
±(t)
+
¡
hd(t)
............................
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.......
.......
.......
.......
.......
.......
..
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.......
.......
.......
.......
..
...............................................................................................................................................................
.......
.......
......
.......
........................................................................
........................................................................
..........................................................
.......
.......
.......
.......
²
²C
R
+
¡
±(t)
+
¡
hb(t)
What is the relationship between ha(t) and hb(t) ? Between ha(t) and hd(t) ?
Homework. Find the step and impulse responses of the series RLC circuit that is critically
damped: when ® = !0 and the characteristic quadratic polynomaial has a single, repeated
root.
53
Evaluation of an Important Integral via the Laplace Transform
¡0:4¡0:20.0
0.2
0.4
0.6
0.8
1.0
sin(x)
x
¡6¼ ¡4¼ ¡2¼ 0 6¼4¼2¼
x
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sinc(x) , sin(x)
x=1
x
1Xn=0
(¡1)n x2n+1
(2n+ 1)!=1
x
·x¡ x
3
3!+x5
5!¡ : : :
¸=
·1¡ x
2
3!+x4
5!¡ : : :
¸(1)
sinc(0) = limx!0
sin(x)
x= 1 (2)
sinc(¡x) = sinc(x) is even (3)
We want to show that 1Z¡1
sin(x)
xdx = ¼: (4)
Since sinc(x) is even, observe that
1Z¡1
sin(x)
xdx = 2
1Z0
sin(x)
xdx: (5)
54
Our approach is to ¯nd the Laplace transform of the function
Si(t) ,tZ0
sin(x)
xdx; (6)
called the \sine integral." Our desired integral (half of it!) is the ¯nal value of Si(t), that
is Si(1). Perhaps it is helpful to write Si(t) using ¿ as the dummy variable of integration
Si(t) ,tZ0
sin(¿)
¿d¿: (7)
We will use the following Laplace transform properties. If L[f(t)] = F (s), then
L·f(t)
t
¸=
1Zs
F (¸) d¸ (8)
L24 tZ0
f(¿) d¿
35 = F (s)
s(9)
limt!1 f(t) = lim
s!0sF (s) ¯nal value theorem (10)
The Laplace transform of
g(t) = sin(t)u(t) (11)
is
G(s) =1
s2 + 1: (12)
The Laplace transform of
h(t) =g(t)
t= sinc(t)u(t) (13)
is then
H(s) =
1Zs
d¸
¸2 + 1: (14)
This integral can be found in a table (say in your calculus book, or you can use a symbolic
mathematical tool such as MAPLE, or you can use a web-based integrator4). Although
not important for our purposes, let's do it ourselves just for fun.
Consider y = tan(x) so that
dy
dx=d
dx
sin(x)
cos(x)= 1 + tan2(x)
4If you know one, tell the class. If you don't know one, ¯nd one!
55
dy
1 + y2= dxZ
dy
1 + y2=
Zdx = x = tan¡1(y)
1Zs
dy
1 + y2= tan¡1(1)¡ tan¡1(s) = ¼
2¡ tan¡1(s) = tan¡1(1=s)
..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... .......
.......
....................
1
s
μ
¼=2¡ μ
tan¡1(s) + tan¡1(1=s) =¼
2
Therefore the Laplace transform of h(t) = sinc(t)u(t) is
H(s) = tan¡1(1=s): (15)
The Laplace transform of
f(t) = Si(t) =
tZ0
sinc(¿) d¿ (16)
is now
F (s) =tan¡1(1=s)
s(17)
and the ¯nal value theorem gives
Si(1) = f(1) = lims!0
sF (s) = lims!0
tan¡1(1=s) =¼
2: ¥ (18)
56
Proposition (Spectral form of the Dirac-delta). This Fourier integral satis¯es the
de¯nition of the Dirac-delta (distribution or generalized function):
1
2¼
1Z¡1
ej!t d! = ±(t): (1)
Given the integral1Z
¡1
sinx
xdx = ¼; (2)
Dirichlet's discontinuous integral falls out
1Z¡1
sin®x
xdx = ¼ sgn(®) = §¼ (® ? 0): (3)
Integrate the left hand side of (1) from a to b:
bZa
1
2¼
1Z¡1
ej!t d! dt =1
2¼
1Z¡1
bZa
ej!t dt d! =1
2¼
1Z¡1
ejb! ¡ eja!j!
d!
=1
2¼
1Z¡1
·sin b!
!¡ sin a!
!
¸d! =
1
2¼[¼ sgn(b)¡ ¼ sgn(a)] =
½1; a < 0 < b
0; o.w.(4)
This is the sampling property of ±(t) ¤:
57
Fourier Transform
F (!) = F [f(t)] =1Z
¡1f(t)e¡j!t dt F() f(t) = F¡1[F (!)] = 1
2¼
1Z¡1
F (!)ej!t d!
¯̄̄̄¯̄1Z
¡1f(t)e¡j!t dt
¯̄̄̄¯̄ ·
1Z¡1
¯̄f(t)e¡j!t
¯̄dt =
1Z¡1
¯̄f(t)
¯̄dt
Therefore a su±cient (not necessary) condition for the existence of F (!) is the absolute
integrability of f(t)1Z
¡1
¯̄f(t)
¯̄dt <1:
Note that the zero frequency value of the Fourier transform (or spectrum)
F (0) =
1Z¡1
f(t) dt
is the DC or \average value" of the signal f(t).
We can easily demonstrate the legitimacy of the Fourier transform pair by taking the
inverse transform of the forward transform
F¡1nF£f(t)¤o = 1
2¼
1Z¡1
24 1Z¡1
f(¿)e¡j!¿d¿
35| {z }
F (!)
ej!td!
=
1Z¡1
f(¿)1
2¼
1Z¡1
ej!(t¡¿)d!
| {z }±(¿¡t)
d¿
=
1Z¡1
f(¿)±(¿ ¡ t) d¿ = f(t):
58
Fourier Transform Exercises
Find an expression for the Fourier transform of the following time-domain signals, and
sketch BOTH f(t) and F (!):
0. f(t) = e¡atu(t) causal, decaying exponential (a > 0)
1. f(t) = A¦(t=T ) rectangular pulse
2. f(t) = B¤(t=T ) triangular pulse
3. f(t) = C constant
4. f(t) = ±(t¡ t0) delayed Dirac impulse5. f(t) = Ae¡®t
2
Gaussian signal \and it is a trick"
6. f(t) = e¡®jtj
7. f(t) =
½cos
sin
¾(!0t)
8. f(t) = u(t) Heaviside unit-step function (why is this one so hard?)
hint: write u(t) = lima!0
e¡atu(t) and note lima!0
a
a2 + !2= ¼±(!)
9. f(t) = sgn(t) signum (or signature or sign) function
10. f(t) = m(t) cos(!0t) AM modulated signal
i.e. !0=2¼ = 1250 (kHz) and m(t) = \Your Cheatin' Heart"
11. f(t) = sinc(®t)
12. f(t) = ¦(t=T )~ sgn(t)13. f(t) = t¦(t=T )
14. f(t) = D combT (t) = D1P
n=¡1±(t¡ nT ) Woodward's comb function
15. f(t) =1P
k=¡1g(t¡ kT ) periodic replication where F [g(t)] = G(!)
16. f(t) = A1P
k=¡1¦[(t¡ kT )=¿ ] rectangular pulse train
59
Fourier Series
f(t) = a0 +
1Xn=1
[an cosn!0t+ bn sinn!0t] =
1Xn=¡1
cnejn!0t (!0 = 2¼=T )
a0 =1
T
t0+TZt0
f(t) dt
½anbn
¾=2
T
t0+TZt0
f(t)
½cos
sin
¾(n!0t) dt
cn =1
T
t0+TZt0
f(t)e¡jn!0t dt
Fourier Transform
F (!) = F [f(t)] =1Z
¡1f(t)e¡j!t dt F() f(t) = F¡1[F (!)] = 1
2¼
1Z¡1
F (!)ej!t d!
Properties Signal Transform
1. superposition c1f1(t) + c2f2(t) c1F1(!) + c2F2(!)
2. time delay f(t¡ t0) e¡j!t0F (!)
3. time scale f(®t)1
j®jF³!®
´4. duality F (t) 2¼f(¡!)5. frequency translation f(t)ej!0t F (! ¡ !0)6. convolution f(t)~ g(t) F (!)G(!)
7. multiplication f(t)g(t)1
2¼F (!)~G(!)
8. time di®erentiationdf(t)
dtj!F (!)
9. frequency di®erentiation ¡jtf(t) d
d!F (!)
10. integrationtR
¡1f(¿) d¿ ¼F (0)±(!) +
1
j!F (!)
60
1Z¡1
jf(t)j2 dt = 1
2¼
1Z¡1
jF (!)j2 d!
u(t)F() 1
j!+ ¼±(!)
¦(t=¿)F() ¿
sin(!¿=2)
!¿=2= ¿ sinc(!¿=2)
sinc(¯t)F() ¼
¯¦(!=2¯)
combT (t)F() 2¼
Tcomb2¼=T (!)
Z-Transform
Zfx[n]g = X(z) =1X
n=¡1x[n]z¡n
Properties Signal Transform
1. superposition c1x1[n] + c2x2[n] c1X1(z) + c2X2(z)
2. time delay x[n¡m] z¡mX(z)
3. modulation x[n]³n X(z=³)
4. convolution x[n]~ y[n] X(z)Y (z)
5. frequency di®erentiation nx[n] ¡ z ddzX(z)
6. accumulationnP
k=¡1x[k] X(z)
1
1¡ z¡1
61
Lff(t)g = F (s) =1Z0
f(t)e¡st dt is the transform pair f(t)L, F (s)
property t-domain s-domain
linearity ®f(t) + ¯g(t) ®F (s) + ¯G(s)
time delay f(t¡ t0)u(t¡ t0) e¡st0F (s) (t0 ¸ 0)time scaling f(®t) 1
®F (s=®) (® > 0)
exponential modulation e¡®t f(t) F (s+ ®)
time di®erentiation df(t)=dt sF (s)¡ f(0)reprise d2f(t)=dt2 s2F (s)¡ sf(0)¡ f 0(0)time integration
R t0f(¿) d¿ F (s)=s
time integrationR t¡1 f(¿) d¿ F (s)=s+ 1
s
R 0¡1 f(t) dt
s-domain di®erentiation tf(t) ¡dF (s)=dsdivision by t f(t)=t
R1sF (¸) d¸
t-domain convolution f(t)~ g(t) F (s)G(s)
Initial value theorem limt!0
f(t) = lims!1 sF (s)
Final value theorem limt!1 f(t) = lim
s!0sF (s)
Short Table of Laplace Transforms (s = ¾ + j!)
±(t) 1 (8s)u(t)
1
s(¾ > 0)
e¡atu(t)1
s+ a(¾ > ¡Re[a])
te¡atu(t)1
(s+ a)2(¾ > ¡Re[a])
tnu(t)n!
sn+1(¾ > 0)
cos(®t)u(t)s
s2 + ®2(¾ > 0)
sin(®t)u(t)®
s2 + ®2(¾ > 0)
62
Fourier Transform of the Gaussian
If f(t) = exp[¡®t2] then F (!) =
r¼
®exp[¡!2=4®].
This one is a bit involved to show, especially without complex variable theory. But it's
so pervasive in signal analysis and communications that we use it almost daily, so here is
a classical derivation that uses nothing more than integration-by-parts and a simple ¯rst
order di®erential equation.
A preliminary integral
I =
1Z¡1
e¡x2
dx =p¼
is required. The trick5 to derive this is to consider
I2 =
1Z¡1
e¡x2
dx
1Z¡1
e¡y2
dy =
1Z¡1
1Z¡1
e¡(x2+y2)dx dy:
Conversion to polar coordinates via
x = r cosÁ and y = r sinÁ
gives
I2 =
2¼Z0
1Z0
e¡r2
r dr dÁ = (2¼)
μ1
2
¶= ¼ and so I =
p¼:
Our desired Fourier integral is
F (!) =
1Z¡1
e¡®t2
e¡j!tdt = 2Z 1
0
e¡®t2
cos!t dt
since f(¡t) = f(t) is even. Di®erentiation w.r.t. ! gives
dF (!)
d!= ¡2
1Z0
te¡®t2
sin!t dt:
Integration by parts Zudv = uv ¡
Zv du
5\And it is a trick!"
63
with
u = sin!t du = ! cos!t dt
dv = ¡2te¡®t2 v =e¡®t
2
®
givesdF (!)
d!=1
®e¡®t
2
sin!t
¯̄̄̄1t=0
¡ !®
Z 1
0
e¡®t2
cos!t dt
ordF (!)
d!= ¡ !
2®F (!)
dF (!)
d!F (!)
= ¡ !
2®
d
d!lnF (!) = ¡ !
2®
lnF (!) = ¡!2
4®+ C
F (!) = eCe¡!2=4®
eC = F (0) =
1Z¡1
e¡®t2
dt =
r¼
®
F (!) =
r¼
®e¡!
2=4®
solid curves: ® = 1, dashed curves: ® = 2
¡3 ¡2 ¡1 0 1 2 3
t
0.0
0.2
0.4
0.6
0.8
1.0
f(t)
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..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
¡8 ¡6 ¡4 ¡2 0 2 4 6 8
!
0.0
0.5
1.0
1.5
2.0
F (!)
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64
Fourier Transform of the Heaviside Unit Step Function
F (!) = F [f(t)] =1Z
¡1f(t)e¡j!t dt F() f(t) = F¡1[F (!)] = 1
2¼
1Z¡1
F (!)ej!t d!
Recall that a su±cient (but not necessary) condition for the Fourier transform F (!) of a
signal f(t) to exist is that the signal be absolutely integrable, that is
1Z¡1
jf(t)j dt <1:
The Heaviside unit step function is not absolutely integrable
1Z¡1
ju(t)j dt =1Z0
dt ¥1
and therefore we might anticipate that its Fourier transform will be both di±cult to eval-
uate and exhibit singular behavior. Recall or note that the periodic signal cos(!0t) is not
absolutely integrable, and its spectrum consists of Dirac-deltas
F [cos(!0t)] =1Z
¡1
ej!0t + e¡j!0t
2e¡j!tdt =
1
2
24 1Z¡1
e¡j(!¡!0)tdt+
1Z¡1
e¡j(!+!0)tdt
35= ¼
£±(! ¡ !0) + ±(! + !0)
¤:
Two preliminary formulae or ideas:
First: lim®!0
®=¼
®2 + !2= ±(!) (1)
This is the Cauchy or Lorentz Dirac-delta sequence (that we already studied back on page
7). Clearly
lim®!0
®=¼
®2 + !2= 0 if !6= 0.
The integral
®
¼
1Z¡1
d!
®2 + !2=1
¼
1Z¡1
dx
1 + x2=1
¼tan¡1(x)
¯̄̄̄1x=¡1
=¼=2¡ (¡¼=2)
¼= 1
gives unity independently of ®, so the truth of (1) is veri¯ed.
65
Second: lim®!0
¡j!®2 + !2
=1
j!(2)
Rewrite it as
lim®!0
1
j![1 + (®=!)2]=1
j!
which is clearly true if ! 6= 0. The only other possibility is that our limit also has a
Dirac-delta at the origin of strength or weight K
lim®!0
1
j![1 + (®=!)2]=1
j!+K±(!):
Integrate both sides of this from ! = ¡² to ! = +²
1
j
²Z¡²
d!
![1 + (®=!)2]=
1
j
²Z¡²
d!
!+ K
²Z¡²±(!) d!
and observe that the integral on the left side and the ¯rst integral on the right side vanish
because of the odd symmetry of their integrands. Therefore the only possibility is K = 0
and the truth of (2) is established.
One of our ¯rst examples (or exercises) of Fourier integral evaluations was the causal,
decaying exponential
F£e¡®tu(t)¤ = 1
®+ j!(® > 0):
The condition that ® > 0 is necessary to ensure convergence of the integral, by making
the time-domain function decay su±ciently fast as t!1. Although the restriction ® > 0enabled us to evaluate the Fourier transform of the decaying exponential, let's try to get
our desired Fourier transform of the plain Heaviside unit-step by taking the limit of the
above result as ®! 0. That is
F [u(t)] = lim®!0
F£e¡®tu(t)¤ = lim®!0
1
®+ j!:
Rewrite1
®+ j!=
®¡ j!(®+ j!)(®¡ j!) =
®
®2 + !2+
¡j!®2 + !2
so that
F [u(t)] = lim®!0
®
®2 + !2+ lim®!0
¡j!®2 + !2
= ¼±(!) +1
j!(F)
from (1) and (2) above.
66
Since
2u(t) = 1 + sgn(t)
we now have the Fourier transform of the signum function as
F [sgn(t)] = 2F [u(t)]¡F [1] = 2·¼±(!) +
1
j!
¸¡ 2¼±(!) = 2
j!:
With
F [u(t)] = ¼±(!) + 1
j!
then by the time-di®erentiation property of the Fourier transform
F·df(t)
dt
¸= j!F [f(t)]
where
±(t) =du(t)
dt;
we have the correct transform
F [±(t)] = j!·¼±(!) +
1
j!
¸= 1:
Using the convolution property of the Fourier transform
F [f(t)~ g(t)] = F (!)G(!)
and the integral identitytZ
¡1f(¿) d¿ = f(t)~ u(t);
we have the Fourier transform of the integral as
F24 tZ¡1
f(¿) d¿
35 = F (!) ·¼±(!) + 1
j!
¸= ¼F (0)±(!) +
F (!)
j!:
67
Duality Property of the Fourier Transform
Recall our notation for the Fourier transform
F (!) = F [f(t)] =1Z
¡1f(t)e¡j!t dt F() f(t) = F¡1[F (!)] = 1
2¼
1Z¡1
F (!)ej!t d!:
For our duality property, we want the Fourier transform of some signal F (t):
F [F (t)] =1Z
¡1F (t)e¡j!t dt: (1)
The fact that our time domain signal is a \big" F of t means that we interpret our big F
as related to some \little" f via a direct Fourier transform
F (x) =
1Z¡1
f(»)e¡jx»d» (2)
and therefore the particular little f that must have spawned our big F must be
f(») =1
2¼
1Z¡1
F (x)ejx»dx: (3)
According to (2), we can also write
F (t) =
1Z¡1
f(»)e¡jt»d» (4)
and insertion of this into (1) yields
F [F (t)] =1Z
¡1
8<:1Z
¡1f(»)e¡jt»d»
9=; e¡j!tdt: (5)
Reverse the orders of the » and t integrations
F [F (t)] =1Z
¡1f(»)
24 1Z¡1
e¡j(»+!)tdt
35| {z }recognize as 2¼±(»+!)
d» (6)
68
and so
F [F (t)] = 2¼1Z
¡1f(»)±(» + !) d» = 2¼f(¡!): (7)
Our desired duality property of the Fourier transform is here F [F (t)] = 2¼f(¡!) wheref(t)
F, F (!) is the notation for the original Fourier transform pair F [f(t)] = F (!).
Example. Since we already know the Fourier transform pair
¦
μt
T
¶F, T sinc(!T=2); (8)
it follows from duality that
F [T sinc(tT=2)] = 2¼¦³¡!T
´(9)
or
F [sinc(T2 t)] =2¼
T¦³!T
´(10)
by the linearity of the Fourier transform operator, and invoking the even symmetry of the
pulse function ¦. Now T is just a remnant parameter from the original pair (8), so let's
introduce a new (better!) constant ® = T=2 such that the above is
F [sinc(®t)] = ¼
®¦³ !2®
´: (11)
Add this entry in your table of Fourier transforms!
Another View. Consider a function ª(x) related to a given function Á(u) through the
direct Fourier transform operator
ª(x) =
1Z¡1
Á(u)e¡juxdu: (12)
Consider a second function Ã(x) related to our given function Á(u) through the inverse
Fourier transform operator
Ã(x) =1
2¼
1Z¡1
Á(u)ejuxdu: (13)
Observe that ª(x) = 2¼Ã(¡x). If we know the forward transform, then we also know thereverse transform of any function, by inspection.
69
Half-Power Bandwidth or Pulse width
..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
x
f1(x)
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x
f2(x)
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x
f3(x)
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Consider the three functions (or signals) graphed above. The function f1(x) has a clear
maximum and a \function width" or \pulse width" about the maximum where most of its
energy is concentrated. The same observation applies to the signal f2(x), but its \pulse
width" is perhaps not as obvious. The oscillatory and somewhat wild signal f3(x) possesses
no obvious maximum that we see in its graph and it is unclear how to de¯ne any kind of
a pulse width for it.
Often our time-domain signals are one of the \linear" signals in an electric circuit, such as
a voltage v(t) or a current i(t). In a mechanical system, the displacements x(t) or velocities
v(t) are examples of one of the so-called \linear" signals that we might be interested in.
The instantaneous power associated with one of the linear signals f(t) is proportional to
jf(t)j2. If the signal f(t) does possess a clear maximum and most of its energy or power6
6This is an example of hazy (perhaps lazy, too) language. Power p(t) (in Watts) is not equal to energyE(t) (in Joules) but we are approximately thinking that p(t) / E(t) / jf(t)j2.
70
is concentrated around the maximum (such as the functions f1 and f2 above), then we
de¯ne the half-power pulse width as
T1=2 = t2 ¡ t1where
jf(t1;2)j2jfmaxj2 =
1
2or
jf(t1;2)jjfmaxj =
1p2:
When the signal magnitude is given or graphed in units of decibels (dB), the half-power
points t1;2 are also called the \3 (db)" points since
10 log10(1=2) = 20 log10(1=p2) ¼ ¡3:
The half-power pulse width is also then called the \3 (dB) pulse width." This measure of
\signal duration" applies equally well in the frequency domain, where the term bandwidth
is commonly used instead of pulse width. The symbol for bandwidth is typically BW in
communications work, but a simple B1=2 seems ¯ne.
Example: Gaussian pulse. Note in this case that jf(t)j = f(t) and jF (!)j = F (!).
This will not be true in general.
......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
t
f(t) = exp(¡®t2)
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t1 t2
jf jmaxjf jmaxp
2
T1=2............................................... .............. .............................................................
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!
F (!) =
r¼
®exp(¡!2=4®)
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!1 !2
jF jmaxjF jmaxp
2
B1=2.......................................................................... ............................................................ ..............
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exp(¡®t22)1
=1p2
exp(¡!22=4®)1
=1p2
t2 =
rln 2
2®!2 =
p2® ln 2
T1=2 = t2 ¡ t1 = 2t2 =r2 ln 2
®B1=2 = !2 ¡ !1 = 2!2 =
p8® ln 2
The \BT product" B1=2T1=2 = 4 ln 2 is independent of ®.
71
Example: The sinc function. Examine the two graphs of j sinc(x)j = j sin(x)=xj, theone on the top using a linear ordinate scale and the one on the bottom using a logarithmic
(dB) ordinate scale. The abscissa is x, which can serve as a scaled frequency ! for the
spectrum of a time-domain rectangular pulse.
¡10 ¡8 ¡6 ¡4 ¡2 0 2 4 6 8 100.0
0.2
0.4
0.6
0.8
1.0
x
jsinc(x)j
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..
1=p2
x1
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¡10 ¡8 ¡6 ¡4 ¡2 0 2 4 6 8 10¡30
¡25
¡20
¡15
¡10
¡5
0
x
jsinc(x)j(dB)
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x1
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sin(x1)
x1=
1p2
=) x1 =p2 sin(x1) =) x1 = 1:391557378251510
If f(t) = ¦(t=T ) is the standard rectangular pulse of pulse width T , then since the spectrum
is F (!) = T sinc(!T=2), we see that the half-power bandwidth is
B1=2 = 2!1 ¼ 2£ 2£ 1:39T
¼ 5:56
T:
72
Homework
1. Find the pulse width, bandwidth, and BT product of these two signals:
f(t) = exp(¡®jtj) and g(t) = exp(¡®t)u(t):Graph jf(t)j and its amplitude spectrum jF (!)j, and also jg(t)j and jG(!)j. Use at leasttwo values for the constant ®.
2. Compare the half-power pulse width T1=2 described above with the (statistical class)
of pulse width Dt that is used in the mathematical analysis of the Uncertainty Principle
that appears on page 74. If the signal is normalized to have unit energy
1Z¡1
jf(t)j2 dt = 1
then the (square of) pulse width is
D2t =
1Z¡1
t2jf(t)j2 dt:
In other words, simply take Dt as de¯ned here, to be yet another measure of a signal's
pulse width. For a speci¯c signal, use the Gaussian pulse.
73
Uncertainty Principle for the Fourier Integral
To simplify our notation, assume that the signals under consideration are normalized to
have unit energy:1Z
¡1jf(t)j2 dt = 1
2¼
1Z¡1
jF (!)j2 d! = 1:
If we further assume that both f(t) and its Fourier transform F (!) are centered about
t = 0 and ! = 0, respectively, then it is reasonable to de¯ne the t and !-domain durations
by
D2t =
1Z¡1
t2jf(t)j2 dt and D2! =
1Z¡1
!2jF (!)j2 d!:
Our so-called \uncertainty principle" states:
If limt!§1
pjtjf(t) = 0; Then DtD! ¸
r¼
2:
prf: Equality holds in Schwarz's inequality¯̄̄̄¯̄bZa
g1g2 dt
¯̄̄̄¯̄2
·bZa
jg1j2 dtbZa
jg2j2 dt
only if g2(t) = kg1(t). Let g1(t) = tf(t) and g2(t) = f0(t):¯̄̄̄
¯̄1Z
¡1tf(t)
df(t)
dtdt
¯̄̄̄¯̄2
·1Z
¡1jtf(t)j2 dt
1Z¡1
¯̄̄̄df(t)
dt
¯̄̄̄2dt:
Integrate by parts
1Z¡1
tfdf
dtdt =
tf2(t)
2
¯̄̄̄1¡1| {z }
=0 by above
¡12
1Z¡1
jf j2 dt| {z }=1 by above
= ¡12:
From the Fourier transform pair f 0(t)() j!F (!) and Parseval's theorem,
1Z¡1
¯̄̄̄df
dt
¯̄̄̄2dt =
1
2¼
1Z¡1
j!F (!)j2 d!
and so Schwarz's inequality gives
1
4·
1Z¡1
jtf(t)j2 dt 12¼
1Z¡1
j!F (!)j2 d!
74
or¼
2· D2
tD2!: ¥
Equality holds ifdf(t)
dt= ktf(t):
The solution of this di®erential equation is
f(t) = Cekt2=2:
Let k = ¡2®; this is the inescapable Gaussian function
f(t) =
μ2®
¼
¶1=4e¡®t
2 () F (!) =
μ2®
¼
¶1=4r¼
®e¡!
2=4® =
μ2¼
®
¶1=4e¡!
2=4®:
75
Theorem (Riemann-Lebesgue Lemma). Let f(x) 2 L1(¡1;1). That is, f(x) isabsolutely integrable
1Z¡1
jf(x)j dx <1:
Then the integrals1Z
¡1f(x) cos¸xdx;
1Z¡1
f(x) sin¸xdx;
tend to zero as ¸!1.
Proof. Consider the cosine integral. Let ² be a given positive number. Then we can choose
X so large that1ZX
jf(x)j dx < ²;¡XZ¡1
jf(x)j dx < ²:
Hence ¯̄̄̄¯̄1ZX
f(x) cos¸xdx
¯̄̄̄¯̄ < ²;
¯̄̄̄¯̄¡XZ¡1
f(x) cos¸xdx
¯̄̄̄¯̄ < ²
for all values of ¸. Next, we can de¯ne a function Á(x), absolutely continuous in the
interval (¡X;X), such thatXZ
¡Xjf(x)¡ Á(x)j dx < ²:
Then ¯̄̄̄¯̄XZ
¡Xff(x)¡ Á(x)g cos¸xdx
¯̄̄̄¯̄ < ²
for all values of ¸. Finally
XZ¡X
Á(x) cos¸xdx =Á(X) sin¸X
¸+Á(¡X) sin¸X
¸¡ 1
¸
XZ¡X
Á0(x) sin¸xdx;
and (for a ¯xed X) we can choose ¸0 so large that the modulus of this is less than ² for
¸ > ¸0. Then ¯̄̄̄¯̄1Z
¡1f(x) cos¸xdx
¯̄̄̄¯̄ < 4² (¸ > ¸0):
This proves the theorem for the cosine integral; a similar proof applies to the sine integral.
¥
76
The above is the working-class proof of the Riemann-Lebesgue lemma, as given by E.C.
Titchmarsh in Introduction to the Theory of Fourier Integrals, 1937. Perhaps you will
prefer this one, from R.R. Goldberg, Fourier Transforms, Cambridge University Press,
1962.
Definition (the Class Lp). Suppose 1 · p < 1. The function f on (¡1;1) is saidto be of class Lp (written f 2 Lp) if
1R¡1
jf(x)jp dx <1. If f 2 Lp then kfkp is de¯ned tobe 0@ 1Z
¡1jf(x)jp dx
1A1=p
:
The symbol kfkp is read as the Lp norm of f .
Theorem (Riemann-Lebesgue). If f 2 L1 then
lim!!§1F (!) = lim
!!§1
1Z¡1
e¡j!tf(t) dt = 0:
Proof. Since
F (!) =
1Z¡1
e¡j!tf(t) dt; (1)
then
¡F (!) =1Z
¡1e¡j![t+(¼=!)]f(t) dt =
1Z¡1
e¡j!tf(t¡ ¼=!) dt: (2)
Subtracting (2) from (1) we obtain
2F (!) =
1Z¡1
e¡j!t [f(t)¡ f(t¡ ¼=!)] dt:
Hence
2jF (!)j ·1Z
¡1jf(t)¡ f(t¡ ¼=!)j dt: (3)
But since f 2 L1,
lim!!§1
1Z¡1
jf(t)¡ f(t¡ ¼=!)j dt = 0 (4)
by the continuity-in-the-mean theorem. The theorem follows from (3) and (4). ¥
77
Riemann-Lebesgue Lemma
If f 0(t) is absolutely integrable, then F (!)! 0 as ! ! §1.
PROOF: Integrate by parts
F (!) =
1Z¡1
f(t)e¡j!t dt =f(t)e¡j!t
¡j!¯̄̄̄1t=¡1
+1
j!
1Z¡1
f 0(t)e¡j!tdt
If f(§1) is ¯nite, then the boundary terms go to zero as ! ! §1. The integral is theFourier transform of the derivative of f(t)
1Z¡1
f 0(t)e¡j!tdt = F [f 0(t)]
and this Fourier integral exists (has ¯nite magnitude jF [f 0(t)] j < 1) if the derivative isabsolutely integrable
1Z¡1
jf 0(t)j dt · A <1:
If jF [f 0(t)] j <1, then the integral term
F [f 0(t)]j!
¡¡¡¡¡!!!§1
0:
Exercise. Examine the high frequency limiting behavior (! ! §1) of the spectra ofthese signal pairs and interpret in terms of the Riemann-Lebesgue Lemma:
u(t)F() 1
j!+ ¼±(!)
¦(t=T )F() T sinc(!T=2)
sinc(¯t)F() ¼
¯¦(!=2¯)
e¡®tu(t) F() 1
®+ j!
±(t)F() 1
1
t
F() ¼
jsgn(!)
78
Convergence of the Fourier Integral Representation
F (!) = F [f(t)] =1Z
¡1f(t)e¡j!t dt F() f(t) = F¡1[F (!)] = 1
2¼
1Z¡1
F (!)ej!t d!
Assume f 0(t) is absolutely integrable so that the Riemann-Lebesgue lemma guarantees
lim!!1F (!) = 0:
Denote the inverse Fourier transform of the Fourier transform as
ef(t) = lim−!1
1
2¼
−Z¡−
1Z¡1
f(¿)e¡j!¿d¿ ej!t d!
= lim−!1
1Z¡1
f(¿)1
2¼
−Z¡−
ej!(t¡¿)d! d¿
= lim−!1
1Z¡1
f(¿)sin[−(¿ ¡ t)]¼(¿ ¡ t) d¿
= lim−!1
1Z¡1
f(x+ t)sin−x
¼xdx:
If f is continuous at t, then clearly ef(t) = f(t) if we accept the Dirac-delta representationlim−!1
sin−x
¼x= ±(x):
If f su®ers a jump discontinuity at t, in that
limx!0¡
f(t+ x) = f(t¡)6= f(t+) = limx!0+
f(t+ x);
then it is appropriate to break the integral above at x = 0 so that
ef(t) = lim−!1
0Z¡1
f(x+ t)sin−x
¼xdx+ lim
−!1
1Z0
f(x+ t)sin−x
¼xdx
= lim−!1
0Z¡1
f(x+ t)¡ f(t¡)¼x
sin−xdx+ lim−!1
f(t¡)
0Z¡1
sin−x
¼xdx
+ lim−!1
1Z0
f(x+ t)¡ f(t+)¼x
sin−xdx+ lim−!1
f(t+)
1Z0
sin−x
¼xdx
79
The functionf(x+ t)¡ f(t¡)
¼x
is continuous at x = 0¡, and is well-behaved everywhere else by our initial assumption.(We are assuming that f is continuous everywhere except at the particular t of interest. If
f actually has a ¯nite number of such jump discontinuities, then each one can be handled
in a similar fashion.) By the Riemann-Lebesgue lemma, the ¯rst term on the right-hand
side of the above integral vanishes as −!1. The same argument applies to the function
f(x+ t)¡ f(t+)¼x
and its Fourier integral. From the famous integral
1Z¡1
sinu
udu = ¼
and its variation 1Z¡1
sin¯u
udu = ¼ sgn¯;
then for − > 0 we have0Z
¡1
sin−x
¼xdx =
1Z0
sin−x
¼xdx = 1
2 :
Therefore, we have shown
F¡1 fF [f(t)]g = ef(t) = 12
£f(t¡) + f(t+)
¤:
80
System Transfer Function
....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................................................................................... ............. ...................................................................................... .............x(t) y(t)h(t)
The zero-state response (in the time domain) of our linear, time-invariant (LTI) system is
the convolution of the input signal with the system impulse response
y(t) = x(t)~ h(t): (1)
The direct Fourier transform of this equation is
Y (!) = X(!)H(!) (2)
which exhibits the important role of the system transfer function
Ffh(t)g = H(!): (3)
By transforming to the frequency domain, the nontrivial convolution operator has been
replaced by a simple multiplication! Besides algebraic simplicity, working in the frequency
domain also gives us a whole another conceptual view of the input/output dynamics of our
systems. Electrical engineers work so much in the frequency domain that we sometimes
loose touch with the original time domain!
For an arbitrary input signal x(t), the time-domain response y(t) now requires an inverse
Fourier transform
y(t) = F¡1fX(!)H(!)g: (4)
However, the transfer function H(!) immediately gives the response of a LTI system to a
time-harmonic signal. The simplest time-harmonic input signal is the complex exponential
form
x(t) = ej!0t: (5)
(The only other choice would be a real-valued cosine or sine, but as we have seen repeatedly
this semester, the mathematics of signal analysis usually (always?) is much simpler in terms
of the complex exponential form. And in deriving important concepts and principles, we
always take the simpler path.) Note the use of !0 to denote some particular frequency, to
avoid confusion with the Fourier transform frequency variable plain !. The response of
our system to the excitation (5) is
y(t) = h(t)~ x(t) =1Z
¡1h(¿)x(t¡ ¿) d¿ =
1Z¡1
h(¿)ej!0(t¡¿) d¿
= ej!0t1Z
¡1h(¿)e¡j!0¿ d¿ = ej!0tH(!0): (6)
81
Observe that the response (steady-state since the excitation has been on forever) of a
LTI system to a time-harmonic signal is a time-harmonic signal of the same frequency as
the excitation. Only the amplitude and phase of the response have been modi¯ed by the
amplitude and phase of the system transfer function, evaluated (of course) at the excitation
frequency. The amplitude and phase are apparent in the polar representation
H(!) = jH(!)jej£(!): (7)
Note that if the system impulse response h(t) is a real-valued function of time, then we
can recover the separate responses to the real and imaginary parts of a complex input by
inspection:
....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................................................................................... ............. ...................................................................................... .............x(t) = xr(t) + jxi(t) y(t) = xr(t)~ h(t) + jxi(t)~ h(t)h(t)
....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................................................................................... ............. ...................................................................................... .............Refx(t)g = xr(t) yr(t) = Refy(t)gh(t) real
....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................................................................................... ............. ...................................................................................... .............Imfx(t)g = xi(t) yi(t) = Imfy(t)gh(t) real
F Exercise 1. Show that the time-harmonic response of a real LTI system to a cosine
excitation x(t) = A cos(!0t+ Á) of arbitrary amplitude A and phase Á is
y(t) = AjH(!0)j cos[!0t+ Á+£(!0)]where the amplitude and phase of the transfer function are de¯ned in (7).
Hint: write x(t) = RefAejÁej!0tg. In AC circuit analysis, we call AejÁ the complex phasor.
F Exercise 2. Specialize the above result to the case where x(t) = sin(!0t):
F Exercise 3. Now consider the periodic excitation
x(t) =
1Xn=¡1
cnejn!0t
applied to a LTI system having arbitrary (not necessarily real) impulse response h(t). The
transfer function is the generic H(!). What is the form of the response? Think about the
linearity of the system.
82
Superposition Interpretation of System Transfer Function
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............................................................................................................................................................................................................. ..............
§
H(!)
H(!)
H(!)
H(!)
H(!)
x(t) =Pn
X(!n)ej!nt ....................................................................................... ..........
....Pn
H(!n)X(!n)ej!nt = y(t)
X(!2)ej!2t H(!2)X(!2)e
j!2t
X(!1)ej!1t H(!1)X(!1)e
j!1t
X(!n)ej!nt H(!n)X(!n)e
j!nt
........................................................................................................................................................................................................................................................................................................................................
......................................................................................................... .............. ......................................................................................................... .........
.....H(!)x(t) = 12¼
1Z¡1
X(!)ej!td! 12¼
1Z¡1
H(!)X(!)ej!td! = y(t)
83
Transfer Function for Linear Electric Circuits
Example.
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............................
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....................................................................................................................................................................................................
..................................................................
.............................................................................................................................................
.......
......
.......
......
............................................................................................................................................................................
²
²R
L
+
¡
x(t)
+
¡
y(t)
Recall (look back in your notes; no one memorizes such speci¯cs) that the impulse response
of the given RL circuit is
h(t) = ±(t)¡ RLexp
μ¡RLt
¶u(t): (8)
Most of us would agree that the correct derivation of this time-domain result is a nontrivial
exercise. We ¯rst found the step response s(t) and then di®erentiated to get h(t) = s0(t).Fortunately, the Fourier transform allows us to circumvent the time-domain di®erential
equations and work with purely algebraic equations. Temporarily set the input x(t) = vg(t)
and the output y(t) = vL(t) for obvious dimensional consistency and to use signal labels
for voltages that look like voltages. At least this helps my circuit analysis.
............................
............................
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..................................................................................................................................................................................................................................................................................................
....................................................................................................................................................................................................
..................................................................
.............................................................................................................................................
.......
......
.......
......
............................................................................................................................................................................
²
²R
L
+
¡
vL(t)
..................................................................................................................................................................................................................................................................................................
.......
.......
.......
.......
.......
.......
..................................................................
+
¡vg(t)
................................................ ...................
i(t)
Kirchho®'s voltage law is
vg(t) = Ri(t) + Ldi(t)
dt(9)
and the direct Fourier transform of this is
Vg(!) = [R+ j!L]I(!): (10)
Note the natural occurrence of the familiar complex impedance. The ratio of output to
input voltage spectraVL(!)
Vg(!)=
j!L
R+ j!L=
ZL
ZR + ZL(11)
could have been written by inspection as a simple voltage divider. Our transfer function is
H(!) =Y (!)
X(!)=
j!L
R+ j!L=
j!L=R
1 + j!L=R=
j!=!b
1 + j!=!b(12)
where the constant !b = R=L (s¡1) is called the break frequency or half-power frequency
or maybe even cut-o® frequency of this simple high-pass ¯lter.
F Exercise 4. Verify that the Fourier transform of impulse response (8) gives transfer
function (12).
84
¡40
¡35
¡30
¡25
¡20
¡15
¡10
¡5
0
10¡2 10¡1 100 101 102
!=!b
jHj (dB)
.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.........................................................................................................................................................................................................................................................................................................................................
(a) Magnitude response.
10¡2 10¡1 100 101 102
!=!b
argH
0
¼=8
¼=4
3¼=8
¼=2 ............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
(b) Phase response.
Fig. 1. Frequency behavior of single-pole high-pass ¯lter.
85
The magnitude and phase response of the transfer function (12)
H(!) =j!=!b
1 + j!=!b(13)
are graphed versus a logarithmic and normalized frequency abscissa in Fig. 1. We also use
a logarithmic (unnatural or base 10) scale to express the magnitude in decibels according
to
jHj (dB) = 20 log10 jHj: (14)
Note that the low and high frequency approximations
H(!) =j!=!b
1 + j!=!b=
8>><>>:j!=!b; ! ¿ !b
j
1 + j; ! = !b
1; ! À !b
(15)
clearly show that this ¯lter passes the high frequencies unscathed. In the low frequency
regime (! ¿ !b) where this ¯lter has the greatest e®ect between the input and output,
the frequency-domain relationship is
Y (!) ¼ j!X(!)
!b: (16)
Inverse Fourier transformation of (16) to the time domain gives
y(t) ¼ 1
!b
dx(t)
dt(17)
so we see that a high-pass ¯lter acts as a di®erentiator when it is having its biggest e®ect
on the input.
F Exercise 5. Consider the simple circuit having the positions of R and L reversed
relative to the circuit above. Find its transfer function and show that this low-pass ¯lter
acts as an integrator when it is having its greatest e®ect on the input signal.
Homework Problem: Rectangular Pulse Response of Ideal Low-Pass Filter.
Consider an ideal low-pass ¯lter of bandwidth B and subsequent transfer function
H(!) = ¦³ !2B
´:
Find the transient response y(t) of this ¯lter to a rectangular pulse of pulse-width T
x(t) = ¦(t=T ):
Graphical results simplify with normalized time t=T as the independent variable, for various
values of the dimensionless parameter BT . This problem is most readily solved in terms
of a special function, the sine integral7
Si(u) =
uZ0
sin »
»d» Si(¡u) = ¡Si(u) Si(1) = ¼=2:
7M. Abramowitz and I.A. Stegun, Handbook of Mathematical Functions. National Bureau of Standards,1970, pp. 231{232, Table 5.1. (\Big Red"or \Citizen's Handbook"). MATLAB has it built-in, too!
86
Parseval's Theorem for the Fourier Integral
The energy associated with the signal f(t) is
E =1Z
¡1jf(t)j2 dt =
1Z¡1
f(t)f¤(t) dt
=
1Z¡1
24 12¼
1Z¡1
F (!)ej!td!
3524 12¼
1Z¡1
F (−)ej−td−
35¤ dt=
1Z¡1
24 12¼
1Z¡1
F (!)ej!td!
3524 12¼
1Z¡1
F ¤(−)e¡j−td−
35 dt=1
2¼
1Z¡1
F (!)
1Z¡1
F ¤(−)
8<: 1
2¼
1Z¡1
ej(!¡−)tdt
9=; d− d!
=1
2¼
1Z¡1
F (!)
1Z¡1
F ¤(−)±(−¡ !) d− d! = 1
2¼
1Z¡1
F (!)F ¤(!) d!
=1
2¼
1Z¡1
jF (!)j2 d!
F Exercise 6. Derive the generalized Parseval theorem
1Z¡1
f(t)g¤(t) dt =
F Exercise 7. Find a simple evaluation for the integrals
1Z¡1
sinx
xdx
1Z¡1
·sinx
x
¸2dx:
We met the ¯rst one back on page 54. The similarity (more than similar! ) in the numerical
values strikes me as a strange coincidence. What do you think? (You'll need the answers,
¯rst!)
87
Other Conventions for the Fourier Transform Pair
Note that our forward Fourier transform could be scaled by any constant K as long as we
divide by K in performing the inverse:
F (!) = K
1Z¡1
f(t)e¡j!tdt F() f(t) =1
2¼K
1Z¡1
F (!)ej!td!:
The common choice K = 1=p2¼ gives a pleasingly symmetric pair
F (!) =1p2¼
1Z¡1
f(t)e¡j!tdt F() f(t) =1p2¼
1Z¡1
F (!)ej!td!:
One-sided (unilateral) trigonometric forms
Fc(!) =
1Z0
f(t) cos(!t) dtF() f(t) =
2
¼
1Z0
Fc(!) cos(!t) d!
Fs(!) =
1Z0
f(t) sin(!t) dtF() f(t) =
2
¼
1Z0
Fs(!) sin(!t) d!
Note that the sine transform is a natural choice if f(0) = 0, while the cosine transform ¯ts
the case when f 0(0) = 0.
The use of frequency f (Hz=cycle/s) instead of ! (1/s)
Introduce the frequency f (Hz) such that ! = 2¼f , and our usual Fourier transform pair
X(!) =
1Z¡1
x(t)e¡j!tdt F() x(t) =1
2¼
1Z¡1
X(!)ej!td!
becomes
X(f) =
1Z¡1
x(t)e¡j2¼ftdt F() x(t) =
1Z¡1
X(f)ej2¼ftdf:
This form has the advantage that both variables t and f appear symmetrically without
any multiplying factors outside the integrals. Its disadvantage (to me) is the appearance
of a 2¼ in every kernel function exp(§j2¼ft). Its supporters are often experimentalists,because of their natural preference to measure frequency f in (Hz), as opposed to the
natural frequency !. The use of f can also pave the way for a slightly smoother transition
to the discrete-time world. Either way you look at it, the Fourier transform pair has
to have 2¼'s somewhere: The question is, do you place them as harmless scaling factors
outside the integrals, or do you turn them loose as pesky irritants inside the arguments
of the complex exponentials, so that every di®erential and integral operation results in a
factor of 2¼ or (2¼)¡1 ? Seriously, the world is divided and you will inevitably need to
translate between the two forms, but not in our course, where, at least for now, ! rules!
88
ECE 370 QUIZ 7 9 October 1940 Name IgÄor Reiruμof, III
Take-Home: due next class, Wednesday October 31 12:00 noon
Consider a real-valued function f(t) that possesses a de¯nite Fourier transform F (!).
Assume that f(t) is absolutely integrable so that we are assured of the ¯niteness of F (!)
for all frequencies.
1.) If f(t) = f(¡t) is even, what can you deduce about the symmetry (even/odd parity)of F (!)? Is F (!) pure real, pure imaginary, or arbitrarily complex?
2.) If f(t) = ¡f(¡t) is odd, what can you deduce about the symmetry (even/odd parity)of F (!)? Is F (!) pure real, pure imaginary, or arbitrarily complex?
3.) What is the physical (geometric) signi¯cance of F (0)?
F (!) =
1Z¡1
f(t)e¡j!tdt
************************** SOLUTION **************************
F (!) =
1Z¡1
f(t) cos(!t) dt¡ j1Z
¡1f(t) sin(!t) dt = Fr(!)¡ jFi(!)
If f(t) = f(¡t) is even, then Fi(!) = 0 and Fr(!) = 21Z0
f(t) cos(!t) dt = Fr(¡!):
That is, the spectrum of an even, real signal is pure real and is an even function of !.
If f(t) = ¡f(¡t) is odd, then Fr(!) = 0 and Fi(!) = 21Z0
f(t) sin(!t) dt = ¡Fi(¡!):
That is, the spectrum of an odd, real signal is pure imaginary and is an odd function of !.
F (0) =
1Z¡1
f(t) dt is the integral of f(t); the DC value, total area under the curve.
89
Cascade of Two LTI Systems
....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................................................................................... ............. ...................................................................................... .............x(t) y(t)h1(t) h2(t)
....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................................................................................... .............
h(t) = h1(t)~ h2(t)
H(!) = H1(!)H2(!)
Linear Dispersionless Filter
....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................................................................................... ............. ...................................................................................... .............x(t) y(t) = x(t¡ t0)h(t) = ±(t¡ t0)
H(!) = e¡j!t0 \linear phase"
90
Homework Problem: Rectangular Pulse Response of Ideal Low-Pass Filter.
Consider an ideal low-pass ¯lter of bandwidth B and subsequent transfer function
H(!) = ¦³ !2B
´: (1)
Find the transient response y(t) of this ¯lter to a rectangular pulse of pulse-width T
x(t) = ¦(t=T ): (2)
Graphical results simplify with normalized time t=T as the independent variable, for various
values of the dimensionless parameter BT . This problem is most readily solved in terms
of a special function, the sine integral8
Si(z) =
zZ0
sin »
»d» Si(¡z) = ¡Si(z) Si(1) = ¼=2: (3)
....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................................................................................... ............. ...................................................................................... .............x(t) y(t)h(t)
The ¯lter impulse response is
h(t) = F¡1[H(!)] = B
¼sinc(Bt) (4)
which can be found from your table of Fourier transforms or by direct evaluation of the
inverse transform integral. Similarly, the spectrum of the input signal is
X(!) = F [x(t)] = T sinc(!T=2): (5)
At least three di®erent, but related, approaches to ¯nding the output y(t) are worth
considering.
Time-Domain Convolution.
y(t) = x(t)~ h(t) =1Z
¡1x(¿)h(t¡ ¿) d¿ =
T=2Z¡T=2
h(t¡ ¿) d¿ (6)
=B
¼
T=2Z¡T=2
sin[B(t¡ ¿)]B(t¡ ¿) d¿ =
B
¼
T=2Z¡T=2
sin[B(¿ ¡ t)]B(¿ ¡ t) d¿ =
1
¼
B(T=2¡t)ZB(¡T=2¡t)
sin »
»d»
(7)
8M. Abramowitz and I.A. Stegun, Handbook of Mathematical Functions. National Bureau of Standards,1970, pp. 231{232, Table 5.1. (\Big Red"or \Citizen's Handbook"). MATLAB has it built-in, too!
91
y(t) =1
¼
©Si[B(T=2¡ t)]¡ Si[¡B(T=2 + t)]ª = 1
¼
©Si[B(t+ T=2)]¡ Si[B(t¡ T=2)]ª
(8)
=1
¼
½Si
·BT
μt
T+1
2
¶¸¡ Si
·BT
μt
T¡ 12
¶¸¾(9)
Note that y(t ! 1) = 0. Also note that time appears normalized to the incident pulse
width T as in t=T , and that the single (dimensionless!) parameter that characterizes the
¯lter response to the rectangular pulse is the product BT .
Frequency-Domain Multiplication.
y(t) = F¡1[X(!)H(!)] = 1
2¼
1Z¡1
X(!)H(!)ej!td! =1
2¼
BZ¡B
T sinc(!T=2)ej!td!
(10)
=1
¼
BZ¡B
sin(!T=2)
!ej!td! only the even part of ej!t contributes (11)
=1
¼
BZ¡B
sin(!T=2)
!cos(!t) d! =
2
¼
BZ0
sin(!T=2)
!cos(!t) d! (12)
Use the trig identity 2 sinX cosY = sin(X ¡ Y ) + sin(X + Y ) so that
y(t) =1
¼
BZ0
©sin[!(T=2¡ t)] + sin[!(T=2 + t)]ª d!
!: (13)
Let » = !(T=2¨ t) so thatBZ0
sin[!(T=2¨ t)] d!!=
B(T=2¨t)Z0
sin »
»d» = Si[B(T=2¨ t)] (14)
whereupon (13) is
y(t) =1
¼
©Si[B(T=2¡ t)] + Si[B(T=2 + t)]ª
=1
¼
©Si[B(t+ T=2)]¡ Si[B(t¡ T=2)]: (15)
92
Linear Combination of Step-Responses. Recognize that our particular input can be
written as
x(t) = ¦(t=T ) = u(t+ T=2)¡ u(t¡ T=2) (16)
so that the response is
y(t) = s(t+ T=2)¡ s(t¡ T=2) (17)
where the step response is
s(t) =
tZ¡1
h(¿) d¿ =B
¼
tZ¡1
sinc(B¿) d¿ =1
¼
BtZ¡1
sin »
»d» (18)
=1
¼
24 0Z¡1
sin »
»d» +
BtZ0
sin »
»d»
35 = 1
¼
h¼2+ Si(Bt)
i=1
2+1
¼Si(Bt): (19)
The linear combination (17) of shifted step responses now yields
y(t) =1
¼
©Si[B(t+ T=2)]¡ Si[B(t¡ T=2)]: (20)
The MATLAB routine sinint.m calls the symbolic toolbox, built upon MAPLE, and does
an extremely slow and ine±cient numerical integration to compute the function Si(x).
It doesn't work at all on my computer, so here is an alternate computation that uses
MATLAB's routine expint.m for the exponential integral, a di®erent but related special
function.
E1(z) =
1Zz
e¡t
tdt
Si(x) =¼
2+ ImfE1(jx)g
The routine for E1(jx) will not work for (real) values of x · 0, but that's ok since we knowthat Si(0) = 0 and it is an odd function Si(¡x) = ¡Si(x). Therefore Si(x) = Si(jxj) sgn(x).
function f=sinintRWS(x)
% f=sinintRWS(x) sine integral Si(x) for real x
f=zeros(size(x));
k=find(x»=0);f(k)=(pi/2+imag(expint(i*abs(x(k))))).*sign(x(k));
93
¡2:0 ¡1:5 ¡1:0 ¡0:5 0.0 0.5 1.0 1.5 2.0¡0:2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
t=T
y(t)
dashed curve: BT = 10
solid curve: BT = 50
...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
..........................................................................................................................................................................................
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..................................................................................................................
.................................................................................................................................................................................................................................................................................................................................................................................
..........................................................................................................................................................................................................................................................................................................................................................................
.........................................................................................................................
...........................................................................
Fig. 1. Rectangular pulse response of ideal low-pass ¯lter.
94
Ideal Band-Pass Filter
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!......................................................................................................................................
H(!)
..............................
¡!0 !00
B..................................................................................................... ....................................................................................... ..........
.... B ....................................................................................... ...................................................................................................................
H(!) = ¦
μ! ¡ !0B
¶+¦
μ! + !0
B
¶
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......
f(t) F (!)
B
2¼sinc
μBt
2
¶¦³ !B
´
ej!0tf(t) F (! ¡ !0)
h(t) =B
2¼sinc(Bt=2)
£ej!0t + e¡j!0t
¤=B
¼sinc(Bt=2) cos(!0t)
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......
t
h(t)
!0 = 100, B = 20
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95
Complex Phasors for Time-Harmonic Circuit Analysis. Consider the sinusoidal-
steady-state response of the given RLC circuit, that is ¯nd vo(t) if vi(t) = Vp cos(!0t+Á).
The time-domain integro-di®erential equation (Kirchho®'s voltage law) that is appropriate
for this network is
Ldi(t)
dt+1
C
tZ¡1
i(¿) d¿ +Ri(t) = vi(t)
and the output voltage is simply a scaled version of the mesh current, that is vo(t) = Ri(t).
.......
........................................................................................................................................................................................................................................
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................................
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................................
..................................................................i(t)
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..
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²
²
..........................................................
..........................................................
..........................................................
.
+
¡
vi(t)
+
¡
vo(t)
LC
R...............................................................................................................................................................................
(1) Solve for vo(t) by representing the time-domain signals in terms of their Fourier
transforms.
(2) Show that the Fourier transform analysis is equivalent to, and is therefore the mo-
tivation for, the (simple and compact!) complex phasor analysis of time-harmonic
circuits such as this one. Recall that the time-harmonic signal i(t) is represented
in terms of the complex phasor I via
i(t) = RefIe+j!tg:
The Fourier transform of the input or generator voltage is
Vi(!) = F [vi(t)] = F [Vp cos(!0t+ Á)] = F½Vp
2
£ej(!0t+Á) + e¡j(!0t+Á)
¤¾=Vp
2
£2¼ejÁ±(! ¡ !0) + 2¼e¡jÁ±(! ¡ !0)
¤:
The Fourier transform of the entire (KVL) equation above (both the LHS and RHS) is
j!LI(!) +1
C
·I(!)
j!+ ¼I(0)±(!)
¸+RI(!) = ¼Vp
£ejÁ±(! ¡ !0) + e¡jÁ±(! + !0)
¤:
Physically argue that I(0) = 0 here, since we expect zero steady-state DC current through
the capacitor, whereuponμR+ j!L+
1
j!C
¶| {z }
Z(!)
I(!) = ¼Vp£ejÁ±(! ¡ !0) + e¡jÁ±(! + !0)
¤:
96
Now solve this algebraic equation for the unknown spectrum
I(!) = ¼VpejÁ±(! ¡ !0) + e¡jÁ±(! + !0)
Z(!):
Fourier inversion proceeds as
i(t) =1
2¼
1Z¡1
I(!)ej!td!
=Vp
2
24ejÁ 1Z¡1
±(! ¡ !0)Z(!)
ej!td! + e¡jÁ1Z
¡1
±(! + !0)
Z(!)ej!td!
35=Vp
2
·ejÁ
ej!0t
Z(!0)+ e¡jÁ
e¡j!0t
Z(¡!0)¸
note Z(¡!0) = Z¤(+!0)
=1
2
·Vpe
jÁ
Z(!0)ej!0t +
Vpe¡jÁ
Z¤(!0)e¡j!0t
¸=1
2
·Vpe
jÁ
Z(!0)ej!0t + c.c.
¸an expression plus its complex conjugate
= Re·Vpe
jÁ
Z(!0)ej!0t
¸= Re
£Iej!0t
¤where the complex phasor
I =Vpe
jÁ
Z(!0)=
Vi
Z(!0)
introduced itself! If you still want it,
Vo = RI and vo(t) = Ri(t):
It looks like the two steps alluded to in the problem statement are more naturally done
together.
97
Homework Problem on Uncertainty Principle
Consider the signal
x(t) =1pT¦(t=T )
where T > 0 and the total energy in x(t) is
Ex =1Z
¡1jx(t)j2 dt = 1:
(The multiplying factor T¡1=2 ensures that Ex is independent of T .) If x(t) is the inputvoltage to the given electric circuit, with ¯xed values of R and L, consider the energy in
the output voltage y(t)
Ey =1Z
¡1jy(t)j2 dt:
Again, note that changing T has no e®ect on the total input energy. If T is increased, will
the output energy Ey increase, decrease, or remain the same? Explain.
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........
.......
.......
.......
.
²
²L
R
+
¡
x(t)
+
¡
y(t)
98
Approximate Analysis of Dispersion
x(t) y(t)H(!)
............................................................................................................................................................................................................................................................................................................................................................................
........................................................... ............. ........................................................... .............
Fig. 1. System Block Diagram
If the modulating signal or envelope of a high frequency sinusoid of frequency !0 is
the baseband signal a(t), having bandwidth considerably less than !0 (say 10%), then the
spectrum of the the modulated carrier
x(t) = a(t) cos(!0t) (1)
is
X(!) = 12 [A(! ¡ !0) +A(! + !0)]: (2)
An example spectrum is shown in Fig. 2.
...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
0.............
¡!0.............
!0
!
jX(!)j
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Fig. 2. Typical Narrow-Band Spectrum
For real signals x(t) and h(t), the spectra satisfy X(¡!) = X¤(!) and H(¡!) = H¤(!).The response of the system is the inverse Fourier transform ofX(!)H(!), with the positive
and negative frequency contributions written separately as
99
y(t) =1
2¼
1Z0
X(!)H(!)ej!td! +1
2¼
0Z¡1
X(!)H(!)ej!td!
=1
2¼
1Z0
X(!)H(!)ej!td! +1
2¼
1Z0
X(¡−)H(¡−)e¡j−td−
=1
2¼
1Z0
X(!)H(!)ej!td! +1
2¼
1Z0
£X(−)H(−)ej−t
¤¤d−
=1
2¼
1Z0
X(!)H(!)ej!td! + c.c.
= 12 [y+(t) + y
¤+(t)] (3)
The signal y+(t) is called the pre-envelope of y(t), and consists of positive frequencies only
y+(t) =1
2¼
1Z0
[A(! ¡ !0) +A(! + !0)]H(!)ej!td!
¼ 1
2¼
1Z¡1
A(! ¡ !0)H(!)ej!td! = 1
2¼
1Z¡1
A(−)H(−+ !0)ej(−+!0)td− (4)
where A(−) is the original, baseband or low-pass signal.
The phase or argument of the transfer function H(!) in a neighborhood of the carrier
frequency !0 is of most interest. Write
H(!) = exp[j£(!)]: (5)
A Taylor series about !0 of the phase function
£(!0 + −) = £(!0) + −£0(!0) + 1
2−2£00(!0) + : : : (6)
gives the approximate transfer function
H(−+ !0) ¼ ej£(!0)ej−£0(!0): (7)
An approximation for the pre-envelope (4) is therefore
y+(t) ¼ 1
2¼ej[!0t+£(!0)]
1Z¡1
A(−)ej−[£0(!0)+t]d−
= ej[!0t+£(!0)]a[t+£0(!0)] (8)
and so the ¯nal approximation for the output of the dispersive ¯lter (5) is
y(t) ¼ a[t+£0(!0)] cos[!0t+£(!0)]: (9)
The phase delay (of the carrier) is due to £(!0) and the group delay (of the envelope) is
due to £0(!0).
100
Numerical Example. The phase function of a length of waveguide transmission line is
£(!) = ¡p!2 ¡ !2c t0
where !c is the cuto® frequency of the guide and t0 = d=c where d is the length and c is
the free-space phase velocity.
If !c = 0, then £0(!0) = ¡t0.
If !c = 25, !0 = 30, t0 = 3, then
£0(!0) = ¡ (30)(3)p302 ¡ 252 = ¡5:43
The following ¯gures are the result of an inverse FFT approximation for the signal
x(t) = e¡t2
cos(!0t)
applied to the input of the above dispersive waveguide section. The 4096 point FFT uses
¢! = 0:3.
101
0 1 2 3 4 5 6 7 8 9 10
t
¡1:0
¡0:5
0.0
0.5
1.0
y(t)
!c = 0, t0 = 3, !0 = 30
........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................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0 1 2 3 4 5 6 7 8 9 10
t
¡1:0
¡0:5
0.0
0.5
1.0
y(t)
!c = 25, t0 = 3, !0 = 30
........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................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Fig. 3. Dispersive Filter Output
102
Fourier Integral Solution of Potential Problem
Laplace's equation
r2Ã(x; y) =μ@2
@x2+@2
@y2
¶Ã(x; y) = 0
boundary conditions
Ã(x; 0) =
½1; jxj < 10; jxj > 1
separate variables
Ã(x; y) = X(x)Y (y)
X 00(x)X(x)| {z }=¡k2
+Y 00(y)Y (y)| {z }=+k2
= 0
solution form
Ã(x; y) = cos(kx)e¡kjyj
Ã(x; y) =
1Z0
A(k) cos(kx)e¡kjyj dk
Ã(x; 0) =
1Z0
A(k) cos(kx) dk
inverse Fourier transform
A(k) =2
¼
1Z0
Ã(x; 0) cos(kx) dx =2
¼
1Z0
cos(kx) dx =2
¼
sin k
k
Ã(x; y) =2
¼
1Z0
sin k cos kx
ke¡kjyjdk
=1
¼
1Z0
sin[k(x+ 1)]¡ sin[k(x¡ 1)]k
e¡kjyjdx
=1
¼
·tan¡1
μx+ 1
jyj¶¡ tan¡1
μx¡ 1jyj
¶¸using G&R 3.941(1).
103
−3 −2 −1 0 1 2 30
1
2
3
4
5
6
Fig. 1. Equipotentials
104
Fourier Integral Solution of Potential Problem
M.J. Ablowitz and A.S. Fokas, Complex Variables, Cambridge, 1997, page 60 prob 9.
Laplace's equation
r2Ã(x; y) = 0boundary conditions
Ã(x; 0) = sgn(x) = §1 (x ? 0)
separate variables
Ã(x; y) = X(x)Y (y)
X 00(x)X(x)| {z }=¡k2
+Y 00(y)Y (y)| {z }=+k2
= 0
solution form
Ã(x; y) = sin(kx)e¡kjyj
Ã(x; y) =
1Z0
B(k) sin(kx)e¡kjyj dk
Ã(x; 0) =
1Z0
B(k) sin(kx) dk
inverse Fourier transform
B(k) =2
¼
1Z0
Ã(x; 0) sin(kx) dx =2
¼
1Z0
sin(kx) dx =2
¼k
a bit tricky! recall ECE 370 and sgn(x) = u(x)¡ u(¡x)
Ã(x; y) =2
¼
1Z0
sin kx
ke¡kjyjdk
=2
¼tan¡1
μx
jyj¶= 1¡ 2
¼tan¡1
μ jyjx
¶using G&R 3.941(1).
105
The Fourier Transform of combT (t) is2¼Tcomb2¼=T (!)
Recall our notation F (!) for the Fourier transform
F (!) = F [f(t)] =1Z
¡1f(t)e¡j!t dt F() f(t) = F¡1[F (!)] = 1
2¼
1Z¡1
F (!)ej!t d!: (1)
The Fourier transform of Woodward's comb function, or impulse train
f(t) = combT (t) =
1Xn=¡1
±(t¡ nT ) (2)
is therefore
F [combT (t)] =1X
n=¡1e¡jn!T = F (!) = F (! + 2¼=T ); (3)
periodic in ! with period 2¼=T . Let's integrate F (!) over the range a < ! < b, where
¡¼=T < a < b < ¼=T . The minimum and maximum restrictions cover one period, centeredabout the origin (! = 0). We will make use of the discontinuous sum
1Xn=1
sin(nx)
n=¼ ¡ jxj2
sgn(x) (¡2¼ < x < 2¼) (4)
where, in this context, we de¯ne
sgn(x) =
8><>:¡1; x < 0
0; x = 0
+1; x > 0
(5)
in order to ensure the vanishing of the sum for x = 0. The integral of interest is
bZa
1Xn=¡1
e¡jn!Td! = b¡ a+ 21Xn=1
bZa
cos(n!T ) d!
= b¡ a+ 2
T
1Xn=1
1
n[sin(nbT )¡ sin(naT )]
= b¡ a+³ ¼T¡ jbj
´sgn(b)¡
³ ¼T¡ jaj
´sgn(a)
=
½2¼=T if ¡ ¼=T < a < 0 < b < ¼=T0 if a < b < 0 or 0 < a < b
: (6)
Therefore, we see that F [combT (t)] = 2¼Tcomb2¼=T (!): ¥
Now we are in a position to derive the Fourier series from the Fourier integral. And recall
that we essentially derived the Fourier integral from the \spectral representation of the
Dirac-delta" function. We are in excellent shape to now do some real damage.
106
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t
combT (t)
¡2T ¡T 0 T 2T
: : : : : :
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!
¡4¼T
¡2¼T
02¼
T
4¼
T
: : : : : :
2¼
Tcomb2¼
T
(!)F()
Fourier Transform of a Periodic Signal
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¡T 0 T 2T
t
g(t)
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...
¡T 0 T 2T
t
: : : : : :
f(t) = f(t+ T )
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Consider the arbitrary periodic signal f(t) = f(t+T ), and let's focus our attention on the
particular period 0 · t · T . Any period would work, but since all periods are equivalentwe might as well work with the selected one. (The function f does not have to go to zero
at t = nT ; it was simply easier to draw one like this. And of course the function f does
not have to be positive, or even real-valued.) Introduce the new aperiodic function g(t),
also de¯ned for all t according to
g(t) =
½f(t); 0 · t · T0; t < 0 or t > T:
(1)
We can easily reproduce the entire periodic signal f(t) by replicating g(t) all up and down
the t-axis. That is, we write f(t) as the periodic extension of g(t) in the form
f(t) =
1Xn=¡1
g(t¡ nT ): (2)
107
Recognize that (2) can be written as
f(t) = g(t)~1X
n=¡1±(t¡ nT ) = g(t)~ combT (t): (3)
The Fourier transform of (3) is
F (!) = G(!) ¢ 2¼Tcomb2¼
T
(!)
=2¼
TG(!)
1Xn=¡1
±(! ¡ n!0) where !0 ,2¼
T
=2¼
T
1Xn=¡1
G(n!0)±(! ¡ n!0): (4)
The spectrum F (!) of the periodic signal f(t) consists of discrete spectral lines at integer
multiples n!0 of the fundamental frequency !0 = 2¼=T . The nth multiple is called the
nth harmonic. Note that negative n are just as important as positive n.
!
!0¡10 10
¡8 8¡6 6
4¡4 2¡2 0..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
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................................................................................................
108
The inverse Fourier transform of F (!) should give us our original periodic f(t). It does.
In fact, it gives us a whole new representation for a periodic signal
f(t) =1
2¼
1Z¡1
F (!)ej!td! =
1Xn=¡1
G(n!0)
T
1Z¡1
±(! ¡ n!0)ej!td! =1X
n=¡1
G(n!0)
Tejn!0t
(5)
where
G(n!0) = G(!)¯̄!=n!0
=
1Z¡1
g(t)e¡jn!0tdt =
TZ0
g(t)e¡jn!0tdt =
TZ0
f(t)e¡jn!0tdt: (6)
We usually call the Fourier coe±cients cn instead of G(n!0)=T and write the whole
package as
f(t) =
1Xn=¡1
cnejn!0t cn =
1
T
TZ0
f(t)e¡jn!0tdt: (7)
Of course any period will work for the integration limits, since the total integrand is
periodic with period T :
cn =1
T
t0+TZt0
f(t)e¡jn!0tdt: (8)
You now have the complete and uncluttered derivation of the Fourier series representation
of a periodic signal, starting from the Fourier integral. The two key components are the
correspondingly appropriate spectral representations for the aperiodic Dirac-delta function
and for its periodic version, Woodward's comb function. All that was required on our part
was a reasonable °uency with the Fourier transform properties and a few fundamental
signals.
F Exercise 8. Find an interested (or at least willing) fellow student or more who are at
or above your present level with respect to Fourier analysis, and explain to them, in detail,
the above derivation starting from the top of three pages back. If you explain it correctly
and with some degree of coherence, and your entire audience has not deserted you, buy the
survivor(s) a Coke. (They can split it to ¯t any ¯nancial boundary conditions, applied, of
course, to your total solution.) You deserve one, too. Otherwise go back and do any of the
homework exercises (including enough of the 16 Fourier transform pairs on page 59) that
strike your fancy. If none ¯t that description, see me for more.
109
Periodic Signals and Fourier Series
The period of a periodic signal is the smallest positive number T such that f(t) = f(t+T )
for all t. It will be frequently convenient to de¯ne !0 , 2¼=T to be the fundamental
frequency of the periodic signal, with n!0 the nth harmonic frequency for integer n =
0;§1;§2; : : : . An important property of a periodic signal is that the integralt0+TZt0
f(t) dt =
TZ0
f(t) dt =
T=2Z¡T=2
f(t) dt =
3T=4Z¡T=4
f(t) dt = : : :
is independent of the time t0. That is, integration over any complete period yields the same
result. Another useful observation is that if f(t) = f(t + T ), then the product f(t)ejn!0t
is also periodic with period T , so that the integral
t0+TZt0
f(t)ejn!0t dt
is also independent of the time t0. F Be sure you can show/verify this important property
for yourself.
DEFN: Orthogonality. Two functions (generally complex) are said to be orthogonal on
the domain a · t · b if the inner product
hf(t); g(t)i¯̄̄ba,
bZa
f(t)g¤(t) dt = 0:
In any given situation, let's assume the interval of interest a · t · b is known and ¯xed, sothat we can condense the notation on the inner product angle brackets. The inner product
of a signal and itself is sometimes called the energy in the signal
hf(t); f(t)i ,bZa
jf(t)j2 dt = E
and the square-root of this necessarily non negative quantity is called the norm of the
signal
kf(t)k = +phf(t); f(t)i =
24 bZa
jf(t)j2 dt351=2 :
When dealing with periodic signals so that the interval of interest is any period t0 · t ·t0 + T , the e®ective or root-mean-square value is
frms =
24 1T
t0+TZt0
jf(t)j2 dt351=2 :
110
Kronecker delta. This simple function is a function of two integers m and n and is
de¯ned as
±m;n =
½1; m = n
0; m6= n:It is unrelated to the Dirac-delta function ±(t) that is explicitly written as a function of
the continuous time variable t. The operational advantage of introducing this notational
convenience will be clear as we progress.
Orthogonality of the Fourier basis.
hejn!0t; ejm!0tiT =t0+TZt0
ej(n¡m)!0tdt = T±n;m (n;m = 0;§1;§2; : : : )
hcos(n!0t); cos(m!0t)iT =t0+TZt0
cos(n!0t) cos(m!0t) dt =T
2±n;m (n;m = 1; 2; : : : )
hsin(n!0t); sin(m!0t)iT =t0+TZt0
sin(n!0t) sin(m!0t) dt =T
2±n;m (n;m = 1; 2; : : : )
hcos(n!0t); sin(m!0t)iT =t0+TZt0
cos(n!0t) sin(m!0t) dt = 0 (n;m = 0; 1; 2; : : : )
When n = 0 the nth cosine is unity, and we have
h1; cos(m!0t)iT =t0+TZt0
cos(m!0t) dt = T±0;m (m = 0; 1; 2; : : : )
h1; sin(m!0t)iT =t0+TZt0
sin(m!0t) dt = 0 (m = 1; 2; : : : )
Note that sin(m!0t) is trivial, and therefore not used, when m = 0.
F Homework: Verify/derive each of the above orthogonality relations. You will need to
use trig identities to simplify the products in the integrands.
111
Derivation of Fourier Coe±cients by Minimizing the Mean Square Error
The concept of the Fourier series arose naturally back on page 108 when we wrote a
periodic signal as the inverse Fourier integral of its Fourier transform. In doing so, the
Fourier coe±cients appeared naturally. Now let's start over by assuming that the periodic
signal f(t) = f(t+ T ) can be represented by a Fourier series, call it
ef(t) = 1Xn=¡1
cnejn!0t
to temporarily avoid any premature claims that ef(t) = f(t). One approach to derive theexpression for the Fourier coe±cients cn is to ask the question: What values of the cnminimize the mean square error between the original signal f(t) and its proposed Fourier
series representation ef(t) ?Firstly, write the mean square error
E2 = 1
T
t0+TZt0
jf(t)¡ ef(t)j2 dt = 1
T
t0+TZt0
£f(t)¡ ef(t)¤£f¤(t)¡ ef¤(t)¤ dt
=1
T
t0+TZt0
£jf(t)j2 + j ef(t)j2 ¡ f(t) ef¤(t)¡ f¤(t) ef(t)¤ dt:We need the following component integrals.
1
T
t0+TZt0
j ef(t)j2 dt = 1
T
t0+TZt0
1Xn=¡1
cnejn!0t
| {z }ef(t)
1Xm=¡1
c¤me¡jm!0t
| {z }ef¤(t)dt
=
1Xn=¡1
cn
1Xm=¡1
c¤m1
T
t0+TZt0
ej(n¡m)!0tdt
=
1Xn=¡1
cn
1Xm=¡1
c¤m±n;m
=
1Xn=¡1
cnc¤n
112
1
T
t0+TZt0
f¤(t) ef(t) dt = 1Xn=¡1
cn1
T
t0+TZt0
f¤(t)ejn!0tdt
1
T
t0+TZt0
f(t) ef¤(t) dt = 1Xn=¡1
c¤n1
T
t0+TZt0
f(t)e¡jn!0tdt
Now the mean square error is
E2 = 1
T
t0+TZt0
jf(t)j2 dt +1X
n=¡1cnc
¤n ¡
1Xn=¡1
cn1
T
t0+TZt0
f¤(t)ejn!0tdt
¡1X
n=¡1c¤n1
T
t0+TZt0
f(t)e¡jn!0tdt
To minimize E2 with respect to the Fourier coe±cients cn, set to zero the partial derivativewith respect to each of the ck (for all integer k)
@
@ckE2 = c¤k ¡
1
T
t0+TZt0
f¤(t)ejk!0tdt = 0
or
c¤k =1
T
t0+TZt0
f¤(t)ejk!0tdt
the conjugate of which gives the required Fourier coe±cient formula
ck =1
T
t0+TZt0
f(t)e¡jk!0tdt
where k can now be replaced by n
cn =1
T
t0+TZt0
f(t)e¡jn!0tdt:
113
Fourier Coe±cients by a Direct Inner Product
Write
f(t) =
1Xn=¡1
cnejn!0t
and then take the inner product of both sides with ejm!0t
t0+TZt0
f(t)e¡jm!0tdt =1X
n=¡1cn
t0+TZt0
ej(n¡m)!0tdt
=
1Xn=¡1
cnT±n;m = Tcm
and therefore
cm =1
T
t0+TZt0
f(t)e¡jm!0tdt:
Trigonometric Form of the Fourier Series
f(t) = a0 +
1Xn=1
£an cos(n!0t) + bn sin(n!0t)
¤To ¯nd the trig coe±cients an and bn (n = 0; 1; 2; : : : ) in terms of the complex exponential
coe±cients cn (n = : : : ;¡1; 0; 1; 2; : : : ) set1X
n=¡1cne
jn!0t = a0 +
1Xn=1
£an cos(n!0t) + bn sin(n!0t)
¤
c0 +
1Xn=1
hcne
jn!0t + c¡ne¡jn!0ti= a0 +
1Xn=1
£an cos(n!0t) + bn sin(n!0t)
¤
c0 +
1Xn=1
hcn©cos(n!0t) + j sin(n!0t)
ª+ c¡n
©cos(n!0t)¡ j sin(n!0t)
ªi= a0 +
1Xn=1
£an cos(n!0t) + bn sin(n!0t)
¤
c0+
1Xn=1
h(cn+c¡n) cos(n!0t)+j(cn¡c¡n) sin(n!0t)
i= a0+
1Xn=1
£an cos(n!0t)+bn sin(n!0t)
¤114
Therefore
a0 = c0 =1
T
t0+TZt0
f(t) dt
an = cn + c¡n =2
T
t0+TZt0
f(t) cos(n!0t) dt (n = 1; 2; : : : )
bn = j(cn ¡ c¡n) = 2
T
t0+TZt0
f(t) sin(n!0t) dt (n = 1; 2; : : : )
F Homework. Derive the trig Fourier coe±cients by taking the direct inner product of
f(t) = a0 +
1Xn=1
£an cos(n!0t) + bn sin(n!0t)
¤with each of the trig Fourier basis functions cos(m!0t), sin(m!0t), and unity.
115
Two Standard Notations for the Trigonometric Fourier Series
f(t) = a0 +
1Xn=1
£an cos(n!0t) + bn sin(n!0t)
¤a0 =
1
T
t0+TZt0
f(t) dt
an =2
T
t0+TZt0
f(t) cos(n!0t) dt (n = 1; 2; : : : )
bn =2
T
t0+TZt0
f(t) sin(n!0t) dt (n = 1; 2; : : : )
f(t) =a0
2+
1Xn=1
£an cos(n!0t) + bn sin(n!0t)
¤an =
2
T
t0+TZt0
f(t) cos(n!0t) dt (n = 0; 1; 2; : : : )
bn =2
T
t0+TZt0
f(t) sin(n!0t) dt (n = 1; 2; : : : )
The second form uses the general form for an to also calculate the DC term a0, avoiding
the need for a separate formula for a0.
It is silly, but observe that this captures the spirit:
f(t) =|7+
1Xn=1
£an cos(n!0t) + bn sin(n!0t)
¤| = 7
T
t0+TZt0
f(t) dt
F Homework. In some applications, the Fourier series is written in the form
f(t) = A0 +
1Xn=1
An cos(n!0t+ Án):
Find an expression for An and Án in terms of an and bn.
116
Riemann-Lebesgue Lemma - Fourier Coe±cients
f(t) = f(t+ T ) =
1Xn=¡1
cnejn!0t; !0 = 2¼=T
Tcn =
t0+TZt0
f(t)e¡jn!0tdt
=f(t)e¡jn!0t
¡jn!0
¯̄̄̄¯t0+T
t=t0
+1
jn!0
t0+TZt0
f 0(t)e¡jn!0tdt
by integration-by-parts. The boundary terms are zero by the periodicity of the integrand.
The remaining integral has¯̄̄̄¯̄t0+TZt0
f 0(t)e¡jn!0tdt
¯̄̄̄¯̄ ·
t0+TZt0
jf 0(t)j dt · K <1
if f 0(t) is absolutlely integrable. Therefore
jcnj · K
2n¼:
If f 0(t) is absolutlely integrable, then cn ! 0 as n!1.
Corollary.
If f (k)(t) is absolutely integrable, but f (k+1)(t) is not, then cn = O(n¡k) as n!1.
117
Gibb's Phenomenon - Example Square wave
0.0 0.5 1.0
t=T
¡1:0
¡0:5
0.0
0.5
1.0
f(t)
S5(t)
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Fourier series of the above square wave is
f(t) =4
¼
1Xk=1
sin[(2k ¡ 1)!0t]2k ¡ 1 ; !0 = 2¼=T
Truncated Fourier series
SN (t) =4
¼
NXk=1
sin[(2k ¡ 1)!0t]2k ¡ 1
Di®erentiate to ¯nd its extrema
S0N (t) =8
T
NXk=1
cos[(2k ¡ 1)!0t]
write
cos[(2k ¡ 1)!0t] = sin(2k!0t)¡ sin[(2k ¡ 1)!0t]2 sin!0t
All terms cancel except
S0N (t) =4
T
sin(2N!0t)
sin!0t
118
Smallest zero is
t0 =¼
2N!0
SN (t0) =4
¼
NXk=1
sinh(2k¡1)¼2N
i2k ¡ 1 =
4
¼
NXk=1
¼
2N
sinh(2k¡1)¼2N
i(2k¡1)¼2N
Riemann sumaZ0
sinx
xdx ¼
NXk=1
sin£2k¡12 ¢x
¤2k¡12 ¢x
¢x
where N¢x = a. Here ¢x = ¼=N so a = ¼.
SN (t0) ¼ 2
¼
¼Z0
sinx
xdx =
2
¼Si(¼) ¼ 1:18
overshoot is1:18¡ 1
total jump of 2£ 100% ¼ 9%
Pointwise Convergence of the Fourier Series
The Dirichlet kernel is
DN (x) =
NXn=¡N
ejnx = e¡jNx2NXn=0
ejnx = e¡jNx1¡ ej(2N+1)x1¡ ejx =
sin[(N + 1=2)x]
sin(x=2);
an even function DN (¡x) = DN (x). Consider a periodic function f(t) = f(t+ T ) havinga single jump discontinuity at the point ¡T=2 < t < T=2. De¯ne the right and left-handlimits
f(t+) = lim²!0+
f(t+ ²) and f(t¡) = lim²!0+
f(t¡ ²):
With the Fourier coe±cients
cn =1
T
T=2Z¡T=2
f(¿)e¡j2n¼¿=T d¿;
the truncated Fourier series efN (t) = NXn=¡N
cnej2n¼t=T
119
is
efN (t) = T=2Z¡T=2
f(¿)1
T
NXn=¡N
ej2n¼(t¡¿)=Td¿ =
T=2Z¡T=2
f(¿)1
TDN [2¼(t¡ ¿)=T ] d¿
=
tZ¡T=2
f(¿)1
TDN [2¼(¿ ¡ t)=T ] d¿
| {z }I¡
+
T=2Zt
f(¿)1
TDN [2¼(¿ ¡ t)=T ] d¿| {z }
I+
I¡ =
tZ¡T=2
[f(¿)¡ f(t¡)] 1TDN [2¼(¿ ¡ t)=T ] d¿ + f(t¡)
tZ¡T=2
1
TDN [2¼(¿ ¡ t)=T ] d¿
Introduce the change-of-integration variable
x =2¼
T(N + 1=2)(¿ ¡ t)
such that1
TDN [2¼(¿ ¡ t)=T ] d¿ = sinx
sin
μx
2N + 1
¶ dx
(2N + 1)¼
¿ = ¡T=2 =) x = ¡(2N + 1)¼(1=2 + t=T ) = ¡X¿ = t =) x = 0
¿ ¡ t = T
(2N + 1)¼
I¡ =1
¼
0Z¡X
½f
·T
(2N + 1)¼x+ t
¸¡ f(t¡)
¾sinx
(2N + 1) sin
μx
2N + 1
¶dx
+f(t¡)¼
0Z¡X
sinx
(2N + 1) sin
μx
2N + 1
¶dxSince ¡X < x < 0 , as N !1,
f
·T
(2N + 1)¼x+ t
¸¡¡¡¡!N!1
f(t¡)
120
so that the ¯rst integral above vanishes. The denominator becomes
(2N + 1) sin
μx
2N + 1
¶¡¡¡¡!N!1
x
and with X !1, the second integral gives
I¡ ¡¡¡¡!N!1
f(t¡)¼
0Z¡1
sinx
xdx =
f(t¡)2
:
The same philosophy yields
I+ ¡¡¡¡!N!1
f(t+)
¼
1Z0
sinx
xdx =
f(t+)
2
so thatNX
n=¡Ncne
j2n¼t=T ¡¡¡¡!N!1
f(t+) + f(t¡)2
: ¥
F Homework Consider the simpler case where f(t) is continuous everywhere, and there-
fore demonstrate thatNX
n=¡Ncne
j2n¼t=T ¡¡¡¡!N!1
f(t):
121
ECE 370 QUIZ 3 16 January 1978 NAME
Take Home - Due Wednesday January 18
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t
f(t)
.......
.......
.....
¡T...................
¡T=2 0...................
T=2...................
T...................
3T=2
................... 1
: : : : : :
1.) Find a Fourier series representation of the signal f(t).
2.) Use MATLAB to graph the Nth partial sum of this Fourier series. For example, if
f(t) =a0
2+
1Xn=1
an cos(n!0t);
then let
fN (t) =a0
2+
NXn=1
an cos(n!0t):
Pick several N to demonstrate (pictorially) the convergence of this Fourier series.
There are no sines (bn ´ 0) in this even function, and the DC or average value is
a0
2=1
T
TZ0
f(t) dt =1
2; (1)
as the graph shows (perhaps after a little thought). The cosine coe±cients (for n = 1; 2; : : : )
are
an =2
T
t0+TZt0
f(t) cos(n!0t) dt =2
T
T=2Z¡T=2
f(t) cos(n!0t) dt =2
T¢ 2
T=2Z0
f(t) cos(n!0t) dt (2)
122
by the even symmetry of the product of f(¡t) = f(t) and the nth cosine. The last integralabove requires an expression for f(t) in the range 0 · t · T=2, which is the line of slope2=T through the origin: f(t) = 2t=T . Required now is
an =2
T¢ 2
T=2Z0
2t
Tcos(n!0t) dt =
8
T 2
T=2Z0
t cos(n!0t) dt: (3)
In terms of di®erent variables, we need the integral
I =
dZ0
x cos(®x) dx: (4)
It is perfectly acceptable to use a table of integrals (perhaps in your calculus book, or
a table such as the CRC). But here are two di®erent techniques to evaluate it. The
¯rst method is based on the observation that the parameter ® is independent from the
integration variable x, so that we can write
I =
dZ0
x cos(®x) dx =d
d®
dZ0
sin(®x) dx =d
d®
1¡ cos(®d)®
=d sin(®d)
®¡ 1¡ cos(®d)
®2: (5)
The second method is integration by parts:
d(uv) = u dv + v du =)bZa
udv = uv¯̄̄ba¡
bZa
v du:
The choice
u = x
dv = cos(®x) dx=)
du = dx
v =1
®sin(®x)
gives
I =
dZ0
x cos(®x) dx =x
®sin(®x)
¯̄̄dx=0
¡ 1
®
dZ0
sin(®x) dx
=d
®sin(®d) +
1
®2cos(®x)
¯̄̄dx=0
=d
®sin(®d) +
cos(®d)¡ 1®2
; (6)
123
in agreement with the ¯rst result (5). The needed integral in (3) has upper limit d = T=2
and frequency (the factor multiplying the integration variable t in the argument of the
cosine) ® = n!0, whereupon ®d = n!0T=2 = n¼ (recall !0T = 2¼). Note that the values
of the trig functions are thus sin(n¼) = 0 and cos(n¼) = (¡1)n. The Fourier coe±cients(3) are now
an = ¡ 8
T 21¡ (¡1)n(n!0)2
= ¡ 2
(n¼)2[1¡ (¡1)n] =
8<: ¡μ2
¼
¶21
n2; n odd
0; n even:
(7)
Together with the DC value (1), the entire trigonometric Fourier series for our even, peri-
odic function f(t) is
f(t) =1
2¡μ2
¼
¶2 1Xn=1;3;5;:::
1
n2cos(2n¼t=T ): (8)
The sum is reasonably clear (at least to me), but it would more commonly be written in
the (probably nicer looking) completely equivalent form
f(t) =1
2¡μ2
¼
¶2 1Xn=1
1
(2n¡ 1)2 cos[(2n¡ 1)2¼t=T ]: (9)
And it would be perfectly acceptable to write it in terms of the fundamental frequency as
f(t) =1
2¡μ2
¼
¶2 1Xn=1
1
(2n¡ 1)2 cos[(2n¡ 1)!0t];
as long as it is perfectly clear that we have de¯ned !0 = 2¼=T .
This series converges nicely, since janj » n¡2 as n!1 (in fact, these an are 1=n2 for all
odd n. In fact, Fourier series representations of known functions are used to sum certain
series. In this case, note that f(T=2) = 1 and so (9) gives
1 =1
2¡μ2
¼
¶2 1Xn=1
1
(2n¡ 1)2 cos[(2n¡ 1)¼]
or 1Xn=1
1
(2n¡ 1)2 =¼2
8:
Note that the sum over both odd and even integers is
1Xn=1
1
n2=
1Xn=1
1
(2n¡ 1)2 +1Xn=1
1
(2n)2
=
1Xn=1
1
(2n¡ 1)2 +1
4
1Xn=1
1
n2
124
or
3
4
1Xn=1
1
n2=
1Xn=1
1
(2n¡ 1)2
=¼2
8
and therefore 1Xn=1
1
n2=¼2
6:
This is the Riemann zeta function
³(x) =
1Xn=1
1
nx
of argument x = 2.
125
On the Convergence and Summation of Some Series
The basic form to consider is
S =
1Xn=1
an = a1 + a2 + a3 + : : :
where the indexing can just as easily start at n = 0
1Xn=0
an = a0 + a1 + a2 + : : :
and negative indices are also commonly encountered
1Xn=¡1
an = a0 +
1Xn=1
an +
1Xn=1
a¡n:
If the series1Pn=1
janj converges, then we say that1Pn=1
an is absolutely convergent. Any series
that is absolutely convergent is convergent, since¯̄̄̄¯1Xn=1
an
¯̄̄̄¯ ·
1Xn=1
janj:
A convergent series that is not absolutely convergent is called conditionally convergent.
Here is an important class of series: With p real
1Xn=1
1
npconverges i® p > 1.
This can be shown by using the integral argument (Do it!) on the next page.
126
If all of the terms of the original series are positive (true for exp(z) if z > 0), then a related
integral can provide error bounds by the following reasoning.
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x
g(x)........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
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............................................
N+1 N+2 N+3 : : :
1Xn=N+1
g(n) >
1ZN+1
g(x) dx
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N N+1 N+2 N+3 : : :
1Xn=N+1
g(n) <
1ZN
g(x) dx
Therefore, we have both a lower and an upper bound, so that the error or \tail" sum is
1ZN+1
g(x) dx <
1Xn=N+1
g(n) <
1ZN
g(x) dx:
127
If the series1Pn=1
janj converges, then we must have limn!1 an = 0: The converse is not true,
as we demonstrate by example, since the series
1Xn=1
1
np
is divergent for p = 1=2 but1pn¡¡¡!n!1
0:
Comparison Test. Let1Pn=1
an and1Pn=1
bn be two series with nonnegative terms. If there
exists an index N such that an · bn for all n > N , then the convergence of the series1Pn=1
bn implies the convergence of the series1Pn=1
an, and the divergence of the series1Pn=1
an
implies the divergence of1Pn=1
bn.
Weierstrass M-Test for Absolute Convergence. Let1Pn=1
an and1Pn=1
bn be series.
Suppose there exists an index N such that janj · bn for all n > N . Then a su±cient
condition for absolute convergence of the series1Pn=1
an is that the series1Pn=1
bn converges.
Ratio (d'Alembert's) Test. Suppose that the limit limn!1
¯̄̄̄an+1
an
¯̄̄̄= ® exists for the
series1Pn=1
an.
(1) if ® < 1 then the series1Pn=1
an converges absolutely
(2) if ® > 1 then the series1Pn=1
an diverges
(3) there exist both absolutely convergent and divergent series for which ® = 1.
128
Acceleration of series convergence by subtracting out the asymptotic form
(Kummer transform):
Consider the series1Pn=1
f(n) where the large n form of f(n) is
f(n) ¡¡¡!n!1
fasy(n):
If the original terms f(n) decay so slowly with n as to be numerically disagreeable, but
the asymptotic form can be summed by some alternate method, then write
1Xn=1
f(n) =
1Xn=1
[f(n)¡ fasy(n)] +1Xn=1
fasy(n):
The terms of the di®erence f(n)¡fasy(n) will approach zero faster than the original f(n),so that truncation of the di®erence sum can be made at a smaller n to achieve a given
level of accuracy.
Example. One series that can be summed exactly using the Poisson sum formula (on
pages 150-151) is1X
n=¡1
1
n2 + b2=¼
bcoth¼b:
We choose an example series where we know the exact value of the sum simply to be able
to check our result. Let's get this series into the form1Pn=1
f(n), where f(¡n) = f(n) is aneven function of the summation index,
1Xn=¡1
1
n2 + b2= 2
1Xn=1
1
n2 + b2+1
b2
and therefore
S =
1Xn=1
1
n2 + b2=¼b coth¼b¡ 1
2b2:
Now let's compare the brute-force summation of S as it stands, with a simple Kummer
transform. Having the exact value of S allows us to gauge our accuracy. The asymptotic
form of the summand f(n) is fasy(n) = n¡2 so that the Kummer transform is
S =
1Xn=1
·1
n2 + b2¡ 1
n2
¸+
1Xn=1
1
n2:
The asymptotic or \tail" sum is the Riemann zeta function of argument 2 (a well-known,
actually famous sum) that we encountered back on page 124
³(2) =
1Xn=1
1
n2=¼2
6:
129
The di®erence terms
1
n2 + b2¡ 1
n2=n2 ¡ (n2 + b2)n2(n2 + b2)
=¡b2
n2(n2 + b2)= O(n¡4)
decay much faster than the original terms of O(n¡2). The Kummer transform of the
original sum S is now
S = ¡b21Xn=1
1
n2(n2 + b2)+¼2
6:
To do the comparison, label the truncated brute-force sum
S0(N; b) =
NXn=1
1
n2 + b2
and the truncated Kummer-accelerated sum
SK(N; b) = ¡b2NXn=1
1
n2(n2 + b2)+¼2
6:
With the exact value
S(b) =¼b coth¼b¡ 1
2b2
let's compare the numerical performance of
S0(N; b)
S(b)with
SK(N; b)
S(b)
as a function of the truncation index N for one or more values of the parameter b.
NS0(N; 10)
S(10)
SK(N; 10)
S(10)
101 0:499726021056655 1:125492496191760
102 0:934787298579118 1:000214644029801
103 0:993428003029169 1:000000218841745
104 0:999342482848786 1:000000000219156
Exercise. Using ³(4) =1Pn=1
n¡4 = ¼4=90, perform a second Kummer transform on the
series in the example and compare its convergence to the data above.
130
Homework Problem: Convergence of Fourier Series. Consider the three periodic
functions x(t), y(t), and z(t) as graphed below. Find appropriate Fourier series expansions
for each of the three functions. Study and discuss the convergence of each series, including
graphical display.
For example, say the odd function y(t) is most conveniently represented by the Fourier
sine series
y(t) =
1Xn=1
bn sin(2n¼t=T )
where the coe±cients bn are known. Truncation of this Fourier series yields the approxi-
mation
yN (t) =
NXn=1
bn sin(2n¼t=T ):
Compare these truncated series yN (t) with the original function y(t) for several (you decide
which range of values has signi¯cance to your convergence study) values of the ¯nal or
truncation index N . Certainly, the rate of decay of jbnj as n ! 1 is the important
feature.
Note: How are x(t), y(t), and z(t) related? How then are their Fourier series related?
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......
................
¡T=2 0 T=2 Tt
................... +1
x(t)
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t
+1
¡1
y(t)
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t.........................................................................
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...Nz(t)
(+1)
(¡1)
131
Fourier Series of Full-Wave Recti¯ed Sine Wave
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¡T 0 T
t
sin( 12!0t) = sin(¼t=T )
.......
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.....
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.......
.....
.......
.......
.......
.....
¡T 0 T
t
f(t) = j sin(12!0t)j = j sin(¼t=T )j
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The even function f(t) = f(¡t) is expanded in a Fourier series
f(t) =a0
2+
1Xn=1
an cos(n!0t) (1)
where its fundamental frequency is !0 = 2¼=T , in terms of its period T . Note that the
period of the sine wave prior to recti¯cation is 2T . Equivalently, the frequency of the
un-recti¯ed sine wave is half the frequency of the recti¯ed signal f(t).
The DC or average value of f(t) is
a0
2=1
T
TZ0
f(t) dt =1
T
TZ0
sin³!02t´dt =
2
!0Tcos³!02t´¯̄̄̄¯0
t=T
=2
2¼[1¡ cos¼] = 2
¼: (2)
For n = 1; 2; : : :
an =2
T
t0+TZt0
f(t) cos(n!0t) dt =2
T¢ 2
T=2Z0
sin³!02t´cos(n!0t) dt (3)
132
since
f(t) =¯̄̄sin³!02t´¯̄̄= sin
³!02t´
for 0 · t · T=2: (4)
Use of the trig identity
2 sinA cosB = sin(A¡B) + sin(A+B)yields
an =2
T
T=2Z0
½sin£(n+ 1
2 )!0t¤¡ sin£(n¡ 1
2 )!0t¤¾dt: (5)
Let's economize the algebra and do both forms at once by considering
2
T
T=2Z0
sin£(n§ 1
2)!0t¤dt =
2
T
1
(n§ 12 )!0
cos£(n§ 1
2 )!0t¤¯̄̄̄¯0
t=T=2
=1¡ cos£(n§ 1
2 )!0T=2¤
(n§ 12 )!0T=2
=1¡ cos£(n§ 1
2 )¼¤
(n§ 12 )¼
=1
(n§ 12)¼
:
(6)
Insertion of (6) into (5) gives
an =1
¼
·1
n+ 12
¡ 1
n¡ 12
¸=1
¼
(n¡ 12 )¡ (n+ 1
2 )
(n+ 12 )(n¡ 1
2)= ¡ 1
¼(n2 ¡ 14 ): (7)
Together with the DC term (2), the full Fourier series for f(t) is now
f(t) =2
¼¡ 1
¼
1Xn=1
1
n2 ¡ 14
cos(2n¼t=T ); (8)
and its truncated version (Nth partial sum) is
fN (t) =2
¼¡ 1
¼
NXn=1
1
n2 ¡ 14
cos(2n¼t=T ): (9)
..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.......
.......
.....
0 1
t=T
solid curve: N = 3
dashed curve: N = 20
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133
Parseval's Theorem for the Fourier Series
Consider a periodic signal f(t) = f(t+ T ) and its Fourier series representation
f(t) =
1Xn=¡1
cnejn!0t (!0 = 2¼=T ):
The average power (per period) \in" or \of" f(t) is
Pav =1
T
t0+TZt0
jf(t)j2 dt
=1
T
t0+TZt0
f(t)f¤(t) dt =1
T
t0+TZt0
1Xn=¡1
cnejn!0t
1Xm=¡1
c¤me¡jm!0tdt
=
1Xn=¡1
cn
1Xm=¡1
c¤m1
T
t0+TZt0
ej(n¡m)!0tdt =1X
n=¡1cn
1Xm=¡1
c¤m ±mn =1X
n=¡1cnc
¤n
=
1Xn=¡1
jcnj2
F Homework: Find the comparable expression in terms of the trigonometric Fourier
series coe±cients.
134
Homework Problem: Power Supply Performance.
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¡T 0 T
t
sin(12!0t) = sin(¼t=T )
.......
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.....
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.....
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.....
¡T 0 T
t
x(t) = j sin(12!0t)j = j sin(¼t=T )j
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................................................................................................................
x(t) =2
¼¡ 1
¼
1Xn=1
1
n2 ¡ 14
cos(2n¼t=T )
Consider a real power supply, consisting of a full-wave recti¯er followed by a simple low-
pass ¯lter. Find an expression for, and graph, the time-domain output y(t) of the power
supply for several values of the important parameter !b=!0. De¯ne the percent ripple as
the ratio of average power in the unwanted harmonics to the desired DC power. Graph
percent ripple versus !b=!0. Recall Parseval's theorem for Fourier series.
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............................................
²
²R
C
+
¡
x(t)
+
¡
y(t)
135
Solution. The transfer function of the low-pass ¯lter is
H(!) =
1
j!C
R+1
j!C
=1
1 + j!RC=
1
1 + j!=!bwith !b ,
1
RC:
In polar form
H(!) = jH(!)jej£(!)
the magnitude of H(!) is
jH(!)j = 1p1 + (!=!b)2
and the phase of H(!) is
£(!) = ¡ tan¡1(!=!b):Note that the ¯lter impulse response is necessarily real, since the ¯lter is an actual, physical
electric circuit. Also note that H(0) = 1 is real; that is, £(0) = 0. With the ¯lter input
written as the particular Fourier series
x(t) =
1Xn=0
an cos(n!0t);
the response of the ¯lter is
y(t) =
1Xn=0
anjH(n!0)j cos[n!0t+£(n!0)]:
Note that the DC term is now called a0, instead of a0=2. Recall that !0 , 2¼=T is the
fundamental frequency of the recti¯ed sine, which is twice the frequency of the pure sine
wave that is the initial input to the full-wave recti¯er. Speci¯cally, the above expression is
y(t) =2
¼¡ 1
¼
1Xn=1
1
n2 ¡ 14
1p1 + (n!0=!b)2
cos£n!0t¡ tan¡1(n!0=!b)
¤:
Note that the single (dimensionless!) numerical parameter required to characterize the
complete power supply output is !0=!b. Time t is always multiplied by !0 = 2¼=T , so
that in terms of normalized time t=T , we have
y
μt
T
¶=2
¼¡ 1
¼
1Xn=1
1
n2 ¡ 14
1p1 + (n!0=!b)2
cos
·2n¼
t
T¡ tan¡1(n!0=!b)
¸: (1)
The form of the Fourier series for y(t) is, with An = anjH(n!0)j and Án = £(n!0),
y(t) =
1Xn=0
An cos(n!0t+ Án)
136
and the total average power in the output y(t) is
Ptot =1
T
t0+TZt0
jy(t)j2 dt
=1
T
t0+TZt0
" 1Xn=0
An cos(n!0t+ Án)
#" 1Xm=0
Am cos(m!0t+ Ám)
#¤dt
=
1Xn=0
1Xm=0
AnA¤m
1
T
t0+TZt0
cos(n!0t+ Án) cos(m!0t+ Ám) dt:
The DC terms (n = 0 and/or m = 0) are a little di®erent, but I included them in the series
(starting at n = 0 instead of starting at n = 1) for notational convenience. Recall or note
that Á0 ´ 0. The inner products of interest are:
1
T
t0+TZt0
1 ¢ 1 dt = 1 (n = m = 0)
1
T
t0+TZt0
1 ¢ cos(m!0t+ Ám) dt = 0 (n = 0;m = 1; 2; : : : )
1
T
t0+TZt0
cos(n!0t+ Án) cos(m!0t+ Ám) dt =1
2±nm (n;m = 1; 2; : : : ):
Therefore, the total average power in the signal y(t) is
Ptot = jA0j2 + 12
1Xn=1
jAnj2;
which is the form of Parseval's theorem that applies for our particular representation of
y(t). The DC (n = 0) term is our desired power supply output, while all of the higher
harmonics (n = 1; 2; : : : ) give rise to the unwanted \ripple." Therefore, one de¯nition of
percent ripple is
r , unwanted power in the ripples
total power output£ 100%
or
r =
12
1Pn=1
jAnj2
jA0j2 + 12
1Pn=1
jAnj2£ 100%: (2)
137
However, the problem statement evidently de¯ned percent ripple as
r =
12
1Pn=1
jAnj2
jA0j2 £ 100%:
These two r expressions will be nearly identical when the majority of the output power is
in the DC component, but for a very poor power supply I suppose some marketing/sales
people would prefer to use (2). The second form might be greater than 100%!
¡0:75 ¡0:50 ¡0:25 0.00 0.25 0.50 0.75
t=T
¡0:2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
y(t)
!i=!b = 0:1
10
1
input frequency of unrecti¯ed sine is !i = !0=2
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0 1 2 3 4 5
!i=!b = freq of unrecti¯ed sine/¯lter break
0
2
4
6
8
10
12
14
16
18
20
r (%)
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138
Periodic Excitation of a Simple GLC Electric Circuit
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................ig(t)
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L
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...
² ²
² ²
G C
v(t)
.................................................................................
If the source signal ig(t) = ig(t + T ) is a periodic signal, then the steady-state solution
for all of the linear signals (common voltage v(t), all three branch currents) will also be
periodic with the same fundamental frequency !0 = 2¼=T of the excitation. If we know
the waveform ig(t), then we can ¯nd its Fourier series representation
ig(t) =
1Xn=¡1
cnejn!0t: (1)
Here is a case where it is more convenient and natural to use the complex exponential
form of the Fourier series: The trigonometric form would su±ce in principle, but carrying
around two sets of coe±cients an and bn is more awkward. Plus di®erentiating exp(jn!0t)
is much cleaner than di®erentiating cos(n!0t) and sin(n!0t).
Kirchho®'s current law supplies the equation-of-motion for the unknown voltage signal
v(t) in our circuit
Gv(t) + Cdv(t)
dt+1
L
tZ¡1
v(¿) d¿ = ig(t): (2)
This integrodi®erential equation is converted to a pure di®erential equation by di®erenti-
ating once to eliminate the integral operator in the inductive current
Cd2v(t)
dt2+G
dv(t)
dt+1
Lv(t) = i0g(t): (3)
The order of the terms was rearranged to put the di®erential equation in standard form,
and the prime on the generator current denotes di®erentiation with respect to its argument,
which is our independent variable t. The periodic voltage signal is written as a Fourier
series
v(t) =
1Xn=¡1
°nejn!0t (4)
139
where the Fourier coe±cients °n are unknown, so far. Insertion of both series (1) and (4)
into the di®erential equation (3) gives
1Xn=¡1
°n
·(jn!0)
2C + (jn!0)G+1
L
¸ejn!0t =
1Xn=¡1
(jn!0)cnejn!0t: (5)
Since the above two Fourier series in (5) are equal for all t and they are written in terms
of the same Fourier basis functions exp(jn!0t), the corresponding coe±cients are equal
°n
·(jn!0)
2C + (jn!0)G+1
L
¸= (jn!0)cn: (6)
Therefore the °n are given explicitly in terms of the known cn
°n =jn!0
(jn!0)2C + (jn!0)G+1
L
cn =jn!0L
1¡ (n!0)2LC + jn!0LG cn; (7)
and our particular solution (4) for the voltage is
v(t) =
1Xn=¡1
jn!0L
1¡ (n!0)2LC + jn!0LG cnejn!0t: (8)
Note that the (SI or mksC) units of the numerator are those of !0L, which are (H/s) or
(−). All three terms in the denominator are dimensionless (check for yourself), and the
Fourier coe±cients cn de¯ned in (1) are in (A). The units of the °n are therefore (V), as
required.
Problems
1. Find an expression for the average power dissipated as heat in this circuit.
2. Give a physical explanation for the absence (zero value) of the n = 0 term in the voltage.
3. If the lossy element is removed from the circuit (G = 0 or G¡1 = R =1), what is thephysical signi¯cance of a harmonic having n!0 = 1=
pLC ?
4. Write down an approximation for the higher-order terms in the series for v(t), that
is as n ! +1. (The terms with negative indices behave similarly.) Interpret this highfrequency approximation in terms of the circuit.
5. If f(t) is a real function of time, show that the coe±cients cn in its complex exponential
Fourier series
f(t) =
1Xn=¡1
cnejn!0t
must satisfy the requirement cn = c¤¡n. One way to see this is to examine f
¤(t) = f(t).
6. If ig(t) is a real-valued source current, show that the expression (8) is also real.
7. Graph the signal v(t) when the current source is a square wave, say of amplitude §I0.Show the e®ect of varying the circuit element values. Use normalized parameters, as much
as possible, to simplify the results.
140
Simple Low-Pass and Hi-Pass Filter Response to a Periodic Input
Problem: Find expressions for and graph the signals ya(t) and yb(t) for several values of
the important (nondimensional) parameters that appear naturally.
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¡T 0 T 2Tt
................... Ax(t)
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..
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²
²R
C
+
¡
x(t)
+
¡
ya(t)
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..
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..
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......
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........
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......
.......
..
²
²C
R
+
¡
x(t)
+
¡
yb(t)
Note that the excitation x(t) has both even and odd parts, but the even part is simply
A=2. The graph of ex(t) = x(t)¡A=2 is clearly an antisymmetric (odd) function, so that itsFourier series is most logically written in the trigonometric form with only sine functions.
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.........................................................................................¡T 0 T 2Tt
...................A=2
...................¡A=2
ex(t) = x(t)¡A=2
ex(t) = ¡ex(¡t) ex(t) = Aμ tT¡ 12
¶(0 < t < T=2)
ex(t) = 1Xn=1
bn sin(n!0t) (!0 , 2¼=T )
bn =2
T
T=2Z¡T=2
ex(t) sin(n!0t) dt = 4
T
T=2Z0
ex(t) sin(n!0t) dt=4A
T
T=2Z0
μt
T¡ 12
¶sin(n!0t) dt
141
T=2Z0
sin(n!0t) dt =cos(n!0t)
n!0
¯̄̄̄0t=T=2
=1¡ cos(n!0T=2)
n!0=1¡ cos(n¼)
n!0=1¡ (¡1)nn!0
=
8<:2
n!0n odd
0 n even
(but this form is not as useful as it might appear)
Zu dv = uv¡
Zv du u = t du = dt dv = sin(n!0t) dt v = ¡cos(n!0t)
n!0
T=2Z0
t sin(n!0t) dt =t cos(n!0t)
n!0
¯̄̄̄0t=T=2
+1
n!0
T=2Z0
cos(n!0t) dt
= ¡ T
2n!0cos(n!0T=2) +
sin(n!0t)
(n!0)2
¯̄̄̄T=2t=0
= ¡ T
2n!0cos(n¼) +
sin(n¼)
(n!0)2
=¡T (¡1)n2n!0
Now go back and assemble the required Fourier coe±cient
bn =4A
T
·¡(¡1)n2n!0
¡ 1¡ (¡1)n
2n!0
¸=
¡4A2n!0T
=¡An¼:
The Fourier series representation of the input signal is therefore
x(t) =A
2¡ A¼
1Xn=1
sin(n!0t)
n:
The transfer functions of our two ¯lters, from a simple frequency domain voltage divider,
are
Ha(!) =
1
j!C
R+1
j!C
=1
1 + j!RC= jHa(!)jej£a(!)
Hb(!) =R
R+1
j!C
=j!RC
1 + j!RC= jHb(!)jej£b(!):
Let's not write out the polar representations explicitly: MATLAB will compute those for
us. That is, assume the magnitude function jHa(!)j and £a(!) are known (they are!), andobviously the same for the b circuit. Note that Ha(0) = 1 and Hb(0) = 0.
142
The Fourier series for the two output signals are therefore
ya(t) =A
2¡ A¼
1Xn=1
jHa(n!0)jn
sin£n!0t+£a(n!0)
¤
yb(t) = ¡A¼
1Xn=1
jHb(n!0)jn
sin£n!0t+£b(n!0)
¤Let's use normalized time t=T and normalize by the amplitude A of the input to write
ya(t=T )
A=1
2¡ 1
¼
1Xn=1
jHa(n!0)jn
sin£2n¼t=T +£a(n!0)
¤yb(t=T )
A= ¡ 1
¼
1Xn=1
jHb(n!0)jn
sin£2n¼t=T +£b(n!0)
¤We need
Ha(n!0) =1
1 + jn!0RC=
1
1 + jn!0=!b¡¡¡!n!1
!b
jn!0
Hb(n!0) =jn!0RC
1 + jn!0RC=
jn!0=!b
1 + jn!0=!b¡¡¡!n!1
1
where
!b ,1
RC
is the break frequency (or cut-o® frequency or half-power frequency) of (both of) the ¯lters.
Therefore, the important nondimensional parameter of our problem is !0=!b = !0RC.
Recall that !0 = 2¼=T is the fundamental frequency of the periodic excitation.
As n!1, the terms in the series for the low-pass ¯lter output ya(t) vary like n¡2 whilethe higher-order terms in the high-pass ¯lter output yb(t) vary like n
¡1. Therefore, thenumerical convergence of the signal ya(t) is much faster. That is, truncation of the ya(t)
series at some ¯nite value of n will give a much better approximation for the in¯nite series
than will the same truncation of the series for yb(t). Note that the outputs ya(t) and yb(t)
are intimately related since
Ha(!) +Hb(!) = 1 =) ya(t) + yb(t) = x(t):
Therefore, we can numerically evaluate the series for ya(t) (because its series converges
faster), and then avoid computing the poorly convergent series for yb(t) simply by using
the known relationship
yb(t) = x(t)¡ ya(t):
143
...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................t=T
!0=!b = 0:1
ya(t)
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yb(t)
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144
...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................t=T
!0=!b = 1
ya(t)
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yb(t)
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145
...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................t=T
!0=!b = 10
ya(t)
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yb(t)
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146
The Vibrating String
This is one of the most celebrated problems of mathematical physics. And since the early
1950's, the unchallenged lead instrument in the sound track of Western culture has been
constructed around a set of six such steel strings. Let's see what our Fourier analysis can
tell us about this simple and important wave equation.
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x
y(x; t0)
0 a
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Fig. 1. Snapshot of string displacement at time t = t0.
Fig. 1 shows the pro¯le (at some particular time t0) of the vertical de°ection of a string, of
total length a, about its equilibrium position y = 0. The string is ¯xed at each end; hence
y(0; t) = y(a; t) = 0 (1)
are the boundary conditions imposed on our unknown function y(x; t), for all t.
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T
T
μ
μ0
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x x+¢x
Fig. 2. Free-body diagram of elemental string length.
147
Consider the free-body diagram (Fig. 2) of a di®erential length of string. If the entire
string is subject to a tensile force T (N), then the sum of the forces in the y-direction,
according to Newton's second law, must be
T sin μ0 ¡ T sin μ = ½¢x@2y
@t2(2)
where ½ is the lineal mass density (kg/m) of the string. For small displacements about
equilibrium
sin μ0 ¼ tan μ0 = @
@xy(x+¢x) and sin μ ¼ tan μ = @
@xy(x) (3)
resulting in@
@xy(x+¢x; t)¡ @
@xy(x; t)
¢x=½
T
@2
@t2y(x; t): (4)
In the limit as ¢x! 0, we have the wave equation·@2
@x2¡ 1
c2@2
@t2
¸y(x; t) = 0 (5)
where c =pT=½ will be the wave speed in (m/s).
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0 a
x
Á(x)
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Fig. 3. Initial displacement of string.
Let's excite our string by holding it in the initial rest position
y(x; 0) = Á(x) (one possibility shown in Fig. 3) (6)
@
@ty(x; 0) = 0 (initially at rest) (7)
and releasing it. We can solve our partial di®erential equation (5) by expanding y(x; t)
into a Fourier series in x, which is equivalent to periodically extending the function y(x)
beyond our original physical domain 0 · x · a. Since y vanishes at the end points x = 0
148
and x = a, we can use a sine series, which is the odd periodic extension of period 2a. So
the form of our sought-after solution is
y(x; t) =
1Xn=1
Bn(t) sin(n¼x=a); (8)
where the functions Bn(t) (Fourier coe±cients for the x-expansion) are unknown at this
point. We need the appropriate Fourier series representation of our initial displacement
(6); write it as
Á(x) =
1Xn=1
bn sin(n¼x=a) (9)
where the bn are known, in principle and in reality, since we know the initial displacement
Á(x):
bn =2
a
aZ0
Á(x) sin(n¼x=a) dx: (10)
Insertion of our solution form (8) into the wave equation (5) yields
1Xn=1
·¡³n¼a
´2¡ 1
c2d2
dt2
¸Bn(t) sin(n¼x=a) = 0: (11)
This is itself a Fourier series representation of zero, so each coe±cient must be zero; that
is ·d2
dt2+³n¼ca
´2 ¸Bn(t) = 0: (12)
The solution of this homogeneous di®erential equation is a linear combination of½cos
sin
¾μn¼ct
a
¶:
We select the cosine (discard the sine) in accordance with the initial condition (7). Now
we have Bn(t) = ¯n cos(n¼ct=a) in terms of (unknown) constants ¯n, which gives
y(x; t) =
1Xn=1
¯n cos(n¼ct=a) sin(n¼x=a): (13)
Evaluation at t = 0, according to the initial condition (6), provides ¯n = bn, and so
y(x; t) =
1Xn=1
bn cos(n¼ct=a) sin(n¼x=a): (14)
149
The physical interpretation of our solution(!) (this is it!!!) is much clearer upon rewriting
it (via a trig identity) as
y(x; t) =1
2
1Xn=1
bn
nsinhn¼a(x¡ ct)
i+ sin
hn¼a(x+ ct)
io(15)
or, ¯nally,
y(x; t) =1
2
h~Á(x¡ ct) + ~Á(x+ ct)
i(16)
where ~Á(¢) means the odd periodic extension (Fig. 4) of the initial displacement Á(x) asgiven in the sine series (9):
~Á(x) =
1Xn=¡1
Á(x¡ 2na)¡1X
n=¡1Á[¡x¡ (2n+ 1)a]: (17)
In (17) we are taking Á(») to be zero outside of the original domain 0 · » · a. Also notethat ~Á(x¨ ct) is a wave propagating with speed c in the §x-direction.
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...
¡a ..........
0..........
a..........
2ax
~Á(x)
Fig. 4. Odd periodic extension of initial displacement of string.
Problems
1.) Verify that any functions f(x + ct) and g(x ¡ ct) satisfy the wave equation (5).Certainly, then, (16) is a solution. Such forms are called the d'Alembert solution.
2.) Verify that (16) satis¯es the initial conditions (6) and (7), and the homogeneous
boundary conditions (1).
3.) Why do we say that f(x + ct) and g(x ¡ ct) are propagating in the ¨x-directions,respectively, with speed c?
150
Poisson Sum Formula
Let f(t) be any function that is Fourier transformable. Then its periodic extension can be
represented by a Fourier series:
g(t) =
1Xn=¡1
f(t+ nT ) =
1Xk=¡1
ckej2k¼t=T
with Fourier coe±cients
ck =1
T
TZ0
g(t)e¡j2k¼t=Tdt
=1
T
1Xn=¡1
TZ0
f(t+ nT )e¡j2k¼t=T dt
with ¿ = t+ nT
=1
T
1Xn=¡1
(n+1)TZnT
f(¿)e¡j2k¼¿=T ej2kn¼d¿
=1
T
1Z¡1
f(¿)e¡j2k¼¿=Td¿ =1
TF
μ2k¼
T
¶:
The resulting Poisson sum formula
1Xn=¡1
f(t+ nT ) =1
T
1Xk=¡1
F
μ2k¼
T
¶ej2k¼t=T
is often written in the common form, with t = 0 and T = 1 such that
1Xn=¡1
f(n) =
1Xk=¡1
F (2k¼):
F Homework. Sum the slowly converging series
1Xn=¡1
1
n2 + b2=¼
bcoth¼b:
151
Solution. Recall that the Fourier transform of g(t) = exp(¡®jtj) with ® > 0 is
G(!) =
0Z¡1
e(®¡j!)tdt+
1Z0
e¡(®+j!)tdt =1
®¡ j! +1
®+ j!=
2®
®2 + !2:
Duality says that the Fourier transform pair
g(t)F() G(!) also gives G(t)
F() 2¼g(¡!):so that the Fourier transform of
f(t) =1
t2 + ®2
is
F (!) =¼
®e¡®j!j:
We will also use the geometric series
1Xk=1
zk =z
1¡ z (jzj < 1):
The Poisson sum formula 1Xn=¡1
f(n) =
1Xk=¡1
F (2k¼)
therefore gives (assume b > 0 and note that the desired sum is even in b)
1Xn=¡1
1
n2 + b2=¼
b
1Xk=¡1
e¡2jkj¼b =¼
b
"1 + 2
1Xk=1
e¡2¼bk#=¼
b
·1 +
2e¡2¼b
1¡ e¡2¼b¸
=¼
b
1 + e¡2¼b
1¡ e¡2¼b =¼
b
e¼b + e¡¼b
e¼b ¡ e¡¼b =¼
bcoth(¼b): ~
It is interesting to look at the limiting case of very small b. If jbj ¿ 1, then
1Xn=¡1
1
n2 + b2¡¡¡!b!0
1
b2+ 2
1Xn=1
1
n2=1
b2+ 2³(2) =
1
b2+¼2
3:
Formula 4.5.67 of Big Red shows that, for jzj < ¼,
coth z =1
z+z
3¡ z3
45+ : : :
so that¼
bcoth(¼b) ¡¡¡!
b!0
¼
b
·1
¼b+¼b
3¡ (¼b)
3
45+ : : :
¸=1
b2+¼2
3+O(b2):
How can you see the behavior of the sum for very large b, that is for b!1 ?
152
Sampling and Reconstruction
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combT (t)
x(t) xs(t)h(t)
y(t)
Consider a continuous-time (analog) signal x(t) that is uniformly sampled at the times
tn = nT (n = : : : ;¡2;¡1; 0; 1; 2; 3; : : : ) so that the sampling frequency is
!s =2¼
T(1)
in terms of the sampling period or interval T . Once x(t) has been sampled, all we have is
a table (list) of numbers x(nT ). But a useful representation of the sampled signal xs(t) in
continuous-time notation is
xs(t) = x(t) combT (t) =
1Xn=¡1
x(nT )±(t¡ nT ): (2)
From our table of Fourier transform properties we need
F£f(t)g(t)¤ = 1
2¼F (!)~G(!): (3)
The easiest way to get this is to take the inverse Fourier transform of the frequency-domain
convolution
F¡1£F (!)~G(!)¤ = 1
2¼
1Z¡1
1Z¡1
F (−)G(! ¡ −) d− ej!t d!
=
1Z¡1
F (−)1
2¼
1Z¡1
G(! ¡ −)ej!td! d−:
The inner integral (in !) is
1
2¼
1Z¡1
G(! ¡ −)ej!td! = 1
2¼
1Z¡1
G(¸)ej(¸+−)td¸ = ej−tg(t)
and therefore
F¡1£F (!)~G(!)¤ = g(t) 1Z¡1
F (−)ej−td− = 2¼f(t)g(t)
which is identical to (3).
Typeset by AMS-TEX
153
The Fourier fransform of the time-domain comb function is
F£combT (t)¤ = 2¼
Tcomb2¼
T
(!) (4)
and thus the spectrum of the sampled signal xs(t) is
Xs(!) =1
2¼X(!)~ 2¼
Tcomb2¼
T
(!) =1
TX(!)~
1Xk=¡1
±(! ¡ k!s)
=1
T
1Xk=¡1
X(! ¡ k!s): (5)
If x(t) is a \low-pass" signal, then
X(!) = 0 for j!j > !h: (6)
This means that the spectrum of x(t) has a highest frequency !h. An example low-pass
spectrum appears in the ¯rst row of the ¯gure below.
X(!)
Xs(!) when !s=2 > !h
Xs(!) when !s=2 < !h
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¡2!s...........
¡!s...........
0...........
2!s
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!s.............
¡!s=2.............
!s=2
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¡!s...........
!s...........
0.............
¡!s=2.............
!s=2
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0...........
¡!h...........
!h
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The middle curve showsXs(!) for the case !s > 2!h and the bottom curve is for !s < 2!h.
If we pass xs(t) through an ideal low-pass ¯lter having transfer function
H(!) = T¦
μ!
!s
¶; (7)
154
then we see that, for all !,
Y (!) = X(!) if !s > 2!h: (8)
In this case, y(t) = x(t) and we have exactly recovered the low-pass signal x(t) from its
sample values! The minimum sampling frequency is called the Nyquist rate or Nyquist
frequency 2!h. If we sample at too low a frequency, as shown in the bottom curve of the
previous ¯gure, then
Y (!)6= X(!) if !s < 2!h (9)
and of course then y(t)6= x(t). The error or distortion that results from the overlap in the
adjacent shifted transforms X(! ¡ k!s) is called aliasing. The whole concept is usuallycalled:
Shannon's sampling theorem. If the low-pass signal x(t) having spectrum X(!) = 0
for j!j > !h is sampled uniformly at the rate !s > 2!h (the Nyquist rate), then x(t) is
exactly recoverable by passing the sampled signal through the ideal reconstruction ¯lter
having transfer function H(!) = T¦(!=!s).
The frequency domain renders this derivation clear and simple. Now let's examine the
output y(t) in the time-domain to get another perspective. The impulse response of the
ideal reconstruction ¯lter is
h(t) = F¡1£T¦(!=!s)¤ = sinc(!st=2) = sinc(¼t=T ); (10)
and this ideal low-pass ¯lter is also called the sinc interpolation ¯lter. Independent of x(t)
being a low-pass signal or not, and for an arbitrary sampling frequency (that is, ignoring
both Nyquist and Shannon), the output of the ¯lter in response to the sampled signal (2)
is always
y(t) = xs(t)~ h(t) =1X
n=¡1x(nT )±(t¡ nT )~ h(t) =
1Xn=¡1
x(nT )h(t¡ nT )
=
1Xn=¡1
x(nT ) sinc
·¼
μt
T¡ n
¶¸: (11)
If k is an integer, observe that
sinc(k¼) =sin(k¼)
k¼=
½1; k = 0
0; k6= 0 = ±k0; (12)
in terms of the Kronecker-delta. Therefore, we see that at the sample times t = mT , the
output is always equal to the input in that
y(mT ) =
1Xn=¡1
x(nT ) sinc [¼ (m¡ n)] =1X
n=¡1x(nT )±mn = x(mT ): (13)
155
Dissection of the Time-Domain Output of the sinc Interpolation Filter: Consider
a small number (four) of samples x[n] = x(nT ) for n = 0; 1; 2; 3. If all other (n = 4; 5; : : :
and n = ¡1;¡2; : : : ) of the x[n] are zero (or absent), then the output (11) of the sincinterpolation ¯lter is
y(t) = x[0] sinc
μ¼t
T
¶+ x[1] sinc
·¼
μt
T¡ 1¶¸
+ x[2] sinc
·¼
μt
T¡ 2¶¸+ x[3] sinc
·¼
μt
T¡ 3¶¸:
Each of these individual terms are graphed below for the example data
x[0] = 0:5 x[1] = 1:0 x[2] = 0:8 x[3] = 0:6
along with the sum of all four terms. Do you see which curve is which?
¡2 ¡1 0 1 2 3 4 5
t=T
¡0:4
¡0:2
0.0
0.2
0.4
0.6
0.8
1.0
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²
²
²
²
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156
DISCRETE-TIME SIGNALS, SYSTEMS, AND TRANSFORMS
A common source for a discrete-time signal is to uniformly sample a continuous-time signal,
as in
x[n] = x(nT ) n = : : : ;¡2;¡1; 0; 1; 2; 3; : : :The discrete-time signal x[n] is also called a sequence; it is a function of the integers
n. Usually we are not concerned with any underlying sample period T that was used to
obtain our sequence from some original x(t). Once we're in the discrete-time world, the
independent time variable is simply the integers n. Our sequence x[n] is simply a list or
table of ordered numbers. Note that the sequence x[n] can take on real or complex values.
Many of our familiar continuous-time signals have discrete-time analogues or cousins, but
be aware that their behavior is often (usually? always?) quite di®erent. For starters, there
is no such thing as continuity in discrete-time. The continuous-time calculus operators of
di®erentiation and integration have analogous operators of di®erencing and accumulation
in discrete-time. Sometimes the discrete-time version of a common signal, for example
the Heaviside unit-step function u[n], goes by the same name and similar symbol as its
continuous-time counterpart. In other cases, di®erent names are used. I suppose the
Heaviside unit-step sequence is a better name that draws attention to its di®erence with
the continuous-time signal u(t).
Heaviside unit-step sequence.
u[n] ,½0; n = ¡1;¡2; : : :1; n = 0; 1; 2; : : :
..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................² ² ²
² ² ² ² ²
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.
n
u[n]: : : : : :
¡3 ¡2 ¡1 0 1 2 3 4
Unlike u(t) that is not well-de¯ned at t = 0, the unit-step sequence clearly takes on the
value u[0] = 1. Note, however that u[0:5] is complete notational nonsense.
Kronecker delta sequence. (also called unit impulse sequence)
±[n] ,½0; n = §1;§2;§3; : : :1; n = 0
..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................² ² ²
²
² ² ² ²................................................................
n
±[n]: : : : : :
¡3 ¡2 ¡1 0 1 2 3 4
Unlike the continuous-time Dirac-delta function ±(t) that is illegal at t = 0, the discrete-
time Kronecker-delta sequence clearly takes on the value ±[0] = 1. The Kronecker-delta
possesses the sampling property
x[n] =
1Xk=¡1
x[k]±[k ¡ n]:
Relationship.
±[n] = u[n]¡ u[n¡ 1] u[n] =
nXk=¡1
±[k]
157
Complex exponential. First, recall that for any positive real number ! (called the
frequency), the continuous-time complex exponential
x(t) = ej!t
is always periodic in time t with period T = 2¼=!, since
x(t+ T ) = ej!(t+T ) = ej!T ej!t = ej!t = x(t)
whenever !T = 2¼, since e2¼j = 1.
The discrete-time complex exponential is substantially di®erent. Consider the discrete-
time sequence
x[n] = ej−n
where the frequency − is any real number: Take it to be positive for starters. Is x[n]
periodic in the independent time variable n ? Let's denote its candidate period by the
speci¯c integer N , and examine
x[n+N ] = ej−(n+N) = ej−Nej−n:
If x[n] is periodic with the period N , then x[n+N ] = x[n] which requires that
−N = 2k¼
where k is any integer. This only happens if
− =k
N2¼ = a rational multiple of 2¼:
However, the discrete-time complex exponential is a periodic function of the frequency
variable −, since
ejn(−+2¼) = e2n¼jejn− = ejn−:
Therefore, in discrete-time work, we only have to consider frequencies − in the primary
range ¡¼ < − · ¼, for example. The highest frequency possible is − = ¼ (or equivalently,¡¼), and the lowest frequency is − = 0 (which is identical with − = 2¼).
158
Exercises.
1. Write an expression, in terms of n, N ,
and u[¢], for the given sequence x[n].
.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................² ² ² ²
²
²
²
²
²
²
² ²
1
2
3
N¡1
N
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n
x[n]
: : : : : :
¡3 ¡2 ¡1 0 1 2 3 : : : N¡1 N N+1
2. Graph the exponential (or geometric) sequence x[n] = anu[n].
Look at the following speci¯c cases:
a = 1=2; a = ¡1=2; a = 2; a = rej− where r = 1=2 and − = 1
Discrete-Time Systems
....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................................................................................... ............. ...................................................................................... .............x[n] y[n]
The equation-of-motion, that describes the relationship between the input x[n] and the
output y[n], for a discrete-time system is often a di®erence equation. An example of a
second-order, constant-coe±cient di®erence equation is
c0y[n] + c1y[n¡ 1] + c2y[n¡ 2] = b0x[n] + b1x[n¡ 1]:When accompanied by two (the di®erence equation is second order) initial conditions of
the form
y[¡1] = y¡1 and y[¡2] = y¡2;we have a discrete-time initial value problem. Classical solution techniques are analogous
to the classical methods used to solve ordinary di®erential equations in continuous-time.
Di®erence equations can sometimes be solved recursively:
y[n] = ¡c1c0y[n¡ 1]¡ c2
c0y[n¡ 2] + b0
c0x[n] +
b1
c0x[n¡ 1]:
The discrete-time version of the Laplace transform (or Fourier transform) is called the
Z-transform, and it, too, provides an attractive path to solve discrete-time di®erenceequations by converting them into purely algebraic equations in the transform (z) domain.
The inverse Z-transform gets our solution back to the time [n] domain.
159
Linear Time-Invariant Discrete-Time Systems
....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................................................................................... ............. ...................................................................................... .............x[n] y[n] = Tfx[n]g
Assume that the zero-state response sequence y[n] to the exciting sequence x[n] is given
by the operator or system transformation rule y[n] = Tfx[n]g. If the operator T satis¯esthe superposition property
Tfc1x1[n] + c2x2[n]g = c1Tfx1[n]g+ c2Tfx2[n]g
for any values of the constants c1 and c2 and for arbitrary sequences x1[n] and x2[n], then
the discrete-time system is linear.
Given that the zero-state response to the input sequence x[n] is y[n], if the response to the
delayed input x[n¡n0] is the delayed response y[n¡n0], then the system is time-invariant.
....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................................................................................... ............. ...................................................................................... .............x[n] y[n] = Tfx[n]g
....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................................................................................... ............. ...................................................................................... .............x[n¡ n0] y[n¡ n0] = Tfx[n¡ n0]g
Input/Output relationship. Using the sampling property of the Kronecker-delta se-
quence, write the input sequence as
x[n] =
1Xk=¡1
x[k]±[n¡ k]:
The system operator T operates only on functions of the running time variable n, treating
all other variables as constants, so that the T operator can be moved inside the summation
in
y[n] = Tfx[n]g = T( 1Xk=¡1
x[k]±[n¡ k])=
1Xk=¡1
x[k]Tf±[n¡ k]g:
The response to a Kronecker-delta or unit impulse sequence is the system impulse response
h[n] = Tf±[n]g. By the (assumed or de¯ned) time-invariance of the linear system, the
160
response due to a delayed unit impulse ±[n¡ k] is the delayed h[n¡ k] = Tf±[n¡ k]g, andtherefore the zero-state response is
y[n] =
1Xk=¡1
x[k]h[n¡ k] = x[n]~ h[n]:
This is the discrete-time convolution sum between the two signals x[n] and h[n].
....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................................................................................... ............. ...................................................................................... .............x[n] y[n] = x[n]~ h[n]h[n]
Example of a LTI Discrete-Time System. Let's look at one particular discrete-time
system in detail and study its input/output dynamics from several perspectives. One
common way to specify a discrete-time system is via a system block diagram. We need to
¯rst introduce several basic components of such system block diagrams.
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Pf [n]
g[n]
f [n] + g[n]summer
.....................................................................................
............................
................................................................
.............................................................................. .............. .............................................................................. ..........
....®f [n] ®f [n]ampli¯er/attenuator
............................................................................................................................................................................................................................
.............................................................................. .............. .............................................................................. ..........
....z¡1f [n] f [n¡ 1]unit delay
The reason that the unit delay is represented by the symbol z¡1 will be clear when westudy the Z-transform.
161
.......
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.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
z¡1®
Px[n] y[n]²
The output of the summer is
y[n] = x[n] + ®y[n¡ 1]
and so the discrete-time system depicted by the block diagram above is governed by an
equation-of-motion that is a linear, constant coe±cient, ¯rst-order di®erence equation.
The impulse response h[n] is the response y[n] to the input sequence x[n] = ±[n], and being
a zero-state response, its initial condition is h[¡1] = 0. Sometimes di®erence equations
can be solved recursively, and this is one
h[n] = ®h[n¡ 1] + ±[n]h[¡1] = 0h[0] = ®h[¡1] + 1 = 1h[1] = ®h[0] + 0 = ®
h[2] = ®h[1] = ®2
h[3] = ®h[2] = ®3
...
h[n] = ®nu[n]
With the unit impulse response h[n] in hand, the zero-state response due to any input
sequence is readily computed via the convolution sum. For example, the unit step response
is
s[n] = u[n]~ h[n] =1X
k=¡1h[k]u[n¡ k] =
nXk=¡1
h[k] =
nXk=¡1
®ku[k] =
nXk=0
®k
=1¡ ®n+11¡ ® u[n]:
The required ¯nite geometric series is detailed two pages ahead.
162
The Z-Transform
In order to derive the Z-transform from the Laplace transform, we need to represent our
sequence x[n] in continuous-time notation: Take a step backwards, so to speak, and write
a sampled (discretized) continuous-time signal as
xd(t) = x(t) combT (t) =
1Xn=¡1
x(nT )±(t¡ nT ) =1X
n=¡1x[n]±(t¡ nT ):
t...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
²
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²
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²..............
²
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²
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²
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..
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..
T............................................. ................ .............................................................
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The bilateral Laplace transform of a continuous-time signal f(t) is de¯ned as the integral
Lff(t)g = F (s) =1Z
¡1f(t)e¡stdt
where s = ¾+j! is a complex frequency variable. The Laplace-transform of the discretized
signal xd(t) is therefore
Xd(s) =
1Xn=¡1
x[n]
1Z¡1
±(t¡ nT )e¡stdt =1X
n=¡1x[n]e¡nsT :
Introduce the transformed variable
z = esT
so that the above Laplace-transform is now called the Z-transform
Zfx[n]g = X(z) =1X
n=¡1x[n]z¡n:
The Z-transform converts a function of the integers n (a sequence) to a function of a
complex variable z.
163
Geometric series. Firstly, consider the ¯nite geometric series
S =
NXn=0
an
where a is an arbitrary complex number. There is no question about the convergence
of this series, since it is simply the sum of a ¯nite number of terms. We can obtain a
closed-form expression for the sum S by this neat trick:
S = 1+a+ a2 + a3 + ¢ ¢ ¢+ aN¡1 + aNaS = a+ a2 + a3 + ¢ ¢ ¢+ aN¡1 + aN + aN+1
S ¡ aS = 1¡aN+1
S(1¡ a) = 1¡ aN+1 =) S =
NXn=0
an =1¡ aN+11¡ a
Note that when a = 1, we have S = N +1, which is the limit as a! 1 of the above result.
Similarly, the expression for the in¯nite geometric series
1Xn=0
an =1
1¡ a (jaj < 1)
explicitly displays its region-of-convergence jaj < 1.
Z-transform example: causal exponential sequence. Consider the sequence
x[n] = anu[n]
where a is an arbitrary complex number. Then the Z-transform of x[n] is
X(z) =
1Xn=0
³az
´n=
1
1¡ az
=z
z ¡ a (jzj > jaj):
The region-of-convergence in the complex z-plane for X(z) to be a valid transform of x[n]
is the exterior of the circle of radius jzj = jaj.
164
Fibonacci Sequence - Di®erence Equation and Z-Transform
f [n] = f [n¡ 1] + f [n¡ 2] initial conditions: f [¡2] = 0; f [¡1] = 1
Zff [n]g = F (z) =1Xn=0
f [n]z¡n
Zff [n¡ 1]g = f [¡1] + z¡1F (z); Zff [n¡ 2]g = f [¡2] + z¡1f [¡1] + z¡2F (z)F (z) = 1 + z¡1F (z) + z¡1 + z¡2F (z)
F (z) =1 + z¡1
1¡ z¡1 ¡ z¡2 =z(z + 1)
z2 ¡ z ¡ 1
F (z)
z=
z + 1
z2 ¡ z ¡ 1 =z + 1³
z ¡ 1+p5
2
´³z ¡ 1¡p5
2
´=
12
³1 + 3p
5
´z ¡ 1+
p5
2
+
12
³1¡ 3p
5
´z ¡ 1¡p5
2
via partial fraction expansion
f [n] =1
2
μ1 +
3p5
¶Ã1 +
p5
2
!n+1
2
μ1¡ 3p
5
¶Ã1¡p52
!nf = 0; 1; 1; 2; 3; 5; 8; 13; 21; 34; 55; 89; 144; : : : for n = ¡2;¡1; 0; 1; 2; : : :
f [n+ 1]
f [n]¡¡¡!n!1
1 +p5
2
165
Convolution Theorem of the Z-Transform
Zff [n]~ g[n]g =1X
n=¡1
( 1Xk=¡1
f [k]g[n¡ k])z¡n
=
1Xk=¡1
f [k]
" 1Xn=¡1
g[n¡ k]z¡n#
(let m = n¡ k) =
1Xk=¡1
f [k]
" 1Xm=¡1
g[m]z¡(m+k)#
=
1Xk=¡1
f [k]z¡k1X
m=¡1g[m]z¡m
= F (z)G(z)
Z-Transform of a Delayed Sequence.
Zff [n¡ n0]g =1X
n=¡1f [n¡ n0]z¡n =
1Xm=¡1
f [m]z¡(m+n0) = z¡n01X
m=¡1f [m]z¡m
= z¡n0F (z)
System Transfer Function.
y[n] = x[n]~ h[n] Z() Y (z) = X(z)H(z)
H(z) =Y (z)
X(z)
Recall the system block diagram for the example four pages back. The Z-transform of thegoverning di®erence equation
y[n]¡ ®y[n¡ 1] = x[n]
is
(1¡ ®z¡1)Y (z) = X(z)and therefore the system transfer function is
Zfh[n]g = H(z) = 1
1¡ ®z¡1 =z
z ¡ ®:
166
Discrete-Time Fourier Transform. With z written in polar form as z = rej−, if we
restrict z to lie on the unit-circle (jzj = r = 1), then we have the discrete-time Fourier
transform
X(z)¯̄̄jzj=1
= X(ej−) =
1Xn=¡1
x[n]e¡jn−:
The discrete-time Fourier transform (DTFT) is a mapping from a function of the integers
n to a periodic function of the continuous real variable −, with period 2¼. Observe that
the DTFT is essentially a Fourier series representation of the periodic function X(ej−),
where x[n] are the Fourier coe±cients! Therefore, the inverse DTFT is simply obtained
by applying the operator
1
2¼
¼Z¡¼
d− ejm−
to the DTFT above
1
2¼
¼Z¡¼X(ej−)ejm−d− =
1Xn=¡1
x[n]1
2¼
¼Z¡¼
ej(m¡n)−d−
| {z }±mn
= x[m]:
The discrete-time Fourier transform pair is therefore
DTFTfx[n]g = X(ej−) =1X
n=¡1x[n]e¡jn−
DTFT¡1fX(ej−)g = x[n] = 1
2¼
¼Z¡¼X(ej−)ejn−d−:
Discrete Fourier Transform. Consider a ¯nite or partial sequence x[n], where our
attention is centered on the indices n = 0; 1; 2; : : : ; N ¡ 1. Its DTFT is
X(ej−) =
N¡1Xn=0
x[n]e¡jn−:
If we now sample the DTFT at N values of −
−k =2k¼
N(k = 0; 1; 2; : : : ; N ¡ 1)
that are uniformly distributed around the unit circle, we have
X(ej−k) =
N¡1Xn=0
x[n]e¡j2nk¼=N :
167
This is a mapping from a function x[n] of the integers n = 0; 1; 2; : : : ; N ¡ 1 to anotherfunction of the same integers k = 0; 1; 2; : : : ; N¡1. Rather than write it as the cumbersomeX(ej−k), we focus our attention on the discrete frequency parameter k and write it as the
discrete Fourier transform or DFT
X[k] =
N¡1Xn=0
x[n]e¡j2nk¼=N (k = 0; 1; 2; : : : ; N ¡ 1):
The inverse DFT is most easily derived by ¯rst looking at the comparable discrete-time
spectral representation of the Kronecker-delta sequence
±[n] =1
N
N¡1Xk=0
e§j2nk¼=N :
The truth of this last equation can be established in a number of ways. Observe that
the sum is a ¯nite geometric series (page 163): The result follows at once. Therefore,
application of the operator
1
N
N¡1Xk=0
ej2mk¼=N (m = 0; 1; 2; : : : ; N ¡ 1)
to the DFT above gives
1
N
N¡1Xk=0
X[k]ej2mk¼=N =
N¡1Xn=0
x[n]1
N
N¡1Xk=0
ej2(m¡n)k=N| {z }±mn
= x[m]:
Hence, the DFT transform pair is
DFTfx[n]g = X[k] =N¡1Xn=0
x[n]e¡j2nk¼=N (k = 0; 1; 2; : : : ; N ¡ 1)
DFT¡1fX[k]g = x[n] = 1
N
N¡1Xk=0
X[k]ej2nk¼=N (n = 0; 1; 2; : : : ; N ¡ 1):
The DFT (FFT is the fast algorithm that e±ciently computes the DFT) is therefore a
transformation that converts N numbers into N other numbers. But each of the sets of
N numbers, x[n] and X[k], are a single period of periodic sequences, each with period N .
To see this, observe that
x[n+N ] =1
N
N¡1Xk=0
X[k]ej2(n+N)k¼=N =1
N
N¡1Xk=0
X[k]ej2nk¼=Nei2k¼
=1
N
N¡1Xk=0
X[k]ej2nk¼=N = x[n]
168
and
X[k +N ] =
N¡1Xn=0
x[n]e¡j2(k+N)n¼=N =N¡1Xn=0
x[n]e¡j2nk¼=Ne¡j2n¼
=
N¡1Xn=0
x[n]e¡j2nk¼=N = X[k]:
Exercises.
1. Find the Z-transform and its region-of-convergence for the ¯nite, rectangular pulse
sequence x[n] = u[n]¡ u[n¡N ], where N > 0.
2. Given the sequence x[n] = ±[n] (n = 0; 1; 2; : : : ; N ¡ 1), ¯nd its DFT X[k].
3. Given the sequence x[n] = 1 (n = 0; 1; 2; : : : ; N ¡ 1), ¯nd its DFT X[k].
4. The Bessel function of order n and argument t is de¯ned as the integral
Jn(t) =1
2¼
2¼Z0
ej(nÁ¡t sinÁ)dÁ:
Approximate this integral using the DFT. Use the fft algorithm in MATLAB to generate
values of Jn(1) and Jn(50) for n = 0; 1; 2; : : : ; 20. Compare with Table 9.4 of Big Red:
M. Abramowitz and I.A. Stegun, Handbook of Mathematical Functions with Formulas,
Graphs, and Mathematical Tables. National Bureau of Standards (also reprinted by Dover).
5. Recall from page 165 that the Z-domain transfer function of the system having block
diagram on page 161 is
H(z) =1
1¡ ®z¡1 :
Graph the magnitude and phase response H(ej−) as a function of − for these values of the
constant ®:
(a) ® = 1 and ® = ¡1 (graph on the same axes)(b) ® = 1=2 and ® = ¡1=2 (also graph on common axes)
Graph 20 log10fjH(j−)jg (dB) for the magnitude characteristic. Explain the frequencybehavior with respect to the sign of ®.
169
DFT APPROXIMATION OF THE CONTINUOUS-TIME
FOURIER TRANSFORM
F (!) =
1Z¡1
f(t)e¡j!t dt ¼1X
n=¡1f(n¢t)e¡j!n¢t¢t ¼ ¢t
N=2Xn=¡N=2+1
f(n¢t)e¡jn!¢t;
subject to appropriate restrictions. Evaluation at speci¯c frequencies gives
F
μk2¼
N ¢t
¶¼ ¢t
N=2Xn=¡N=2+1
f(n¢t)e¡j2¼nk=N :
De¯ne the periodic sequence
x[n] =
½f(n¢t); n = 0; 1; : : : ; N=2
f((n¡N)¢t); n = N=2 + 1; : : : ; N ¡ 1
where x[n +N ] = x[n]. Also note that exp(¡j2¼nk=N) is periodic in both n and k withperiod N . The approximate Fourier transform above is now
F
μk2¼
N ¢t
¶¼ ¢t
N¡1Xn=0
x[n]e¡j2¼nk=N ¼ ¢tX[k]
in terms of the DFT X[k] of the sequence x[n]. If F (!k) is desired for negative k, employ
the periodicity of X[k] = X[k +N ]. Also note that the range of available frequencies is
¡ ¼
¢t
μ1¡ 2
N
¶· ! · ¼
¢t:
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t
170
Numerical Example of Using the FFT to Evaluate the Fourier Integral:
Time-Domain Response of an RC Filter to a Gaussian Pulse
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.
²
²
CR
+
¡
x(t)
+
¡
y(t)
y(t) = x(t)~ h(t) F() Y (!) = X(!)H(!) (1)
The time-domain input to the simple RC ¯lter is a Gaussian pulse
x(t) = e¡®t2
(2)
having spectrum
X(!) =
r¼
®e¡!
2=4®: (3)
The circuit transfer function is
H(!) =Y (!)
X(!)=
R
R+1
j!C
=j!RC
1 + j!RC=
j!=!b
1 + j!=!b(4)
where the ¯lter break frequency is de¯ned to be
!b =1
RC: (5)
The time-domain response is the inverse Fourier transform
y(t) =1
2¼
1Z¡1
H(!)X(!)ej!td! =1
2¼
1Z¡1
j!=!b
1 + j!=!b
r¼
®e¡!
2=4®ej!td!: (6)
Let ® = ¿¡2 so that (6) becomes a little clearer
y(t) =1
2p¼
1Z¡1
j!¿
!b¿
1 +j!¿
!b¿
exp£¡1
4(!¿)2¤exp
·j!¿
t
¿
¸¿ d!: (7)
171
Now introduce the normalized variables
z = !¿; zb = !b¿; v =t
¿(8)
so that the circuit input and output signals are
x(v) = e¡v2
(9)
y(v) =1
2p¼
1Z¡1
jz=zb
1 + jz=zbe¡z
2=4 ejzv dz: (10)
Normalized time is v = t=¿ and the single parameter requiring a numerical value is
zb = !b¿ =¿
RC=
1p®RC
: (11)
This parameter is essentially the ratio of the input \pulse width" to the circuit time-
constant. Note that as the cut-o® frequency of the high-pass ¯lter approaces zero, the
output approaches the input
y(v) ¡¡¡!zb!0
1
2p¼
1Z¡1
e¡z2=4 ejzv dz = x(v): (12)
x(v) =1
2p¼
1Z¡1
e¡z2=4 ejzv dz ¼ 1
2p¼
N=2Xn=¡N=2+1
e¡(n¢z)2=4ejn¢z v¢z
Let
vk =2¼k
N¢z(k = ¡N=2 + 1; : : : ; N=2)
x
μ2¼k
N¢z
¶¼ ¢z
2p¼
N=2Xn=¡N=2+1
e¡(n¢z)2=4ej2¼nk=N
x[k] =N¢z
2p¼FFT¡1 fX[n]g
with
x[n] = e¡(n¢z)2=4:
172
¡4 ¡3 ¡2 ¡1 0 1 2 3 4¡0:50
¡0:25
0.00
0.25
0.50
0.75
1.00
t=¿
¿=RC = 10
1
0:1
x(t)
y(t)
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Fig. 1. Response of RC High-Pass Filter to a Gaussian Pulse x(t) = exp[¡(t=¿)2]Cases: ¿=RC = 0:1; 1; 10
173
Example: DFT Approximation of a Fourier Sine Integral
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0 T=2 T N¢t
t
...................1
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f(t) f(t) =
8>>>>><>>>>>:
0; t < 0
2t
T; 0 · t < T=2
2
T(T ¡ t); T=2 · t < T
0; t ¸ T
exact Fourier sine transform for comparison (never start with one you don't know!)
Fs(!) =
1Z0
f(t) sin(!t) dt =2T
(!T )2
£2 sin(!T=2)¡ sin(!T )¤
Fs(!) ¼N¡1Xn=0
f(n¢t) sin(n!¢t)¢t
Fs
μ2¼k
N¢t
¶¼ ¢t
N¡1Xn=0
f(n¢t) sin(2¼nk=N)
=j¢t
2
N¡1Xn=0
f(n¢t)£e¡j2¼nk=N ¡ ej2¼nk=N¤
discrete Fourier transform (DFT) of a sequence x[n] is the sequence
X[k] =
N¡1Xn=0
x[n]e¡j2¼nk=N (k = 0; 1; 2; : : : ; N ¡ 1)
both sequences are periodic with period N , that is X[k +N ] = X[k]
let x[n] = f(n¢t) for n = 0; 1; 2; : : : ; N ¡ 1
Fs
μ2¼k
N¢t
¶¼ j¢t
2
nX[k]¡X[N ¡ k]
ohopefully good for k < N=2 or slightly less
174
In this particular run, T = 1, ¢t = 0:02, and N = 1024. In the range 0 < ! < 20
the approximate and exact curves are indistinguishable to the naked eye. Therefore, the
percentage error
E(!) = Fap(!)¡ Fex(!)Fex(!)
£ 100%
is also graphed.
0 2 4 6 8 10 12 14 16 18 20¡0:1
0.0
0.1
0.2
0.3
0.4
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!
Fs(!)
0 2 4 6 8 10 12 14 16 18 200.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
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!
E(!)
175
% DFTsineFT.m DFT for a sine Fourier Transform
% 9 November 2008 R.W.Scharstein
% Practice for the Army 3-dim slotted cylinder problem
% and a good example for ECE 370.
%
% Original f(t) is causal, and an isosoles triangular waveform of total
% pulse width T.
T=1; N=1024;dt=0.02;
n=0:(N-1); t=n*dt;
x=(2*t/T).*(t<=(T/2))+(2*(T-t)/T).*(((T/2)<t)&(t<=T))+0*(t>T);
%plot(t,x)
X=fft(x); Y=[X(2:end) X(1)]; XX=fliplr(Y);
%[n' X.' XX.']
Fs=(i*dt/2)*(X-XX); w=2*pi*n/(N*dt);
% Fexact has a divide by zero at w=0, but so what!
Fexact=2*T*(2*sin(w*T/2)-sin(w*T))./(w*T).̂ 2;
% look at the first N/4 data points
m=1:N/4; wm=w(1:N/4); Fsm=Fs(1:N/4);
Fexactm=[0 Fexact(2:N/4)];
[m' wm' Fsm.' Fexactm.']
figure
plot(wm,Fexactm,wm,real(Fsm))
axis([0 20 -0.1 0.5])
error=100*(Fsm-Fexactm)./Fexactm;
figure
plot(wm,error)
axis([0 20 0 2])
176
Discrete-Time Approximation for the Step Response
of an Underdamped RLC Series Circuit
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+
¡V0u(t)
R L
C
+
¡vC(t)
....................................................................................................................................
i(t)
Exact continuous-time solution for veri¯cation:
Kirchho® voltage law gives the equation of motion as the integro-di®erential equation
Ldi(t)
dt+Ri(t) +
1
C
tZ¡1
i(¿) d¿ = V0u(t):
The inductor forces the initial condition i(0) = 0. For t > 0 application of L¡1d=dt toeliminate the integral yields the homogeneous second-order di®erential equation
d2i(t)
dt2+R
L
di(t)
dt+
1
LCi(t) = 0 (t > 0)
A second initial condition i0(0) = V0=L comes from the ¯rst integro-di®erential equation
evaluated at t = 0+ with the capacitor initially uncharged. A nice, clean statement of the
pertinent initial value problem is now
d2i(t)
dt2+ 2®
di(t)
dt+ !20i(t) = 0
with initial conditions
i(0) = 0 and i0(0) = V0=L;
where the introduction of
® , R
2Land !20 ,
1
LC
simpli¯es the mathematics by clarifying the physically-important combinations of the orig-
inal R, L, and C circuit values. The exact solution to this well-posed initial value problem
is
i(t) =V0
¯Le¡®t sin¯t
where
¯ ,q!20 ¡ ®2:
This form is most natural for the subject underdamped case where R < 2pL=C.
177
Discrete-time approximation:
The discretization of the derivative derives from the de¯ning limit as
df(t)
dt, lim
¢t!0
f(t)¡ f(t¡¢t)¢t
=) f(nT )¡ f((n¡ 1)T )T
=) f [n]¡ f [n¡ 1]T
:
A successive application of this gives the second-derivative
d
dt
df(t)
dt=)
f [n]¡ f [n¡ 1]T
¡ f [n¡ 1]¡ f [n¡ 2]T
T=f [n]¡ 2f [n¡ 1] + f [n¡ 2]
T 2:
The discretized forms of the initial conditions i(0) = 0 and i0(0) = V0=L are thus
i[0] = 0 and i[¡1] = ¡V0T=L:
The di®erence equation that approximates the continuous-time di®erential equation is
similarly ©1 + 2®T + (!0T )
2ªi[n]¡ 2(1 + ®T )i[n¡ 1] + i[n¡ 2] = 0:
Using the two initial conditions at n = ¡1 and n = 0, a recursive solution can be con-
structed for n = 1; 2; : : : according to
i[n] =2(1 + ®T )
1 + 2®T + (!0T )2i[n¡ 1]¡ 1
1 + 2®T + (!0T )2i[n¡ 2]:
Numerical Results:
The circuit values
R = 200 (−) L = 20 (H) C = 40 (¹F)
give
® = 5 (s¡1) and ¯ = 35 (s¡1):
The MATLAB code discreteRLC.m compares the normalized exact solution
¯L
V0i(t) = e¡®t sin¯t
to the recursive solution of the discrete-time problem.
178
0.0 0.1 0.2 0.3 0.4 0.5
t (s)
¡0:5
0.0
0.5
1.0
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Normalized Current
solid curve: exact solutiondashed curve: sampling period T = 0:010 (s)
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0.0 0.1 0.2 0.3 0.4 0.5
t (s)
¡0:5
0.0
0.5
1.0
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Normalized Current
solid curve: exact solutiondashed curve: sampling period T = 0:001 (s)
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179