SS Signals and Systems PRESENTATION

Signals and Systems

Dr Reza Danesfahani

Faculty of EngineeringUniversity of Kashan

Kashan, Iran

28th January 2006

Signals and Systems www.danesfahani.com

Part I

Preamble, Introduction, and Overview


Preamble ISlide 3

Syllabus1 Continuous-time and discrete-time signals: classification

and properties.2 Basic properties of systems: linearity, time-invariance,

causality, stability.3 Linear time-invariant (LTI) systems: convolution,

characterization, impulse response.4 Fourier series and Fourier transform: definition, properties,

frequency response of LTI Introduction to filtering.5 Sampling: impulse train sampling, sampling theorem,

reconstruction of signals, effects of under-sampling.6 Z-transforms: definitions, properties, analysis of LTI

systems, transfer functions.7 Communication systems: Amplitude modulation (AM),

demodulationGrading


Preamble IISlide 4

1 Midterm Exam (20%)2 Final Exam(80%)

Type of Exam: Closed-book. Students may bring one A4sheet of formulae to exam.References

A.V. Oppenheim, A.S. Willsky and S.H. Nawab,Signals & Systems,Prentice-Hall International, second edition, 1997.

G.E. Carlson,Signal and Linear System Analysis.,John Wiley & Sons, second edition, 1998.

R.E. Ziemer, W.H. Tranter and D.R. Fannin,Signals & Systems, Continuous and Discrete,Prentice Hall, fourth edition, 1998.


Preamble IIISlide 5

S. Haykin and B.V. Veen,Signals and Systems,John Wiley & Sons, 1999.

C.L. Phillips and J.M. Parr,Signals, Systems, and Transforms,Prentice Hall, second edition, 1999.

F.J. Taylor,Principles of Signals and Systems,McGraw-Hill, 1994.

H.P. Hsu,Theory and Problems of Signals and Systems,McGraw Hill, 1995.


Preamble IVSlide 6

J.R. Buck, M.M. Daniel and A.C. Singer,Computer Explorations in Signals and Systems UsingMATLAB,Prentice Hall, 1997.

L. Balmer,Signals and Systems, An Introduction,Prentice Hall, 1998.

E.W. Kamen and B.S. Heck,Fundamentals of Signals and Systems using MATLAB,Prentice Hall, 1998.

L.B. Jackson,Signals, Systems, and Transforms,Addison-Wesley, 1991.


Preamble VSlide 7

Z.Z. Karu,Signals and Systems Made Rediculously Simple,ZiZi Press, 2001.

S.T. Karris,Signals and Systems with Matlab Applications,Orchard Publications, second edition, 2003.

B.P. Lathi,Signal Processing & Linear Systems,Oxford University Press, 1998.

S. Haykin,Communication Systems,John Wiley and Sons, fourth edition, 2001.


Part II

Signals and Systems


Signals and Systems ISlide 9

DefinitionA signal is a function representing a physical quantity orvariable, and typically it contains information about the behavioror nature or phenomenon. For instance, in a RC circuit thesignal may represent the voltage across the capacitor or thecurrent flowing in the resistor. Mathematically, a signal isrepresented as a function of an independent variable t . Thus, asignal is denoted by x(t).


Signals and Systems IISlide 10

DefinitionAn energy signal is a signal whose total energy is finite, i.e.those signals for which E∞ <∞. Such a signal must have zeroaverage power, since in the continuous time case, for example,

P∞ = limT→∞

E∞2T

= 0

ExampleAn example of a finite-energy signal is a signal that takes onthe value 1 for 0 ≤ t ≤ 1 and 0 otherwise. In this case E∞ = 1and P∞ = 0.


Signals and Systems IIISlide 11

DefinitionA power signal is a signal whose average power is finite, i.e.those signals for which P∞ <∞. Such a signal must haveinfinite total energy.

Example

The constant signal x [n] = 4 is a power signal. It has infiniteenergy, but average power P∞ = 16.

There are signals for which neither P∞ nor E∞ are finite. Asimple example is the signal x(t) = t .Transformation of the independent variable

1 Time shift: A signal and its time shifted version are identicalin shape, but are displaced or shifted relative to each other.


Signals and Systems IVSlide 12

2 Time reversal: The time reversed version of the signal x(t)is obtained by a reflection about t = 0.

3 Time scale: the independent variable is scaled.

ExampleSuppose that we would like to determine the effect oftransforming the independent variable of a given signal, x(t), toobtain a signal of the form x(αt + β), where α and β are givennumbers. A systematic approach to doing this is to first delay oradvance x(t) in accordance with the value of β, and then toperform time scaling and/or time time reversal on the resultingsignal in accordance with the value of α. The delayed oradvanced signal is linearly stretched if |α| < 1, linearlycompressed if |α| > 1, and reversed in time if α < 0.


Signals and Systems VSlide 13

DefinitionIf a continuous-time signal x(t) can take on any value in thecontinuous interval (a,b), where a may be −∞ and b may be+∞, then the continuous-time signal x(t) is called an analogsignal. If a discrete-time signal x [n] can take on only a finitenumber of distinct values, then we call this signal a digitalsignal.


Signals and Systems VISlide 14

DefinitionA signal x(t) is a real signal if its value is a real number, and asignal x(t) is a complex signal if its value is a complex number.A general complex signal x(t) is a function of the form

x(t) = x1(t) + jx2(t)

where x1(t) and x2(t) are real signals and j =√−1.

DefinitionDeterministic signals are those signals whose values arecompletely specified for any given time. Random signals arethose signals that take random values at any given time andmust be characterized statistically.


Signals and Systems VIISlide 15

DefinitionA signal x(t) or x [n] is referred to as an even signal if

x(−t) = x(t)x [−n] = x [n]

A signal x(t) or x [n] is referred to as an odd signal if

x(−t) = −x(t)x [−n] = −x [n]


Signals and Systems VIIISlide 16

The even part of the signal x(t) is

Ex(t) =12[x(t) + x(−t)]

The odd part of the signal x(t) is

Ox(t) =12[x(t)− x(−t)]

Similarly,

Ex [n] =12[x [n] + x [−n]]

The odd part of the signal x [n] is

Ox [n] =12[x [n]− x [−n]]


Signals and Systems IXSlide 17

Example

The odd and even components of x(t) = ejt are

xe(t) =12(ejt + e−jt)

xo(t) =12(ejt − e−jt)


Signals and Systems XSlide 18

DefinitionA continuous-time signal x(t) is said to be periodic with periodT if there is a positive nonzero value of T for which

x(t) = x(t + T ) all t

Any sequence which is not periodic is called a nonperiodic(or aperiodic) sequence. The fundamental period T0 of x(t) isthe smallest positive value of T for which the above equationholds.


Signals and Systems XISlide 19

DefinitionA discrete-time signal x [n] is periodic with period N, where N isa positive integer, if it is unchanged by a time shift of N, i.e. if

x [n] = x [n + N]

for all values of n. The fundamental period N0 is the smallestpositive value of N for which the above equation holds.

DefinitionThe unit step function is defined as

u(t) ∆

1 t > 00 t < 0


Signals and Systems XIISlide 20

DefinitionDirac delta function, often referred to as the unit impulse(ordelta) function, is a signal that has infinite height andinfinitesimal width. The Dirac delta function δ(x) is defined by

ˆ ∞

−∞w(x)δ(x) dx = w(0)

where w(x) is any function that is continuous at x = 0.

Properties of impulse function properties1

Aδ(−t) = Aδ(t)

2 ˆ ∞

−∞Aδ(t) = A


Signals and Systems XIIISlide 21

3

Aδ(t) = 0 for t 6= 0

4

Aδ(t − t0) + Bδ(t − t0) = (A + B)δ(t − t0)

5

[y(t)][Aδ(t − t0)]Ay(t0)δ(t − t0)

DefinitionA system is a mathematical model of a physical process thatrelates the input signal to the output signal.

interconnections of systems1 Series (cascade) interconnection2 Parallel interconnection3 series-parallel interconnection


Signals and Systems XIVSlide 22

4 Feedback interconnection

DefinitionIf the input and output signals are continuous-time signals, thenthe system is called a continuous-time system. If the input andoutput signals are discrete-time signal, then the system iscalled a discrete-time system.

DefinitionA system is said to be memoryless if the output at any timedepends on only the input at that same time. Otherwise, thesystem is said to have memory.


Signals and Systems XVSlide 23

ExampleThe system specified by the relationship

y [n] = (2x [n]− x2[n])2

is memoryless. An example of a discrete-time system withmemory is accumulator or summer

y [n] =n∑

k=−∞x [k ]


Signals and Systems XVISlide 24

DefinitionA system is said to be invertible if distinct inputs lead to distinctoutputs. For a discrete-time case, if a system is invertible, thenan inverse system exists that, when cascaded with the originalsystem, yields an output equal to the input to the first system.

ExampleAn example of an invertible continuous-time system is

y(t) = 2x(t) for which the inverse system is w(t) =12

y(t). An

example of a noninvertible system is y(t) = x2(t).


Signals and Systems XVIISlide 25

DefinitionA system is called causal if its output y(t) at an arbitrary timet = t0 depends on only the input x(t) for t ≤ t0. That is, theoutput of a causal system at the present time depends on onlythe present and/or past values of the input, not on its futurevalues. A system is called noncausal if it is not causal.

DefinitionA system is called time-invariant if a time shift in the inputsignal causes the same time shift in the output signal.Otherwise, the system is time-varying.


Signals and Systems XVIIISlide 26

ExampleConsider the continuous-time system defined by

y(t) = sin[x(t)]

This system is time invariant.

ExampleThe following discrete-time system is time-varying.

y [n] = nx [n]


Signals and Systems XIXSlide 27

DefinitionA system is linear if the following two properties hold:

1 additivity. The response to x1(t) + x2(t) is y1(t) + y2(t).2 homogeneity or scaling.The response to ax1(t) is ay1(t),

where a is any complex constant.The two properties defining a linear system can be combinedinto a single statement:

continuous time:ax1(t) + bx2(t) → ay1(t) + by2(t)discrete time:ax1[n] + bx2[n] → ay1[n] + by2[n]


Signals and Systems XXSlide 28

Example

The system whose input x(t) and output y(t) are related byy(t) = tx(t) is linear, while the system y(t) = x2(t) in not linear.


Part III

Linear Time-Invariant Systems


Linear Time-Invariant Systems ISlide 30

DefinitionIf a system is linear and also time-invariant, then it is called alinear time-invariant (LTI) system.

DefinitionA system is BIBO stable if and only if every bounded inputresults in a bounded output. For a fixed, linear, system anecessary and sufficient condition for BIBO stability is

ˆ ∞

−∞|h(t)|dt <∞

where h(t) is the impulse response of the system.


Linear Time-Invariant Systems IISlide 31

DefinitionThe impulse response h(t) of a continuous-time LTI system isdefined to be the response of the system when the input is δ(t).

DefinitionThe convolution integral or superposition integral of twocontinuous-time signals x(t) and h(t) is

y(t) = x(t) ∗ h(t) =

ˆ ∞

−∞x(τ)h(t − τ) dτ

The properties of convolution are:1 the commutative property. This property is expressed by

x(t) ∗ h(t) = h(t) ∗ x(t)


Linear Time-Invariant Systems IIISlide 32

2 the distributive property. This property is expressed byx(t) ∗ [h1(t) + h2(t)] = x(t) ∗ h1(t) + x(t) ∗ h2(t).

3 the associative property. This property is expressed by

x(t) ∗ h1(t) ∗ h2(t) = x(t) ∗ h1(t) ∗ h2(t)

4 the shift property. This property states that if

f1(t) ∗ f2(t) = c(t)

thenf1(t) ∗ f2(t − T ) = c(t − T )

f1(t − T ) ∗ f2(t) = c(t − T )

andf1(t − T1) ∗ f2(t − T2) = c(t − T1 − T2)


Linear Time-Invariant Systems IVSlide 33

5 The width property. This property states that if thedurations (widths) of f1(t) and f2(t) are T1 and T2,respectively, then the duration (width) of f1(t) ∗ f2(t) isT1 + T2.

DefinitionThe step response s(t) of a continuous-time LTI system isdefined to be the response of the system when the input is astep function u(t).

Properties of continuous-time LTI systems:1 system with or without memory. For a causal

continuous-time LTI system

h(t) = 0 t < 0

2 causality.


Linear Time-Invariant Systems VSlide 34

3 stability. A continuous-time LTI system is BIBO stable if itsimpulse response is absolutely integrable, that is,

ˆ ∞

−∞|h(τ)|dτ <∞

DefinitionThe impulse response h[n] of a discrete-time LTI system isdefined to be the response of the system when the input is δ[n].


Linear Time-Invariant Systems VISlide 35

DefinitionThe convolution sum or superposition sum of the sequencesx [n] and h[n] is expressed by

y [n] =∞∑

k=−∞x [k ]h[n − k ] = x [n] ∗ h[n]

properties of the convolution sum1 Commutative

x [n] ∗ h[n] = h[n] ∗ x [n]

2 Associative

x [n] ∗ h1[n] ∗ h2[n] = x [n] ∗ h1[n] ∗ h2[n]


Linear Time-Invariant Systems VIISlide 36

3 Distributive

x [n] ∗ h1[n] + h2[n] = x [n] ∗ h1[n] + x [n] ∗ h2[n]

Convolution sum operation

y [n] = h[n] ∗ x [n] =∞∑

k=−∞h[k ]x [n − k ]

1 The impulse response h[k ] is time-reversed to obtain h[−k ]and then shifted by n to form h[n − k ] = h[−(k − n)] whichis a function of k with parameter n.

2 Two sequences x [k ] and h[n − k ] are multiplied together forall values of k with n fixed at some value.

3 The product x [k ]h[n − k ] is summed over all k to produce asingle output sample y [n].


Linear Time-Invariant Systems VIIISlide 37

4 Steps 1 to 3 are repeated as n varies over −∞ to ∞ toproduce the entire output y [n].

System with or without memoryCausality. The causality condition for a discrete-time LTIsystem is

h[n] = 0 n < 0

Stability. A DT LTI system is BIBO stable if its impulseresponse is absolutely summable, that is,

∞∑k=−∞

|h[k ]| <∞


Part IV

Fourier Series


Fourier Series ISlide 39

Fourier series: The exponential Fourier series of a powersignal v(t) with period T0 = 1/f0 is

v(t) =∞∑−∞

cnej2πnf0t n = 0, 1,2, · · ·

The series coefficients are related to v(t) by

cn =1T0

ˆT0

v(t)e−j2πnf0tdt

Since the coefficients are complex quantities in general,they can be expressed in the polar form

cn = |cn|ej arg cn


Fourier Series IISlide 40

where arg cn stands for the angle of cn.The plot of |cn| as a function of f represents the amplitudespectrum, and the plot of arg cn represents the phasespectrum.Properties of continuous-time Fourier series:

1 LinearityAx(t) + By(t) ↔ Aak + Bbk

2 Time shiftingx(t − t0) ↔ ak e−jkω0t0

3 Frequency shifting

x(t)ejMω0t ↔ ak−M

4 Conjugationx∗(t) ↔ a∗−k


Fourier Series IIISlide 41

5 Time reversalx(−t) ↔ a−k

6 Time scalingx(αt), α > 0 ↔ αk

7 Periodic convolutionˆ

Tx(τ)y(t − τ) dτ ↔ Tak bk

8 Multiplication

x(t)y(t) ↔+∞∑

l=−∞

albk−l

9 Differentiation

dx(t)dt

↔ jkω0ak = jk2πT

ak


Fourier Series IVSlide 42

10 Integration ˆ t

−∞x(t) dt ↔

(1

jkω0

)ak

11 Conjugate Symmetry for real signalsak = a∗−k<ak = <ak=ak = −=ak|ak | = |a−k |arg ak = −arg a−k

12 Even and odd decompositionxe(t) ↔ <akxo(t) ↔ j=ak


Fourier Series VSlide 43

13 Parseval’s relation

1T

ˆT|x(t)|2 dt =

+∞∑k=−∞

|ak |2

Convergence of the Fourier series: For a signal to haveFourier series, the following conditions develiped by P.L.Dirichlet must be satisfied

1 Over any period, x(t) must be absolutely integrable, that is,ˆ

T|x(t)| dt <∞

2 In any finite interval of time, x(t) is of bounded variation;that is, there are more than a finite number of maxima andminima during any single period of the signal.


Fourier Series VISlide 44

3 In any finite interval of time, there are only a finite numberof discontinuities. Furthermore, each of thesediscontinuities is finite.

Discrete-time Fourier series pair

x [n] =∑

k=<N>

akejkω0n =∑

k=<N>

akejk(2π/N)n

ak =1N

∑n=<N>

x [n]e−jkω0n =1N

∑n=<N>

x [n]e−jk(2π/N)n

Properties of discrete-time Fourier series:1 Linearity

Ax [n] + By [n] ↔ Aak + Bbk

2 Time shiftingx [n − n0] ↔ ak e−jkω0n0


Fourier Series VIISlide 45


x [n]ejMω0n ↔ ak−M

4 Conjugationx∗[n] ↔ a∗−k

5 Time reversalx [−n] ↔ a−k

6 Time scaling

x(m)[n] =

x [n/m], if n is a multiple of m0, otherwise ↔ 1

mak

7 Periodic convolution∑r=<N>

x [r ]y [n − r ] ↔ Nak bk


Fourier Series VIIISlide 46

8 Multiplicationx [n]y [n] ↔

∑l=<N>

albk−l

9 First difference

x [n]− x [n − 1] ↔ (1− e−jkω0)ak

10 Running sum

n∑k=−∞

x [k ] ↔(

11− e−jkω0

)ak


Fourier Series IXSlide 47

11 Conjugate Symmetry for real signalsak = a∗−k<ak = <ak=ak = −=ak|ak | = |a−k |arg ak = −arg a−k

12 Even and odd decompositionxe[n] ↔ <akxo[n] ↔ j=ak


1N

∑n=<N>

|x [n]|2 =∑

k=<N>

|ak |2


Fourier Series XSlide 48

Fourier series and LTI systems The output of an LTIsystem with a period input signal x(t) is

y(t) =∞∑

k=−∞akH(ejkω0)ejkω0t

where ak are the Fourier series coefficients of x(t).Similarly, for a period signal

y [n] =∑

k=<N>

akH(ejkω0)ejkω0n


Part V

Fourier Transform


Fourier Transform ISlide 50

Fourier transform

X (f ) = F [x(t)] =

ˆ ∞

−∞x(t)e−j2πft dt

The Fourier transform for periodic signals is of the form ofa linear combination of impulses equally spaced infrequency, that is,

X (jω) =+∞∑

k=−∞2πakδ(ω − kω0)

where ak are the Fourier series coefficients of x(t).Convergence of Fourier Transform: For a signal to haveFourier transform, the following conditions developed byP.L. Dirichlet must be satisfied


Fourier Transform IISlide 51


T|x(t)| dt <∞




Fourier Transform IIISlide 52

The Fourier transform for periodic signals

X (jω) =+∞∑

k=−∞2πakδ(ω − kω0)

Example

The signal x(t) = sinω0t has nonzero Fourier seriescoefficients a1 = 1/2j and a−1 = −1/2j . Therefore

X (jω) = π/jδ(ω − ω0)− π/jδ(ω + ω0)

Fourier transform pairs: Some of the common Fouriertransform pairs are as follows:


Fourier Transform IVSlide 53

1 Rectangular ∏(tτ

)↔ τsincf τ

2 Triangular ∧ (tτ

)↔ τsinc2f τ

3 Gaussiane−π(bt)2

↔ (1/b)e−π(f/b)2

4 Symmetric exponential

e−b|t| ↔ 2bb2 + (2πf )2

5 Sinc

sinc2Wt ↔ 12W

∏(f

2W

)Signals and Systems www.danesfahani.com

Fourier Transform VSlide 54

6 Sinc squared

sinc22Wt ↔ 12W

∧ (f

2W

)7 Constant

1 ↔ δ(f )

8 Phasorej(ωc t+φ) ↔ ejφδ(f − fc)

9 Sinusoid

cos(ωc t + φ) ↔ 12

[ejφδ(f − fc) + e−jφδ(f + fc)

]10 Impulse

δ(t − td ) ↔ e−jωtd


Fourier Transform VISlide 55

11 Sampling

∞∑k=−∞

δ(t − kTs) ↔ fs∞∑

n=−∞δ(f − nfs)

12 Signumsgn t ↔ 1/jπf

13 Step

u(t) ↔ 1j2πf

+12δ(f )

Fourier transform properties: The Fourier transformproperties are as follows

1 Linearity

ax(t) + by(t) ↔ aX (jω) + bY (jω)


Fourier Transform VIISlide 56

2 Time shiftingx(t − t0) ↔ e−jωt0X (jω)


x(t)ejω0t ↔ X (j(ω − ω0))

4 Conjugationx∗(t) ↔ X ∗(−jω)

5 Time reversalx(−t) ↔ X (−jω)

6 Time and frequency scaling

x(αt) ↔ 1|α|

X(

jωα

)7 Convolution

x(t) ∗ y(t) ↔ X (jω)Y (jω)


Fourier Transform VIIISlide 57

8 Multiplication

x(t)y(t) ↔ 12π

X (jω) ∗ Y (jω)

9 Differentiation in time

dx(t)dt

↔ jωX (jω)

10 Integration

ˆ t

−∞x(t) dt ↔ 1

jωX (jω) + πX (0)δ(ω)

11 Differentiation in frequency

tx(t) ↔ jd

dωX (jω)


Fourier Transform IXSlide 58

12 Conjugate Symmetry for real signalsX (jω) = X ∗(−jω)<X (jω) = <X (−jω)=X (jω) = −=X (−jω)|X (jω)| = |X (−jω)|arg X (jω) = −arg X (−jω)

13 Symmetry

x(t) real and even ↔ X (jω) real and even

x(t) real and odd ↔ X (jω) imaginary and odd

14 Even and odd decomposition for real signals

xe(t) ↔ <X (jω)

xo(t) ↔ j=X (jω)


Fourier Transform XSlide 59

15 Parseval’s relationˆ +∞

−∞|x(t)|2 dt =

12π

ˆ +∞

−∞|X (jω)|2 dω

Convergence of Fourier transform: The sufficientconditions for the convergence are:


T|x(t)| dt <∞




Fourier Transform XISlide 60

Duality theorem: This theorem states that if v(t) and V (f )constitute a known transform pair, and if there exists a timefunction z(t) related to the function V (f ) by

z(t) = V (f )

thenF [z(t)] = v(−f )

where v(−f ) equals v(t) with t = −f .


Part VI

Discrete-Time Fourier Transform


Discrete-Time Fourier Transform ISlide 62

Synthesis equation

x [n] =1

2π

ˆ2π

X (ejω)ejωn dω

Analysis equation

X (ejω) =+∞∑

n=−∞x [n]e−jωn


Discrete-Time Fourier Transform IISlide 63

A periodic sequence x [n] with period N and with theFourier series representation

x [n] =∑

k=<N>

akejk(2π/N)n

has the Fourier transform

X (ejω) =+∞∑

k=−∞2πakδ

(ω − 2πk

N

)

Example

The DTFT of x [n] = cosω0n is

X (ejω) = πδ (ω − ω0) + πδ (ω + ω0)


Discrete-Time Fourier Transform IIISlide 64

Basic properties of the DTFT are the following:1 Periodicity

X (ej(ω+2π)) = X (ejω)

2 Linearity

ax1[n] + bx2[n] ↔ a1X1(ejω) + bX2(ejω)

3 Time shifting

x [n − n0] ↔ e−jωn0X (ejω)


ejω0nx [n] ↔ X (ej(ω−ω0))

5 Conjugationx∗[n] ↔ X ∗(e−jω)

where ∗ denotes the complex conjugate.


Discrete-Time Fourier Transform IVSlide 65

6 Time reversalx [−n] ↔ X (e−jω)

7 Time expansion

x(k)[n] =

x [n/k ], if n = multiple of k0, if n 6= multiple of k ↔ X (ejω)

8 Convolution

x [n] ∗ y [n] ↔ X (e−jω)Y (e−jω)

9 Multiplication

x [n]y [n] ↔ 12π

ˆ2π

X (ejθ)Y (ej(ω−θ)) dθ

10 Differencing in time

x [n]− x [n − 1] ↔ (1− e−jω)X (ejω)


Discrete-Time Fourier Transform VSlide 66

11 Accumulation

n∑k=−∞

x [n] ↔ 11− e−jω X (ejω) + πX (ej0)

+∞∑k=−∞

δ(ω − 2πk)

12 Differentiation in frequency

nx [n] ↔ dX (ejω)

dω

13 Conjugate Symmetry for real signalsX (jω) = X ∗(e−jω)<X (ejω) = <X (e−jω)=X (ejω) = −=X (e−jω)|X (ejω)| = |X (e−jω)|arg X (ejω) = −arg X (e−jω)


Discrete-Time Fourier Transform VISlide 67

14 Symmetry for real, even signals

x [n] real and even ↔ X (ejω) real and even

15 Symmetry for real, odd signals

x [n] real and odd ↔ X (ejω) purely imaginary and odd

16 even-odd decomposition

xe[n] ↔ <X (ejω)xo[n] ↔ j=X (ejω)


+∞∑n=−∞

|x [n]|2 =1

2π

ˆ2π

|X (ejω)|2dω


Discrete-Time Fourier Transform VIISlide 68

Basic discrete-time Fourier transform pairs1 ∑

k=<N>

ak ej(2π/N)n ↔ 2π+∞∑

k=−∞

akδ

(ω − 2πk

N

)2

ejω0n ↔ 2π+∞∑

l=−∞

δ(ω − ω0 − 2πl)

3

cosω0n ↔ π+∞∑

l=−∞

δ(ω − ω0 − 2πl) + δ(ω + ω0 − 2πl)


Discrete-Time Fourier Transform VIIISlide 69

4

sinω0n ↔ π

j

+∞∑l=−∞

δ(ω − ω0 − 2πl)− δ(ω + ω0 − 2πl)

5

x [n] = 1 ↔ 2π+∞∑

l=−∞

δ(ω − 2πl)

6 periodic square wave with period N

x [n] =

1, |n| ≤ N10, N1 < |n| ≤ N/2 ↔ 2π

+∞∑k=−∞

akδ

(ω − 2πk

N

)7

+∞∑k=−∞

δ[n − kN] ↔ 2πN

+∞∑k=−∞

δ

(ω − 2πk

N

)Signals and Systems www.danesfahani.com

Discrete-Time Fourier Transform IXSlide 70

8

anu[n], |a| < 1 ↔ 11− ae−jω

9

x [n] =

1, |n| ≤ N10, |n| > N1

↔sin[ω(N1 +

12

)]

sin(ω/2)

10

sin Wnπn

=Wπ

sinc(

Wnπ

)↔ X (ω) =

1, 0 ≤ |ω| ≤ W0, W < |ω| ≤ π

0 < W < π, X (ω) periodic with period 2π.11

δ[n] ↔ 1


Discrete-Time Fourier Transform XSlide 71

12

u[n] ↔ 11− e−jω +

+∞∑k=−∞

πδ(ω − 2πk)

13

δ[n − n0] ↔ e−jωn0

14

(n + 1)anu[n], |a| < 1 ↔ 1(1− ae−jω)2

15

(n + r − 1)!

n!(r − 1)!anu[n], |a| < 1 ↔ 1

(1− ae−jω)r


Discrete-Time Fourier Transform XISlide 72

Example

Find the convolution of h[n] = αnu[n] and x [n]βnu[n].

H(ejω) =1

1− αe−jω

X (ejω) =1

1− βe−jω

SoY (ejω) =

1(1− αe−jω)(1− βe−jω)

By PFE and taking the inverse transform

y [n] =1

α− β[αn+1u[n]− βn+1u[n]]


Discrete-Time Fourier Transform XIISlide 73

ExampleFind the impulse response of the LTI system characterized by

y [n]− 34

y [n − 1] +18

y [n − 2] = 2x [n]

H(ejω) =2

1− 34

e−jω +18

e−j2ω

By performing PFE on H(ejω) and by inspection

h[n] = 4(

12

)n

u[n]− 2(

14

)n

u[n]


Part VII

Sampling


Sampling ISlide 75

Sampling theorem: A continuous-time signal x(t) can beuniquely reconstructed from its samples xs(t) with twoconditions:

1 x(t) must be band-limited with a maximum frequency ωM .2 Sampling frequency ωs of xs(t) must be greater than 2ωM ,

i.e. ωs > 2ωM .

The second condition is also known as Nyquist criterion.ωs is referred to as Nyquist frequency, i.e., the smallestpossible sampling frequency in order to recover the originalanalog signal from its samples.Sampling of a continuous-time signal x(t) can be done byobtaining its values at periodic times x(kT ), where T is thesampling period.


Sampling IISlide 76

Sampling is done by multiplying x(t) by a train of impulseswith a period T

xp(t) = x(t)p(t)

where

p(t) =+∞∑

n=−∞δ(t − nT )

xp(t) is an impulse train with the amplitude of the impulsesequal to the samples of x(t) at intervals spaces by T ; thatis,

xp(t) =+∞∑

n=−∞x(nT )δ(t − nT )

From the multiplication property

Xp(jω) =1

2π[X (jω)] ∗ P(jω)


Sampling IIISlide 77

As

P(jω) =2πT

+∞∑k=−∞

δ(ω − kωs)

Since convolution with an impulse simply shifts a signal, itfollows that

Xp(jω) =1T

+∞∑k=−∞

X (j(ω − kωs))

That is, Xp(jω) is a periodic function of ω consisting ofsuperposition of shifted replicas of X (jω), scaled by 1/T .


Sampling IVSlide 78

Reconstruction of a signal from its samples usinginterpolation

xr (t) = xp(t) ∗ h(t)

where

xp(t) =+∞∑

n=−∞x(nT )δ(t − nT )

Hence,

xr (t) =+∞∑

n=−∞x(nT )h(t − nT )

For an ideal LPF

h(t) =ωcT sin(ωc t)

πωc t


Sampling VSlide 79

So

xr (t) =+∞∑

n=−∞x(nT )

ωcTπ

sin(ωc(t − nT ))

ωc(t − nT )


Part VIII

Discrete Fourier Transform


Discrete Fourier Transform ISlide 81

Discrete Fourier transform: Let x [n] be a finite-lengthsequence of length N, that is,

x [n] = 0 outside the range 0 ≤ n ≤ N − 1

The DFT of x [n], denoted as X [k ], is defined by

X [k ] =N−1∑n=0

x [k ]W knN k = 0,1, · · · ,N − 1

where WN is the N th root of unity given by

WN = e−j(2π/N)


Discrete Fourier Transform IISlide 82

The inverse DFT (IDFT) is given by

x [n] =1N

N−1∑n=0

X [k ]W−knN n = 0,1, · · · ,N − 1

Properties of the DFT: Basic properties of the DFT are thefollowing:

1 Linearity

a1x1[n] + a2x2[n] ↔ a1X1[k ] + a2X2[k ]

2 Time shifting

x [n − n0]modN ↔ W kn0N X [k ] WN = e−j2π/N


W−kn0N x [n] ↔ X [k − k0]modN


Discrete Fourier Transform IIISlide 83

4 Conjugationx∗[n] ↔ X ∗[−k ]modN

where ∗ denotes the complex conjugate.5 Time reversal

x [−n]modN ↔ X [−k ]modN

6 DualityX [n] ↔ Nx [−k ]modN

7 Circular convolution

x1[n]⊗ x2[n] ↔ X1[k ]X2[k ]

where

x1[n]⊗ x2[n] =N−1∑i=0

x1[i]x2[n − i]modN


Discrete Fourier Transform IVSlide 84

8 Multiplication

x1[n]x2[n] ↔ 1N

X1[k ]⊗ X2[k ]

where

X1[k ]⊗ X2[k ] =N−1∑i=1

X1[i]X2[k − i]modN


N−1∑n=0

|x [n]|2 =1N

∑n=0

N − 1|X [k ]|2


Part IX

The z-Transform


The z-Transform ISlide 86

DefinitionFor a general discrete-time signal x [n], the bilateral (ortwo-sided) z-transform X (z) is

X (z) =∞∑

n=−∞x [n]z−n

The variable z is generally complex-valued and is expressed inpolar form as

z = rejΩ

where r is the magnitude of z and Ω is the angle of z.


The z-Transform IISlide 87

DefinitionThe range of values of the complex variable z for which thez-transform converges is called the region ofconvergence (ROC).


The z-Transform IIISlide 88

Example

The z-transform of x [n] = anu[n] is

X (z) =∞∑

n=0

(az−1)n

The ROC is the range of values of z for which |z| > |a|.Therefore,

X (z) =1

1− az−1 =z

z − a, |z| > |a|


The z-Transform IVSlide 89

Example

The z-transform of x [n] = −anu[−n − 1] is

X (z) =1

1− az−1 =z

z − a, |z| < |a|

Example

The z-transform of x [n] = 7(

13

)u[n]− 6

(12

)u[n] is

X (z) =7

1− 13

z−1− 6

1− 12

z−1, |z| > 1

2


The z-Transform VSlide 90

Example

The z-transform of x [n] =

(13

)n

sin(π

4n)

u[n] is

X (z) =

13√

2z

(z − 13

ejπ/4)(z − 13

e−jπ/4)

, |z| > 1/3

The properties of the region of convergence (ROC) for thez-tansform are as follows:

1 The ROC of X (z) consists of a ring in the z-plane centeredabout the origin.

2 The ROC does not contain any poles.


The z-Transform VISlide 91

3 If x [n] is of finite duration, then the ROC is the entirez-plane, except possibly z = 0 and/or z = ∞.

Example

The ROC of δ[n] ↔ 1 consists of the entire z-plane. The ROC ofδ[n − 1] ↔ z−1 consists of the entire z-plane, including z = ∞ butexcluding z = 0. The ROC of δ[n + 1] ↔ z consists of the entirez-plane (including z = 0).

4 If x [n] is a right-sided sequence, and if the circle |z| = r0 isin the ROC, then all finite values of z = 0 for which |z| > r0will also be in the ROC.

5 If x [n] is a left-sided sequence, and if the circe |z| = r0 is inthe ROC, then all values of z for which 0 < |z| < r0 will be inthe ROC.


The z-Transform VIISlide 92

6 If x [n] is two-sided, and if the circle |z| = r0 is in the ROC,then the ROC will consist of a ring in the z-plane thatincludes the circle |z| = r0.

Example

The z-transform of x [n] = b|n|, b > 0 is

X (z) =1

1− bz−1 −1

1− b−1z−1 , b < |z| < 1b

7 If the z-transform X (z) of x [n] is rational, then its ROC isbounded by the poles or extends to infinity.

8 If the z-transform X (z) of x [n] is rational, and if x [n] is rightsided, then the ROC is the region in the z-plane outside theoutermost pole – i.e., outside the circle of radius equal tothe largest magnitude of the poles of X (z). Furthermore, ifx [n] is causal (i.e., if it is right sided and equal to 0 forn < 0), then the ROC also include z = ∞.


The z-Transform VIIISlide 93

9 If the z-transform X (z) of x [n] is rational, and if x [n] is leftsided, then the ROC is the region in the z-plane inside theinnermost nonzero pole – i.e., inside the circle of radiusequal to the smallest magnitude of the poles of X (z) otherthan any at z = 0 and extending inward to and possibleincluding z = 0. In particular, if x [n] is anticausal (i.e., if it isleft sided and equal to 0 for n > 0), then the ROC alsoincludes z = 0.

Inverse z-transform:1 Inversion formula

x [n] =1

2πj

‰X (z)zn−1dz

where

denotes integration around a counterclockwiseclosed circular contour centered at the origin and withradius r .


The z-Transform IXSlide 94

2 Use of table of z-transform pairs. In this method, X (z) isexpressed as a sum

X (z) = X1(z) + · · ·+ Xn(z)

where X1(z) + · · ·+ Xn(z) are functions with know inversetransforms x1[n] + · · ·+ xn[n].

3 Power series expanding.4 Partial fraction expansion. This method provides the most

generally useful inverse z-transform, especially when X (z)is a rational function of z.


The z-Transform XSlide 95

ExampleConsider the z-transform

X (z) =3− 5

6z−1

(1− 14

z−1)(1− 13

z−1)

, |z| > 13

The inverse z-transform is

x [n] =

(14

)n

u[n] + 2(

13

)n

u[n]


The z-Transform XISlide 96


X (z) =3− 5

6z−1

(1− 14

z−1)(1− 13

z−1)

,14< |z| < 1

3


x [n] =

(14

)n

u[n]− 2(

13

)n

u[−n − 1]


The z-Transform XIISlide 97


X (z) =3− 5

6z−1

(1− 14

z−1)(1− 13

z−1)

, |z| < 14


x [n] = −(

14

)n

u[−n − 1]− 2(

13

)n

u[−n − 1]


The z-Transform XIIISlide 98


X (z) = 4z2 + 2 + 3z−1. 0 < |z| <∞

From the power-series definition of the z-transform, we candetermine the inverse transform of X (z) by inspection:

x [n] = 4δ[n + 2] + 2δ[n] + 3δ[n − 1]

Properties of the z-transform are as follows:1 Linearity

ax1[n] + bx2[n] ↔ aX1(z) + bX2(z), ROC: R1 ∩ R2


The z-Transform XIVSlide 99

2 Time shiftingx [n − n0] ↔ z−n0X (z)

ROC: R, except for the possible addition or deletion of the region.

3 scaling in the z-domain

zn0 x [n] ↔ X

(zz0

),ROC : |z0|R

4 Time reversal

x [−n] ↔ X (z−1), ROC:1R

5 Time expansion

xk [n] ↔ X (zk ),ROC : R1/k


The z-Transform XVSlide 100

6 Conjugationx∗[n] ↔ X ∗(z∗), ROC: R

7 Differentiation in the z-domain

nx [n] ↔ −zdX (z)

dz, ROC: R


The z-Transform XVISlide 101

Example

Consider the following z-transform

X (z) =az−1

(1− az−1)2 , |z| > |a|

Fromanu[n] ↔ 1

1− az−1 , |z| > |a|

Hence

nanu[n] ↔ −zddz

(1

1− az−1

)=

az−1

(1− az−1)2 , |z| > |a|

8 Convolution

x1[n] ∗ x2[n] ↔ X1(z)X2(z), ROC: R1 ∩ R2


The z-Transform XVIISlide 102

9 First difference

x [n]− x [n − 1] ↔ (1− z−1)X (z), ROC: R ∩ |z| > 0

10 Accumulation

n∑k=−∞

x [k ] ↔ (1− z−1)−1X (z), ROC: R ∩ |z| > 1

11 Initial value theorem. If x [n] = 0 for n < 0, thenx [0] = lim

z→∞X (z)

Some common z-transform pairs are as follows:1

δ[n] ↔ 1, ROC: All z


The z-Transform XVIIISlide 103

2

u[n] ↔ 11− z−1 , ROC: |z| > 1

3

−u[−n − 1] ↔ 11− z−1 , ROC: |z| < 1

4

δ[n−m] ↔ z−m, ROC: All z except 0 (if m > 0) or ∞ (if m < 0)

5

αnu[n] ↔ 11− αz−1 , ROC: |z| > |α|

6

−αnu[−n − 1] ↔ 11− αz−1 , ROC: |z| < |α|


The z-Transform XIXSlide 104

7

nαnu[n] ↔ αz−1

(1− αz−1)2 , ROC: |z| > |α|

8

−nαnu[−n − 1] ↔ αz−1

(1− αz−1)2 , ROC: |z| < |α|

9

[cosω0n]u[n] ↔ 1− [cosω0]z−1

1− [2 cosω0]z−1 + z−2 , ROC: |z| > 1

10

[sinω0n]u[n] ↔ [sinω0]z−1

1− [2 cosω0]z−1 + z−2 , ROC: |z| > 1


The z-Transform XXSlide 105

11

[rn cosω0n]u[n] ↔ 1− [r cosω0]z−1

1− [2r cosω0]z−1 + r2z−2 , ROC: |z| > r

12

[rn sinω0n]u[n] ↔ [r sinω0]z−1

1− [2r cosω0]z−1 + r2z−2 , ROC: |z| > r

DefinitionA discrete-time LTI system is causal if and only if the ROC of itssystem function is the exterior of a circle, including infinity.


The z-Transform XXISlide 106

DefinitionA discrete-time LTI system with rational system function H(z) iscausal if and only if (a) the ROC is the exterior of a circleoutside the outermost pole; and (b) with H(z) expressed as aratio of polynomials in z, the order of the numerator cannot begreater than the order of the denominator.

ExampleThe system with the following system function is not causal,because the numerator of H(z) is of higher order than thedenominator.

H(z) =z3 − 2z2 + z

z2 +14

z +18


The z-Transform XXIISlide 107

ExampleConsider a system with system function

H(z) =1

1− 12

z−1+

11− 2z−1 , |z| > 2

The impulse response of this system is

h[n] =

[(12

)n

+ 2n]

u[n]

Since h[n] = 0 for n < 0, we conclude that the system is causal.


The z-Transform XXIIISlide 108

DefinitionAn LTI system is stable if and only if the ROC of its systemfunction H(z) includes the unit circle, |z| = 1.


The z-Transform XXIVSlide 109


H(z) =1

1− 12

z−1+

11− 2z−1 , 1/2 < |z| < 2


h[n] =

(12

)n

u[n]− 2nu[−n − 1]

Since h[n] is absolutely summable, we conclude that thesystem is stable. Further, it is seen that the system isnoncausal.


The z-Transform XXVSlide 110


H(z) =1

1− 12

z−1+

11− 2z−1 , |z| < 1/2


h[n] = −[(

12

)n

+ 2n]

u[−n − 1]

Since h[n] is neither stable nor causal.


The z-Transform XXVISlide 111

DefinitionA causal LTI system with rational system function H(z) is stableif and only if all of the poles of H(z) lie inside the unit circle –i.e., they must all have magnitude smaller than 1.

ExampleConsider a causal system with system function

H(z) =1

1− az−1

which has a pole at z = a. For this system to be stable, its polemust be inside the unit circle, i.e., we must have |a| < 1.


The z-Transform XXVIISlide 112

For system characterized by linear constant-coefficientdifference equation, the properties of the z-transformprovide a particularly convenient procedure for obtainingthe system function, frequency response, or time-domainresponse of the system.


The z-Transform XXVIIISlide 113

Example

Consider an LTI system for which the input x [n] and output y [n]satisfy the linear constant-coefficient difference equation

y [n]− 12

y [n − 1] = x [n] +13

x [n − 1]

Applying the z-trasform

H(z) =Y (z)

X (z)=

1 +13

z−1

1− 12

z−1


Part X

Laplace Transform


Laplace Tranform ISlide 115

The basic inverse Laplace transform equation is

x(t) =1

2πj

ˆ σ+jω

σ−jωX (s)est ds

This equation states that x(t) can be represented as aweighted integral of complex exponentials. The contour ofintegration in this equation is the straight line in the s-planecorresponding to all points s satisfying <s = σ. This line isparallel to the jω-axis. Furthermore, we can choose anysuch line in the ROC–i.e., we can choose any value σ suchthat X (σ + jω) converges.Some common Laplace transform pairs are as follows

1

δ(t) ↔ 1


Laplace Tranform IISlide 116

2

u(t) ↔ 1s

3

tn

n!u(t) ↔ 1

sn+1

4

e−atu(t) ↔ 1s + a

5

tne−at

n!u(t) ↔ 1

(s + a)n+1

6

sin(ω0t)u(t) ↔ ω0

s2 + ω20


Laplace Tranform IIISlide 117

7

cos(ω0t)u(t) ↔ ss2 + ω2

0

8

t sin(ω0t)u(t) ↔ 2ω0s(s2 + ω2

0)2

9

t cos(ω0t)u(t) ↔(s2 − ω2

0)

(s2 + ω20)2

10

e−at sin(ω0t)u(t) ↔ ω0

(s + a)2 + ω20

11

e−at cos(ω0t)u(t) ↔ s + a(s + a)2 + ω2

0


Laplace Tranform IVSlide 118

The Laplace transform of a general continuous-time signalx(t) is defines as

X (s) ∆

ˆ +∞

−∞x(t)e−st dt

The independent variable s corresponds to a complexvariable in the exponent of e−st . The complex variable scan be written as s = σ + jω, with σ and ω the real andimaginary parts, respectively.Properties of the Laplace transform are as follows:

1 Linearity

ax1(t) + bx2(t) ↔ aX1(s) + bX2(s), ROC contains R1 ∩ R2


Laplace Tranform VSlide 119

2 Time shifting

x(t − t0) ↔ est0X (s), ROC = R

3 Shifting in the s-domain

es0tx(t) ↔ X (s − s0), ROC = R + <s0

4 Time scaling

x(at) ↔ 1|a|

X(s

a

), ROC =

Ra

5 Conjugationx∗(t) ↔ X ∗(s∗), ROC=R

6 Convolution property

x1(t) ∗ x2(t) ↔ X1(s)X2(s),R1 ∩ R2


Laplace Tranform VISlide 120

7 Differentiation in the time domain

x(t)dt

↔ sX (s), ROC = R

8 Differentiation in the s-domain

−tx(t) ↔ dX (s)

ds, ROC = R

9 Integration in the time domain

ˆ t

−∞x(τ) dτ ↔ 1

sX (s), ROC contains R ∩ <s > 0

10 Initial- and final-value theorems state that if x(t) = 0 fort < 0 and x(t) contains no impulses or higher-ordersingularities at t = 0, then x(0+) = lim

x→∞sX (s) and

limt→∞

x(t) = lims→0

sX (s).


Laplace Tranform VIISlide 121

The properties of the region of convergence (ROC) forLaplace transform are as follows:

1 The ROC of X (s) consists of strips parallel to the jω-axis inthe s-plane.

2 For rational Laplace transforms, the ROC does not containany poles.

3 If x(t) is of finite duration and is absolutely integrable, thenthe ROC is the entire s-plane.

4 If x(t) is right sided, and if the line <s = σ0 is in the ROC,then all values of s for which <s > σ0 will also be in theROC.

5 If x(t) is left sided, and if the line <s = σ0 is in the ROC,then all values of s for which <s < σ0 will also be in theROC.

6 If x(t) is two sided, and if the line <s = σ0 is in the ROC,then the ROC will consist of a strip in the s-plane thatincludes the line <s = σ0.


Laplace Tranform VIIISlide 122

7 If the Laplace transform X (s) of x(t) is rational, then itsROC is bounded by poles or extends to infinity. In addition,no poles of X (s) are contained in the ROC.

8 If the Laplace transform X (s) of x(t) is rational, then if x(t)is right sided, the ROC is the region in the s-plane to theright of the rightmost pole. If x(t) is left sided, the ROC isthe region in the s-plane to the left of the leftmost pole.


Part XI

Amplitude (Linear) ModulationDouble Sideband-Suppressed Carrier


Double Sideband (DSB) ISlide 124

Most of the common signalling techniques consist ofmodulating an analog or digital baseband signal onto acarrier.All can be represented by the generic bandpass form

s(t) = g(t) cos(2πfc t)

ors(t) = <g(t) exp(j2πfc t)

where fc is the carrier frequency.The function g(t) is referred to as the envelope of themodulated signal. The desired type of modulating signal isobtained by selecting the appropriate modulation functiong[m(t)] where m(t) is the analog or digital basebandmessage signal.


Double Sideband (DSB) IISlide 125

We always assume that m(t) is real but the function g(t) iscomplex or real. Either the frequency, phase or amplitudeof the carrier is varied in proportion to the baseband signalm(t).Modulation is a process that causes a shift in the range offrequencies in a signal.The term baseband is used to designate the band offrequencies of the signal delivered by the source or theinput transducer.In telephony, the baseband is 0 to 3.5 kHz.In television, the baseband is 0 to 4.3 MHz.Baseband communication does not use modulation.

Example: Long-distance PCM over optical fibers.Carrier Communication uses modulation.


Double Sideband (DSB) IIISlide 126

Example: Long-haul communication over a radio link.One of the basic parameters (amplitude, frequency, orphase) of a sinusoidal carrier of high frequency ωc is variedin proportion to m(t). This results in AM, FM, or PM,respectively.FM and PM belong to the class of modulation known asangle modulation.


Double Sideband (DSB) IVSlide 127

Figure: Transmitter.


Double Sideband (DSB) VSlide 128

Figure: receiver.


Double Sideband (DSB) VISlide 129

The carrier amplitude A is made directly proportional to themodulating signal m(t). The DSB signal is

m(t) cosωc t ↔ 12[M(ω + ωc) + M(ω − ωc)]

That is, the DSB modulation translates or shifts thefrequency spectrum to the left and the right by ωc .If the bandwidth of m(t) is B Hz, then the bandwidth of themodulated signal is 2B Hz.Lower sideband (LSB) is a portion of DSB spectrum whichlies below ωc .Upper sideband (USB) is a portion of DSB spectrum whichlies above ωc .


Double Sideband (DSB) VIISlide 130

DSB signal does not contain a discrete component of thecarrier frequency ωc . For this reason DSB is calledDSB-SC.

Demodulation, or detection, is the process of recoveringthe signal from the modulated signal.

in Synchronous or coherent detection the local oscillatorhas the same frequency and phase as the carrier used formodulation.Modulation can be obtained by multiplier modulators,nonlinear modulators, and switching modulators.


Double Sideband (DSB) VIIISlide 131

1 Multiplier Modulator: Here modulation is achieved directlyby multiplying m(t) by cosωc t using an analog multiplierwhose output is proportional to the product of two inputsignal. Another way to multiply two signals is throughlogarithmic amplifiers. Here the basic components are alogarithmic and an antilogarithmic amplifier with outputsproportional to the log and antilog of their inputs,respectively. Using two logarithmic amplifiers, we generateand add the logarithms of the two signals to be multiplied.The sum is then applied to an antilogarithmic amplifier toobtain the desired product. That is,

AB = antilog[log(AB)]

= antilog[log A + log B]

It is rather difficult to maintain linearity in this kind ofamplifier, and they tend to be rather expensive.


Double Sideband (DSB) IXSlide 132

2 Nonlinear Modulator: Modulation can be achieved by usingnonlinear devices, such as a semiconductor diode or atransistor.

3 Switching Modulator: The multiplication operation requiredfor modulation can be replaced by a simpler switchingoperation in which a modulated signal can be obtained bymultiplying m(t) not only by a pure sinusoidal but by anyperiodic signal w(t) of the fundamental radian frequencyωc . For example with square pulse train

w(t) =12

+2π

(cosωc t − 1

3cos 3ωc t +

15

cos 5ωc t − · · ·)

or with a square waveform

w(t) =4π

(cosωc t − 1

3cos 3ωc t +

15

cos 5ωc t − · · ·)


Double Sideband (DSB) XSlide 133

Frequency mixing or frequency conversion is used tochange the carrier frequency of a modulated signalm(t) cosωc t from ωc to some other frequency ωI . This canbe done by multiplying m(t) cosωc t by 2 cosωmixt , whereωmix = ωc + ωI for up-conversion or ωmix = ωc − ωI fordown-conversion and then bandpass-filtering the product.


Double Sideband (DSB) XISlide 134

Figure: Block diagram of product modulator.


Double Sideband (DSB) XIISlide 135

Figure: DSB waveforms.


Double Sideband (DSB) XIIISlide 136

Figure: Spectrum of baseband signal.


Double Sideband (DSB) XIVSlide 137

Figure: Spectrum of DSB-SC modulated wave.


Double Sideband (DSB) XVSlide 138

Figure: Coherent detector for demodulating DSB-SC modulatedwave.


Double Sideband (DSB) XVISlide 139

Figure: Spectrum of a product modulator output with a DSB-SCmodulated wave as input.


Double Sideband (DSB) XVIISlide 140

Problem

Consider a multiplex system in which four input signals m1(t),m2(t), m3(t), and m4(t) are respectively multiplied by thecarrier waves

[cos(2πfat) + cos(2πfbt)][cos(2πfat + α1) + cos(2πfbt + β1)]

[cos(2πfat + α2) + cos(2πfbt + β2)]

[cos(2πfat + α3) + cos(2πfbt + β3)]

and the resulting DSB-SC signals are summed and thentransmitted over a common channel. In the receiver,demodulation is achieved by multiplying the sum of theDSB-SC signals by the four carrier waves separately and thenusing filtering to remove the unwanted components.


Double Sideband (DSB) XVIIISlide 141

(a) Determine the conditions that the phase angles α1, α2, α3and β1, β2, β3 must satisfy in order that the output of thek th demodulator is mk (t), where k = 1, 2, 3, 4.

(b) Determine the minimum separation of carrier frequenciesfa and fb in relation to the bandwidth of the input signals soas to ensure a satisfactory operation of the system.


Part XII

Amplitude (Linear) ModulationOrdinary Amplitude Modulation


Amplitude Modulation (AM) ISlide 143

AM = DSB-SC + Carrier.Time domain expression

ϕAM(t) = A cosωc t + m(t) cosωc t= [A + m(t)] cosωc t

Spectrum

ϕAM(t) ↔ 12[M(ω+ωc)+M(ω−ωc)]+πA[δ(ω+ωc)+δ(ω−ωc)]

That is, the AM spectrum is simply a translated version ofthe modulated signal (both positive and negative withpower equally distributed between the two) plus deltafunctions at the carrier line spectral component.


Amplitude Modulation (AM) IISlide 144

Larger power at the transmitter, which makes it ratherexpensive.Less expensive receivers.The envelope of AM has the information about themessage m(t) only if the AM signal [A + m(t)] cosωc tsatisfies the condition A + m(t) ≥ 0 for all t . This meansthat

A ≥ mp

where mp is the peak amplitude (positive or negative) ofm(t).Modulation index

µ =mp

A


Amplitude Modulation (AM) IIISlide 145

Because A ≥ mp and because there is no upper bound onA, it follows that

0 ≤ 1

as the required condition for the viability of demodulation ofAM by an envelope detector.Overmodulation results if µ > 1 or A < mp.

DefinitionPower efficiency

η =useful powertotal power

=Ps

Pc + Ps

where Pc the carrier power, and Ps is the power in sidebands.


Amplitude Modulation (AM) IVSlide 146

Example

For tone modulation m(t) = µA cosωmt , Ps =(µA)2

2, and

Pc =A2

2. Hence

η =µ2

2 + µ2 100%

With maximum modulation index, that is µ = 1, η = 33.3%,which means that only 33.3% of the power is used to transmitthe sidebands.


Amplitude Modulation (AM) VSlide 147

DefinitionThe percentage of positive modulation on an AM signal is

(Amax − Ac)/(Ac)× 100 = max[µm(t)]× 100

DefinitionThe percentage of negative modulation is

(Ac − Amin)/(Ac)× 100 = min[µm(t)]× 100


Amplitude Modulation (AM) VISlide 148

DefinitionThe overall modulation percentage is

(Amax − Amin)/(2Ac)× 100

or(max[µm(t)]−min[µm(t)])/(2)× 100

So, the positive modulation refers to the maximum amountthe amplitude of the modulated carrier is increased overthe maximum amplitude of the unmodulated carrier.And, the negative modulation refers to the maximumamount the amplitude of the modulated carrier isdecreased in comparison with the unmodulated carrier.


Amplitude Modulation (AM) VIISlide 149

Note that for any message signal which is symmetric (atleast in terms of maximum and minimum values) about thex axis, the percentage of positive and negative modulation(and therefore the overall modulation) will always be equal.The percentage of modulation can be greater than 100%,in which case Amin has a negative valueGeneration of AM signal

vbb′(t) = [c cosωc t + m(t)]w(t)

where

w(t) =

[12

+2π

(cosωc t − 1

3cos 3ωc t +

15

cos 5ωc t − · · ·)]


Amplitude Modulation (AM) VIIISlide 150

Hence

vbb′(t) =c2

cosωc t +2π

m(t) cosωc︸︷︷︸AM

+ other terms︸︷︷︸suppressed by bandpass filter

Demodulation of AM signals1 Rectifier Detector:

vR = [A + m(t)] cosωc tw(t)

where

w(t) =

[12

+2π

(cosωc t − 1

3cos 3ωc t +

15

cos 5ωc t − · · ·)]

Hence

vR =1π

[A + m(t)] + other terms of higher frequencies


Amplitude Modulation (AM) IXSlide 151

2 Envelope Detector: In an envelope detector, the output ofthe detector follows the envelope of the modulated signal.The time constant RC must be large compared to 1/ωc butshould be small compared to 1/2πB, where ωc and B andradian frequency of carrier and bandwidth of the messageto be recovered.

Compared with an AM signal, a DSBSC signal has infinitepercentage modulation because there is no carrier linecomponent.The modulation efficiency of a DSBSC signal is 100%since no power is wasted in a discrete carrier.However, a product detector is always required for aDSBSC signal (more expensive than an envelopedetector).


Amplitude Modulation (AM) XSlide 152

Figure: AM waveforms (a) Message.


Amplitude Modulation (AM) XISlide 153

Figure: AM waveforms (b) AM wave with µ < 1.


Amplitude Modulation (AM) XIISlide 154


Amplitude Modulation (AM) XIIISlide 155

Figure: AM waveforms (c) AM wave with µ > 1.


Amplitude Modulation (AM) XIVSlide 156

Figure: Spectrum of baseband signal.


Amplitude Modulation (AM) XVSlide 157

Figure: Spectrum of AM wave.


Amplitude Modulation (AM) XVISlide 158

AKA quadrature multiplexingThe DSB signals occupy twice the bandwidth required forthe baseband. This disadvantage can be overcome bytransmitting two DSB signals using carriers of the samefrequency but in phase quadrature.Modulation

ϕQAM(t) = m1(t) cosωc t + m2(t) sinωc t

demodulation

2ϕQAM(t) cosωc t = 2[m1(t) cosωc t + m2(t) sinωc t ] cosωc t= m1(t) + m1(t) cos 2ωc t + m2(t) sin 2ωc t︸︷︷︸

suppressed by lowpass filtering


Amplitude Modulation (AM) XVIISlide 159

Phase error results in crosstalk

2ϕQAM(t) cos(ωc t + θ)


Amplitude Modulation (AM) XVIIISlide 160

Figure: Transmitter.


Amplitude Modulation (AM) XIXSlide 161

Figure: Receiver.


Part XIII

Amplitude (Linear) ModulationSingle Sideband Modulation


Single Sideband (SSB) ISlide 163

The DSB spectrum has two sidebands; the uppersideband (USB) and the lower sideband (LSB), bothcontaining the complete information of the basebandsignal. A scheme in which only one sideband is transmittedis known as sideband (SSB) transmission, which requiresonly one-half the bandwidth of the DSB signal.

Time-domain representation

M(ω) = M+(ω) + M−(ω)

m(t) = m+(t) + m−(t)


Single Sideband (SSB) IISlide 164

|M+(ω)| and |M−(ω)| are not even functions of ω.Therefore, m+(t) and m−(t) cannot be real; they arecomplex.

m+(t) =12[m(t) + jmh(t)]

andm−(t) =

12[m(t)− jmh(t)]

To determine mh(t), we note that

M+(ω) = M(ω)u(ω)

=12

M(ω)[1 + sgn(ω)]

=12

M(ω) +12

M(ω)sgn(ω)


Single Sideband (SSB) IIISlide 165

It can be seen that

jmh(t) ↔ M(ω)sgn(ω)

HenceMh(ω) = −jM(ω)sgn(ω)

From the table of Fourier transform pairs

1/πt ↔ −jsgn(ω)

Thereforemh(t) = m(t) ∗ 1/πt

or

mh(t) =1π

ˆ ∞

−∞

m(α)

t − αdα


Single Sideband (SSB) IVSlide 166

which is known as the Hilbert transform of m(t). TheHilbert transformer has the following frequency response

H(ω) = −jsgn(ω)

It follows that |H(ω)| = 1 and θh(ω) = −π/2 for ω > 0 andπ/2 for ω < 0. Thus, if we delay the phase of everycomponent of m(t) by π/2 (without changing its amplitude),the resulting signal is mh(t), the Hilbert transform of m(t).

ExampleFor the simple case of a tone modulation, that is, when themodulating signal is a sinusoid m(t) = cosωmt , the Hilberttransform is

mh(t) = cos(ωmt − π

2

)= sinωmt


Single Sideband (SSB) VSlide 167

USB Spectrum

ΦUSB(ω) = M+(ω − ωc) + M−(ω + ωc)

USB signal

φUSB(t) = m+(t)ejωc t + m−(t)e−jωc t

Substituting for m+(t) and m−(t)

φSSB(t) = m(t) cosωc t ∓mh(t) sinωc t

where the minus sign applies to USB and the plus signaapplies to LSB.Generation of SSB signals


Single Sideband (SSB) VISlide 168

1 Selective-Filtering Method: A DSB-SC signal is passedthrough a bandpass filter to eliminate the undesiredsideband. This method is used in speech processing.

2 Phase-Shift Method:

φSSB(t) = m(t) cosωc t ∓mh(t) sinωc t

Demodulation of SSB signals1 Synchronous Method: φSSB(t) is multiplied by cosωc t and

is then passed through a low-pass filter.


Single Sideband (SSB) VIISlide 169

2 Envelope Detection Method:

φSSB+C = A cosωc t + [m(t) cosωc t + mh(t) sinωc t ]

where the envelope E(t) is given by

E(t) = [A + m(t)]2 + m2h(t)]1/2

= A[1 +

2m(t)A

+m2(t)

A2 +m2

h(t)A2

]1/2

With A |m(t)| and A |mh(t)|

E(t) ' A[1 +

2m(t)A

]1/2


Single Sideband (SSB) VIIISlide 170

Using binomial expansion and discarding higher orderterms

E(t) ' A[1 +

m(t)A

]= A + m(t)


Single Sideband (SSB) IXSlide 171

Figure: (a) Spectrum of a message signal m(t) with an energy gap ofwidth 2fa centered on the origin.


Single Sideband (SSB) XSlide 172

Figure: (b) Spectrum of corresponding SSB signal containing theupper sideband.


Single Sideband (SSB) XISlide 173

ProblemConsider the modulated waveform

s(t) = Ac cos(2πfc t) + m(t) cos(2πfc t)− m(t) sin(2πfc t)

which represents a carrier plus an SSB signal, with m(t)denoting the message and m(t) its Hilbert transform.Determine the conditions for which an ideal envelope detector,with s(t) as input, would produce a good approximation to themessage signal m(t).

CCITT standardgroup: 12 channelssupergroup: 5 groupsmastergroup: 10 supergroups


Single Sideband (SSB) XIISlide 174

Figure: Modulation steps in an FDM system.


Part XIV

Amplitude (Linear) ModulationVestigial Sideband Modulation


Vestigial Sideband (VSB) ISlide 176

AKA asymmetric sideband system is a compromisebetween DSB and SSB.VSB inherits the advantages of DSB and SSB but avoidstheir disadvantages at a small cost.VSB signals are relatively easy to generate, and, at thesame time, their bandwidth is only (typically 25%) greaterthan that of SSB signal.In VBS, instead of rejecting one sideband completely (as inSSB), a gradual cutoff of one sideband, is accepted.Demodulation Methods:

1 Synchronous detection.2 Envelope (or rectifier) detector.


Vestigial Sideband (VSB) IISlide 177

If the vestigial shaping filter that produces VSB from DSBis Hi(ω), then the resulting VSB signal spectrum is

ΦVSB(ω) = [M(ω + ωc) + M(ω − ωc)]Hi(ω)

This shaping filter allows the transmission of one sideband,but suppresses the other sideband, not completely, butgradually.

The product e(t) of multiplying the incoming VSB signalφVSB(t) by 2 cosωc t is

e(t) = 2ΦVSB(t) cosωc t ↔ [ΦVSB(ω + ωc) + ΦVSB(ω − ωc)]


Vestigial Sideband (VSB) IIISlide 178

The signal e(t) is further passed through a low-passequalizer filter of transfer function H0(ω) whose output isrequired to be m(t). Hence,

M(ω) = [ΦVSB(ω + ωc) + ΦVSB(ω − ωc)]

By substitution of ΦVSB(ω) in this equation and eliminatingthe spectra ±2ωc by a low-pass filter H0(ω), we obtain

H0(ω) =1

Hi(ω + ωc) + Hi(ω − ωc)|ω| ≤ 2πB

Use of VSB in broadcast television: The baseband videosignal of television occupies an enormous bandwidth of4.5 MHz.


Vestigial Sideband (VSB) IVSlide 179

The vestigial shaping filter Hi(ω) cuts off the lowersideband of the DSB spectrum of a television signalgradually at 0.75 MHz to 1.25 MHz below the carrierfrequency fc . The resulting VSB spectrum bandwidth is6 MHz.

Linearity of amplitude modulation: Principle ofsuperposition applies in all amplitude modulationschemes (AM, DSB, SSB, VSB).


Vestigial Sideband (VSB) VSlide 180

Figure: Filtering scheme for the generation of VSB modulated wave.


Vestigial Sideband (VSB) VISlide 181

Figure: Magnitude response of VSB filter; only the positive-frequencyportion is shown.


Vestigial Sideband (VSB) VIISlide 182

Problem

The single tone modulating signal m(t) = Am cos(2πfmt) isused to generate the VSB signal

s(t) =12

aAmAc cos[2π(fc +fm)t ]+12

AmAc(1−a) cos[2π(fc−fm)t ]

where a is a constant, less than unity, representating theattenuation of the upper side frequency.

(a) Find the quadrature component of the VSB signal s(t).(b) The VSB signal, plus the carrier Ac cos(2πfc t), is passed

through an envelope detector. Determine the distortionproduced by the quadrature component.

(c) What is the value of constant a for which this distortionreaches its worst possible condition?


Part XV

Angle (Exponential) Modulation


Angle (Exponential) Modulation ISlide 184

generalized sinusoidal signal φ(t)

φ(t) = A cos θ(t)

where θ(t) is the generalized angle and is a function of t .

φ(t) = A cos(ωc t + θ0)

instantaneous frequency and phase

ωi(t) =dθdt

θ(t) =

ˆ t

−∞ωi(α) dα


Angle (Exponential) Modulation IISlide 185

ExampleThe instantaneous frequency in hertz ofcos 200πt cos(5 sin 2πt) + sin 200πt sin(5 sin 2πt) is found bynoticing that the phase is θ(t) = 200πt − 5 sin 2πt . Hence,ωi = 200π − 10π cos 2πt .

angle modulation possibilities: phase modulation (PM) andfrequency modulation (FM).In PM, the angle θ(t) is varied linearly with m(t):

θ(t) = ωc t + θ0 + kpm(t)

where kp is a constant and ωc is the carrier frequency.Assuming θ0 = 0, without loss of generality

θ(t) = ωc t + kpm(t)


Angle (Exponential) Modulation IIISlide 186

PM signalφPM(t) = A cos[ωc t + kpm(t)]

Instantaneous frequency ωi(t) in PM

ωi(t) =dθdt

= ωc + kpm(t)

Hence, in PM, the instantaneous frequency ωi varieslinearly with the derivative of the modulating signal.In FM the instantaneous frequency ωi is varied linearly withthe modulating signal.

ωi(t) = ωc + kf m(t)

where kf is a constant.


Angle (Exponential) Modulation IVSlide 187

In FM

θ(t) =

ˆ t

−∞[ωc + kf m(α)] dα

= ωc t + kf

ˆ t

−∞m(α) dα

FM signal

φFM(t) = A cos[ωc t + kf

ˆ t

−∞m(α) dα

]


Angle (Exponential) Modulation VSlide 188

The generalized angle-modulated carrier φEM can beexpressed as

φEM(t) = A cos[ωc t + ψ(t)]

= A cos[ωc t +

ˆ t

−∞m(α)h(t − α) dα

]FM results if h(t) = kpδ(t), and PM results if h(t) = kf u(t)

Power of an angle-modulated wave is A2/2.


Angle (Exponential) Modulation VISlide 189

Figure: Illustrative AM, FM, and PM waveforms.


Angle (Exponential) Modulation VIISlide 190

Narrow-band FM

φFM(t) ' A[cosωc t − kf a(t) sinωc t ]

Narrow-band PM

φPM(t) ' A[cosωc t − kpm(t) sinωc t ]

Carson’s rule for FM

BFM = 2B(β + 1)

where B is the message BW, and the deviation ratio β is

β =∆fB


Angle (Exponential) Modulation VIIISlide 191

where the frequency deviation ∆f is

∆f =kf mp

2π

where kf and mp are the FM modulator sensitivity and thepeak of the message, respectively.Carson’s rule for PM

BPM = 2(∆f + B)

where the frequency deviation ∆f is

∆f =kpm′

p

2π

where kp and m′p are the PM modulator sensitivity and the

peak of the derivative of the message, respectively.Signals and Systems www.danesfahani.com

Angle (Exponential) Modulation IXSlide 192

ExampleFor

xc(t) = 10 cos[(108)πt + 5 sin 2π(103)t ]

the phase θ(t) = ωc t + φ(t). Hence, φ(t) = 5 sin 2π(103)t andφ′(t) = 5(2π)(103) cos 2π(103)t . Therefore, |φ(t)|max = 5 rad,and ∆ω = |φ′(t)|max = 5(2π)(103) rad/s or ∆f = 5 kHz.

ExampleGiven the angle-modulated signalxc(t) = 10 cos(2π108t + 200 cos 2π103t), the instantaneousfrequency is ωi = 2π(108)− 4π(105) sin 2π(103)t . So

∆ω = 4π(105), ωm = 2π(103), and β =∆ω

ωm= 200. The BW is

WB = 2(β + 1)ωm = 8.04π(105) rad/s


Angle (Exponential) Modulation XSlide 193

Immunity of angle modulation to nonlinearities. Forexample, consider a second-order nonlinear device whoseinput x(t) and output y(t) are related by

y(t)a1x(t) + a2x2(t)

Ifx(t) = cos[ωc t + ψ(t)]

then

y(t) =a2

2+ a1 cos[ωc t + ψ(t)] +

a2

2cos[2ωc t + 2ψ(t)]


Angle (Exponential) Modulation XISlide 194

A similar nonlinearity in AM not only causes unwantedmodulation with carrier frequencies nωc but also causesdistortion of the desired signal. For instance, if a DSB-SCsignal m(t) cosωc t passes through a nonlinearityy(t) = ax(t) + bx3(t), the output is

y(t) =

[am(t) +

3b4

m3(t)]

cosωc t +b4

m3(t) cos 3ωc t

Passing this signal through a bandpass filter yields[am(t) + (3b/4)m3(t)] cosωc t . The distortion component(3b/4)m3(t) is present along with the desired signal am(t).In telephone systems, several channels are multiplexedusing SSB signals. The multiplexed signal is frequencymodulated and transmitted over a microwave radio relaysystem with many links in tandem.


Angle (Exponential) Modulation XIISlide 195

Generation of FM waves by indirect method of Armstrong:In this method the NBFM is converted to WBFM by usingfrequency multipliers. Thus, if we want a 12-fold increasein the frequency deviation, we can use a 12th-ordernonlinear device or two second-order and one third-orderdevice in cascade.Demodulation of FM: If we apply ϕFM(t) to an idealdifferentiator, the output is

ϕFM(t) = A[ωc + kf m(t)] sin[ωc t + kf

ˆ t

−∞m(α)d(α)

]The signal ϕFM(t) is both amplitude and frequencymodulated, the envelope being A[ωc + kf m(t)]. Themessage can be obtained by envelope detection of ϕFM(t).


Angle (Exponential) Modulation XIIISlide 196

Figure: (a) Scheme for generating an FM wave by using a phasemodulator.


Angle (Exponential) Modulation XIVSlide 197

Figure: (b) Scheme for generating a PM wave by using a frequencymodulator.


Angle (Exponential) Modulation XVSlide 198

Figure: Narrowband FM.


Angle (Exponential) Modulation XVISlide 199

Figure: Plots of Bessel functions of the first kind for varying order.


Angle (Exponential) Modulation XVIISlide 200

Figure: Discrete amplitude spectra of an FM signal, normalized withrespect to the carrier amplitude, for the case of sinusoidal modulationof fixed frequency and varying amplitude. Only the spectra forpositive frequencies are shown.


Angle (Exponential) Modulation XVIIISlide 201



Angle (Exponential) Modulation XIXSlide 202



Angle (Exponential) Modulation XXSlide 203

Figure: Discrete amplitude spectra of an FM signal, normalized withrespect to the carrier amplitude, for the case of sinusoidal modulationof varying frequency and fixed amplitude. Only the spectra forpositive frequencies are shown.


Angle (Exponential) Modulation XXISlide 204



Angle (Exponential) Modulation XXIISlide 205



Angle (Exponential) Modulation XXIIISlide 206

Figure: Block diagram of the indirect method of generating awideband FM signal.


Angle (Exponential) Modulation XXIVSlide 207

Figure: Block diagram of frequency multiplier.


Angle (Exponential) Modulation XXVSlide 208

Problem

Consider a narrow-band FM signal approximately defined by

s(t) ' Ac cos(2πfc t)− βAc sin(2πfc t) sin(2πfmt)

(a) Determine the envelope of this modulated signal. What isthe ratio of the maximum to the minimum value of thisenvelope? Plot this ratio versus β, assuming that β isrestricted to the interval 0 ≤ β ≤ 0.3.

(b) Determine the average power of the narrow-band FMsignal, expressed as a percentage of the average power ofthe unmodulated carrier wave. Plot this result versus β,assuming that β is restricted to the interval 0 ≤ β ≤ 0.3.


Angle (Exponential) Modulation XXVISlide 209

(c) By expanding the angle θi(t) of the narrow-band FM signals(t) in the form of a power series, and restricting themodulation index β to a maximum value of 0.3 radians,show that

θi(t) ' 2πfc t + β sin(2πfmt)− β3

3sin3(2πfmt)

What is the value of the harmonic distortion for β = 0.3?

Problem

The sinusoidal modulating wave

m(t) = Am cos(2πfmt)

is applied to a phase modulator with phase sensitivity kp. Theunmodulated carrier wave have frequency fc and amplitude Ac .


Angle (Exponential) Modulation XXVIISlide 210

(a) Determine the spectrum of the resulting phase-modulatedsignal, assuming that the maximum phase deviationβp = kpAm does not exceed 0.3 radians.

(b) Not included.

Problem

Suppose that the phase-modulated signal of the previous problem

has an arbitrary value for the maximum phase deviation βp.This modulated signal is applied to an ideal band-pass filterwith mid-band frequency fc and a passband externding fromfc − 1.5fm to fc + 1.5fm. Determine the envelope, phase, andinstantaneous frequency of the modulated signal at the filteroutput as functions of time.


Angle (Exponential) Modulation XXVIIISlide 211

ProblemAn FM signal with modulation index β = 1 is transmittedthrough and ideal band-pass filter with mid-band frequency fcand bandwidth 5fm, where fc is the carrier frequency and fm isthe frequency of the sinusoidal modulating wave. Determinethe amplitude spectrum of the filter output.

Problem

Consider a wide-band PM signal produced by a sinusoidalmodulating wave Am cos(2πfmt), using a modulator with aphase sensitivity equal to kp radian per volt.(a) Show that if the maximum phase deviation of the PM signal

is large compared with one radian, the bandwidth of thePM signal varies linearly with the modulation frequency fm.


Angle (Exponential) Modulation XXIXSlide 212

(b) Compare this characteristic of a wide-band PM signal withthat of wide-band FM signal.

ProblemAn FM signal with a frequency deviation of 10 kHz at amodulation frequency of 5 kHz is applied to two frequencymultipliers connected in cascade. The first multiplier doublesthe frequency and the second multiplier triples the frequency.Determine the frequency deviation and the modulation index ofthe FM signal obtained at the second multiplier output. What isthe frequency separation of the adjacent side frequencies ofthis FM signal?


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