ELEC 361 LECTURE NOTES — WEEK 7: Chapter 5

Representation of Aperiodic Signals: The Discrete-Time FourierTransform (DTFT)

• The DTFT is used for aperiodic discrete-time signals.

• The development of DTFT for discrete-time aperiodic signals fol-lows exactly the same as that of the Fourier transform for continuous-time aperiodic signals.

• That is, let x[n] be the aperiodic signal of interest. We construct aperiodic signal x[n] for which x[n] is one period. In this case, x[n] iscomprised of infinite number of replicas of x[n] where each replicais centered an an integer multiple of N, where N is the period ofx[n].

• Example: Consider the following figure which illustrates an ex-ample of x[n] and the construction of x[n]. Clearly, x[n] is definedbetween −N1 and N2. Consequently, N has to be chosen such thatN > N1 +N2 + 1 so that adjacent replicas do not overlap.

n

N−

][nx

N

1N− 2N

… …

][~ nx

1N− 2N

Fig. 7.1. Example of constructing an periodic signal from anaperiodic signal.

1

• Clearly, as we let N →∞, x[n] = x[n], as desired.

• Let us now examine the FS representation of x[n] :

x[n] =P<N>

akejk(2π/N)n

whereak=

1

N

P<N>

x[n]e−jk(2π/N)n

• Since x[n] is defined between −N1 and N2, ak in the above expressionsimplifies to

ak =1

N

N2Pn=−N1

x[n]e−jk(2π/N)n

=1

N

∞Pn=−∞

x[n]e−jk(2π/N)n

• Now defining the function

X(ejw) =∞P

n=−∞x[n]e−jwn

• From above we can see that the coefficients ak are related to X(ejw)

as followsak =

1

NX(ejkw0)

where w0 = 2π/N is the spacing of the samples in the frequencydomain.

• Therefore,

x[n] =P<N>

1

NX(ejkw0)ejkw0n

=1

2π

P<N>

X(ejkw0)ejkw0nw0

• As N increases w0 decreases, and as N → ∞ the above equationbecomes an integral.

• One important observation here is that the function X(ejw) is pe-riodic in w with period 2π. So is the function ejwn. Therefore, asN →∞, x[n] = x[n].

2

• This leads us to the following pair of equations:

x[n] =1

2π

R2π

X(ejw)ejwndw Synthesis equation

X(ejw) =∞P

n=−∞x[n]e−jwn Analysis equation

• Example: Let x[n] = δ[n]→ X(ejw) = 1.

• Example: Letx[n] =

½1, |n| ≤ 50, e.w.

X(ejw) =5P

n=−5e−jwn

=sin [w(11/2)]

sin [w/2]

-10 -8 -6 -4 -2 0 2 4 6 8 10-2

-1

0

1

2

n

x[n]

-15 -10 -5 0 5 10 15-5

0

5

10

15

w

X(ejw)

3

The Fourier Transform of Periodic Signals• Consider the continuous time signal

x(t) = ejw0t

This signal is periodic. Furthermore, the Fourier series represen-tation of this signal is just an impulse of weight one centered atw = w0.

• Now consider this signalx[n] = ejw0n

It is also periodic, and there is one impulse per period. However,the separation between adjacent impulses is 2π, which agrees withthe properties of DTFT. In particular, the DTFT for this signalis

X(ejw) =∞P

l=−∞2πδ(w − w0 − 2πl).

Example: Letx[n] = cosw0n with w0 =

2π

5

This signal can be expressed as

x[n] =1

2

¡ejw0n + e−jw0n

¢We can immediately write

X(ejw) =∞P

l=−∞πδ(w − 2π

5− 2πl) +

∞Pl=−∞

πδ(w +2π

5− 2πl)

Or equivalently,

X(ejw) = πδ(w − 2π5) + πδ(w +

2π

5) − π ≤ w < π

where X(ejw) is periodic with period 2π, as depicted below.

π2− π2

02 w−− π 02 w+− π

π π

02 w−π 02 w+π w

π

X(ejw)

0w− 0w

… …

Fig. 7.2. DTFT of x[n] = cosw0n.4

Properties of the DTFT

• PeriodicityX(ej(w+2π)) = X(ejw)

• Linearity: If

x1[n] F←→ X1(ejw)

x2[n] F←→ X2(ejw)

Then⇒ ax1[n] + bx2[n] F←→ aX1(e

jw) + bX2(ejw)

• Time-Shifting: Ifx[n] F←→ X(ejw)

Thenx[n− n0] F←→ e−jwn0X(ejw)

• Frequency Shifting: Ifx[n] F←→ X(ejw)

Thene−jw0nx[n] F←→ X(ej(w−w0))

• Conjugation and Conjugate Symmetry

x[n] F←→ X(ejw)

x∗[n] F←→ X∗(e−jw)

For real-valued signals,

x∗[n] = x[n]⇒ X(ejw) = X∗(e−jw).

For real-valued and even signals, the Fourier transform is real andeven.

For real-valued and odd signals, the Fourier transform is purelyimaginary and odd.

• Differencingx[n]− x[n− 1] F←→

¡1− e−jw

¢X(ejw)

5

• AccumulationnP

m=−∞x[m] F←→

1

1− e−jwX(ejw) + πX(ej0)

∞Pm=−∞

δ(w − 2πk)

where the impulse train on the right-hand side of the above equa-tion reflects the average value (or dc component) that may resultfrom the summation.

• Time Reversal: Ifx[n] F←→ X(ejw)

Thenx[−n] F←→ X(e−jw)

• Time Expansion: Let k be a positive integer. Define

x(k)[n] =

½x[n/k], if n is a multiple of k0, otherwise

Now ifx[n] F←→ X(ejw)

thenx(k)[n] F←→ X(ejkw)

see the example below.

0 2 4 60

2

4

6

w

X (2)(ejw)

-10 -5 0 5 100

1

2

n

x[n]

0 2 4 60

2

4

6

w

X(ejw)

-10 -5 0 5 100

1

2

x (2)[n]

n

-10 -5 0 5 100

1

2

n

x (3)[n]

0 2 4 60

2

4

6

w

X (3)(ejw)

6

• Differentiation in Frequency: If

x[n] F←→ X(ejw)

thennx[n] F←→ j

dX(ejw)

dw

• Parseval’s RelationnP

n=−∞|x[n]|2 = 1

2π

R2π

¯X(ejw)

¯2dw

• Convolution: Lety[n] = x[n] ∗ h[n]

where ∗ denotes convolution. Then

Y (ejw) = X(ejw)H(ejw)

Example: Consider the following system

][][ nubnx n= ??][ =ny

][][ nuanh n=

From the convolution property, we have

Y (ejw) = X(ejw)H(ejw)

=1

(1− ae−jw) (1− be−jw)

=A

1− ae−jw+

B

1− be−jw

where, by using partial fraction expansion, one can show thatA = a

a−b and B = − ba−b .

Therefore,

y[n] =1

a− b

£an+1u[n]− bn+1u[n]

¤=

an+1 − bn+1

a− bu[n].

7

• Multiplication: Lety[n] = x1[n] · x2[n]

thenY (ejw) =

1

2π

R2π

X1(ejθ)X2(e

j(w−θ))dθ

Systems Characterized by Linear Constant-Coefficient DifferenceEquations

• Ageneral linear constant-coefficient difference equation for an LTIsystem with input x[n] and output y[n] is of the form

NPk=0

aky[n− k] =MPk=0

bkx[n− k].

• Now applying the Fourier transform to both sides of the aboveequation, we have

NPk=0

ake−jkwY (ejw) =

MPk=0

bke−jkwX(ejw).

• But we know that the input and the output are related to eachother through the impulse response of the system, denoted by h[n],

i.e.,y[n] = x[n] ∗ h[n].

• Applying the convolution propertyY (ejw) = X(ejw)H(ejw)

Or

H(ejw) =Y (ejw)

X(ejw)

=

MPk=0

bke−jkw

NPk=0

ake−jkw

• Therefore, if one is given a difference equation corresponding tosome system, the Fourier transform of the impulse response ofthe system (also called the frequency response) can found directlyfrom the difference equation by applying the Fourier transform.

8

• By applying the inverse Fourier transform to the frequency re-sponse obtained, one can obtain the impulse response itself.

• Example: Consider the causal LTI system that us characterizedby the difference equation

y[n]− ay[n− 1] = x[n]

with |a| < 1. From the above discussion, it is easy to see that thefrequency response of the system is

H(ejw) =1

1− ae−jw

From tables (or by applying inverse Fourier transform), one caneasily find that

h[n] = anu[n].

9

ELEC 361 LECTURE NOTES — WEEK 8: Chapter 7

Representation of A Continuous-Time Signal by Its Samples

• Suppose that we have a continuous-time signal denoted by x(t).

Sampling x(t) at a certain rate results in a sampled (or a discrete-time) version of it, denoted by x[n].

• Intuitively, the signal x[n] tells us partial information about x(t) inthe sense that x[n] is defined only at certain time instants and notat all t, as is x(t).

• The question now is: Is it possible to recover x(t) from x[n]? Ourintuition may tell us that there is no way to recover x(t) fromx[n] because simply there could be many different signals that, ifthey were sampled, they would result in the same discrete-timesequence x[n].

• Well, fortunately, if x(t) satisfies certain conditions and is sampledproperly, the resulting x[n] uniquely specifies x(t). In other words,the relation between x[n] and x(t) is one-to-one. Therefore, a signalin one form (discrete or continuous) can be obtained from theother form very easily and perfectly.

• The theory that governs this result is called the sampling theorem.

Impulse-Train Sampling

• One way of sampling a continuous-time signal x(t) is bymultiplyingit in the time domain by an impulse train, denoted by p(t), asdepicted below.

1

ELEC 361 LECTURE NOTES — WEEK 10: Chapter 9

Chapter 9 (cont’d)The inverse Laplace Transform

• In the pervious sections we learned how to compute the Laplacetransform of signals. In this section, we will learn how to computethe inverse Laplace transform.

• Let X(s) be the Laplace transform of a signal x(t). The inverseLaplace transform is given by

x(t) =1

2πj

σ+jωRσ−jω

X(s)estds

for all values of s in the ROC.• A very useful technique in finding the inverse Laplace transformis to expand X(s) in the form

X(s) =mPi=1

Ai

s+ ai

From the ROC of X(s), one can find the ROC for each individualterm in the above expression. The inverse Laplace transform canthen be obtained for every term separately very easily.

• Example: LetX(s) =

1

(s+ 1)(s+ 2), for Res > −1.

The above expression can be expanded using partial fraction ex-pansion as follows

X(s) =1

(s+ 1)− 1

(s+ 2)

with Res > −1 for the first term and Res > −2 for the secondterm. Clearly, the signal is right-sided. Applying inverse Laplacetransform to each term individually yields

x(t) = e−tu(t)− e−2tu(t).

• Example: Consider X(s) given in the above example, but now withthe following ROC

ROC: − 2 < Res < −11

Clearly, the term corresponding to the pole s = −2 is right-sidedand the other is left-sided. therefore,

x(t) = −e−tu(−t)− e−2tu(t).

• Example: Now, if the ROC changes in the above example toROC: Res < −2

Obviously, both terms are now left-sided. Hence, we get

x(t) = −e−tu(−t) + e−2tu(−t).

• It is clear from the above three examples that the ROC plays amajor role in computing the inverse Laplace transform. Once theROC is determined, x(t) can be uniquely found.

Properties of the Laplace Transform

• Linearity:x1(t) L←→ X1(s) ROC = R1

x2(t) L←→ X2(s) ROC = R2

Thenax1(t) + bx2(t) L←→ aX1(s) + bX2(s) ROC = R1 ∩R2

• Time Shifting:x(t) L←→ X(s) ROC = R

⇒ x(t− t0) L←→ e−st0X(s) ROC = R

• Shifting in the s−Domainx(t) L←→ X(s) ROC = R

⇒ es0tx(t) L←→ X(s− s0) ROC = R+Res0

• Time Scalingx(t) L←→ X(s) ROC = R

⇒ x(at) L←→1

|a|X(s

a) ROC = aR

2

— Example: Let

x1(t) = e−tu(t) L←→1

s+ 1ROC: Res > −1

Now letx2(t) = e−2tu(2t) L←→

1

|2|X(s

2) =

1

|2|1

s2+ 1

=1

s+ 2

with ROC: Res > −2.

• Conjugationx(t) L←→ X(s) ROC = R

⇒ x∗(t) L←→ X∗(s∗) ROC = R

• Convolutionx1(t) L←→ X1(s) ROC = R1

x2(t) L←→ X2(s) ROC = R2

Thenx1(t) ∗ x2(t) = X1(s)X2(s) ROC contains R1 ∩R2

• Differentiation in the Time-Domainx(t) L←→ X(s) ROC = R

⇒ dx(t)

dtL←→ sX(s) ROC contains R

— Example:

x(t) = u(t) L←→1

sROC: Res > 0

⇒ δ(t) =du(t)

dtL←→ 1 ROC contains the entire s− plane

• Differentiation in the s−Domainx(t) L←→ X(s) ROC = R

⇒−tx(t) L←→dX(s)

dsROC = R

• Integration in the Time-Domainx(t) L←→ X(s) ROC = R

⇒tR−∞

x(τ)dτ L←→1

sX(s) ROC contains R ∩ Res > 0

3

• The Initial-Value Theorem

x(0+) = lims→∞

sX(s)

— Example:u(t) L←→

1

s

→ u(0+) = lims→∞

s1s= 1.

• The Final-Value Theorem: If x(t) has a finite limit as t→∞, then:

limt→∞

x(t) = lims→0

sX(s)

— Example:x(t) = e−2tu(t)←→ 1

s+ 2, Res > −2

→ limt→∞

x(t) = 0 OR lims→0

ss+2

= 0.

— Example:x(t) = e2tu(t)←→ 1

s+ 2, Res > 2

→ lims→0

ss+2

= 0 but limt→∞

x(t) =∞ 6= 0. (the final-value Theoremholdsonly if x(t) has a finite value as t→∞.)

• Example: x(t) = tu(t)

X(s) =∞R0

te−stdt

=1

s2ROC: Res > 0

Now using the differentiation in the s−domain property yields

u(t) L←→1

sROC: Res > 0

⇒ tu(t) L←→ − d

ds

µ1

s

¶=1

s2ROC: Res > 0.

• Example:x(t) = cos(w0t)u(t) =

·ejw0t + e−jw0t

2

¸u(t)

4

→ X(s) =1

2

∞R0

ejw0te−stdt+1

2

∞R0

e−jw0te−stdt, Res > 0

=1

2

1

s− jw0+1

2

1

s+ jw0

=s

s2 + w20Res > 0

• Example:x(t) = sin(w0t)u(t) =

·ejw0t − e−jw0t

2j

¸u(t)

→ X(s) =1

2j

1

s− jw0+1

2j

1

s+ jw0

=w0

s2 + w20, Res > 0

• Example:cos(w0t)|t=0+ = 1 and sin(w0t)|t=0+ = 0

→ lims→∞

ss

s2 + w20= 1 and lim

s→∞s

w0s2 + w20

= 0.

• Example:£e−αt cos(w0t)

¤u(t)←→ (s+ α)

(s+ α)2 + w20, Res > −α.

• Example:tu(t)←→ 1

s2, Res > 0

→ t2u(t)←→− d

ds

µ1

s2

¶=2

s3, Res > 0.

5

Analysis and Characterization of LTI Systems Using the LaplaceTransform

• Let x(t) be an input to some LTI system whose impulse responseis h(t), the output y(t) of the system is given by

y(t) = h(t) ∗ x(t)where ∗ denotes convolution. In the s−domain, the above expres-sion becomes

Y (s) = H(s)X(s)

where Y (s) is the Laplace transform (LT) of y(t), X(s) is the LT ofx(t), and H(s) is the LT of h(t).

• It is customary to refer to H(s) as the transfer function of the LTIsystem.

• Causality:

— The ROC associated with the transfer function of a causalsystem is a right-half plane.

— A system with rational transfer function is causal iff the ROCis the right-half plane to the right of the rightmost pole.

• Example: Consider an LTI system whose H(s) is given by

H(s) =s− 1

(s+ 1)(s− 2)=

2/3

s+ 1+1/3

s− 2Note that since the ROC is not specified and there are two poles,then there are three possible solutions:

1. Res < −1 : Then the solution is left-sided, and is given by

h(t) = −23e−tu(−t)− 1

3e2tu(−t).

Clearly this system is non causal.2. −2 < Res < −1 : Then the solution must have a term that isleft-sided and another that is right-sided; that is,

h(t) =2

3e−tu(t)− 1

3e2tu(−t).

The system now is non causal.6

3. Res > −1 : The solution here is given by

h(t) =2

3e−tu(t) +

1

3e2tu(t)

which is right-sided. In addition the ROC is extending towards∞. Thus, the system is causal.

• Example: Consider now a system whose H(s) is given by

H(s) =es

s+ 1, Res > −1

Applying inverse Laplace transform yields

h(t) = e−(t+1)u(t+ 1)

Note that although the ROC is the right-half plane of the right-most pole, the system is not causal. This is simply because thetransfer function is not rational.

• Stability

— An LTI system is stable iff the ROC of its transfer functionincludes the jω axis.

— Clearly, a causal system with rational transfer function is sta-ble iff all of the poles of the transfer function lie in the left-halfplane.

• Example: LetH(s) =

1

(s+ 1)(s− 2)As mentioned above, there are three possible solutions, which aredisplayed in the following figure.

7

- unstable - causal

2

2 -1

Im

Re

ROC

-1

Im

Re

ROC

2 -1

Im

Re

ROC

- stable - non causal - unstable

- non causal

Fig. 10.1. Three possible solutions, depending on the ROC.

LTI Systems Characterized by Linear Constant-Coefficient

• One of the great things about Laplace transform is that it can beused to solve fairly complicated linear differential equations veryeasily.

• Consider for example the following differential equation:dy(t)

dt+ 3y(t) = x(t).

Applying Laplace transform to both sides yieldssY (s) + 3Y (s) = X(s)

And the transfer function can be obtained as

H(s) =Y (s)

X(s)

=1

s+ 3.

Since this system has only one pole, there are two possible solu-tions:

8

1. Res > −1 : The system in this case is causal and is given byh(t) = e−3tu(t)

2. Res < −1 : This results in the non-causal solution, namely,h(t) = −e−3tu(−t)

• Of course, the same procedure can be used to obtain H(s) fromany differential equations with constant coefficients.

• A general linear constant-coefficient differential equation is of theform

NPk=0

akdky(t)

dtk=

MPk=0

bkdky(t)

dtk

Applying Laplace transform to both sides yieldsµNPk=0

aksk

¶Y (s) =

µMPk=0

bksk

¶X(s),

or

H(s) =

MPk=0

bksk

NPk=0

aksk

Clearly, the zeros of the transfer function are the solutions ofMPk=0

bksk = 0

and the poles are the solutions ofNPk=0

aksk= 0.

• We remark here that the transfer function doesn’t tell additionalinformation about the ROC of the system. The ROC is normallyspecified by additional information such as knowledge about thestability and causality of the system.

• Example: Suppose we know if the input to an LTI system isx(t) = e−3tu(t),

then the output isy(t) =

£e−t − e−2t

¤u(t)

9

Applying Laplace transform to x(t) and y(t) yields

X(s) =1

s+ 3, Res > −3

andY (s) =

1

(s+ 1)(s+ 2), Res > −1.

Accordingly, the transfer function H(s) is

H(s) =Y (s)

X(s)

=s+ 3

(s+ 1)(s+ 2)

=s+ 3

s2 + 3s+ 2

Now to determine the ROC of H(s), we know from the convolutionproperty that the ROC of Y (s) must include at least the intersec-tion of the ROCs of X(s) and H(s). Since H(s) has two poles, thenthere are three choices for the ROC. However, since we have someknowledge about the ROC of Y (s), this limits our choices for theROC of H(s) to one. That is, Res > −1. This means that H(s) isstable and causal.

— From H(s), one can obtain differential equation that relates X(s)and Y (s). It is easy to see that

d2y(t)

dt2+3

dy(t)

dt+2y(t) =

dx(t)

dt+3x(t).

• Example: Suppose that we are given the following informationabout an LTI system:

1. The system is causal.2. The system is rational and has only two poles, at s = −2 and

s = 4.

3. If x(t) = 1, then y(t) = 0.

4. The value of the impulse response at t = 0+ is 4.

From the above information, find H(s).

10

)( Tx −

0

)(tx

t

0

)(txp)(tx

)(tp

T−T2−T3− T3T2 t

1

)(tp

T

… …

0

)(txp

t

)0(x

Fig. 8.1. Impulse-train sampling.

• The periodic impulse train p(t) is referred to as the sampling func-tion, the period T as the sampling period, and the fundamentalfrequency of p(t), ws = 2π/T, as the sampling frequency.

• In the time domain,xp(t) = x(t)p(t),

wherep(t) =

∞Pn=−∞

δ(t− nT )

• But we know from previous discussion that xp(t) can be expressed2

asxp(t) =

∞Pn=−∞

x(nT )δ(t− nT ).

• Note that xp(t) here is a discrete-time signal. We used to refer tothis signal as x[n].

• Multiplication the time domain corresponds to convolution in thefrequency domain, thus

Xp(jw) =1

2π

∞R−∞

X(jθ)P (jw − θ)dθ

where X(jw) and P (jw) are the Fourier transform of x(t) and p(t),

respectively.

• Knowing that the Fourier transform of an impulse train is animpulse train, the P (jw) is expressed as

P (jw) =2π

T

∞Pn=−∞

δ(w − nws).

Therefore,Xp(jw) =

1

T

∞Pn=−∞

X(j (w − nws)).

• The above equation suggests that Xp(jw) (which is the Fouriertransform of the sampled sequence x[n]) is periodic in w consist-ing of a superposition of shifted replicas of X(jw), scaled by by1/T.

• Example: Let X(jw) be the Fourier transform of some CT signalx(t), P (jw) be the Fourier transform of an impulse train p(t) usedto sample x(t), and Xp(jw) be the Fourier transform of the sampledsignal x[n], which are depicted below.

3

Mw Ms ww −

)( jwX p

sw−sw2−sw3− sw sw2

Mw− Mw0

)( jwX

w

0 sw3 w

Tπ2

)( jwP

… …

sw−sw2−sw3− sw sw20 sw3 w

T1

… …

Fig. 8.2. Illustration of the sampling process in the frequency domain.

• It can be seen from the figure that adjacent replicas of X(jw) arenot overlapping provided that

ws > 2wM

which means if the sampling frequency is greater than twice thehighest frequency component of the signal, the original CT signalcan be completely recovered from its sampled sequence. We willshow shortly how this can be done.

• Otherwise, the adjacent replicas of X(jw) will overlap, and, as aresult, recovering x(t) from x[n] becomes not possible. The followingfigure illustrates when this happens.

4

)( jwX p

MwMs ww −

sw− sw w

T1

… …

this is what happens when Ms ww 2<

Fig. 8.3. Illustration of aliasing (overlapping) which happens when the signal x(t) is undersampled.

• This leads us to the sampling theorem.

• The Sampling Theorem: Let x(t) be a band-limited signal withX(jw) = 0 for |w| > wM .Then x(t) is uniquely determined by its samplesx(nT ), n = 0,±1,±2, . . . , if

ws > 2wM

wherews =

2π

T.

• The reconstruction of x(t) from x[n] is accomplished by processingthe latter by an ideal low pass filter with gain T and cutoff fre-quency greater than wM and less than ws−wM . The resulting signalwill exactly equal x(t). This is illustrated in the following figure.

5

1

this is what happens when Ms ww 2>

)(txr)(tx p)(tx

)(tp

)( jwH

Mw Ms ww −

)( jwX p

Mw− Mw0

)( jwX

w

sw−sw2−sw3− sw sw20 sw3 w

T1

… …

cw− cw0

)( jwH

w

T )( MscM wwww −<<

Mw− Mw0

)( jwXr

w

Fig. 8.4. Illustration of constructing x(t) from x[n].

6

Sampling with a Zero-Order Hold

• As was explained above, the impulse-train sampling approach re-quires generating a perfect impulse train p(t). However, generatingsuch a signal is not possible in real life.

• As an alternative, the sampling signal is modified such that itbecomes realizable in practice.

• This modification involves generating a sampled signal in a formreferred to as a zero-order hold. Specifically, this system samplesx(t) at a given instant and holds that value until the next instantat which a sample is taken, which can be accomplished by, again,a low pass filter.

• This new sampling process is illustrated in the following figure.

)(tx p

)(0 th

0

1

T t

0

)(tx

t

)(tx

)(tp

0

)(tx p

t

)(0 tx

0

)(0 tx

t

Fig. 8.5. Illustration of the zero-order hold process (or sample-and-hold process).

7

• From the above figure, x0(t) is obtained from xp(t) by simply passingthe latter through an LTI system with impulse response h0(t).

• Now to reconstruct x(t) from x0(t), we process x0(t) by an LTI systemwith impulse response hr(t) such that the output of this system r(t)

is identical to the original CT signal x(t).

• This happens only if the cascade of h0(t) and hr(t) is equivalent to theideal low pass filter shown in Fig. 8.4 whose frequency response isH(jw). Before proceeding further, we illustrate this reconstructionprocess in the following diagram.

)( jwH

)(tr)(txp

)(0 th

0

1

T t

)(tx

)(tp

)(0 tx

)(

)(

jwH

thr

Fig. 8.6. Cascade of the representation of a zero-order hold with a reconstruction filter.

• It is clear from the above figure that the following must hold toperfectly reconstruct x(t)

H(jw) = H0(jw)Hr(jw)

or equivalentlyHr(jw) =

H(jw)

H0(jw).

• But H0(jw) can be expressed as

H0(jw) = e−jwT/2·2 sin(wT/2)

w

¸hence,

Hr(jw) =ejwT/2H(jw)2 sin(wT/2)

w

.

8

Reconstruction of a Signal from Its Samples Using Interpolation

• We mentioned before that, provided that the sampling conditionsare satisfied, the original signal can be recovered exactly fromits samples simply by passing the sampled sequence through areconstruction filter.

• We illustrated that in the frequency domain and everything lookednice and simple. However, the question now is: What happens inthe time domain?

• The reconstruction process in the time domain is referred to asinterpolation. Interpolation is simply a process by which the val-ues of the original signal x(t) between adjacent samples in xp(t) aredetermined by using a linear combination of the values of thesesamples.

• In our elaboration on interpolation, we assume ideal samplingand the reconstruction filter is an ideal low pass filter. As wasmentioned before, the output of the reconstruction filter is simplythe convolution of the input xp(t) and the impulse response of thatfilter h(t), i.e.,

xr(t) = xp(t) ∗ h(t)where xp(t) is the discrete-time version of x(t) consisting of a trainof weighted impulses.

• As such, xr(t) can be expressed as

xr(t) =∞P

n=−∞x(nT )h(t− nT )

whereh(t) =

wcT sin(wct)

πwct

consequently,xr(t) =

wcT

π

∞Pn=−∞

x(nT )sin(wc (t− nT ))

wc (t− nT )

• The above equation suggests that the output of the reconstructionfilter (sometimes referred to as interpolation filter) is a weighted-sum of shifted versions of a sinc function where:

9

— Weighted-Sum: this is because the nth sinc function in thesummation above is weighted by the values x(nT ).

— Sinc Function: the sinc function is the impulse response of theideal low pass filter.

— Shifted Versions of a Sinc Function: they are shifted becausethe input to the low pass filter is a sequence of impulses. Theoutput of the filter due to each impulse is a sinc function butshifted depending on the location of the impulse.

The Effect of Undersampling:Aliasing

• As mentioned before, when the signal x(t) is undersampled, i.e.,when the sampling frequency ws < 2wM , the signal replicas in thefrequency domain overlap, resulting in altering x(t) beyond recov-erability.

• This overlap is usually referred to as aliasing. The consequencesof aliasing are large and undesirable.

• To illustrate this point further, we now consider a simple example.

• Example: Letx(t) = cosw0t.

Now we use various sampling frequencies, specifically:

— ws= 6w0

— ws= 3w0

— ws=32w0

— ws=65w0

The effect of sampling x(t) in the frequency domain for all ofthese sampling frequencies is depicted in the following figure.

10

0w− 0w

w

π

)( jwX

0w− 0w

sw− sww

)( jwX p

… …

2sw ( )0wws −

06wws =

0w− 0wsw− sw w

)( jwX p

… …

2sw

( )0wws −

03wws =

0wsw− sw w

)( jwX p

… …

2sw( )0wws −

023 wws =

Aliasing

No Aliasing

No Aliasing

sww0sw−w

)( jwX p

… …

2sw( )0wws −

056 wws =

Aliasing

Fig.8.7.The effect of oversampling and undersampling in the frequency domain.11

ELEC 361 LECTURE NOTES — WEEK 9: Chapters 7&9

Chapter 7 (cont’d)Discrete-Time Processing of Continuous-Time Signals

• It is often advantageous to convert a continuous-time signal intoa discrete-time signal so that processing is done in the discrete-time domain using a discrete-time processor such as a computeror a microcomputer. Nevertheless, after processing, it is alwaysthe case that the discrete-time signal is converted back into acontinuous-time signal. The ‘recovered’ continuous-time signalcould be similar to the original signal, or an altered version of it,depending on the application.

• As you may guess, the above process consists of three parts: (1)sampling, (2) discrete-time processing, and (3) converting theprocessed signal into a continuous-time signal.

• This three-step process in illustrated in the following figure.

)(tyc][nyd][nxd)(txc conversion to DT

discrete-time system

conversion to CT

Fig. 9.1. DT processing of CT signals.

• Form the above figure, we can see that the output of the sampleris xc(nT ) where T is the sampling period, and n is integer. Forconvenience, we represent xc(nT ) by

xd[n] = xc(nT )

so mathematically we can say that xd[n] is a discrete-time signal.

• The output of the DT system is also discrete, denoted by yd[n].

• yd[n] is then converted back into a CT signal, denoted by yc(t).

• The conversion of xc(t) into xd[n] is called continuous-to-discrete(D/C), and the conversion of yd[n] into yc(t) is called discrete-to-continuous (C/D).

1

• We elaborate further on the relationship between the signals abovein the following block diagram.

T T

)(tyc)(][ nTyny cd =)(][ nTxnx cd =)(txc C/D conversion

discrete-time system

D/C conversion

Fig. 9.2. Notation for C/D conversion and D/C conversion.

• Note if we consider the sampled sequence xc(nT ), the spacing be-tween adjacent samples is T, whereas the spacing between adjacentsamples in the sequence xd[n] is unity although both sequences cor-respond to the same signal.

• This is because xc(nT ) is plotted against time (the x-axis), whereasxd[n] is plotted against n, which is an integer and the spacing be-tween two consecutive integers is 1. This is illustrated in thefollowing figure.

C/D conversion

][nxd

0

)(tx

t

)(tx p)(tx

)(tp

)(txp

t

Conversion of impulse train to

discrete-time sequence

-3 -2 -1 0 1 2 3

][nxd

n

-3T -2T -T 0 T 2T 3T

1TT =

)(txp

-1 0 1

][nxd

n

-T 0 T

12TT =

Fig. 9.3. Sampling with a periodic impulse train followed by conversion to a discrete-time sequence.2

• Now we examine the processing stages described above in thefrequency domain.

• Let Xp(jw) be the Fourier transform (FT) of xp(t), which can beexpressed in terms of the sample values of xc(t) as

xp(t) =∞P

n=−∞xc(nT )δ(t− nT ) (9.1)

which is simply an impulse train except that the nth impulse isweighted by xc(nT ).

• Recall that the FT of an impulse train is an impulse train, and inthis case it is

Xp(jw) =∞P

n=−∞xc(nT )e

−jwnt (9.2)

where this follows from the fact that the FT of δ(t− nT ) is e−jwnt.

• We now consider the FT of xd[n] which is given by

Xd(ejΩ) =

∞Pn=−∞

xd[n]e−jΩn (9.3)

or equivalently,Xd(e

jΩ) =∞P

n=−∞xc(nT )e

−jΩn. (9.4)

• By comparing equations (9.2) and (9.4), we observe that Xp(jw)

and Xd(ejΩ) are related through

Xd(ejΩ) = Xp(j

Ω

T).

• We also know that Xp(jw) simply consists of an infinite number ofreplicas of Xc(jw) centered an integer multiple of ws, i.e.,

Xp(jw) =∞P

n=−∞Xc(j(w − nws)).

Consequently,Xd(e

jΩ) =∞P

n=−∞Xc(j(

Ω− 2πnT

)).

• The relationship between Xc(jw), Xp(jw) and Xd(ejΩ) is illustrated in

the following figure.3

)( Ωjd eX

1

)( jwX p

Mw− Mw0

)( jwX

w

1

2Tπ

−1

2Tπ0 w

1

1T

1TT =

)( Ωjd eX

π2− π20 Ω

1

1T

1TT =

)( jwX p

2

2Tπ

−2

2Tπ0 w

2

1T

12 2TTT ==

π2− π20 Ω

2

1T

12 2TTT ==

Fig. 9.4. Relationship between Xc(jw), Xp(jw), and Xd(ejΩ) for different sampling rates.

4

Chapter NineThe Laplace Transform

• It was mentioned in an earlier chapter that the response of an LTIsystem with impulse response h(t) to a complex exponential inputof the form est is

y(t) = H(s)est

whereH(s) =

∞R−∞

h(t)e−stdt. (9.5)

• If we let s = jw (pure imaginary), the integral in (9.5) is essentiallythe Fourier transform of h(t). For arbitrary values of the complexvariable s, this expression is referred to as the Laplace transformof h(t).

• Therefore, the Laplace transform of a general signal x(t) is definedas

X(s) =∞R−∞

x(t)e−stdt. (9.6)

• Note that s is a complex variable, which can be expressed in gen-eral as

s = σ + jw.

when s = jw (9.6) becomes

X(jw) =∞R−∞

x(t)e−jwtdt

which is the Fourier transform of x(t). Therefore, the Fourier trans-form is a special case of the Laplace transform.

• Equation (9.6) can also be expressed as

X(σ + jw) =∞R−∞

x(t)e−(σ+jw)tdt

=∞R−∞

£x(t)e−σt

¤e−jwtdt

which is essentially the Fourier transform of the signal x(t)e−σt.

• Example: Letx(t) = e−atu(t)

5

The Fourier transform X(jw), with a > 0, is

X(jw) =∞R−∞

e−atu(t)e−jwtdt

=∞R0

e−ate−jwtdt

=1

jw + a

On the other hand, the Laplace transform of x(t) is

X(s) =∞R−∞

e−atu(t)e−stdt

=∞R0

e−(s+a)tdt

orX(σ + jw) =

∞R0

e−(a+σ)te−jwtdt

=1

jw + a+ σwhere a+ σ > 0

Since s = σ + jw, the last equation becomes

X(s) =1

s+ aRe s > −a.

Conclusion:e−atu(t) L←→

1

s+ a, Re s > −a.

• We conclude from the above example that the Laplace transformexists for this particular x(t) only if

Re s > −a.• The region in the complex plane in which the Laplace transformexists (or converges) is called region of convergence (ROC). TheROC for the above example is given in the following figure.

-a

Im

Re

ROC

6

• Example: Letx(t) = −e−atu(−t)

Then

X(s) = −∞R−∞

e−atu(−t)e−stdt

= −0R−∞

e−ate−stdt

=1

s+ a

which converges if Re s+ a < 0 ⇒ Re s < −a, which is illustratedbelow.

-a

Im

Re

ROC

• Example: Letx(t) = 3e−2tu(t)− 2e−tu(t)

Applying the Laplace transform to x(t) yields

X(s) = −∞R−∞

£3e−2tu(t)− 2e−tu(t)¤ e−stdt

= 3∞R−∞

e−2tu(t)e−stdt− 2∞R−∞

e−tu(t)e−stdt

=3

s+ 2− 2

s+ 1

where for these integrals to converge we must have Re s > −2 forthe first term and Re s > −1 for the second term. Therefore, theROC is the intersection of the ROCs for the individual terms, i.e.,the overall ROC is Re s > −1.

7

• Example: Let

x(t) = e−2tu(t) + e−t cos (3t)u(t)

=

·e−2t +

1

2e−(1−3j)t +

1

2e−(1+3j)t

¸u(t)

⇒

X(s) =1

s+ 2+1

2

µ1

s+ (1− 3j)¶+1

2

µ1

s+ (1 + 3j)

¶=

2s2 + 5s+ 12

(s2 + 2s+ 10) (s+ 2)

with the condition that

Re s > −2, for the first term and

Re s > −1 for the second term.

Therefore, the ROC is the region where Re s > −1, which is theintersection of the individual regions.

• Example: Letx(t) = δ(t)

⇒X(s) = 1

and the ROC is the entire s−plane.

8

The Region of Convergence for Laplace Transform

Let X(s) be the Laplace transform of some signal x(t). The ROC ofX(s), in general, has the following characteristics:

1. The ROC of X(s) consists of strips parallel to the jw-axis in thes-plane.

2. For rational Laplace transforms, the ROC doesn’t contain anypoles.

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

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

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

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

7. If the Laplace transform X(s) of x(t) is rational, then the ROC isbounded by poles or extends to infinity. In addition, no poles ofX(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 the right of therightmost pole. If x(t) is left sided, the ROC is the region in thes-plane to the left of the leftmost pole.

• Example: LetX(s) =

1

(s+ 1) (s+ 2)

Clearly, there are two poles: s = −1 and s = −2. This yields threepossibilities for the ROC where each possibility corresponds to adifferent signal. These possibilities are:

1. Re s > −1 =⇒ The signal must be right-sided.2. Re s < −2 =⇒ The signal must be left-sided.3. −2 < Re s < −1 =⇒ The signal must be two-sided.

9

ELEC 361 LECTURE NOTES — WEEK 11: Chapter 10

The Z Transform

• In the previous chapter, we introduced the Laplace transform asan extension of the continuous-time Fourier transform. One of theseveral advantages of Laplace transform is that it allows us to per-form analysis of continuous-time signals whose Fourier transformdoesn’t exist.

• In this chapter, we extend this transformation to discrete-timesignals, but now the transformation is called the z−transform.

• Let h[n] be the impulse response of a LTI system. The response ofthis system to a complex exponential input of the form zn is

y[n] = H(z)zn (11.1)

whereH(z) =

∞Pn=−∞

h[n]z−n. (11.2)

• The Fourier transform of h[n] can be obtained by evaluating thez−transform given in (11.2) at z = ejw with w real, i.e.,

H(ejw) =∞P

n=−∞h[n]e−jwn. (11.3)

• The expression in (11.2) is referred to as the z−transform of h[n]

where z is a complex variable.

• Now we consider important relationships between the discrete-time Fourier transform and the z−transform. We first express thecomplex variable z in polar form as

z = rejw,

where r is the magnitude of z and w is the phase of z. Representingz as such, (11.2) can be expressed as

H(rejw) =∞P

n=−∞h[n]

¡rejw

¢−nor equivalently,

H(rejw) =∞P

n=−∞

¡h[n]r−n

¢e−jwn. (11.4)

1

• By comparing equations (11.3) and (11.4), we can see that theexpression in (11.4) is essentially the Fourier transform of thesequence x[n] multiplied by a real exponential r−n. Of course theexponential r−n may be decaying or growing with increasing n,

depending on whether r is greater than or less than 1.

• If we let r = 1, thenX(z)|z=ejw = X(ejw)

which suggests that the z−transform reduces to the Fourier trans-form on the contour in the complex z−plane corresponding to acircle with a radius of unity, as depicted in the figure shown below.

ωjez =

1 Re

Im

Unit Circle

z-plane

Fig. 11.1. Representation of the unit circle in the z-plane.

• From the above discussion we see that, for convergence of thez−transform, we require that the Fourier transform of x[n]r−n con-verge. Clearly, for any specific sequence x[n], we would expect thisconvergence for some values of r and not for others, as we Laplacetransform.

• The range of values for which the z−transform converges is referredto as the region of convergence (ROC).

• If the ROC includes the unit circle, then the Fourier transformconverges.

• Example: Consider the signal x[n] = anu[n].

2

— The z−transform of x[n] is

X(z) =∞P

n=−∞anu[n]z−n

=∞Pn=0

¡az−1

¢nFor X(z) to converge, we require that

∞Pn=0

|az−1|n <∞. Therefore,the ROC is the range of values of z for which |az−1| < 1, orequivalently, |z| > |a|.

— ThenX(z) =

∞Pn=0

¡az−1

¢n=

1

1− az−1

=z

z − a, |z| > |a| (11.5)

— Therefore, the z−transform exists for any value of a, with aROC determined by the magnitude of a according to the aboveequation.

— When a = 1, x[n] is the unit step sequence with z−transform

X(z) =1

1− 1z−1 , |z| > 1

— The z−transform in (11.5) is a rational function. Therefore,the z−transform is characterized by its zeros and poles. Forthe above example, the there is one zero at z = 0 and one poleat z = a. The pole zero plot for this function is depicted in thefollowing figure.

a 1 Re

Im

Unit Circle

z-plane

ROC

Fig. 11.2. Pole-zero plot for the above example for a < 1 and real.3

• Example: Now let x[n] = −anu[−n− 1]. Then

X(z) = −∞P

n=−∞anu[−n− 1]z−n

= −−1P

n=−∞anz−n

= 1−∞Pn=0

¡a−1z

¢−nThis sum converges if |a−1z| < 1 or |z| < |a−1| . Consequently,

X(z) = 1− 1

1− a−1z

=1

1− az−1

=z

z − a, , |z| < |a|

The pole-zero plot for this example is shown below.

a 1 Re

Im

Unit Circle

z-plane

ROC

Fig. 11.3. Pole-zero plot and ROC for the above example.

4

• Example: Now let

x[n] = 7

µ1

3

¶n

u[n]− 6µ1

2

¶n

u[n].

The z−transform is given by

X(z) =∞P

n=−∞

µ7

µ1

3

¶n

u[n]− 6µ1

2

¶n

u[n]

¶z−n

= 7∞Pn=0

µ1

3z−1¶n

− 6∞Pn=0

µ1

2z−1¶n

=7

1− 13z−1− 6

1− 12z−1

=z(z − 3

2)¡

z − 13

¢ ¡z − 1

2

¢provided that

¯13z−1¯< 1 and

¯12z−1¯< 1, or equivalently, |z| > 1

3and

|z| > 12. Therefore, the ROC is |z| > 1

2. The pole-aero plot of the

above example is given below. In the figure, we first find the ROCfor each term individually and then find the ROC of both termscombined, similar to what we used to do in Laplace transform.

1/3 1 Re

Im

Unit Circle

z-plane

ROC

1/2 1 Re

Im

Unit Circle

z-plane

ROC

1/2 1 Re

Im

Unit Circle

z-plane

ROC

1/3 3/2

Fig. 11.4. Pole-zero plot for the above example. The top figure (left) corresponds to the termwith pole z = 1/3, and the one on the right is for the term with pole z = 1/2. The bottom plot

is for both terms combined.

5

The Region of Convergence for the Z−Transform

In the following we list some of the important properties of the ROCof the z−transform:

1. The ROC of X(z) consists of a ring in the z−plane centered aboutthe origin.

2. The ROC does not contain any poles.

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

• Example: Let x[n] = δ[n]. Then

X(z) =∞P

n=−∞δ[n]z−n

= 1

which suggests that the ROC is the entire z−plane, includingz = 0 and z =∞.

• Example: Now consider x[n] = δ[n − n0] where n0 6= 0. X(z) thenbecomes

X(z) =∞P

n=−∞δ[n− n0]z

−n

= z−n0

If n0 > 0 then the ROC contains the entire z−plane except atz = 0. But if n0 < 0, the ROC contains the entire z−plane exceptat z =∞.

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

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

• Properties 4 and 5 above parallel the corresponding propertiesfor Laplace transform.

6

6. If x[n] is a two-sided sequence, and if the circle |z| = r0 is in theROC, then all finite values of z for which |z| = r0 will also be in theROC.

• Since the sequence is two sided, then it can decomposed into atleast one left-sided sequence and one right-sided sequence. Forthe right-sided sequence, the ROC is bounded on the insideby a circle and extending outward to infinity. For the left-sided sequence, the ROC is bounded on the outside by a circleand extending to the origin. The ROC for both sequencescombined is the intersection of both individual ROCs.

• Example: Consider the signal

x[n] =

½an, 0 ≤ n ≤ N − 1, a > 00, otherwise

Then

X(z) =N−1Pn=0

anz−n

=N−1Pn=0

¡az−1

¢n=

1− (az−1)N1− az−1

=1

zN−1zN − aN

z − a.

— Since the system has a finite impulse response, then weshould expect, according to Property 3 above, the ROC toinclude the entire z−plane except possibly at z = 0 and/orz =∞.

— It is clear from the above expression for X(z) that there is apole at z = 0 of order N − 1.

— Also, there seems to be a pole at z = a. Well, it is true, butthere is also a zero at z = a (from the numerator). Therefore,the pole and zero at z = a cancel out. What is left is apolynomial in the numerator of degree N−1, suggesting thatthere are N − 1 zeros. These zeros are

zk = aej(2πk/N), k = 1, . . . , N − 17

— The pole-zero plot for the above example is depicted in thefollowing figure.

a

1 Re

Im

Unit Circle

z-plane

(N-1)st order pole

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

8. If the z−transform X(z) of x[n] is rational, and if x[n] is right-sided,then the ROC is the region in the z−plane outside the outermostpole (i.e., outside the circle of radius equal to the largest magni-tude of the poles of X(z).Also, if x[n] is causal (i.e., if it is right-sidedand equal to 0 for n < 0), then the ROC also includes z =∞.

9. If the z−transform X(z) of x[n] is rational, and if x[n] is left-sided,then the ROC is the region in the z−plane inside the innermostpole (i.e., outside the circle of radius equal to the smallest mag-nitude of the poles of X(z) other than any at z = 0 and extendinginward to and possibly including z = 0. In particular, if x[n] is an-ticausal (i.e., if it is left-sided and equal to 0 for n > 0), then theROC also includes z = 0.

8

• Example: Consider the following z−transform

X(z) =1¡

1− 13z−1¢(1− 2z−1)

Since there are two poles, then there are three possibilities for theROC:

— ROC: |z| > 2 =⇒ the ROC is extending outward from the out-ermost pole, suggesting that the sequence x[n] is right-sided.The corresponding pole-zero plot is shown in the following fig-ure.

1/3 1 Re

Im

Unit Circle

z-plane

ROC

2

ROC for right-sided squence.

— ROC: 1/3 < |z| < 2 =⇒ the ROC is bounded between two poles,suggesting that the sequence x[n] is two-sided. The correspond-ing pole-zero plot is shown in the following figure.

1/3 1 Re

Im

Unit Circle

z-plane ROC

2

ROC for two-sided sequence.

9

— ROC: |z| < 1/3 =⇒ the ROC is inward from the innermost pole,suggesting that the sequence x[n] is left-sided. The correspond-ing pole-zero plot is shown in the following figure.

1/3 1 Re

Im

Unit Circle

z-plane

ROC

2

ROC for left-sided sequence.

10

1

ELEC 361 LECTURE NOTES – WEEK 12: Sections 10.3-10.7

Concordia University

The inverse z-transform

- The equation for the inverse z-transform is a contour integration in the z-plane

(counterclockwise) as follows:

∫ −= dzzzXj

nx n 1)(21][π

- This equation can be obtained by writing the z-transform of ][nx as the Fourier

transform of the signal nrnx −][ as was discussed before, and taking the inverse

Fourier transform.

- In general, one can find the partial fraction expansion for the z-transform expressions

that are rational functions of z. We will have:

∑=

−−=

m

i i

i

zaAzX

111

)(

- One can then take the inverse z-transform of each individual term very easily.

- Example 12.1: Find the inverse z-transform of the following expression:

)311)(

411(

653

)(11

1

−−

−

−−

−=

zz

zzX

with the following regions of convergence:

(a) 31

>z

(b) 31

41

<< z

- Solution:

The partial fraction expansion for this expression in 1−z is given below:

11

311

2

411

1)(−− −

+−

=zz

zX

2

(a) Since the ROC of )(zX is outside the outermost pole, the ROC corresponding to

each term must be outside the pole associated with each one. This means that the

signal corresponding to each term must be right-sided. Therefore, we will have:

][][][ 21 nxnxnx +=

31,

311

2][

41,

411

1][

1

Z2

1

Z1

>−

→←

>−

→←

−

−

zz

nx

zz

nx

So:

][312][

41][ nununx

nn

+

=

(b) Since the ROC of )(zX is the intersection of 31

<z and 41

>z (in the form of a

ring), the signal corresponding to the pole 31

=z is a left-sided signal and the

signal corresponding to the pole 41

=z is a right-sided signal. Therefore, we will

have:

][][][ 21 nxnxnx +=

31,

311

2][

41,

411

1][

1

Z2

1

Z1

<−

→←

>−

→←

−

−

zz

nx

zz

nx

So:

]1[312][

41][ −−

−

= nununx

nn

- One can also use long division to find a power series for a rational z-transform.

- Example 12.2: Using long division, find the inverse z-transform of the following

3

expression:

111)( −−

=az

zX

with the following regions of convergence:

(a) az >

(b) az <

- Solution:

(a) We have:

L+++

−

−

−

−−

−

−−

−

−

−

221

22

221

1

1

11

111

zaaz

zazaaz

azaz

az

This means that:

L+++=−

−−−

2211 1

11 zaazaz

Since az > , we have 11 <−az and the series given by (12.1) converges. So, by

comparing this series with the general equation for the z-transform of a signal we

will have:

L,]2[,]1[,1]0[,0for0][ 2axaxxnnx ===<=

and in general:

][][ nuanx n= .

(b) For az < the series given by (12.1) does not converge as 11 >−az . Therefore,

one can use the following form of long division:

L−−−

−

+−

−−

−

−

−

221

1

1

1

111

zaza

zaza

az

(12.1)

4

This means that:

L−−−=−

−−−

22111

1 zazaaz

Since az < , we have 11 <− za and the series given by (12.2) converges. So, by

comparing this series with the general equation for the z-transform of a signal we

will have:

L,]2[,]1[,0for0][ 21 −− −=−−=−≥= axaxnnx

and in general:

]1[][ −−−= nuanx n .

- One can find the inverse z-transform of non-rational expressions of z, by writing that

expression as a power series (for example by using Taylor series expansion).

- Example 12.3: Using the power series expansion, find the inverse z-transform of the

following expression:

azezX za >= ,)( )/(

- Solution: With az > or equivalently 11 <−az , we can use the Taylor series

expansion for )/( zae as follows:

1,!2

1 122

1)( 1

<+++= −−

−−

azzaaze az L

This implies that:

<

≥=0,0

0,!][

n

nna

nxn

Geometric evaluation of the Fourier transform from the pole-zero plot

- We know that the z-transform reduces to the Fourier transform for 1=z (i.e., for the

values of the complex variable z on the unit circle, provided that the ROC of the z-

transform includes the unit circle. As a result, one can use the pole-zero plot of a

transfer function to find the frequency response of the system by evaluating the

magnitude and phase of the system on the unit circle in the complex plane.

- Consider the following transfer function:

(12.2)

5

)(

)()(

1

1

j

n

j

i

m

i

pz

zzzH

−Π

−Π=

=

=

where iz ’s and jp ’s represent the zeros and poles of the transfer function,

respectively. We have:

jj

n

j

ij

m

ij

pe

zeeH

−Π

−Π=

=

=

ω

ω

ω

1

1)(

)()()(11∑∑==

−∠−−∠=∠n

jj

jm

ii

jj pezeeH ωωω

Note that xe j −ω and )( xe j −∠ ω represent the magnitude and phase of the vector v

from the point x to the point ωje (which is a point on the unit circle with the phase

ω ) in the complex plane, respectively. This is shown in the following figure:

We will use this result to find the frequency response of any discrete-time LTI system

with rational transfer function given by (12.3). At any frequency ω , find the

magnitude and phase of the vectors drawn from the poles and zeros to the point ωje (

a point on the unit circle with angle ω ). The magnitude of the frequency response at

ω is equal to the product of the magnitudes of all vectors associated with the zeros

divided by the product of the magnitudes of all vectors associated with the poles.

Similarly, the phase of the frequency response at ω is equal to the summation of the

angles of all vectors associated with the zeros minus the summation of the angles of

ωje

Imz

Rez

z-plane

x

vω

(12.3)

6

all vectors associated with the poles.

- For example, consider an LTI system with a impulse response ][nh and assume that

the transfer function of the system (the z-transform of the impulse response) is a

rational function of z whose pole-zero configuration is given as follows:

- From this plot, it can be concluded that the z-transform of the impulse response is:

))(()()(

21

1

pzpzzzKzH−−

−=

- The amplitude and angle of the vectors 1v , 2v , and 3v depend on the frequency ω .

- The magnitude of the frequency response of this system is proportional to 32

1

vvv

, and

is shown in the following figure:

ωje

Imz

Rez

z-plane

1v

ω3v

2v

P2

z1

p1

7

- The phase of the frequency response of this system is equal to 321 vvv ∠−∠−∠ , and

is shown in the following figure:

- The magnitude of the frequency response is large at those frequencies that correspond

to the points on the unit circle which are close to the poles and far from the zeros.

Similarly, the magnitude of the frequency response is small at those frequencies that

correspond to the points on the unit circle which are close to the zeros and far from

the poles.

Properties of the continuous-time Fourier transform

- The properties of the z-transform are very similar to those of the Laplace transform.

The following properties can easily been shown using the definition of the z-

transform.

- Linearity:

21Z

2Z

1Z

RRcontain ROC),()(][][

RROC),(][

RROC),(][

IszbYzaXnbynaxzYnyzXnx

+→←+⇒

=→←

=→←

- Time shifting:

8

infinity.or origin theof deletionor addition possible

for theexcept ,R ROC),(][

RROC),(][0Z

0

Z

=→←−⇒

=→←− zXznnx

zXnxn

- For example, the z-transform of the unit impulse ][nδ is equal to 1 and the ROC is

the entire z-plane. The z-transform of ]1[ −nδ is equal to z and the ROC is the entire

z-plane except for the infinity. The z-transform of ]1[ +nδ is equal to 1−z and the

ROC is the entire z-plane except for the origin.

- Scaling in the z-domain: For any complex number 0z we have:

R ROC),(][

RROC),(][

00

Z0

Z

zzzXnxz

zXnx

n =→←⇒

=→←

- Time reversal:

RzXnx

zXnx1 ROC),1(][

RROC),(][

Z

Z

=→←−⇒

=→←

- Conjugation:

RzXnxzXnx

=→←⇒

=→←

ROC),(][RROC),(][

**Z*

Z

- The convolution property:

2121Z

21

22Z

2

11Z

1

RRcontain ROC),()(][][

RROC),(][

RROC),(][

IszXzXnxnxzXnxzXnx

→←∗⇒

=→←

=→←

- Differentiation in the z-domain:

RdzzdXznnx

zXnx

=−→←⇒

=→←

ROC,)(][

RROC),(][

Z

Z

- The initial-value theorem: If 0][ =nx for 0<n , then:

)(lim]0[ zXxz ∞→

=

9

This property can be easily seen from the definition of the z-transform.

- The final-value theorem: If 0][ =nx for 0<n and if ][nx has a finite value as

∞→n , then:

)]()1[(lim][lim 1

1zXznx

zn

−

→∞→−=

Analysis and characterization of LTI systems using z-transforms

- Consider a discrete-time LTI system with the impulse response ][nh and the transfer

function (system function) )(zH . From the convolution property we have:

)()()(][][][ zXzHzYnxnhny =⇒∗=

- Causality: A discrete-time LTI system is causal if and only if the ROC of its transfer

function is the exterior of a circle, including infinity.

- A discrete-time LTI system with rational transfer function )(zH is causal if and only

if:

(a) the ROC is the exterior of a circle outside the outermost pole;

(b) with )(zH expressed as a ratio of polynomials in z, the order of the numerator

cannot be greater than the order of the denominator.

- Stability: A discrete-time LTI system is stable if and only if the ROC of its transfer

function )(zH includes the unit circle, 1=z .

- A causal discrete-time LTI system with rational transfer function )(zH is stable if

and only if all of the poles of )(zH lie inside the unit circle (i.e., they must all have

magnitude smaller than 1).

- Systems characterized by linear constant coefficients differential equations: Consider

an LTI system whose input and output are related through the following difference

equation:

∑∑==

−=−M

kk

N

kk knxbknya

00][][

- Taking the z-transform of both sides of this equation, we will have:

10

∑

∑

∑∑

=

−

=

−

=

−

=

−

==⇒

=

N

k

kk

M

k

kk

M

k

kk

N

k

kk

za

zb

zXzYzH

zbzXzazY

0

0

00

)()()(

)()(

- Example 12.4: Find the z-transform of a discrete-time system whose input ][nx and

output ][ny are related through the following differential equation:

]1[31][]1[

21][ −+=−− nxnxnyny

- Solution: We have:

1

1

11

211

311

)()()(

)(31)()(

21)(

−

−

−−

−

+==⇒

+=−

z

z

zXzYzH

zXzzXzYzzY

There are two different choices for the impulse response of the system with the given

difference equation, as we have two different choices for the ROC: 21

>z and

21

<z .

(a) For 21

>z we have:

]1[)21(

31][)

21(][

211

131

211

1

211

311

1

1

1

11

1

−+=⇒−

+−

=−

+−

−

−

−−

−

nununhz

zzz

znn

(b) For 21

<z we have:

][)21(

31]1[)

21(]1)1([)

21(

31]1[)

21(][ 11 nununununh nnnn −−−−−=−−−−−−−= −−