1497-chapter5

Chapter 5

Transform Analysis of Linear Time-Invariant Systems

Der-Feng Tseng

Department of Electrical EngineeringNational Taiwan University of Science and Technology

(through the courtesy of Prof. Peng-Hua Wang of National Taipei University)

February 19, 2015

Der-Feng Tseng (NTUST) DSP Chapter 5 February 19, 2015 1 / 75

http://homepage.ntust.edu.tw/dtseng/

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Outline

1 5.1 Frequency Response

2 5.2 System Functions

3 5.3 Frequency Response

4 5.4 Magnitude and Phase

5 5.5 All-Pass System

6 5.6 Minimum-Phase System

7 5.7 Generalized Linear Phase



About The figures

All the figures are from Discrete-Time Signal Processing, 2e, byOppenheim, Schafer, and Buck,Prentice Hall, Inc.



5.1 Frequency Response



Ideal Lowpass Filter

The frequency response of the ideal lowpass filter is

Hlp(ejω) =

{

1, |ω| < ωc

0, ωc < |ω| ≤ π.

The corresponding impulse response is

hlp[n] =sinωcn

πn, −∞ < n < ∞.

The ideal lowpass filter is noncausal. It is NOT computational

realizable.



Ideal Highpass Filter

The frequency response of the ideal highpass filter is

Hhp(ejω) =

{

0, |ω| < ωc

1, ωc < |ω| ≤ π.

Since Hhp(ejω) = 1−Hlp(e

jω), the corresponding impulse responseis

hhp[n] = δ[n]− hlp[n] = δ[n]−sinωcn

πn, −∞ < n < ∞.



Delay

The impulse response and the frequency response of the ideal delayfilter is

hid[n] = δ[n− nd], Hid(ejω) = e−jωnd .

The frequency response of the ideal lowpass filter with delay ndis

Hlp(ejω) =

{

e−jωnd , |ω| < ωc

0, ωc < |ω| ≤ π.

The corresponding impulse response is

hlp[n] =sinωc(n− nd)

π(n− nd), −∞ < n < ∞.

Ideal delay ⇒ Linear phase



Group Delay

The linearity of the phase response can be measured by the groupdelay defined by

τ(ω) = grd[H(ejω)] = −d

dω∢H(ejω).



Example 5.1



5.2 System Functions



Linear Difference Equations

The linear constant-coefficient difference equations

N∑

k=0

aky[n− k] =

M∑

m=0

bmx[n−m]

We can calculate the output recursively by

y[n] =1

a0

(

−

N∑

k=1

aky[n− k] +

M∑

m=0

bmx[n−m]

)

We want to characterize the LTI systems by the linear differenceequation.Take z-transform, we have

H(z) =Y (z)

X(z)=

M∑

m=0

bmz−m

N∑

k=0

akz−k



Example 5.2

Suppose that the system function of a LTI system is

H(z) =(1 + z−1)2

(1− 12z

−1)(1 + 34z

−1),

find the difference equation that is satisfied by the input and output ofthis system.

Solution. Since

H(z) =1 + 2z−1 + z−2

1 + 14z

−1 − 38z

−2=

Y (z)

X(z),

we have

y[n] + 14y[n− 1]− 3

8y[n− 2] = x[n] + 2x[n− 1] + x[n− 2]



ROC

Causal ⇒ ROC is outside the outermost pole

Anti-causal ⇒ ROC is inside the innermost pole.

Stable ⇒ ROC include the unit circle.



Example 5.3

Consider the LTI system with input and output related through thedifference equation

y[n]− 52y[n− 1] + y[n− 2] = x[n].

Discuss the stability, causality and ROC.



Example 5.3

Solution.

H(z) =1

1− 52z

−1 + z−2=

1

(1− 12z

−1)(1 − 2z−1)

ROC = {z : |z| > 2}: causal, not stable.

ROC = {z : |z| < 12}: not causal, not stable.

ROC = {z : 12 < |z| < 2}: not causal, stable.



Inverse System

Given H(z), the inverse system Hi(z) satisfies

H(z)Hi(z) = 1

⇒ h[n] ∗ hi[n] = δ[n]⇒ Regions of convergence of H(z) and Hi(z) must

overlap.



Example 5.4

Let H(z) be

H(z) =1− 0.5z−1

1− 0.9z−1.

Find the inverse system Hi(z) and discuss the ROC.

Solution.

Hi(z) =1− 0.9z−1

1− 0.5z−1

ROC of Hi(z) is either |z| > 0.5 or |z| < 0.5.

If the ROC of H(z) is |z| > 0.9, the ROC of Hi(z) is |z| > 0.5.

If the ROC of H(z) is |z| < 0.9, we have two inverse systems.



Example 5.5

Let H(z) be

H(z) =z−1 − 0.5

1− 0.9z−1, |z| > 0.9

Find the inverse system Hi(z) and discuss the ROC.

Solution.

Hi(z) =1− 0.9zz−1

z−1 − 0.5=

−2 + 1.8z−1

1− z−1

ROC of Hi(z) is either |z| > 0.5 or |z| < 0.5.

Both ROCs of |z| > 2 and |z| < 2 overlap with |z| > 0.9, so both arevalid inverse system.

If the ROC of Hi(z) is |z| < 2,hi1[n] = 2n+1u[−n− 1]− 1.8 · 2n−1u[−n]. (stable, noncausal)If the ROC of Hi(z) is |z| > 2,hi2[n] = −2n+1u[n] + 1.8 · 2n−1u[n− 1]. (unstable, causal)



Impulse Response

Suppose a causal system function can be expressed in the form

H(z) =

M−N∑

r=0

Brz−r +

N∑

k=1

Ak

1− dkz−1

Its impulse response is

h[n] =

M−N∑

r=0

Brδ[n− r] +

N∑

k=1

Akdnku[n]

If at least one nonzero pole of H(z) is not canceled by a zero, oneterms of the form Akd

nku[n] exists. ⇒ infinite impulse response (IIR)

system.

H(z) has no poles except at z = 0 ⇒ finite impulse response (FIR)system.



Example 5.6

Consider a causal system whose input and output satisfy the differenceequation

y[n]− ay[n− 1] = x[n]

The system function is

H(z) =1

1− az−1

The ROC is |z| > |a|, thecondition for stability is |a| < 1.

h[n] = anu[n]



Example 5.7

Consider the following impulse response and transfer function

h[n] =

{

an, 0 ≤ n ≤ M

0, otherwise,H(z) =

1− aM+1z−(M+1)

1− az−1

The zeros areaej2πk/(M+1), k = 0, 1, . . . ,M.

The pole at z = a is canceled bya zero.

The difference equation isy[n] =

∑Mk=0 a

kx[n− k]

An equivalent equation isy[n]− ay[n− 1] =x[n]− aM+1x[n−M − 1]



5.3 Frequency Response




The frequency response to a system function

H(z) =

M∑

m=0

bmz−m

N∑

k=0

akz−k

is H(ejω) =

M∑

m=0

bme−jωm

N∑

k=0

e−jωk

The frequency response can be factored as

H(ejω) =

(

b0a0

)

M∏

m=1

(1− cme−jω)

N∏

k=1

(1 − dke−jω)

Magnitude response: |H(ejω)|

Gain in dB = 20 log10 |H(ejω)|.

Attenuation in dB = −20 log10 |H(ejω)| = − Gain in dB.



Phase Response

Phase response

∢H(ejω) = ∢[b0/a0] +

M∑

m=1

∢[1− cme−jω]−

N∑

k=1

∢[1− dke−jω ]

Group delay

grd[H(ejω)] =

M∑

m=1

d

dωarg[1− cme−jω ]−

N∑

k=1

d

dωarg[1− dke

−jω ]

=

M∑

m=1

|dm|2 −ℜ{dme−jω}

1 + |dm|2 − 2ℜ{dme−jω}−

N∑

k=1

|ck|2 −ℜ{cke

−jω}

1 + |ck|2 − 2ℜ{cke−jω}

where arg[·] represents the continuous phase.



Phase Response

Principle value: −π < ARG[H(ejω)] ≤ π

continuous phase response

∢H(ejω) = ARG[H(ejω)] + 2πr(ω)

where r(ω) is a positive or negative integer.



Single Zero

H(ejω) = 1− rejθe−jω

|H(ejω)|2 = 1 + r2 − 2r cos(ω − θ)

Principle value of phase response

ARG[H(ejω)] = arctan

[

r sin(ω − θ)

1− r cos(ω − θ)

]

Group delay

grd[H(ejω)] =r2 − r cos(ω − θ)

1 + r2 − 2r cos(ω − θ)



Single Zero



Example 5.8

Consider the following second order system

H(z) =1

(1− rejθz−1)(1 − re−jθz−1)=

1

1− 2r cos θz−1 + r2z−2



Example 5.8



Example 5.10

Consider the following second order system

H(z) =0.05634(1+ z−1)(1 − 1.0166z−1 + z−2)

(1− 0.683z−1)(1− 1.4461z−1 + 0.7957z−2)



Example 5.10



5.4 Magnitude and Phase



Example 5.11

Consider the following two systems

H1(z) =2(1− z−1)(1 + 0.5z−1)

(1− 0.8ejπ/4z−1)(1 − 0.8e−jπ/4z−1)

H2(z) =(1− z−1)(1 + 2z−1)

(1− 0.8ejπ/4z−1)(1 − 0.8e−jπ/4z−1)

We can show that C1(z) = C2(z). That is, the magnitude responses ofboth systems are the same.



Example 5.12

Suppose we are given the pole-zero plot for C(z) in the figure below. Wewant to determine a causal stable H(z).



Example 5.12

3 poles and 3 zeros for H(z)

Poles: (P1, P2, P3) for causality and stability

Zeros:

Real coefficient: (z3, z1, z2) or (z6, z1, z2) or (z3, z4, z5) or (z6, z4, z5)⇒ 4 choices.Complex coefficient: additional (z3, z1, z5) or (z3, z2, z4) or (z6, z1, z5)or (z6, z2, z4) ⇒ 8 choices.

If the number of poles and zeros are not restricted, we have infinitechoice of H(z). For example, if

H(z) = H1(z)z−1 − a∗

1− az−1,

then H(z) and H1(z) have the same C(z).



5.5 All-Pass System



All-Pass System

A first order all-pass system function is

Hap(z) =z−1 − a∗

1− az−1or Hap(z) =

a∗ + z−1

1 + az−1

A general all-pass system function is

Hap(z) =a∗n + a∗n−1z

−1 + a∗n−2z−2 + · · ·+ a∗0z

−n

a0 + a1z−1 + a2z−2 + · · ·+ anz−n

where a0 = 1.



Example 5.13



Properties of All-Pass System

Any all pass system can be factored as a cascade of first order all-passsystems.

Hap(z) = A

Mr∏

k=1

z−1 − dk1− dkz−1

Mc∏

k=1

z−1 − e∗k1− ekz−1

z−1 − ek1− e∗kz

−1

The group delay of causal and stable all-pass system is positive.

grd

[

e−jω − re−jθ

1− rejθe−jω

]

=1− r2

|1− rejθe−jω |2≥ 0

The phase response of causal and stable all-pass system is negative.

arg[Hap(ejω)] = −

∫ ω

0

grd[Hap(ejφ)]dφ + arg[Hap(e

j0)] ≤ 0

since Hap(ej0) = A, arg[Hap(e

j0)] = 0.



Frequency response



5.6 Minimum-Phase System



Minimum-Phase System

A minimum-phase has all its poles and zeros inside the unit circle.

Any stable and causal system can be factored as a cascade of aminimum-phase system and an all-pass system.

H(z) = Hmin(z)Hap(z)

Put all zeros and poles of H(z) inside the unit circle into H ′

min(z).Suppose 1/c∗, |c| < 1 is a zero of H(z) outside the unit circle. Put1/c∗ to Hap(z). Put c as a pole of Hap(z) and as a zero of Hmin(z).

H(z) = H ′

min(z)(z−1 − c∗) = [H ′

min(z)(1− cz−1)]z−1 − c∗

1− cz−1



Example 5.14

H1(z) =1 + 3z−1

1 + 12z

−1=

(

31 + 1

3z−1

1 + 12z

−1

)(

z−1 + 13

1 + 13z

−1

)

H2(z) =(1 + 3

2e+jπ/4z−1)(1 + 3

2e−jπ/4z−1)

1− 13z

−1

=9

4

(z−1 + 23e

−jπ/4)(z−1 + 23e

jπ/4)

1− 13z

−1

=

[

9

4

(1 + 23e

−jπ/4z−1)(1 + 23e

jπ/4z−1)

1− 13z

−1

]

×

[

(z−1 + 23e

−jπ/4)(z−1 + 23e

jπ/4)

(1 + 23e

−jπ/4z−1)(1 + 23e

jπ/4z−1)

]



System Compensation

Suppose Hd(z) is a distorting system, we want to compensate it bycascading a causal and stable system Hc(z).

Perfect compensation is possible only if Hd(z) is a minimum-phasesystem.

Suppose Hd(z) = Hdmin(z)Hap(z), let

Hc(z) =1

Hdmin(z),

then G(z) = Hap(z).

Magnitude is exactly compensated, but phase is modified.



Example 5.15

Hd(z) = (1− 0.9ej0.6πz−1)(1 − 0.9e−j0.6πz−1)

× (1− 1.25ej0.8πz−1)(1 − 1.25e−j0.8πz−1)

= (1− 0.9ej0.6πz−1)(1 − 0.9e−j0.6πz−1)(1.25)2

× (z−1 − 0.8ej0.8π)(z−1 − 0.8e−j0.8π)

⇒ Hmin(z) = (1.25)2(1 − 0.9ej0.6πz−1)(1 − 0.9e−j0.6πz−1)

× (1− 0.8ej0.8πz−1)(1− 0.8e−j0.8πz−1)

Hap(z) =(z−1 − 0.8ej0.8π)(z−1 − 0.8e−j0.8π)

(1− 0.8ej0.8πz−1)(1 − 0.8e−j0.8πz−1)



Example 5.15



Minimum Phase-Lag

Minimum phase-lag property

arg[H(ejω)] = arg[Hmin(ejω)] + arg[Hap(e

jω)] ≤ arg[Hmin(ejω)]

Among all system functions that have the same magnitude response,the minimum-phase system has the maximal phase response.The phase-lag function is the negative of the phase response. Thus,the minimum-phase system has the minimum phase-lag response. Thisis the origin of the minimum-phase system.

Minimum group-delay property

grd[H(ejω)] = grd[Hmin(ejω)] + grd[Hap(e

jω)] ≥ grd[Hmin(ejω)]



Minimum Energy-Delay




|h[0]| ≤ |hmin[0]|

Total energy

∞∑

n=0

|h[n]|2 =1

2π

∫ π

π

|H(ejω)|2dω

=1

2π

∫ π

π

|Hmin(ejω)|2dω =

∞∑

n=0

|hmin[n]|2

Partial energym∑

n=0

|h[n]|2 ≤

m∑

n=0

|hmin[n]|2



5.7 Generalized Linear Phase



Pure Delay

Consider Hid(ejω) = e−jωα, |ω| < π, we have

|Hid(ejω)| = 1, ∢Hid(e

jω) = −αω, grd[Hid(ejω)] = α

The impulse response is

Hid[n] =sinπ(n− α)

π(n− α), −∞ < n < ∞



Non-Integer Delay

Consider H(ejω) = |H(ejω)|e−jωα, |ω| < π

For example, a lowpass filter

Hlp(ejω) =

{

e−jωα, |ω| < ωc,

0, ωc < |ω| < π.

Its impulse response is

Hlp[n] =sinωc(n− α)

ωc(n− α), −∞ < n < ∞



Example 5.16

Hlp[n] =sinωc(n− α)

ωc(n− α), −∞ < n < ∞

If α = nd is an integer, then Hlp[2nd − n] = Hlp[n] → The impulseresponse is symmetric about n = nd



Example 5.16



Generalized Linear Phase

Generalized linear-phase system

H(ejω) = A(ejω)e−jαω+jβ

grd[H(ejω)] = α, arg[H(ejω)] = β − αω, 0 < ω < π

A necessary condition for generalized linear phase

∞∑

n=−∞

h[n] sin[ω(n− α) + β] = 0

β = 0, π, 2α = M =an integer, h[2α− n] = h[n]β = π/2, 3π/2, 2α = M =an integer, h[2α− n] = −h[n]



Causal G Linear Phase

If the system is causal, then

∞∑

n=0

h[n] sin[ω(n− α) + β] = 0

Since h[2α − n] = ±h[n], we have h[n] = 0 for n < 0 and n > M .That is

h[n] = ±h[M − n]



Type I Linear Phase System

M is even, h[n] = h[M − n]

Frequency response

H(ejω) =M∑

n=0

h[n]e−jωn = e−jωM/2

M/2∑

k=0

a[k] cosωk

where

a[0] = h[M/2]

a[k] = 2h[M/2− k], k = 1, 2, . . . ,M/2.



Example 5.17



Type II Linear Phase System

M is odd, h[n] = h[M − n]

Frequency response

H(ejω) =M∑

n=0

h[n]e−jωn = e−jωM/2

(M+1)/2∑

k=0

b[k] cos[ω(k − 12 )]

whereb[k] = 2h[(M + 1)/2− k] k = 1, 2, . . . , (M + 1)/2.



Example 5.18



Type III Linear Phase System

M is even, h[n] = −h[M − n]

Frequency response

H(ejω) =M∑

n=0

h[n]e−jωn = je−jωM/2

M/2∑

k=1

c[k] sinωk

wherec[k] = 2h[M/2− k] k = 1, 2, . . . ,M/2.



Example 5.19



Type IV Linear Phase System

M is odd, h[n] = −h[M − n]

Frequency response

H(ejω) =M∑

n=0

h[n]e−jωn = je−jωM/2

(M+1)/2∑

k=1

d[k] sin[ω(k − 12 )]

whered[k] = 2h[(M + 1)/2− k] k = 1, 2, . . . , (M + 1)/2.



Example 5.20



FIR Linear Phase System



Pole-Zero plot



1497-chapter5

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