ec 2314 digital signal processing by dr. k. udhayakumar

EC 2314 Digital Signal Processing

By

Dr. K. Udhayakumar

The z-Transform

Dr. K. Udhayakumar

Content

Introduction z-Transform Zeros and PolesRegion of Convergence Important z-Transform Pairs Inverse z-Transform z-Transform Theorems and Properties System Function

The z-Transform

Introduction

Why z-Transform?

A generalization of Fourier transformWhy generalize it?

– FT does not converge on all sequence– Notation good for analysis– Bring the power of complex variable theory deal with

the discrete-time signals and systems

The z-Transform

z-Transform

Definition The z-transform of sequence x(n) is defined by

n

nznxzX )()(

Let z = ej.

( ) ( )j j n

n

X e x n e

Fourier Transform

z-Plane

Re

Im

z = ej

n

nznxzX )()(

( ) ( )j j n

n

X e x n e

Fourier Transform is to evaluate z-transform on a unit circle.

Fourier Transform is to evaluate z-transform on a unit circle.

z-Plane

Re

Im

X(z)

Re

Im

z = ej

Periodic Property of FT

Re

Im

X(z)

X(ej)

Can you say why Fourier Transform is a periodic function with period 2?

Can you say why Fourier Transform is a periodic function with period 2?

The z-Transform

Zeros and Poles

Definition

Give a sequence, the set of values of z for which the z-transform converges, i.e., |X(z)|<, is called the region of convergence.

n

n

n

n znxznxzX |||)(|)(|)(|

ROC is centered on origin and consists of a set of rings.

ROC is centered on origin and consists of a set of rings.

Example: Region of Convergence

Re

Im

n

n

n

n znxznxzX |||)(|)(|)(|

ROC is an annual ring centered on the origin.

ROC is an annual ring centered on the origin.

xx RzR ||

r

}|{ xx

j RrRrezROC

Stable Systems

Re

Im

1

A stable system requires that its Fourier transform is uniformly convergent.

Fact: Fourier transform is to evaluate z-transform on a unit circle.

A stable system requires the ROC of z-transform to include the unit circle.

Example: A right sided Sequence

)()( nuanx n )()( nuanx n

1 2 3 4 5 6 7 8 9 10-1-2-3-4-5-6-7-8

n

x(n)

. . .

Example: A right sided Sequence

)()( nuanx n )()( nuanx n

n

n

n znuazX

)()(

0n

nn za

0

1)(n

naz

For convergence of X(z), we require that

0

1 ||n

az 1|| 1 az

|||| az

az

z

azazzX

n

n

10

1

1

1)()(

|||| az

aa

Example: A right sided Sequence ROC for x(n)=anu(n)

|||| ,)( azaz

zzX

|||| ,)( az

az

zzX

Re

Im

1aa

Re

Im

1

Which one is stable?Which one is stable?

Example: A left sided Sequence

)1()( nuanx n )1()( nuanx n

1 2 3 4 5 6 7 8 9 10-1-2-3-4-5-6-7-8n

x(n)

. . .

Example: A left sided Sequence

)1()( nuanx n )1()( nuanx n

n

n

n znuazX

)1()(

For convergence of X(z), we require that

0

1 ||n

za 1|| 1 za

|||| az

az

z

zazazX

n

n

10

1

1

11)(1)(

|||| az

n

n

n za

1

n

n

n za

1

n

n

n za

0

1

aa

Example: A left sided Sequence ROC for x(n)=anu( n1)

|||| ,)( azaz

zzX

|||| ,)( az

az

zzX

Re

Im

1aa

Re

Im

1

Which one is stable?Which one is stable?

The z-Transform

Region of Convergence

Represent z-transform as a Rational Function

)(

)()(

zQ

zPzX where P(z) and Q(z) are

polynomials in z.

Zeros: The values of z’s such that X(z) = 0

Poles: The values of z’s such that X(z) =

Example: A right sided Sequence

)()( nuanx n |||| ,)( azaz

zzX

Re

Im

a

ROC is bounded by the pole and is the exterior of a circle.

Example: A left sided Sequence

)1()( nuanx n|||| ,)( az

az

zzX

Re

Im

a

ROC is bounded by the pole and is the interior of a circle.

Example: Sum of Two Right Sided Sequences

)()()()()( 31

21 nununx nn

31

21

)(

z

z

z

zzX

Re

Im

1/2

))((

)(2

31

21

121

zz

zz

1/3

1/12

ROC is bounded by poles and is the exterior of a circle.

ROC does not include any pole.

Example: A Two Sided Sequence

)1()()()()( 21

31 nununx nn

21

31

)(

z

z

z

zzX

Re

Im

1/2

))((

)(2

21

31

121

zz

zz

1/3

1/12

ROC is bounded by poles and is a ring.

ROC does not include any pole.

Example: A Finite Sequence

10 ,)( Nnanx n

nN

n

nN

n

n zazazX )()( 11

0

1

0

Re

Im

ROC: 0 < z <

ROC does not include any pole.

1

1

1

)(1

az

az N

az

az

z

NN

N

1

1

N-1 poles

N-1 zeros

Always StableAlways Stable

Properties of ROC

A ring or disk in the z-plane centered at the origin. The Fourier Transform of x(n) is converge absolutely iff the ROC

includes the unit circle. The ROC cannot include any poles Finite Duration Sequences: The ROC is the entire z-plane except

possibly z=0 or z=. Right sided sequences: The ROC extends outward from the outermost

finite pole in X(z) to z=. Left sided sequences: The ROC extends inward from the innermost

nonzero pole in X(z) to z=0.

More on Rational z-Transform

Re

Im

a b c

Consider the rational z-transform with the pole pattern:

Find the possible ROC’s

Find the possible ROC’s

More on Rational z-Transform

Re

Im

a b c

Consider the rational z-transform with the pole pattern:

Case 1: A right sided Sequence.

More on Rational z-Transform

Re

Im

a b c

Consider the rational z-transform with the pole pattern:

Case 2: A left sided Sequence.

More on Rational z-Transform

Re

Im

a b c

Consider the rational z-transform with the pole pattern:

Case 3: A two sided Sequence.

More on Rational z-Transform

Re

Im

a b c

Consider the rational z-transform with the pole pattern:

Case 4: Another two sided Sequence.

0 5 10 15 200

0.2

0.4

0.6

0.8

1

Bounded Signals

-5

0

5a=0.4

-5

0

5a=0.9

-5

0

5a=1.2

0 5 10-5

0

5a=-0.4

0 5 10-5

0

5a=-0.9

0 5 10-5

0

5a=-1.2

0 10 20 30 40 50 60 70-1

-0.5

0

0.5

1

0 2 4 6 8-1

-0.5

0

0.5

1

BIBO Stability

Bounded Input Bounded Output Stability– If the Input is bounded, we want the Output is

bounded, too– If the Input is unbounded, it’s okay for the Output to

be unbounded

For some computing systems, the output is intrinsically bounded (constrained), but limit cycle may happen

The z-Transform

Important

z-Transform Pairs

Z-Transform Pairs

Sequence z-Transform ROC

)(n 1 All z

)( mn mz All z except 0 (if m>0)or (if m<0)

)(nu 11

1 z

1|| z

)1( nu 11

1 z

1|| z

)(nuan 11

1 az

|||| az

)1( nuan 11

1 az

|||| az

Z-Transform Pairs

Sequence z-Transform ROC

)(][cos 0 nun 210

10

]cos2[1

][cos1

zz

z1|| z

)(][sin 0 nun 210

10

]cos2[1

][sin

zz

z1|| z

)(]cos[ 0 nunr n 2210

10

]cos2[1

]cos[1

zrzr

zrrz ||

)(]sin[ 0 nunr n 2210

10

]cos2[1

]sin[

zrzr

zrrz ||

otherwise0

10 Nnan

11

1

az

za NN

0|| z

Signal Type ROCFinite-Duration Signals

Infinite-Duration Signals

Causal

Anticausal

Two-sided

Causal

Anticausal

Two-sided

Entire z-planeExcept z = 0

Entire z-planeExcept z = infinity

Entire z-planeExcept z = 0And z = infinity

|z| < r1

|z| > r2

r2 < |z| < r1

Some Common z-Transform Pairs

Sequence Sequence Transform Transform ROCROC

1. 1. [n] [n] 1 1 all zall z

2. u[n] 2. u[n] z/(z-1) z/(z-1) |z|>1|z|>1

3. -u[-n-1] 3. -u[-n-1] z/(z-1) z/(z-1) |z|<1|z|<1

4. 4. [n-m] [n-m] zz-m-m all z except 0 if m>0 or all z except 0 if m>0 or ฅฅif m<0if m<0

5. a5. annu[n] u[n] z/(z-a) z/(z-a) |z|>|a||z|>|a|

6. -a6. -annu[-n-1] u[-n-1] z/(z-a) z/(z-a) |z|<|a||z|<|a|

7. na7. nannu[n] u[n] az/(z-a)az/(z-a)22 |z|>|a||z|>|a|

8. -na8. -nannu[-n-1] u[-n-1] az/(z-a)az/(z-a)22 |z|<|a||z|<|a|

9. [cos9. [cos00n]u[n] n]u[n] (z(z22-[cos-[cos00]z)/(z]z)/(z22-[2cos-[2cos00]z+1) ]z+1) |z|>1|z|>1

10. [sin10. [sin00n]u[n] n]u[n] [sin[sin00]z)/(z]z)/(z22-[2cos-[2cos00]z+1) ]z+1) |z|>1|z|>1

11. [r11. [rnncoscos00n]u[n] n]u[n] (z(z22-[rcos-[rcos00]z)/(z]z)/(z22-[2rcos-[2rcos00]z+r]z+r22) ) |z|>r|z|>r

12. [r12. [rnnsinsin00n]u[n] n]u[n] [rsin[rsin00]z)/(z]z)/(z22-[2rcos-[2rcos00]z+r]z+r22) ) |z|>r|z|>r

13. a13. annu[n] - au[n] - annu[n-N] u[n-N] (z(zNN-a-aNN)/z)/zN-1N-1(z-a) (z-a) |z|>0|z|>0

The z-Transform

Inverse z-Transform

Inverse Z-Transform by Partial Fraction Expansion

Assume that a given z-transform can be expressed as

Apply partial fractional expansion First term exist only if M>N

– Br is obtained by long division

Second term represents all first order poles Third term represents an order s pole

– There will be a similar term for every high-order pole Each term can be inverse transformed by inspection

N

k

kk

M

k

kk

za

zbzX

0

0

s

1mm1

i

mN

ik,1k1

k

kNM

0r

rr

zd1

Czd1

AzBzX

Partial Fractional Expression

Coefficients are given as

Easier to understand with examples

s

1mm1

i

mN

ik,1k1

k

kNM

0r

rr

zd1

Czd1

AzBzX

kdz

1kk zXzd1A

1idw

1sims

ms

msi

m wXwd1dwd

d!ms

1C

Example: 2nd Order Z-Transform

– Order of nominator is smaller than denominator (in terms of z-1)– No higher order pole

2

1z :ROC

2

11

4

11

1

11

zz

zX

1

2

1

1

z21

1

A

z41

1

AzX

1

41

21

1

1zXz

41

1A1

41

z

11

2

21

41

1

1zXz

21

1A1

21

z

12

Example Continued

ROC extends to infinity – Indicates right sided sequence

21

z z

21

1

2

z41

1

1zX

11

nu41

-nu21

2nxnn

Example #2

Long division to obtain Bo

1z z1z

21

1

z1

z21

z23

1

zz21zX

11

21

21

21

1z5

2z3z

21z2z1z

23

z21

1

12

1212

11

1

z1z21

1

z512zX

1

2

1

1

z1A

z21

1

A2zX

9zXz21

1A

21

z

11

8zXz1A1z

12

Example #2 Continued

ROC extends to infinity– Indicates right-sides sequence

1z z1

8

z21

1

92zX 1

1

n8u-nu21

9n2nxn

An Example – Complete Solution

386zz1414z3z

limU(z)limc 2

2

zz0

4-z1414z3z

86zz1414z3z

2)(z(z)U

2

2

2

2

2-z1414z3z

86zz1414z3z

4)(z(z)U

2

2

2

4

86zz1414z3z

U(z) 2

2

4z

c2z

ccU(z) 21

0

14-2

1421423(2)Uc

2

21

32-4

1441443(4)Uc

2

42

4z3

2z1

3U(z)

0k,432

0k3,u(k) 1k1k

Inverse Z-Transform by Power Series Expansion

The z-transform is power series

In expanded form

Z-transforms of this form can generally be inversed easily Especially useful for finite-length series Example

n

nz nxzX

2112 2 1 0 1 2 zxzxxzxzxzX

12

1112

z21

1z21

z

z1z1z21

1z zX

1n21

n1n21

2nnx

2n0

1n21

0n1

1n21

2n1

nx

Z-Transform Properties: Linearity

Notation

Linearity

– Note that the ROC of combined sequence may be larger than either ROC– This would happen if some pole/zero cancellation occurs– Example:

Both sequences are right-sided Both sequences have a pole z=a Both have a ROC defined as |z|>|a| In the combined sequence the pole at z=a cancels with a zero at z=a The combined ROC is the entire z plane except z=0

We did make use of this property already, where?

xZ RROC zXnx

21 xx21

Z21 RRROC zbXzaXnbxnax

N-nua-nuanx nn

Z-Transform Properties: Time Shifting

Here no is an integer– If positive the sequence is shifted right– If negative the sequence is shifted left

The ROC can change the new term may– Add or remove poles at z=0 or z=

Example

xnZ

o RROC zXznnx o

41

z z

41

1

1z zX

1

1

1-nu41

nx1-n

Z-Transform Properties: Multiplication by Exponential

ROC is scaled by |zo| All pole/zero locations are scaled If zo is a positive real number: z-plane shrinks or expands

If zo is a complex number with unit magnitude it rotates Example: We know the z-transform pair

Let’s find the z-transform of

xooZn

o RzROCzzXnxz /

1z:ROC z-1

1nu 1-

Z

nure21

nure21

nuncosrnxnjnj

on oo

rz zre1

2/1zre1

2/1zX

1j1j oo

Z-Transform Properties: Differentiation

Example: We want the inverse z-transform of

Let’s differentiate to obtain rational expression

Making use of z-transform properties and ROC

x

Z RROC dz

zdXznnx

az az1logzX 1

1

11

2

az11

azdz

zdXz

az1az

dzzdX

1nuaannx 1n

1nuna

1nxn

1n

Z-Transform Properties: Conjugation

Example

x**Z* RROC zXnx

nxZz nxz nxzX

z nxz nxzX

z nxzX

n

n

n

n

n

n

n

n

n

n

Z-Transform Properties: Time Reversal

ROC is inverted Example:

Time reversed version of

x

Z

R1

ROC z/1Xnx

nuanx n

nuan

111-

1-1

az za-1

za-az1

1zX

Z-Transform Properties: Convolution

Convolution in time domain is multiplication in z-domain Example:Let’s calculate the convolution of

Multiplications of z-transforms is

ROC: if |a|<1 ROC is |z|>1 if |a|>1 ROC is |z|>|a| Partial fractional expansion of Y(z)

2x1x21

Z21 RR:ROC zXzXnxnx

nunx and nuanx 2n

1

az:ROC az11

zX 11

1z:ROC z1

1zX 12

1121 z1az11

zXzXzY

1z :ROC asume az11

z11

a11

zY 11

nuanua1

1ny 1n

The z-Transform

z-Transform Theorems and Properties

Linearity

xRzzXnx ),()]([Z

yRzzYny ),()]([Z

yx RRzzbYzaXnbynax ),()()]()([Z

Overlay of the above two

ROC’s

Shift

xRzzXnx ),()]([Z

xn RzzXznnx )()]([ 0

0Z

Multiplication by an Exponential Sequence

xx- RzRzXnx || ),()]([Z

xn RazzaXnxa || )()]([ 1Z

Differentiation of X(z)

xRzzXnx ),()]([Z

xRzdz

zdXznnx

)()]([Z

Conjugation

xRzzXnx ),()]([Z

xRzzXnx *)(*)](*[Z

Reversal

xRzzXnx ),()]([Z

xRzzXnx /1 )()]([ 1 Z

Real and Imaginary Parts

xRzzXnx ),()]([Z

xRzzXzXnxe *)](*)([)]([ 21R

xj RzzXzXnx *)](*)([)]([ 21Im

Initial Value Theorem

0for ,0)( nnx

)(lim)0( zXxz

Convolution of Sequences

xRzzXnx ),()]([Z

yRzzYny ),()]([Z

yx RRzzYzXnynx )()()](*)([Z

Convolution of Sequences

k

knykxnynx )()()(*)(

n

n

k

zknykxnynx )()()](*)([Z

k

n

n

zknykx )()(

k

n

n

k znyzkx )()(

)()( zYzX

The z-Transform

System Function

Signal Characteristics from Z-Transform

If U(z) is a rational function, and

Then Y(z) is a rational function, too

Poles are more important – determine key characteristics of y(k)

m)u(kb...1)u(kbn)y(ka...1)y(kay(k) m1n1

m

1jj

n

1ii

)p(z

)z(z

D(z)N(z)

Y(z)

zeros

poles

Why are poles important?

m

1j j

j0m

1jj

n

1ii

pz

cc

)p(z

)z(z

D(z)N(z)

Y(z)

m

1j

1-kjjimpulse0 pc(k)ucY(k)

Z-1

Z domain

Time domain

poles

components

Various pole values (1)

-1 0 1 2 3 4 5 6 7 8 90

0.5

1

1.5

2

2.5

-1 0 1 2 3 4 5 6 7 8 90

0.2

0.4

0.6

0.8

1

-1 0 1 2 3 4 5 6 7 8 90

0.2

0.4

0.6

0.8

1

-1 0 1 2 3 4 5 6 7 8 9-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

-1 0 1 2 3 4 5 6 7 8 9-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-1 0 1 2 3 4 5 6 7 8 9-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

p=1.1

p=1

p=0.9

p=-1.1

p=-1

p=-0.9

Various pole values (2)

-1 0 1 2 3 4 5 6 7 8 90

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

-1 0 1 2 3 4 5 6 7 8 90

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

-1 0 1 2 3 4 5 6 7 8 90

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

p=0.9

p=0.6

p=0.3

-1 0 1 2 3 4 5 6 7 8 9-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-1 0 1 2 3 4 5 6 7 8 9-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-1 0 1 2 3 4 5 6 7 8 9-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

p=-0.9

p=-0.6

p=-0.3

Conclusion for Real Poles

If and only if all poles’ absolute values are smaller than 1, y(k) converges to 0

The smaller the poles are, the faster the corresponding component in y(k) converges

A negative pole’s corresponding component is oscillating, while a positive pole’s corresponding component is monotonous

How fast does it converge?

U(k)=ak, consider u(k)≈0 when the absolute value of u(k) is smaller than or equal to 2% of u(0)’s absolute value

|a|ln4

k

3.912ln0.02|a|kln

0.02|a| k

110.36

4|0.7|ln

4k

0.7a

Remember

This!

0 2 4 6 8 10 120

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

y(k)=0.7k

y(11)=0.0198

When There Are Complex Poles …

U(z)za...za1

zb...zbY(z)

nn

11

mm

11

c)...bz(az2

2a

4acbbz

2

0,4acb2 )

2a4acbb

)(z2a

4acbba(zcbzaz

222

0,4acb2 )2a

b4acib)(z

2ab4acib

a(zcbzaz22

2

If

If

Or in polar coordinates,

)irr)(zirra(zcbzaz2 θθθθ sincossincos

What If Poles Are Complex

If Y(z)=N(z)/D(z), and coefficients of both D(z) and N(z) are all real numbers, if p is a pole, then p’s complex conjugate must also be a pole

– Complex poles appear in pairs

l

1j22

j

j0

l

1j j

j0

r)z(2rz)rdz(zbzr

pz

cc

irrzc'

irrzc

pz

ccY(z)

θ

θθ

θθθθ

cos

cossin

sincossincos

coskθdrsinkθbrpc(k)ucy(k) kkm

1j

1-kjjimpulse0

Z-1

Time domain

An Example

0 2 4 6 8 10 12 14 16 18 20-1

-0.5

0

0.5

1

1.5

2

)3

kπcos(0.8)

3kπ

sin(0.82y(k)

0.640.8zzzz

Y(z)

kk

2

2

Z-Domain: Complex Poles

Time-Domain:Exponentially Modulated Sin/Cos

Poles Everywhere

Observations

Using poles to characterize a signal– The smaller is |r|, the faster converges the signal

|r| < 1, converge |r| > 1, does not converge, unbounded |r|=1?

– When the angle increase from 0 to pi, the frequency of oscillation increases Extremes – 0, does not oscillate, pi, oscillate at the maximum frequency

Change Angles

0.9-0.9 Re

Im

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10 12 14-6

-4

-2

0

2

4

6

8

10

12

Changing Absolute Value

Im

Re1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10 12 14-3

-2

-1

0

1

2

3

4

Conclusion for Complex Poles

A complex pole appears in pair with its complex conjugate

The Z-1-transform generates a combination of exponentially modulated sin and cos terms

The exponential base is the absolute value of the complex pole

The frequency of the sinusoid is the angle of the complex pole (divided by 2π)

Steady-State Analysis

If a signal finally converges, what value does it converge to? When it does not converge

– Any |pj| is greater than 1– Any |r| is greater than or equal to 1

When it does converge– If all |pj|’s and |r|’s are smaller than 1, it converges to 0– If only one pj is 1, then the signal converges to cj

If more than one real pole is 1, the signal does not converge … (e.g. the ramp signal)

kdrkbr kk cossin

m

1j

1-kjjimpulse0 pc(k)ucy(k) 21

-1

)z(1z

An Example

kk 0.9)(30.52u(k)0.9z

3z0.5zz

1z2z

U(z)

0 10 20 30 40 50 60-1

0

1

2

3

4

5

6

converge to 2

Final Value Theorem

Enable us to decide whether a system has a steady state error (yss-rss)

Final Value Theorem

1

Theorem: If all of the poles of (1 ) ( ) lie within the unit circle, then

lim ( ) lim ( 1) ( )k z

z Y z

y k z Y z

AAAAAAAAAAAAAAAAAAAAAAAAAAAA

2

1 1

0.11 0.11( )

1.6 0.6 ( 1)( 0.6)

0.11( 1) ( ) | | 0.275

0.6z z

z zY z

z z z z

zz Y z

z

0 5 10 15

-0.35

-0.3

-0.25

-0.2

-0.15

-0.1

-0.05

0

k

y(k)

If any pole of (1-z)Y(z) lies out of or ON the unit circle, y(k) does not converge!

What Can We Infer from TF?

Almost everything we want to know– Stability– Steady-State– Transients

Settling timeOvershoot

– …

Shift-Invariant System

h(n)h(n)

x(n) y(n)=x(n)*h(n)

X(z) Y(z)=X(z)H(z)H(z)

Shift-Invariant System

H(z)H(z)X(z) Y(z)

)(

)()(

zX

zYzH

)(

)()(

zX

zYzH

Nth-Order Difference Equation

M

rr

N

kk rnxbknya

00

)()(

M

rr

N

kk rnxbknya

00

)()(

M

r

rr

N

k

kk zbzXzazY

00

)()(

N

k

kk

M

r

rr zazbzH

00)(

N

k

kk

M

r

rr zazbzH

00)(

Representation in Factored Form

N

kr

M

rr

zd

zcAzH

1

1

1

1

)1(

)1()(

N

kr

M

rr

zd

zcAzH

1

1

1

1

)1(

)1()(

Contributes poles at 0 and zeros at cr

Contributes zeros at 0 and poles at dr

Stable and Causal Systems

N

kr

M

rr

zd

zcAzH

1

1

1

1

)1(

)1()(

N

kr

M

rr

zd

zcAzH

1

1

1

1

)1(

)1()( Re

Im

Causal Systems : ROC extends outward from the outermost pole.

Stable and Causal Systems

N

kr

M

rr

zd

zcAzH

1

1

1

1

)1(

)1()(

N

kr

M

rr

zd

zcAzH

1

1

1

1

)1(

)1()( Re

ImStable Systems : ROC includes the unit circle.

1

Example

Consider the causal system characterized by

)()1()( nxnayny

11

1)(

azzH 11

1)(

azzH

Re

Im

1

a

)()( nuanh n

Determination of Frequency Response from pole-zero pattern

A LTI system is completely characterized by its pole-zero pattern.

))(()(

21

1

pzpz

zzzH

))(()(

21

1

pzpz

zzzH

Example:

))(()(

21

1

00

0

0

pepe

zeeH jj

jj

))(()(

21

1

00

0

0

pepe

zeeH jj

jj

0je

Re

Im

z1

p1

p2

Determination of Frequency Response from pole-zero pattern

A LTI system is completely characterized by its pole-zero pattern.

))(()(

21

1

pzpz

zzzH

))(()(

21

1

pzpz

zzzH

Example:

))(()(

21

1

00

0

0

pepe

zeeH jj

jj

))(()(

21

1

00

0

0

pepe

zeeH jj

jj

0je

Re

Im

z1

p1

p2

|H(ej)|=?|H(ej)|=? H(ej)=?H(ej)=?

Determination of Frequency Response from pole-zero pattern

A LTI system is completely characterized by its pole-zero pattern.

Example:

0je

Re

Im

z1

p1

p2

|H(ej)|=?|H(ej)|=? H(ej)=?H(ej)=?

|H(ej)| =| |

| | | | 1

2

3

H(ej) = 1(2+ 3 )

Example

11

1)(

azzH 11

1)(

azzH

Re

Im

a

0 2 4 6 8-10

0

10

20

0 2 4 6 8-2

-1

0

1

2d

B