The z-Transform · 21) ( 1) 3. x n = − 1. u n − n. u − ... Assume that a given z -transform can be expressed as Apply partial fractional expansion First term exist only if M>N

$Page 1: The z-Transform · 21) ( 1) 3. x n = − 1. u n − n. u − ... Assume that a given z -transform can be expressed as Apply partial fractional expansion First term exist only if M>N$
The z-TransformB Tech VIth Semester EIC- DSP

Content

Introduction

z-Transform

Zeros and Poles

Region of Convergence

Important z-Transform Pairs

Inverse z-Transform

z-Transform Theorems and Properties

System Function

The z-TransformINTRODUCTION

Why z-Transform?A generalization of Fourier transform

Why 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-TransformZ-TRANSFORM

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

∑∞

−∞=

−=n

nznxzX )()(

Let z = e−jω.

( ) ( )j j n

nX e x n eω ω

∞−

=−∞

= ∑

Fourier Transform

z-Plane

Re

Im

z = e−jω

ω

∑∞

−∞=

−=n

nznxzX )()(

( ) ( )j j n

nX e x n eω ω

∞−

=−∞

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

z-Plane

Re

Im

X(z)

Re

Im

z = e−jω

ω

Periodic Property of FT

Re

Im

X(z)

π−π ω

X(ejω)

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

The z-TransformZEROS 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.

Example: Region of Convergence

Re

Im

∞<== ∑∑∞

−∞=

−∞

−∞=

−

n

n

n

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

ROC is an annual ring centered on the origin.

+− << xx RzR ||r

}|{ +−ω <<== xx

j RrRrezROC

Stable Systems

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

Re

Im

1

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=

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

n

x(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 >

azz

azazzX

n

n

−=

−== −

∞

=

−∑ 10

1

11)()(

|||| az >

a−a

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

|||| ,)( azaz

zzX >−

=

Re

Im

1a−a

Re

Im

1

Which one is stable?

Example: A left sided Sequence

)1()( −−−= nuanx n

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

x(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 <

azz

zazazX

n

n

−=

−−=−= −

∞

=

−∑ 10

1

111)(1)(

|||| az <

n

n

n za −−

−∞=∑−=

1

n

n

n za∑∞

=

−−=1

n

n

n za∑∞

=

−−=0

1

a−a

Example: A left sided SequenceROC for x(n)=−anu(− n−1)

|||| ,)( azaz

zzX <−

=

Re

Im

1a−a

Re

Im

1

Which one is stable?

The z-Transform

REGION OF CONVERGENCE

Represent z-transform as a Rational Function

)()()(

zQzPzX = 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) = ∞


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

zzX >−

=

Re

Im

a

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


)1()( −−−= nuanx n |||| ,)( azaz

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

zz

zzX

Re

Im

1/2

))(()(2

31

21

121

+−−

=zz

zz

−1/31/12

ROC is bounded by polesand 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

zz

zzX

Re

Im

1/2

))(()(2

21

31

121

−+−

=zz

zz

−1/31/12

ROC is bounded by polesand is a ring.


Example: A Finite Sequence

10 ,)( −≤≤= Nnanx n

nN

n

nN

n

n zazazX )()( 11

0

1

0

−−

=

−−

=∑∑ ==

Re

Im

ROC: 0 < z < ∞


1

1

1)(1

−

−

−−

=az

az N

azaz

z

NN

N −−

= −11

N-1 poles

N-1 zeros

Always Stable

Properties of ROCA 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


Re

Im

a b c


Case 1: A right sided Sequence.


Re

Im

a b c


Case 2: A left sided Sequence.


Re

Im

a b c


Case 3: A two sided Sequence.


Re

Im

a b c


Case 4: Another two sided Sequence.

The z-Transform

IMPORTANTZ-TRANSFORM PAIRS

Z-Transform PairsSequence z-Transform ROC

)(nδ 1 All z

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

)(nu 111

−− z 1|| >z

)1( −−− nu 111

−− z 1|| <z

)(nuan 111

−− az |||| az >

)1( −−− nuan 111

−− az |||| az <

Z-Transform PairsSequence z-Transform ROC

)(][cos 0 nunω 210

10

]cos2[1][cos1

−−

−

+ω−ω−

zzz

1|| >z

)(][sin 0 nunω 210

10

]cos2[1][sin

−−

−

+ω−ω

zzz

1|| >z

)(]cos[ 0 nunr n ω 2210

10

]cos2[1]cos[1

−−

−

+ω−ω−

zrzrzr

rz >||

)(]sin[ 0 nunr n ω 2210

10

]cos2[1]sin[

−−

−

+ω−ω

zrzrzr

rz >||

−≤≤

otherwise010 Nnan

111

−

−

−−

azza 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 PairsSequence Transform ROC

1. δ[n] 1 all z

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

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

4. δ[n-m] z-m all z except 0 if m>0 or ฅ if m<0

5. anu[n] z/(z-a) |z|>|a|

6. -anu[-n-1] z/(z-a) |z|<|a|

7. nanu[n] az/(z-a)2 |z|>|a|

8. -nanu[-n-1] az/(z-a)2 |z|<|a|

9. [cosω0n]u[n] (z2-[cosω0]z)/(z2-[2cosω0]z+1) |z|>1

10. [sinω0n]u[n] [sinω0]z)/(z2-[2cosω0]z+1) |z|>1

11. [rncosω0n]u[n] (z2-[rcosω0]z)/(z2-[2rcosω0]z+r2) |z|>r

12. [rnsinω0n]u[n] [rsinω0]z)/(z2-[2rcosω0]z+r2) |z|>r

13. anu[n] - anu[n-N] (zN-aN)/zN-1(z-a) |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

zd1C

zd1AzBzX

Partial Fractional Expression

Coefficients are given as

Easier to understand with examples

( ) ( )∑∑∑=

−≠=

−

−

=

−

−+

−+=

s

1mm1

i

mN

ik,1k1

k

kNM

0r

rr

zd1C

zd1AzBzX

( ) ( )kdz

1kk zXzd1A

=−−=

( ) ( )( ) ( )[ ]

1idw

1sims

ms

msi

m wXwd1dwd

d!ms1C

−=

−−

−

−

−−−

=

Example: 2nd Order Z-Transform

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

( )21z :ROC

211

411

111

>

−

−

=−− zz

zX

( )

−

+

−

=−− 1

2

1

1

z211

A

z411

AzX

( ) 1

41

211

1zXz411A

1

41z

11 −=

−

=

−=

−=

−

( ) 2

21

411

1zXz211A

1

21z

12 =

−

=

−=

−=

−

Example Continued

ROC extends to infinity ◦ Indicates right sided sequence

( )21z

z211

2

z411

1zX11

>

−

+

−

−=

−−

[ ] [ ] [ ]nu41-nu

212nx

nn

=

Example #2

Long division to obtain Bo

( ) ( )( )

1z z1z

211

z1

z21z

231

zz21zX11

21

21

21

>−

−

+=

+−

++=

−−

−

−−

−−

1z5

2z3z

21z2z1z

23z

21

1

12

1212

−

+−

+++−

−

−−

−−−−

( )( )11

1

z1z211

z512zX−−

−

−

−

+−+=

( ) 12

1

1

z1A

z211

A2zX −− −

+−

+=

( ) 9zXz211A

21z

11 −=

−=

=

− ( ) ( ) 8zXz1A1z

12 =−=

=

−

Example #2 Continued

ROC extends to infinity◦ Indicates right-sides sequence

( ) 1z z1

8

z211

92zX 11

>−

+−

−= −−

[ ] [ ] [ ] [ ]n8u-nu219n2nx

n

−δ=

An Example – Complete Solution

386zz1414z3zlimU(z)limc 2

2

zz0 =+−+−

==∞→∞→

4-z1414z3z

86zz1414z3z2)(z(z)U

2

2

2

2

+−=

+−+−

−=

2-z1414z3z

86zz1414z3z4)(z(z)U

2

2

2

4

+−=

+−+−

−=

14-2

1421423(2)Uc2

21 =+⋅−⋅

==

86zz1414z3zU(z) 2

2

+−+−

=4z

c2z

ccU(z) 210 −

+−

+=

32-4

1441443(4)Uc2

42 =+⋅−⋅

==

4z3

2z13U(z)

−+

−+=

>⋅+=

= −− 0k,4320k3,

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

z211z

21z

z1z1z211z zX

−

−−−

+−−=

−+

−=

[ ] [ ] [ ] [ ] [ ]1n21n1n

212nnx −δ+δ−+δ−+δ=

[ ]

=

=

=−

−=−

−=

=

2n0

1n21

0n1

1n21

2n1

nx

Z-Transform Properties: LinearityNotation

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 = →←− −

( )41z

z411

1z zX1

1 >

−=

−

−

[ ] [ ]1-nu41nx

1-n

=

Z-Transform Properties: Multiplication by Exponential

C 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 > →←

[ ] ( ) [ ] ( ) [ ] ( ) [ ]nure21nure

21nuncosrnx

njnjo

n oo ω−ω +=ω=

( ) rz zre1

2/1zre1

2/1zX 1j1j oo>

−+

−=

−ω−−ω

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

az11az

dzzdXz

az1az

dzzdX

−−

−

−

+=−⇒

+−

=

[ ] ( ) [ ]1nuaannx 1n −−= −

[ ] ( ) [ ]1nuna1nx

n1n −−= −

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

R1ROC 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 az11zX 11 >

−= − ( ) 1z:ROC

z11zX 12 >−

= −

( ) ( ) ( ) ( )( )1121 z1az11zXzXzY −− −−

==

( ) 1z :ROC asume az11

z11

a11zY 11 >

−−

−−= −−

[ ] [ ] [ ]( )nuanua1

1ny 1n+−−

=

The z-TransformZ-TRANSFORM THEOREMS AND PROPERTIES

Linearity

xRzzXnx ∈= ),()]([Z

yRzzYny ∈= ),()]([Z

yx RRzzbYzaXnbynax ∩∈+=+ ),()()]()([Z

Overlay of the above two

ROC’s

Shift


xn RzzXznnx ∈=+ )()]([ 0

0Z

Multiplication by an Exponential Sequence

+<<= xx- RzRzXnx || ),()]([Z

xn RazzaXnxa ⋅∈= − || )()]([ 1Z

Differentiation of X(z)


xRzdz

zdXznnx ∈−= )()]([Z

Conjugation


xRzzXnx ∈= *)(*)](*[Z

Reversal


xRzzXnx /1 )()]([ 1 ∈=− −Z

Real and Imaginary Parts


xRzzXzXnxe ∈+= *)](*)([)]([ 21R

xj RzzXzXnx ∈−= *)](*)([)]([ 21Im

Initial Value Theorem

0for ,0)( <= nnx

)(lim)0( zXxz ∞→

=

Convolution of Sequences


yRzzYny ∈= ),()]([Z

yx RRzzYzXnynx ∩∈= )()()](*)([Z

Convolution of Sequences

∑∞

−∞=

−=k

knykxnynx )()()(*)(

∑ ∑∞

−∞=

−∞

−∞=

−=

n

n

kzknykxnynx )()()](*)([Z

∑ ∑∞

−∞=

−∞

−∞=

−=k

n

nzknykx )()( ∑ ∑

∞

−∞=

−∞

−∞=

−=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

pzc

c)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 PolesIf 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|ln4k

3.912ln0.02|a|kln0.02|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 ++

2a4acbbz

2 −±−=

0,4acb2 ≥−)

2a4acbb)(z

2a4acbba(zcbzaz

222 −−−

−−+−

−=++

0,4acb2 <− )2a

b4acib)(z2a

b4aciba(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

pzc

c

irrzc'

irrzc

pzc

cY(z)

θθθ

θθθθ

coscossin

sincossincos

coskθdrsinkθbrpc(k)ucy(k) kkm

1j

1-kjjimpulse0 ++×+×= ∑

=

Z-1Time 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)3

kπsin(0.82y(k)

0.640.8zzzzY(z)

kk

2

2

⋅+⋅⋅=

+−+

=

Z-Domain: Complex Poles

Time-Domain:Exponentially Modulated Sin/C

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 PolesA 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)

θθ 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

1z2zU(z)

−⋅++=+

+−

+−

=

0 10 20 30 40 50 60-1

0

1

2

3

4

5

6

converge to 2

Final Value TheoremEnable 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, thenlim ( ) lim ( 1) ( )k z

z Y zy k z Y z∞

−= −

2

1 1

0.11 0.11( )1.6 0.6 ( 1)( 0.6)

0.11( 1) ( ) | | 0.2750.6z z

z zY zz z z z

zz Y zz= =

− −= =

− + − −−

− = = −− 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 time◦ Overshoot

◦ …

Shift-Invariant System

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)X(z) Y(z)

)()()(

zXzYzH =

Nth-Order Difference Equation

∑∑==

−=−M

rr

N

kk rnxbknya

00)()(

∑∑=

−

=

− =M

r

rr

N

k

kk zbzXzazY

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()(

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()( Re

ImCausal Systems : ROC extends outward from the outermost pole.

Stable and Causal Systems

∏

∏

=

−

=

−

−

−= N

kr

M

rr

zd

zcAzH

1

1

1

1

)1(

)1()( Re

ImStable Systems : ROC includes the unit circle.

1

Thanks!Reference : https://akademik.adu.edu.tr/fakulte/muhendislik/personel/uploads/yilmaz.kalkan/dsp3-1577359228.ppt

The z-Transform · 21) ( 1) 3. x n = − 1. u n − n. u − ... Assume that a given z -transform can be expressed as Apply partial fractional expansion First term exist only if M>N

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