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Page 1: The z Transform - UPTshannon.etc.upt.ro/teaching/sp-pi/Course/4_z_Transform_2013.pdf · 1 1 The z Transform The z transform generalizes the Discrete-time Fourier Transform for the

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The z Transform

The z transform generalizes the Discrete-time Fourier Transform for the entire complex plane.For the complex variable is used the notation:

2 2

; ,

arg

jz x j y r e x y

z r x y

z

Ω= + ⋅ = ⋅ ∈

= = +

Ω =

2

notation:

For any discrete-time LTI system:

-the eigen function:

-the eigen value:

The Discrete LTI System Response to a Complex Exponential

[ ] 00 0

j nn nx n z r e Ω= =

System

h[n]

input output

[ ] [ ] [ ] [ ] ( )0 0 0 0 0 0n n k n k n

k ky n h n z h k z z h k z z H z

∞ ∞− −

=−∞ =−∞= ∗ ⋅ = ⋅ ⋅= =∑ ∑

[ ] [ ] ( )0 0 0n ny n h n z z H z= ∗ ⋅=

( ) [ ] k

kH z h k z

∞−

=−∞= ⋅∑

( )0H z0nz

Page 2: The z Transform - UPTshannon.etc.upt.ro/teaching/sp-pi/Course/4_z_Transform_2013.pdf · 1 1 The z Transform The z transform generalizes the Discrete-time Fourier Transform for the

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fast computation of the response of any discrete-time LTI system to linear combinations of complex exponentials of the form:

transfer function, z transform of the impulse response h[n].

[ ] ( )0 0ny n z H z= ⋅

[ ] nk k

kx n c z= ∑

System

h[n]

input output[ ] ( ) n

k k kk

y n c H z z= ∑

( ) [ ] k

kH z h k z

∞−

=−∞= ⋅∑

4

Bilateral z TransformThe bilateral z transform of the signal x[n] :

generalization of the discrete-time Fourier transform

Discrete-time Fourier transform = particular case of z transform on the unit circle |z|=1

[ ] ( ) ( ) [ ] , , 0jn

nZ x n z X z x n z z r e r

∞Ω−

=−∞= = ⋅ = ⋅ ≥∑

[ ] ( ) [ ]( ) [ ] ( ); 0j nn n

nZ x n z x n r e r x n r

∞− Ω− −

=−∞= ⋅ = ⋅ Ω ≥∑ F

[ ] ( ) [ ] ( )1 jz r Z x n e x nΩ= = ⇒ = ΩF|z|=1

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Examples

region of convergence (ROC) = set of values of z for which X(z) is convergent

rational function, zeros = roots of numerator ; poles = roots of denominator.

Pole/zero plot or constellation (PZC).

[ ] [ ]1. , 1nx n a n a= σ <

( ) ( )1

0 0

nn n

n nX z a z az

∞ ∞− −

= == =∑ ∑ 1 1az− <

( ) 11 ,

1X z z a

az−= >

convergent:

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convergent if:

one pole in a.

[ ] [ ]2. 1 , 1nx n a n a= − σ − − <

( ) ( )1 1

11 1 1

1 0

1 1 1n nnn n

n n n nX z a z az

az az az

− − ∞ ∞− −

− − −=−∞ =−∞ = =

⎛ ⎞ ⎛ ⎞= − = − = − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ ∑ ∑ ∑

11 1

a z−<

( ) 11

1

1 1 ; 1 11

X z z aazaz

az

−−

= − = <⎛ ⎞ −−⎜ ⎟⎝ ⎠

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same z transform ; different ROCs

First example Second example

Region of convergence ROC : set of values of z for which theseries X(z) is convergent.

The Z transform exists where the Fourier transform of x[n]r-n

exists, r=|z|

( ) 11

1X z

az−=

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• ROC of a bilateral z transform can not contain any pole • for right sided signals (including causal signals), the

ROC extends outward from the outermost pole, • for left sided signals (including anticausal signals), the

ROC extends inward from the innermost pole• for infinite duration signals, ROC is a ring that doesn’t

include poles – bounded on the interior and exterior by a pole.

• for finite duration signals, the ROC is the entire z-plane, except possibly z=0 or z=∞.

• a causal and stable system’s transfer function has thepoles inside the unit circle. ROC is outside the unit circle.

Properties of the ROC of the Bilateral z Transform

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[ ] [ ] [ ]1 2x n x n x n= +

first component of x[n] is right sided; ROC is outside the circle with radius R-.

second component of x[n] is left sided, ROCis inside the circle with radius R+.

The ROC of X(z)is the intersectionof the ROCs of its components (ring)

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Example[ ]

[ ] [ ] [ ]1 2

3. , 0 1

nx n a a

x n x n x n

= < <

= +

[ ] [ ]

[ ] [ ]

[ ]

1 1

2

1

1 , 1

1

1 1 11 , 11

n

n

x n a n z aaz

x n a n

n za az

a

= σ ⎯⎯→ >−

= σ − −

−⎛ ⎞= σ − − ⎯⎯→ <⎜ ⎟⎝ ⎠ −

( )

2 1 1; 1

n a za a za az a z

a

−⎯⎯→ < <

⎛ ⎞− −⎜ ⎟⎝ ⎠

Right sided (causal) signal

Left sided (anticausal) signal

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Inverse z Transform

• Γ - counterclockwise closed path (contour) included in the region of convergence.

• Γ - encircles the origin and must encircle all poles of X(z).

• integral along a contour. • z transforms in DSP : rational functions. • partial fraction expansion and tables signal – z

transform pairs (knowing the pole/zero plot).

( ) [ ] ( ) [ ]1 11 ; 2

nZ X z n X z z dz x n ROCj

− −Γ

= = Γ ⊂π ∫

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The Bilateral z Transform computationusing its pole/zero plot

X(z) - rational function

( )( )

( )0

1

1

; N

M

kk

pkk

z zX z k z ROC

z z

=

=

−= ∈

poles and zeros ⇒ Z transform without constant k

X(z0) – also known ⇒ Z transform and constant k

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Example

unit circle ⊂ ROC ⇒ discrete-time Fourier transform exists:

A ∈ unit circle; ∠(OA, Ox) = Ψ

the spectrum of signal x[n]

0pole/zero plot 0; 0.5pz z= =( ) 1

1 for 10.5 1 0.5

zX z kz z−

= = =− −

causal signal, ROC: 0.5pz z> =

( ) ( ) 10.5 1 0.5

jj

j jeX X e

e e

ΩΩ

Ω − ΩΩ = = =− −

( ) ( ); OA

XAP

Ω = Φ Ω = ψ −ϕ

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vectors Azpk , Azok: ∠(AZ0k, Ox) = Ψk ; ∠(AZpk, Ox) = ϕk

Magnitude and phase spectrum:

The frequency Ω = length of the arc (of circle) in radians on the unit circle, between its intersection with the positive realaxis and the point A, in trigonometric sense.

General case:

( )0

1

1 1

1

; =

M

M Nkk

k kNk k

pkk

AzX k Argk

Az

=

= =

=

Ω = Φ + ψ − ϕ∏

∑ ∑∏

( )( )

( )0

1

1

; N

M

kk

pkk

z zX z k z ROC

z z

=

=

−= ∈

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The Unilateral z Transform

•For causal signals, unilateral z transform = bilateral ztransform

•ROC: entire complex plane or the outside region of a disc centered in 0.

•Useful for causal systems described by difference equations, with non zero initial conditions.

[ ] ( ) ( ) [ ]0

nu u

nZ x n z X z x n z

∞−

== = ∑

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z Transform Propertiesnotations:

1. Linearity

Proof. Directly, using the definitions. Homework - Prove it.

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

, ;

, ;

u

u

x

y

ZZ

ZZ

x n X z z ROC x n X z

y n Y z z ROC y n Y z

←⎯→ ∈ ←⎯⎯→

←⎯→ ∈ ←⎯⎯→

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

, at leastx y

u u

ax n by n aX z bY z z ROC ROC

ax n by n aX z bY z

+ ←⎯→ + ∈ ∩

+ ←⎯→ +

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2. Time shifting

Proof.

bilateral z transform:

If n0>0, z=0 ∉ ROC. If n0<0, z=∞ ∉ ROC.

unilateral z transform:

( )

( ) [ ]

0

0

0

0

1

0 0

,

, >0

n

n nu

n n

x n n z X z z ROC

x n n z X z x n z n

−− −

=−

− ←⎯→ ∈⎡ ⎤⎣ ⎦⎛ ⎞

− ←⎯→ −⎡ ⎤ ⎜ ⎟⎣ ⎦ ⎜ ⎟⎝ ⎠

[ ]

[ ] [ ]

00

0

0

0

0 00

1

00

, 0

n n m nn mu

n m n

n m m

m m n

Z x n n x n n z x m z z

z x m z x m z n

− =∞ ∞−− −

= =−

∞ −− − −

= =−

− = − =⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎛ ⎞= + >⎜ ⎟⎜ ⎟

⎝ ⎠

∑ ∑

∑ ∑

[ ] ( ) [ ]0

0 00 0

n n m m n nn m

n m mZ x n n x n n z x m z z x m z

− =∞ ∞ ∞− + −− −

=−∞ =−∞ =−∞− = − = =⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦∑ ∑ ∑

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3. Modulation in time

Proof.

More generally:

Homework - Prove it.

[ ] [ ] [ ]( ) (0 0 0 0nj n j n j jn

n nZ e x n e x n z x n e z X e z

∞ ∞ −Ω Ω − Ω − Ω−

=−∞ =−∞= = =∑ ∑

[ ]

[ ]

00 0

00

, n

nu

z zz x n X ROCz z

zz x n Xz

⎛ ⎞←⎯→ ∈⎜ ⎟⎜ ⎟

⎝ ⎠⎛ ⎞

←⎯→ ⎜ ⎟⎜ ⎟⎝ ⎠

[ ] ( )0 0 , j n je x n X e z z ROCΩ − Ω←⎯→ ∈

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4. Time reversal

Proof.

[ ] [ ] [ ] 1 1mm nn

n mZ x n x n z x m X

z z

−∞ ∞=−−

=−∞ =−∞

⎛ ⎞ ⎛ ⎞− = − = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ ∑

[ ] ( )1 1, x n X z ROCz

−− ←⎯→ ∈

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5. Differentiation in time

Proof. direct application of the definitions and previous properties. Homework. Prove these properties.

[ ] [ ] ( ) ( )

[ ] [ ] ( ) ( ) [ ]

1

1

1 1 ;

1 1 1u

x n x n z X z z ROC

x n x n z X z x

− − ←⎯→ − ∈

− − ←⎯→ − − −

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6. Addition in time

Proof.

[ ]( ) [ ]

1

1 ; 11

unk

k

X z x kx k z

z

=−∞−

=−∞

+←⎯→ >

∑∑

[ ] ( ) ( )1 ; 11

n

k

X z zx k X z z ROC ROCzz

∗−

=−∞←⎯→ = ∈ ∩

−−∑

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

11 1 1 ; 1uk

x n y n y n X z z Y z y y x k−

=−∞= − − ←⎯→ = − − − − = ∑

[ ] [ ] [ ] [ ] [ ] ( ) ( ) ( )11 1n

ky n x k x n y n y n X z z Y z−

=−∞= ⇒ = − − ←⎯→ = −∑

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7. Differentiation in z domain

Proof.

By direct application of definitions.

Homework. Prove it.

[ ] ( )

[ ] ( )

;

u

dX znx n z z ROC

dzdX z

nx n zdz

←⎯→− ∈

←⎯→−

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8. Complex conjugation in time domain

Proof.

By direct application of definitions.

Homework. Prove it.

[ ] ( )[ ] ( )

;

u

x n X z z ROC

x n X z

∗ ∗ ∗

∗ ∗ ∗

←⎯→ ∈

←⎯→

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9. Time convolution (convolution theorem)

Proof.

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

[ ] [ ] ( )

[ ] [ ] ( ) ( )

n n

n n k

n kk

n kn k m m k

n k

Z x n y n x n y n z x k y n k z

x k z y n k z

y m z x k z Y z X z

∞ ∞ ∞− −

=−∞ =−∞ =−∞∞ ∞ − −−

=−∞ =−∞∞ ∞− =

− −

=−∞ =−∞

∗ = ∗ = −

= −

= =

∑ ∑ ∑

∑ ∑

∑ ∑

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

,

; at leastx y

u u

x n y n

x n y n X z Y z z ROC ROC

x n y n X z Y z

∗ ←⎯→ ∈ ∩

∗ ←⎯→

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10. Product theorem or convolution theorem in the complex domain

Proof.

[ ] [ ] ( )

[ ] [ ] ( )

1 , 21 ,

2 u u

z dux n y n X u Y ROCj u u

z dux n y n X u Y ROCj u u

Γ

Γ

⎛ ⎞←⎯→ Γ∈⎜ ⎟π ⎝ ⎠⎛ ⎞←⎯→ Γ ⊂⎜ ⎟π ⎝ ⎠

: ; : and

:

y y y yx x x xx y

y yx x

zROC R z R ROC R z R R u R R R

u

ROC R R z R R

− + − + − + − +

− − + +

< < < < < < < <

< <

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

( ) [ ] ( )

112

1 1 2 2

n n n

n nn

n

Z x n y n x n y n z X u u du y n zj

z du z duX u y n X u Yj u u j u u

∞ ∞− − −

Γ=−∞ =−∞

−∞

Γ Γ=−∞

⎛ ⎞= = ⎜ ⎟π⎝ ⎠

⎡ ⎤⎛ ⎞ ⎛ ⎞= =⎢ ⎥⎜ ⎟ ⎜ ⎟π π⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

∑ ∑ ∫

∑∫ ∫

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Particular Cases

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11. The Initial Value Theorem

Proof.

z transform of a causal signal

At the limit :

[ ] ( ) ( )0 lim limuz zx X z X z

→∞ →∞= =

( ) ( ) [ ] [ ] [ ]0

10 1nu

nX z X z x n z x x

z

∞−

== = = + +∑ …

( ) ( ) [ ] [ ] [ ]10 1 0zuX z X z x x xz →∞= = + + ⎯⎯⎯→…

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11. The Final Value Theorem

Proof.

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

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

1 1

0 0 1

0

1 1

0 1 0

n m mn nu u

n n m

mu u

m

x n x n z x n z X z x m z X z

z x m z x X z z X z zx

∞ ∞ ∞+ = − −− −

= = =∞

=

+ − = + − = −

⎛ ⎞= − − = − −⎜ ⎟

⎝ ⎠

∑ ∑ ∑

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

lim lim 1 lim 1uk z zx x k z X z z X z

→∞ → →∞ = = − = −

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

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

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

1 10

0 0

1 1

lim 1 lim 1 0

1 lim 1

lim 1 0

lim lim 1 lim 1

nuz zn

k

kn n

k

uz zk

x n x n z z X z x

x n x n x n x n

x k x

x x k z X z z X z

∞−

→ →=∞

→∞= =

→∞

→ →→∞

+ − = − −

+ − = + −

= + −

⇒ ∞ = = − = −

∑ ∑

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Relation between the z Transform and the Laplace Transform

( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) [ ]

[ ] [ ] ( )

ˆIdeal sampling:

ˆ

Discrete-time signal:

a a

an

snTa a

n n

a s d

n nad d

n n

s

s s

s s s

s

x t X s

x t x nT t nT

x t x nT t nT x nT e

x nT x n

Z x n x n z x nT z

=−∞∞ ∞

=−∞ =−∞

∞ ∞− −

=−∞ =−∞

←⎯→

= δ −

= δ − =

=

⇒ = =

∑ ∑

∑ ∑

L L

( ) ( ) [ ] [ ] ( ); a a sT d dsT

sT

s s

s

x t t Z x n x n x nTe z

z e

δ = ==

=

LLaplace of thesampled analog signal

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Using the z Transform for the Study of the Discrete LTI Systems

• The z transform is useful for studying discrete-time LTI systems (theorem of convolution of discrete-time signals)

• For a causal system, with non zero initial conditions, theunilateral z transform is used.

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The Transfer Function for a Discrete LTI System

•transfer function H(z) - z transform of the impulse response of discrete-LTI system. •describes completely the system in the complex domain.

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Stable system: It has a frequency response. Discrete-time Fourier transform of the impulse response is convergent. The unit circle belongs to the ROC of its transfer function

|z| =1 ⊂ ROCCausal system: Hu(z) = H(z)

ROC is outside of a disc. Causal and stable system:

Unit circle belongs to the ROC: |z| =1 ⊂ ROC. All poles of H(z): inside the unit disc |zp| <1

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Computation of the response of a discreteLTI system using the z transform

• If we know the input signal, x[n] and the system,h[n] ↔ H (z),

• Compute the z transform of the input, X (z). • Compute the product Y(z)=H(z) X (z). • Apply inverse z transform ⇒ the response y[n]. • For a causal input signal and a causal system

with non zero initial conditions, we use theunilateral z transform

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The Computation of the Inverse z Transform

There are three methods that can be used:

1. Direct computation of the integral,2. Partial fraction expansion of the function Y(z),3. Power series expansion of function Y (z).

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2. Partial fraction expansion of the function Y (z)

Y(z) rational function, ratio of polynomials in z-1 or z. We use z-1, denote z-1 = x ( z transforms expressed in

function of z-1 in tables)

( ) ( ) ( )( )

( )( )

( ) ( ) ( )( )

11

1

N z N xY x Y z

D xD z

R xY x I x

D x

−−

−= = =

= +

( ) [ ]1Zk kk k k

k k kI x c x c z c n k

−−= = ←⎯⎯→ δ −∑ ∑ ∑

36

( )( ) ( )1

km ki

im ikm k

s bR x aD x x x x x=

= +− −

∑ ∑∑

( ) ( )( )m m

m

R xa x x

D xx x

⎡ ⎤= −⎢ ⎥⎣ ⎦ = ( ) ( ) ( )

( )1

!

kk

k

s is

s i kk

k

kiR xdb x xD xs i dx x x

⎧ ⎫⎡ ⎤⎪ ⎪= −⎨ ⎬⎢ ⎥− ⎪ ⎪⎣ ⎦⎩ ⎭ =

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Example: 2nd order Z-Transform

( ) ( )( ) ( )

( ) ( )( ) ( )( )

( )

[ ] ( ) ( ) [ ]

[ ] ( ) ( ) [ ]

1

1 1

1 1 1 1

; ROC: 0.50.5 0.25

8 8 8 162 4 2 42 4

8 16 4 42 4 1 0.5 1 0.25

For 0.5 : 4 0.5 0.25

For 0.25 : 4 0.5 0.25 1

For 0.25

u

n n

n n

zY z Y z zz z

z xY zx x x xz z

Y zz z z z

z y n n

z y n n

− −

− − − −

= = >− −

= = = −− − − −− −

= − = −− − − −

⎡ ⎤⇒ > = − σ⎣ ⎦⎡ ⎤< = − − σ − −⎣ ⎦

[ ] ( ) [ ] ( ) [ ]0.5 : 4 0.5 1 4 0.25n nz y n n n< < = − σ − − − σ

Causal signal y[n]

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3. Power series expansion of function Y (z)

Expansion of function Y(z) into a power series

Example #a

( )

[ ]

1

1 1 2

0

, : 0

1 1 1 111! 2! ! !1!

n n

n

z

z

Y z e ROC z

e z z z zn n

y nn

∞− − − − −

=

= >

= + + + + + =

⇒ =

∑… …

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39

Example #b

( )

( ) ( )

[ ] [ ] ( ) [ ]( )

[ ]

2

00 1

, : 0,

1 1 1 111! 2! ! !

1 1 1! !

1 11! 1 !

m m

m

n n

n n

n

z

z

Y z e ROC z z

e z z z zm m

z zn n

y n n n nn n

=−

− −

=−∞ =−∞

= ≥ ≠ ∞

= + + + + + =

= = +− −

⇒ = δ − σ − − = σ −− −

∑ ∑

… …

40

( ) ( )

( ) ( ) [ ] ( )

[ ] ( )

[ ] ( ) [ ]

1

1 1

1

1

ln 1 , :

1 1; 1

initial value theorem:

0 lim ln 1 0

1 1

n nn nn

n

zn

n

Y z az ROC z a

a aY z z y n n

n n

y az

y n a nn

+ +∞−

=

−→∞

= + > ⇒

− −= ⎯⎯→ = ≥

= + =

−= σ −

causal signal

Example #c

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41

Example #d

( ) [ ] [ ]11 ,

1nY z z a y n a n

az−= > ⇒ = σ

42

Example #e, same transform as #d, different ROC

( )

( ) ( ) [ ] [ ]

1

11

1

1 , 1

anticausal signal, z transform contains only powers of z

1m n n n

m n

zY z z az aaz

Y z a z a z y n a n

∞ −− −

= = −∞

= = <−−

⇒ = − = − ⇒ = − σ − −∑ ∑

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43

Discrete LTI Systems Described by Linear Constant Coefficient Difference Equations

[ ] [ ] 00 0

; 0N M

k kk k

a y n k b x n k a= =

− = − ≠∑ ∑

( ) ( ) ( ) ( )( )

0

0 0

0

transfer function

Mk

N M kk k k

k k Nkk k

kk

b z N za z Y z b z X z H z

D za z

− − =

−= =

=

= ⇒ = =∑

∑ ∑∑

Transfer function of a discrete LTI system : rational function in z or z-1. Zeros : roots of numerator N(z); Poles : roots of denominator D(z)

44

Causal and stable system: poles inside unit circle |z|=1

The initial value theorem can be applied:

The degree of the numerator of the transfer function of a

causal and stable system is smaller or equal with the

degree of its denominator, M≤ N

1pkz <

[ ] ( ) ( ) ( )( )

0 lim lim lim - finiteuz z z

N zh H z H z

D z→∞ →∞ →∞= = =

Degree M

Degree N

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45

The Contribution of the Poles of a Causal Discrete LTI System at its

Impulse ResponseConsider two cases: simple poles and double poles.

#1 Simple complex conjugated poles

Contribution of the two poles in the impulse response :

[ ]neaear pp jnjnnp σ⋅⎟

⎠⎞⎜

⎝⎛ ⋅+⋅ Ω−Ω *

( )1 1

and

1 1

j jp p p p

j jp p

p p

p p

z r e z r e

a aH zr e z r e z

Ω − Ω∗

Ω − Ω− −

= =

= + + +− −

… …

46

The pair of poles are on the unit circle.

Partial impulse response = oscillation with fixed amplitude that persists even after the end of the input signal:

System = oscillator, critically stable.

1pr < Partial impulse response decreases in timeNo instability from these poles.

1pr > Partial impulse response increases in timeInstability.

( )sin p pA nΩ +Φ

1pr =

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47

#2. A pair of double complex conjugate poles

Contribution of the two poles in the impulse response :

( )( ) ( )

1 1 2 22 21 1 1 1

and

1 1 1 1

j jp p p p

j j j jp p p p

p p

p p p p

z r e z r e

a a a aH z

r e z r e z r e z r e z

Ω − Ω∗

∗ ∗

Ω − Ω− − Ω − Ω− −

= =

= + + + + +− − − −

… …

( ) ( ) [ ]1 2sin sinn np p p p p pA r n A nr n n⎡ ⎤Ω +Φ + Ω +Ψ σ⎣ ⎦

1pr < Partial impulse response decreases in timeNo instability from these poles.

1pr > Partial impulse response increases in timeInstability.

48

difference equation, non-zero initial conditions ⇒unilateral Z transform

Causal input signal, x[-n]=0 for n >0:

Response of a Discrete LTI System Described by a Difference Equation

[ ] [ ] [ ] [ ] 00 0 0 0

, 0 N M N M

k k k u k uk k k k

a y n k b x n k a a Z y n k b Z x n k= = = =

− = − ≠ ⇒ − = −∑ ∑ ∑ ∑

( ) [ ] ( )0 1 0

N k Mk n k

k u k uk n k

a z Y z y n z b z X z− −

= = =

⎡ ⎤+ − =⎢ ⎥⎣ ⎦

∑ ∑ ∑

( ) [ ] ( ) [ ]1 10 0

N k M kk n k n

u uk kn nk k

a z Y z y n z b z X z x n z− −

= == =

⎡ ⎤ ⎡ ⎤+ − = + −⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦∑ ∑ ∑ ∑

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49

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

( ) ( )( )[ ]

[ ]

[ ] [ ]( )

[ ] .1 , 1 ; 1

.1

11

11

1

11

11

.1 ; 1

1

,01 , , 1

Example

0

0

0

000

0

0

0

0

111

111

111

11

><σ⎟⎟⎠

⎞⎜⎜⎝

−+

−+−=

−+

−−+

−−

−=

=−

−+

−−=

>−

=−+−

≠−σ==−−

Ω

Ω

++

−−Ω−ΩΩ

Ω

−−−Ω

−Ω−

Ω

zanea

keea

kayany

azay

azeaka

zeeake

azay

azzekzY

zze

kzyzYazzY

ynkenxnxnayny

j

nj

j

nn

jjj

j

ju

juu

nj

50

First Order Systems

0pole/zero plot 0; 0.5pz z= = [ ] [ ] [ ]

( )

[ ] [ ]1

1

; 1

, 1 np

y n ay n kx nk kzH z z ROCaz z a

z a z a h n ka n

− − =

⇒ = = ∈− −

> = < ⇒ = σ

( ) ( ) ( )1

0 1, max. freq. response for 0, 1 0, max. for ,

Positive low-pass filterNegative high-pass filter

j jOAH e e

PA PA

aa

aa

Ψ−ϕ Ω−ϕΩ = =

< < Ω =− < < Ω = π

⇒⇒

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51

Second Order Systems[ ] [ ] [ ] [ ] ( )

2

1 2 1 2 21 2 1 2

1 2 1

k kzy n a y n a y n kx n H za z a z z a z a− −+ − + − = ⇒ = =

+ + + +

1,2

21 2

21 2

2 21 2 1

2

1 2

4 complex conjugated poles

4 real poles . The system is considered causal and stable system.

41 magnitude

2The conditions imposed on ,

jp

a a

z e

a a

a a aa

a a

±

< ⇒

=

≥ ⇒

+ −= = <

θρ

ρ

21

2 2

21 2 2 1 2 1

for stability are :Complex conjugated poles

1 , 4

Real poles: 4 0 ; 1 ; 1.

aa a

a a a a a a

< >

− ≥ > − − > −

1.

2.

21 1 2

0 1,2

40 ,

2p

a a az z

− ± −= =

52

( ) ( ).1 : system theof responsefrequency The

pole. real double single a of existence the toscorrespond parabola The

212_____

2_____

1

ϕ−ϕ−Ω=Ω je

APAP

kH

Magnitude: even; Phase: odd

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53

Transfer Function of Equivalent System for Serial or Parallel Interconnections of two

discrete LTI Systems

[ ] [ ] [ ]( ) ( ) ( ).zHzHzH

nhnhnh

e

e

21

21 ; +=

+=[ ] [ ] [ ]( ) ( ) ( ).zHzHzH

nhnhnh

e

e

21

21 ; =

∗=

54

Digital Filters Implementation Forms Obtained Using the z

Transform[ ] [ ] [ ] [ ] [ ]. ,1 ,

100

00∑∑∑∑====

−−−=⇔==−=−N

kk

M

kk

M

kk

N

kk knyaknxbnyNMaknxbknya

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55

Direct form I (N+M adders). First form of implementation using z transform.

Each adder has 2 inputs. One adder with 2N inputs (M=N, a0=1).

56

Direct form II (2N adders M=N). Second form of implementation using z transform.

Each adder has 2 inputs. Two adders with N inputs each (M=N, a0=1).