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Outline Introduction Relation Between LT and ZT Analyzing LTI Systems with ZT Geometric Evaluation Unilateral ZT

Signals and Systems

Lecture 8: Z Transform

Farzaneh Abdollahi

Department of Electrical Engineering

Amirkabir University of Technology

Winter 2012

Farzaneh Abdollahi Signal and Systems Lecture 8 1/29
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Introduction

Relation Between LT and ZTROC PropertiesThe Inverse of ZT

ZT Properties

Analyzing LTI Systems with ZT

Geometric Evaluation

LTI Systems Description

Unilateral ZT

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Z transform (ZT) is extension of DTFT

Like CTFT and DTFT, ZT and LT have similarities and differences.

We had defined x[n] = zn as a basic function for DT LTI systems,s.t.Zn H(z)zn

In Fourier transform z = ej, in other words, |z| = 1

In Z transform z = rej

By ZT we can analyze wider range of systems comparing to FourierTransform

The bilateral ZT is defined:

X(z) =

x[n]zn

X(rej) =

x[n](rej)n =

{x[n]rn}ejn

= F{x[n]rn}

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Region of Convergence (ROC)

Note that: X(z) exists only for a specific region of z which is calledRegion of Convergence (ROC)

ROC: is the z = rej

by which x[n]rn

converges:ROC : {z = rej s.t.

n= |x[n]rn| < }

Roc does not depend on Roc is absolute summability condition of x[n]rn

If r = 1, i,e, z = ejX(z) = F{x[n]}

ROC is shown in z-plane

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Example

Consider x[n] = anu[n]

X(z) =

n= anu[n]zn =

n=0(az1)n

If |z| > |a| X(z) is bounded

X(

z) =

z

za, ROC

: |z|

>|a|

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Example

Consider x[n] = anu[n 1]

X(z) =

n= anu[n 1]zn1

n=0(a1z)n

If |a1z| < 1 |z| < |a|, X(z) is bounded

X(z) = zza

, ROC : |z| < |a|

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In the recent two examples two different signals had similar ZT but withdifferent Roc

To obtain unique x[n] both X(z) and ROC is required

If X(z) = N(z)D(z)

Roots of N(z) zeros of X(z); They make X(z) equal to zero. Roots of D(z) poles of X(z); They make X(z) to be unbounded.


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Relation Between L and ZT In LT: x(t)

LX(s) =

x(t)estdt = L{x(t)}

Now define t = nT:X(s) = limT0

n= x(nT)(esT)n.T =

limT0 T

n= x[n](esT)n

In ZT: x[n]

Z

X(z) =

n= x[n]zn

= Z{x[n]} by taking z = esT ZT is obtained from LT.

j axis in s-plane is changed to unite circle in z-plane

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z-Plane s-Plane

|z| < 1 (insider the unit circle) Re{s} < 0 (LHP)

special case: z = 0 special case: Re{s} = |z| > 1 (outsider the unit circle) Re{s} > 0 (RHP)

special case: z = special case: Re{s} = |z| = cte (a circle) Re{s} = cte (a vertical line)

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ROC Properties

ROC ofX

(z

) is a ring in z-plane centered at origin ROC does not contain any pole

If x[n] is of finite duration, then the ROC is the entire z-plane, exceptpossibly z = 0 and/or z =

X(z) =N2

n=N1x[n]zn

If N1 < 0x[n] has nonzero terms for n < 0, when |z| 0 negative powerof z will be unbounded

If N2 > 0x[n] has nonzero terms for n > 0, when |z| positive powerof z will be unbounded

If N1 0 only negative powers of z exist z = ROC

If N2 0 only positive powers of z exist z = 0 ROC Example:

ZT LT

[n] = 11 ROC: all z (t)1 ROC: all z[n 1]z1 ROC: z= 0 (t T)esT ROC: Re{s} =

[n + 1]z ROC: z= (t + T)esT

ROC: Re{s} = Farzaneh Abdollahi Signal and Systems Lecture 8 10/29

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ROC Properties

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

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 |z| < r0 will also be in the ROC.

If x[n] is a two-sided sequence and if the circle [z] = r0 is in the ROC,then ROC is a ring containing |z| = r0. If X(z) is rational

The ROC is bounded between poles or extends to infinity, no poles of X(s) are contained in ROC

If x[n] is right sided, then ROC is in the out of the outermost pole Ifx[n] is causal and right sided then z = ROC

If x[n] is left sided, then ROC is in the inside of the innermost pole Ifx[n] is anticausal and left sided then z = 0 ROC

If ROC includes |z| = 1 axis then x[n] has FT

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The Inverse of Z Transform (ZT)

By considering r fixed, inverse of ZT can be obtained from inverse of FT:

x[n]rn = 12

2 X(rejz

)ejnd

x[n] = 12

2 X(rej)rne(j)nd

assuming r is fixed dz = jzd

x[n] = 12j

X(z)zn1dz

Methods to obtain Inverse ZT:1. If X(s) is rational , we can use expanding the rational algebraic into a

linear combination of lower order terms and then one may use X(z) = Ai

1aiz1 x[n] = Aiaiu[n] if ROC is out of pole z = ai

X(z) = Ai1aiz

1 x[n] = Aiaiu[n 1] if ROC is inside of z = ai

Do not forget to consider ROC in obtaining inverse of ZT!

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Methods to obtain Inverse ZT:2. If X is nonrational, use Power series expansion of X(z), then apply

[n + n0]zn0

Example: X(z) = 5z2 z + 3z3

x[n] = 5[n + 2] [n + 1] + 3[n 3]

3. If X is rational, power series can be obtained by long division Example: X(z) = 11az1 , |z| > |a|

1 1az1

1+az1+(az1)2+...

1 + az1

az1

az1 + a2z2

...

x[n] = anu[n]


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Methods to obtain Inverse ZT:

Example: X(z) = 11az1

, |z| < |a|

X(z) = a1z( 11a1z

)

1

1a1z

1+a1z+(a1z)2+...1 + a1z

a1z

a1z + az2z2

...

x[n] = 1anu[n 1]


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ZT Properties

Linearity: ax1[n] + bx2[n]aX1(z) + bX2(z) ROC contains: R1

R2

If R1

R2 = it means that Lp does not exit By zeros and poles cancelation ROC can be lager than R1

R2

Time Shifting:x[n n0]zn0 X(z) with ROC=R (maybe 0 or is

added/omited) If n0 > 0z

n0 may provide poles at origin 0 ROC If n0 < 0z

n0 may eliminate infrty from ROC

Time Reversal: x[n]X( 1z

) with ROC= 1R

Scaling in Z domain: zn

0 x[n]X(

z

z0 ) with ROC = |z0|R If X(z) has zero/poles at z = aX( z

z0) has zeros/poles at z = z0a

Differentiation in the z-Domain: nx[n] zdX(z)dz

with ROC = R

Convolution: x1[n] x2[n]X1(z)X2(z) with ROC containing R1 R2

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ZT Properties

Conjugation: x[n]X(z) with ROC = R If x[n] is real

X(z) = X(z) IfX(z) has zeros/poles at z0 it should have zeros/poles at z

0 as well

Initial Value Theorem: If x[n] = 0 for n < 0 then x[0] = limz X(z) limz X(z) = limz

n=0 x[n]z

n = x[0] + limz

n=1 x[n]zn =

x[0] For a casual x[n], if x[0] is bounded it means # of zeros are less than or

equal to # of poles (This is true for CT as well)

Final Value Theorem: If x[n] = 0 for n < 0 and x[n] is bounded whenn then x() = limz1(1 z

1)X(z)


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Analyzing LTI Systems with ZT

ZT of impulse response is H(z) which is named transfer function orsystem function.

Transfer fcn can represent many properties of the system: Casuality: h[n] = 0 for n < 0 It is right sided

ROC of a causal LTI system is out of a circle in z-plane, it includes Note that the converse is not always correct For a system with rational transfer fcn, causality is equivalent to ROC being

outside of the outermost pole (degree of nominator should not be greaterthan degree od denominator)

Stability: h[n] should be absolute summable its FT converges

An LTI system is stable iff its ROC includes unit circle A causal system with rational H(z) is stable iff all the poles of H(z) are

inside the unit circle

DC gain in DT is H(1) and in CT is H(0)

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Geometric Evaluation of FT by Zero/Poles Plot

Consider X1(z) = z a

|X1|: length of X

1 X1: angel of X1

Now consider X2(s) =1

za= 1

X1(s)

|X2| =1

X1(z)| X2 = X1


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For higher order fcns:

X(z) = MR

i=1(zi)Pj=1(zj)

|X(z)| = |M|

Ri=1 |zi|Pj=1 |zj|

X(z) =

M+R

i=1(z i)R

j=1(zj)

Example:H(s) = 1

1+az1

, |z| > |a|, a real h(t) = anu[n] H(ej) = v1

v2 |H(ej)| = 1|v2|

at = 0, |H(ej)| = 11a

is max

0 < < : |H(ej

)| at = , |H(ej)| = 1

1+a

H(ej) = v1 v2 = v2 at = 0,H(ej) = 0 0 < < , H(ej)

at

=,H

(ej

) = 0Farzaneh Abdollahi Signal and Systems Lecture 8 19/29

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First Order Systems

a in Dt first order systems plays a similar role of time constant in CT |a|

|H(ej)| at = 0 Impulse response decays more rapidly Step response settles more quickly

In case of having multiple poles, the poles closer to the origin, decay morerapidly in the impulse response

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Second Order Systems

consider a second order system with poles at z1 = rej, z2 = rej

H(z) = z2

(zz1)(zz2)= 1

12rcosz1+r2z2, 0 < r < 1, 0 < <

h[n] = rnsin(n+1)

sinu[n]

H = 2v1v

2 Starting from = 0 to = :

At first v2 is decreasing v2 is min at = max |H| Then v2 is increasing

If r is closer to unit, then |H| is greater at poles and change ofH issharper

If r is closer to the origin, impulse repose decays more rapidly and stepresponse settles more quickly.

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Bode Plot ofH(j)

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LTI Systems Description

Nk=0 aky[n k] =

Mk=0 bkx[n k]

Nk=0 aks

kY(z) =

Mk=0 bks

kX(z)

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

Mk=0 bkz

k

Nk=0 akz

k

ROC depends on placement of poles boundary conditions (right sided, left sided, two sided,...)

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Feedback Interconnection of two LTI systems: Y(z) = Y1(z) = X2(z) X1(z) = X(z) + Y2(z) = X(z) + H2(z)Y(z) Y(z) = H1(z)X1(z) = H1(z)[X(z) + H2(z)Y(z)]

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

H1(z)1H2(z)H1(z)

ROC: is determined based on roots of 1 H2(z)H1(z)

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Block Diagram Representation for Casual LTI Systems

We can represent a transfer fcn by different methods:

Example: H(z) = 12z1

1 14z1

= ( 11 1

4z1

)(1 2z1)


O li I d i R l i B LT d ZT A l i LTI S i h ZT G i E l i U il l ZT

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Unilateral ZT

It is used to describe casual systems with nonzero initial conditions:X(z) =

0 x[n]zn = UZ{x[n]}

If x[n] = 0 for n < 0 then X(z) = X(z)

Unilateral ZT of x[n] = Bilateral LT of x[n]u[n]

If h[n] is impulse response of a casual LTI system then H(z) = H(z)

ROC is not necessary to be recognized for unilateral ZT since it is alwaysoutside of a circle

For rational X(z), ROC is outside of the outmost pole


O tli I t d ti R l ti B t LT d ZT A l i LTI S t ith ZT G t i E l ti U il t l ZT
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Similar Properties of Unilateral and Bilateral ZT Convolution: Note that for unilateral ZT, If both x1[n] and x2[n] are zero

for t < 0, then X(z) = X1(z)X2(z)

Time Scaling

Time Expansion

Initial and Finite Theorems: they are indeed defined for causal signals

Differentiating in z domain: The main difference between UL and ZT is in time differentiation:

UZ{x[n 1]} =

0 x[n 1]zn = x[1] +

n=1 x[n 1]z

n =x[1] +

m=0 x[m]z

m1

UZ{x[n 1]} = x[1] + z1

X(z) UZ{x[n 2]} = x[2] + z1x[1] + z2X(z) UZ{x[n + 1]} =

n=0 x[n + 1]z

n =

m=1 x[m]zm+1x[0]z

UZ{x[n + 1]} = zX(z) zx[0] UZ{x[n + 2]} = z2X(z) z2x[0] zx[1] Follow the same rule for higher orders


O tline Introd ction Relation Bet een LT and ZT Anal ing LTI S stems ith ZT Geometric E al ation Unilateral ZT
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Example

Consider y[n] + 2y[n 1] = x[n], where y[1] = , x[n] = u[n] Take UZ:

Y(z) + 2[+ z1]Y(z) = X(s)

Y(z) =2

1 + 2z1

ZIR

+X(z)

1 + 2z1

ZSR Zero State Response (ZSR): is a response in absence of initial values

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

Transfer fcn is ZSR ZSR: Y1(z) =

21+2z1

y1[n] = 2(2)n

u[n] Zero Input Response (ZIR): is a response in absence of input (x[n] = 0) ZIR: Y2(z) =

(1+2z1)(1z1)

y[n] = y1[n] + y2[n]

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