lecture on z transform

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1 1 SIGNALS AND SYSTEMS 16 December 2006 2 Transforms are regularly used to make calculation or mathematical analysis simpler. • Examples include: – Logarithms (used to simplify multiplication) slide-rule – Laplace transform. (widely used in linear analysis, circuits & systems including control systems) – Fourier transforms. (Move us from the time to frequency domain) – The Z transform is the primary mathematical tool for the analysis and synthesis of digital filters. • The system function of a digital filter is defined as the z transform of its unit-sample response. TRANSFORMS

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Lecture On Z Transform

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Page 1: Lecture On Z Transform

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SIGNALS AND SYSTEMS

16 December 2006

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Transforms are regularly used to make calculation or

mathematical analysis simpler.

• Examples include:

– Logarithms (used to simplify multiplication) slide-rule

– Laplace transform. (widely used in linear analysis, circuits &

systems including control systems)

– Fourier transforms. (Move us from the time to frequency domain)

– The Z transform is the primary mathematical tool for the analysis

and synthesis of digital filters.

• The system function of a digital filter is defined as the z transform

of its unit-sample response.

TRANSFORMS

Page 2: Lecture On Z Transform

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Z-transforms

For discrete-time systems, z-transforms play the same role of Laplace transforms do in continuous-time systems

[ ]∑∞

−∞=

−=k

kzkhzH ][

Bilateral Forward z-transform

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Region of Convergence

Region of the complex z-plane for which forward z-transform converges

Im{z}

Re{z}Entire plane

Im{z}

Re{z}Complement of a disk

Im{z}

Re{z}Disk

Four possibilities (z=0 is a special case and may or may not be included)

Im{z}

Re{z}

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5

Z complex……….. φjnnn erz =

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z-transform infinite sum , does not converge for all zExample

x[n] = (0.5)nu[n]

X(z) = 1/(1-0.5z-1) for |z|>0.5

Region of Convergence

Page 4: Lecture On Z Transform

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ROC Properties:1)The ROC of X(z) consist of a ring in the z-plane centered about the origin.

2) The ROC does not contains any poles

3) If x[n] is of finite duration, then ROC is the entire z-plane , except possibly z=0 and/or z=infinity

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EXAMPLE:

∑∞

−∞=

− =⇔n

nznn 1][][ δδ

With an ROC consisting of the entire z plane including zero and infinity.

Consider ]1[ −nδ1]1[]1[ −

−∞=

− =−⇔− ∑ zznnn

nδδ

This z transform is well defined except at z=0;

Page 5: Lecture On Z Transform

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4) If x[n] is a right sided sequence then ROC will be outside the outside excluding 0 but including infinity.

5) If x[n] is a left sided sequence then ROC will be a circle including 0 and excluding infinity.

Im{z}

Re{z}Complement of a disk

Im{z}

Re{z}Disk

ROC is BLUE

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6) If x[n] is two sided sequence then ROC will consist of a ring in z-plane

Im{z}

Re{z}

7) If the z-transform X(z) of x[n] is rational, then its ROC is bounded by poles or extends to infinity.

Page 6: Lecture On Z Transform

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x[n] = anu[n] X(z) =1/1-az-1 for |z| > |a|.

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Page 7: Lecture On Z Transform

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What happens if a < 0?x[n] = anu[n] X(z) =1/1- az-1 for |z| > |a|.

Page 8: Lecture On Z Transform

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||||1

1]1[][ 1 αα <

−⇔−−−= − zfor

aznunx n

Anti causal

(left sided signals)

Page 9: Lecture On Z Transform

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))(()(

][][ 1

1||

αααα

α−−

−⇔= −

zzz

nunx nTwo Sided Sequence

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Inverse z-transform1) Using Contour integration……. A difficult

procedure.

[ ] [ ] dzzzFj

kf kjc

jc

1

21 −

∞+

∞−∫=

π

2) Using long Division………..A relatively easier method

3) Decomposition using partial fractions and then applying inverse transform using z transform properties

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21

23

12][

2

2

+−

++=

zz

zzzX

21

21

21

23

1

21][

−−

−−

+−

++=

zz

zzzX

Ratio of polynomial z-domain functions

•Divide through by the highest power of z

EXAMPLE

Page 11: Lecture On Z Transform

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Factor denominator into first-order factors

( )11

21

121

1

21][

−−

−−

++=

zz

zzzX

12

1

10 1

211

][ −− −

+−

+=z

A

z

ABzX

Use partial fraction decomposition to get first-order terms

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Find B0 by polynomial division

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23

2121

23

21

1

12

1212

+−

+++−

−−

−−−−

z

zz

zzzz

Express in terms of B0

( )11

1

1211

512][

−−

+−+=

zz

zzX

Page 12: Lecture On Z Transform

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Solve for A1 and A2

8

21

121

211

21

921

441121

1

1

21

2

21

21

1

1

1

=++

=−

++=

−=−

++=

−++

=

=

−−

=−

−−

z

z

z

zzA

zzz

A

Express X[z] in terms of B0, A1, and A2

( ) 11 1

8

21

1

92 −

− −+

−−=

zzzX

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Use table to obtain inverse z-transform

[ ] [ ] [ ] [ ]kukukkxk

821

9 2 +

−= δ

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Properties of Z transform

Linearity

ax1[n] + bx2[n] X1(z) + bX2(z).⇔X(z) = (ax1[n] + bx2[n])z-n,∑

−∞=n

X(z) = a x1[n]z-n +b x2[n]z-n,∑∞

−∞=n∑

−∞=n

X(z) = a X1(z)+ bX2(z).

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x[n-k] z-k

X(z).⇔

nx[n] -zdX(z)/dz⇔Difference in z Domain

Time shifting

Time reversal

RwithROCzXnxRwithROCzXnx

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

=⇔−=⇔

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Convolution Property

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

2121

22

11

zXzXnxnxzXnxzXnx

⇔⇔⇔ ROC=R1

ROC=R2

ROC=R1 R2I

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Analysis and Characterization of LTI systems using z transform1) Causality:

A discrete time LTI system is causal if and only if ROC of its system function is the exterior of a circle, including infinity

A DT LTI system with rational transfer function is causal if and only if:

a) The ROC is exterior of a circle outside the outermost pole.

b) With H(z) expressed as a ratio of polynomials in z, the order of the numerator can not be greater than the order of the denominator

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Stability:An LTI system is stable if and only if the ROC of its system function H(z) includes the unit circle. |z| = 1

EXAMPLE:

]1[2][)2/1(][ −−−= nununh nn

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Causal and Stable

A causal LTI system with rational system function H(z) is stable if and only if all of its poles lie inside the unit circle I.e. they must all have magnitude less than 1.

Example:11/1)( −−= azzH

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Causality

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Stability

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LTI system characterized by LCCD equations

Find the impulse response of the system followed by following difference equation

y[n] – 1/2y[n-1] = x[n] + 1/3 x[n-1]

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System functions and Block Diagram representations

1) Direct form I

2) Direct form II (canonical form)

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Draw forms for following system

H(z) = 1/(1+1/4z-1 – 1/8z-2)

y[n] + 1/4y[n-1]-1/8y[n-2]=x[n]