Discrete Time System

h[n]

1

Yogananda Isukapalli

Applications of Z-Transform in Signal Processing - I

Discrete Time System

h[n]

Definition:

å¥

-¥=

-=n

nznhzH ][)(

Causal : {h[n]} = 0 for n < 0

å¥

=

-=0

][)(n

nznhzH

Example :

Consider two identical systems :

{h[n]} = (1/2)n , n ³ 0

h1[n]x[n] y[n]

h2[n]

h1[n] = h2[n] = (1/2)n , n ³ 0

2

Causal and Stable systems

Discrete Time System

h[n]

Response :

y[n] = (x[n]* h1[n] )* h2[n]

Taking the z transform we have:

Y(z) = X(z)H1(z) H2(z)

21

12

11

211

)()(

211

1)(

211

1)(

÷øö

çèæ -

=

-=

-=

-

-

-

z

zXzY

zzH

zzH

Suppose Now : x[n] = 1 for n ³ 0

111)( --

=z

zX

Now:

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Discrete Time System

h[n]

Therefore now the response of the system for the given Input would be :

( )

0..)21)(1()

21(24][

)21()

21(

21

4)( ..

)21)(1(

)( ..

2111

1)(

2

2

2

3

211

³+--=

--

--

-=

--=

÷øö

çèæ --

=--

nnny

z

z

z

zzzzYei

zz

zzYei

zzzY

nn

4

Discrete Time System

h[n]

Stability Theorem :

The system is said to be stable if and only if :

å¥

-¥=

¥<n

nh |][|

Recall :

å

å

å

¥

-¥=

¥

-¥=

¥

-¥=

£

-£

¥<<

-=

m

m

m

mhlny

mnxmhny

thenlx

assume

mnxmhny

|][||][|

|][||][||][|

|[.]|:

][][][

5

Discrete Time System

h[n]

Follows : |y[n]| is bounded if and only if :

å¥

-¥=

¥<n

nh |][|

Stable Causal Systems :

Example :

2

2

1

10)(

)}({]}[{

pzC

pzCCzH

zHnh

-+

-+=

«

Poles : p1, p2

Follows : h[n] = C0d[n] + C1p1n + C2p2

n

If h[n] ® 0, as n ®¥ then :

Implies ® if |pi| < 1 for all ih[n] ® 0, as n ®¥

å¥

-¥=

¥<n

nh |][|

6

Discrete Time System

h[n]

p1 p2

|z| = 1

ROC includes the unit circle |z| = 1 for a stable {|pi| < 1} causal systems.

Formal Definition: A causal system with a rational transfer function H(z) is stable if all the poles (pi) of H(z) are located inside the unit circle.

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Discrete Time System

h[n]

Example :

0][][][)(

)(

1

2

®=

-=

-

nhnunpnh

pzzzH

n

if |p| < 1as n ®¥

Stable even for repeated poles if |pi| < 1

Marginally Stable

nnh

zzzH

1][)1(

)(

=

-=

å¥

-¥=

¥<n

nh |][|

is not satisfied.

For finite N is bounded

pole is on |z| = 1

å¥

-¥=nnh |][|

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Discrete Time System

h[n]

Complex Conjugate poles on the unit circle :

å ¥<

=+=

-+

-=

-

---

|][|

][)cos(2][][

)1(1

)1(1

)(

0

11

00

00

nh

nunnheenh

ezezzH

njnj

jj

q

qq

qq

is satisfied for finite N.

Multiple poles repeated on the Unit circle will make the system to be unstable.

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Note: q=wT (rad)

Discrete Time System

h[n] Examples

1. The trapezoidal integration formula can be represented as an IIR digital filter represented by a difference equation given by

y[n] = y[ n – 1 ] + (1/2) { x[n] + x[ n – 1 ] }with y[-1] = 0. Determine the transfer function of the abovefilter.

Soln) Given, the difference equation representing trapezoidal integration

formula as:y[n] = y[ n – 1 ] + (1/2) { x[n] + x[ n – 1 ] }

Taking the z-transform of the above equation gives that:

)1(2

)1( )( ..

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

:is H(z)function transfer Therefore,

]X(z)[1 21 ]1)[( ..

} )( )( { 21 )( )(

1

1

11

11

-

-

--

--

-+

=

=

+=-

++=

zzzHei

zzzYei

zXzzXzYzzY

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Discrete Time System

h[n]

2. Let H(z) be the transfer function of a causal stable LTI discrete-time system. Let G(z) be the transfer function obtained by replacing z-1 in H(z) with α + z-1/ 1+ α z-1. Show that G(1)=H(1) and G(-1)=H(-1).

Soln) Given that the transfer function G(z) is obtained from H(z)by replacing

1-

-11

1-

-11

z1z )()( i.e.

z1zby

aa

aa

++

==++ -- zzHZGz

11

z1z

by z replace

or

zz1

by z replace i.e.

1-

1-

aa

aa

++

++

a) To show G(1) = H(1)

)1()1

1()1(

1, zWhen z1z )()(

HHG

zzHZG

=++

=

=++

==

aa

aa

b) G(-1) = H(-1)

When z = -1,

H(-1)

)1

1( )1(

=--

=-a

aHG

Discrete Time System

h[n]

3. Determine the transfer function of a causal stable LTI discrete-timesystem described by the following difference equation:

y[n] = 5x[n] + 5x[n-1] + 0.4x[n-2] + 0.32x[n-3]- 0.5y[n-1] + 0.34y[n-2] + 0.08y[n-3]

Express the transfer function in a factored form and sketch its pole-zeroplot. Is the system BIBO stable?

Soln)The difference equation of a casual LTI discrete system is:y[n] = 5x[n] + 5x[n-1] + 0.4x[n-2] + 0.32x[n-3]

- 0.5y[n-1] + 0.34y[n-2] + 0.08y[n-3]

Taking z-transform of the above equation:

4)2.01(

7619.4)5.01(

6237.5)8.01(

0256.1)(..

08.034.05.0132.04.055

X(z)Y(z))(

:is H(z)function transfer The

]32.04.055)[(]08.034.05.0Y(z)[1

..

)(08.0)(34.0)(5.0-

)(32.0)(4.0)(5)(5)(

111

321

321

321321

321

321

-+

+-

+--

=

--++++

==

+++=--+

++

+++=

---

---

---

------

---

---

zzzHei

zzzzzzzH

zzzzXzzz

ei

zYzzYzzYz

zXzzXzzXzzXzY

12

Discrete Time System

h[n]

The system is stable since all the poles lie inside the unit circle

The pole zero plot of the transfer function using Matlab:

13

Discrete Time System

h[n]

4. Using z-transform methods, determine the explicit expression For the impulse response h(n) of a causal LTI discrete-time systemwhich develops an output y[n]=4(0.75)n µ[n] for an input x[n]=3(0.25)n µ[n].

Soln)

Given that the system response

for an input

n][4(0.75) y[n] nµ=

n][3(0.25) x[n] nµ=

Taking z transform of each of the equation, we get

)25.01(

13)(X and )75.01(

14 [z] 11 -- -=

-=

zz

zy

Therefore, the transfer function H(z) is:

[ ] ]1[(0.75)25.0][(0.75) 34 h[n]

)75.01(

25.0 - )75.01(

134

)75.01()25.01(.

34

)()( )(

1-nn

1

1

1

1

1

--=

úû

ùêë

é--

=

--

==

-

-

-

-

-

nn

zz

z

zz

zXzYzH

µµ

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