Z - Transform - KSUZ - Transform 1 CEN543, Dr. Ghulam Muhammad King Saud University The z-transform...

Z - Transform

1CEN543, Dr. Ghulam Muhammad

King Saud University

The z-transform is a very important tool in describing and analyzing digital systems.

It offers the techniques for digital filter design and frequency analysis of digital signals.

n

nznxzX ][)(

Definition of z-transform:

For causal sequence, x(n) = 0, n < 0:

Where z is a complex variable

All the values of z that make the summation to exist form a region of convergence.

CEN543, Dr. Ghulam Muhammad King Saud University

2

Example 1

Given the sequence, find the z transform of x(n).

Problem:

Solution:

We know,

Therefore,

111

1

1

1)(

1

z

z

z

zzX

When, 1||1|| 1 zz

Region of convergence (ROC)


3

Example 2

Given the sequence, find the z transform of x(n).

Problem:

Solution:

Therefore,

az

z

z

aazzX

1

1

1

1)(

1

When, azaz ||1|| 1

Region of convergence


4

Z-Transform Table


5

Example 3

Find z-transform of the following sequences.

Problem:

a. b.

Solution:

a. From line 9 of the Table:

b. From line 14 of the Table:


6

Z- Transform Properties (1)

Linearity:

a and b are arbitrary constants.

Example 4

Find z- transform of

Line 3

Line 6

Using z- transform table:

Therefore, we get

Problem:

Solution:


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Shift Theorem:

Verification:

Since x(n) is assumed to be causal:

Then we achieve,

n = m


8

Example 5

Find z- transform of

Problem:

Solution:

Using shift theorem,

Using z- transform table, line 6:


9


Convolution

In time domain,

In z- transform domain,

Verification:

Eq. (1)

Using z- transform in Eq. (1)


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Example 6Problem:

Solution:

Given the sequences,

Find the z-transform of their convolution.

Applying z-transform on the two sequences,

From the table, line 2

Therefore we get,


11

Inverse z- Transform: Examples

Find inverse z-transform of

We get,

Using table,


We get,

Using table,

Example 7

Example 8


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Inverse z- Transform: Examples


Since,

By coefficient matching,

Therefore,


Example 9

Example 10


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Inverse z-Transform: Using Partial Fraction


First eliminate the negative power of z.

Example 11

Dividing both sides by z:

Finding the constants:

Therefore, inverse z-transform is:

Problem:

Solution:


14


Example 12


We first find B:

Next find A:

Problem:

Solution:


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Example 12 – contd.

Using polar form

Now we have:

Therefore, the inverse z-transform is:


16


Example 13


m = 2, p = 0.5

Problem:

Solution:


17


From Table:

Finally we get,


18

Partial Function Expansion Using MATLAB

Example 14

The denominator polynomial can be found using MATLAB:

Therefore,

The solution is:

Problem:

Solution:


19


Example 15Problem:

Solution:


20


Example 16Problem:

Solution:


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Difference Equation Using Z-Transform

The procedure to solve difference equation using z-transform:

1. Apply z-transform to the difference equation.

2. Substitute the initial conditions.

3. Solve for the difference equation in z-transform domain.

4. Find the solution in time domain by applying the inverse z-transform.


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Example 17

Solve the difference equation when the initial condition is

Taking z-transform on both sides:

Substituting the initial condition and z-transform on right hand side using Table:

Arranging Y(z) on left hand side:

Problem:

Solution:


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Solving for A and B:

Therefore,

Taking inverse z-transform, we get the solution:


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Example 18

A DSP system is described by the following differential equation with zero initial condition:

a. Determine the impulse response y(n) due to the impulse sequence x(n) = (n).

b. Determine system response y(n) due to the unit step function excitation, where u(n) = 1 for n 0.

Taking z-transform on both sides:

Problem:

Solution:

Applying on right sidea.


25


We multiply the numerator and denominator by z2

Hnece the impulse response:

Solving for A and B:

Therefore,


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The input is step unit function:

Corresponding z-transform:

b.

[Slide 24]

Do the middle

steps by yourself!


27

Example 19Problem:

Solution:

Determine the z-transform and the ROC of the signal: )()3(4)2(3)( nunx nn

)(4)(3)(

)(3)( and )(2)(

21

21

nxnxnx

nunxnunx nn

Let

Therefore

Applying z-transform: )(4)(3)( 21 zXzXzX

3||:ROC ,31

1)(

2||:ROC ,21

1)(

12

11

zz

zX

zz

zX

But

3|| :ROC ,31

4

21

3)(

11

z

zzzXFinally,

Intersection


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Example 20Problem:

Solution:

Determine the z-transform and the ROC of the signal: )1()()( nubnunx nn

1

1

0

11

0

)()()(l

l

n

n

n

nn

n

nn zbzzbzzX

ROC1: |z| > ||ROC2: |z| < |b|

Case (a). |b| < ||

Case (b). |b| > ||

ROCs do not overlap, so X(z) does not exist.

ROC of X(z) is ||< z < |b|


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A

AAAAAAA

1...)1(.... 232

11

1

1

1

1

bzzb

zb


1

1 )(l

lzb

111 1

1

1

1)(

bzzb

b

bzzzX

Z - Transform - KSUZ - Transform 1 CEN543, Dr. Ghulam Muhammad King Saud University The z-transform...

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Transcript of Z - Transform - KSUZ - Transform 1 CEN543, Dr. Ghulam Muhammad King Saud University The z-transform...