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Z - Transform
1
CEN352, 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
n
z
n
x
z
X ]
[
)
(
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.
CEN352, Dr. Ghulam Muhammad
King Saud University
2
Example 1
Given the sequence, find the z transform of x(n).
Problem:
Solution:
We know,
Therefore,
1
1
1
1
1
1
)
( 1





 
z
z
z
z
z
X
When, 1
|
|
1
|
| 1




z
z
Region of convergence
CEN352, Dr. Ghulam Muhammad
King Saud University
3
Example 2
Given the sequence, find the z transform of x(n).
Problem:
Solution:
Therefore,
a
z
z
z
a
az
z
X





 
1
1
1
1
)
( 1
When, a
z
az 



|
|
1
|
| 1
Region of convergence
CEN352, Dr. Ghulam Muhammad
King Saud University
4
Z-Transform
Table
CEN352, Dr. Ghulam Muhammad
King Saud University
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:
CEN352, Dr. Ghulam Muhammad
King Saud University
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:
CEN352, Dr. Ghulam Muhammad
King Saud University
7
Z- Transform Properties (2)
Shift Theorem:
Verification:
Since x(n) is assumed to be causal:
Then we achieve,
n = m
CEN352, Dr. Ghulam Muhammad
King Saud University
8
Example 5
Find z- transform of
Problem:
Solution:
Using shift theorem,
Using z- transform table, line 6:
CEN352, Dr. Ghulam Muhammad
King Saud University
9
Z- Transform Properties (3)
Convolution
In time domain,
In z- transform domain,
Verification:
Eq. (1)
Using z- transform in Eq. (1)
CEN352, Dr. Ghulam Muhammad
King Saud University
10
Example 6
Problem:
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,
CEN352, Dr. Ghulam Muhammad
King Saud University
11
Inverse z- Transform: Examples
Find inverse z-transform of
We get,
Using table,
Find inverse z-transform of
We get,
Using table,
Example 7
Example 8
CEN352, Dr. Ghulam Muhammad
King Saud University
12
Inverse z- Transform: Examples
Find inverse z-transform of
Since,
By coefficient matching,
Therefore,
Find inverse z-transform of
Example 9
Example 10
CEN352, Dr. Ghulam Muhammad
King Saud University
13
Inverse z-Transform: Using Partial Fraction
Find inverse z-transform of
First eliminate the negative power of z.
Example 11
Dividing both sides by z:
Finding the
constants:
Therefore, inverse z-transform is:
Problem:
Solution:
CEN352, Dr. Ghulam Muhammad
King Saud University
14
Inverse z-Transform: Using Partial Fraction
Example 12
Dividing both sides by z:
We first find B:
Next find A:
Problem:
Solution:
CEN352, Dr. Ghulam Muhammad
King Saud University
15
Example 12 – contd.
Using polar form
Now we have:
Therefore, the inverse z-transform is:
CEN352, Dr. Ghulam Muhammad
King Saud University
16
Inverse z-Transform: Using Partial Fraction
Example 13
Dividing both sides by z:
m = 2, p = 0.5
Problem:
Solution:
CEN352, Dr. Ghulam Muhammad
King Saud University
17
Example 13 – contd.
From Table:
Finally we get,
CEN352, Dr. Ghulam Muhammad
King Saud University
18
Partial Function Expansion Using MATLAB
Example 14
The denominator polynomial can be
found using MATLAB:
Therefore,
The solution is:
Problem:
Solution:
CEN352, Dr. Ghulam Muhammad
King Saud University
19
Partial Function Expansion Using MATLAB
Example 15
Problem:
Solution:
CEN352, Dr. Ghulam Muhammad
King Saud University
20
Partial Function Expansion Using MATLAB
Example 16
Problem:
Solution:
CEN352, Dr. Ghulam Muhammad
King Saud University
21
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.
CEN352, Dr. Ghulam Muhammad
King Saud University
22
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:
CEN352, Dr. Ghulam Muhammad
King Saud University
23
Example 17 – contd.
Solving for A and B:
Therefore,
Taking inverse z-transform, we get the solution:
CEN352, Dr. Ghulam Muhammad
King Saud University
24
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 side
a.
CEN352, Dr. Ghulam Muhammad
King Saud University
25
Example 18 – contd.
We multiply the numerator and denominator by z2
Hnece the impulse
response:
Solving for A and B:
Therefore,
CEN352, Dr. Ghulam Muhammad
King Saud University
26
Example 18 – contd.
The input is step unit function:
Corresponding z-transform:
b.
[Slide 24]
Do the
middle
steps by
yourself!

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Reference for z and inverse z transform

  • 1. Z - Transform 1 CEN352, 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 n z n x z X ] [ ) ( 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.
  • 2. CEN352, Dr. Ghulam Muhammad King Saud University 2 Example 1 Given the sequence, find the z transform of x(n). Problem: Solution: We know, Therefore, 1 1 1 1 1 1 ) ( 1        z z z z z X When, 1 | | 1 | | 1     z z Region of convergence
  • 3. CEN352, Dr. Ghulam Muhammad King Saud University 3 Example 2 Given the sequence, find the z transform of x(n). Problem: Solution: Therefore, a z z z a az z X        1 1 1 1 ) ( 1 When, a z az     | | 1 | | 1 Region of convergence
  • 4. CEN352, Dr. Ghulam Muhammad King Saud University 4 Z-Transform Table
  • 5. CEN352, Dr. Ghulam Muhammad King Saud University 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. CEN352, Dr. Ghulam Muhammad King Saud University 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:
  • 7. CEN352, Dr. Ghulam Muhammad King Saud University 7 Z- Transform Properties (2) Shift Theorem: Verification: Since x(n) is assumed to be causal: Then we achieve, n = m
  • 8. CEN352, Dr. Ghulam Muhammad King Saud University 8 Example 5 Find z- transform of Problem: Solution: Using shift theorem, Using z- transform table, line 6:
  • 9. CEN352, Dr. Ghulam Muhammad King Saud University 9 Z- Transform Properties (3) Convolution In time domain, In z- transform domain, Verification: Eq. (1) Using z- transform in Eq. (1)
  • 10. CEN352, Dr. Ghulam Muhammad King Saud University 10 Example 6 Problem: 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. CEN352, Dr. Ghulam Muhammad King Saud University 11 Inverse z- Transform: Examples Find inverse z-transform of We get, Using table, Find inverse z-transform of We get, Using table, Example 7 Example 8
  • 12. CEN352, Dr. Ghulam Muhammad King Saud University 12 Inverse z- Transform: Examples Find inverse z-transform of Since, By coefficient matching, Therefore, Find inverse z-transform of Example 9 Example 10
  • 13. CEN352, Dr. Ghulam Muhammad King Saud University 13 Inverse z-Transform: Using Partial Fraction Find inverse z-transform of 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. CEN352, Dr. Ghulam Muhammad King Saud University 14 Inverse z-Transform: Using Partial Fraction Example 12 Dividing both sides by z: We first find B: Next find A: Problem: Solution:
  • 15. CEN352, Dr. Ghulam Muhammad King Saud University 15 Example 12 – contd. Using polar form Now we have: Therefore, the inverse z-transform is:
  • 16. CEN352, Dr. Ghulam Muhammad King Saud University 16 Inverse z-Transform: Using Partial Fraction Example 13 Dividing both sides by z: m = 2, p = 0.5 Problem: Solution:
  • 17. CEN352, Dr. Ghulam Muhammad King Saud University 17 Example 13 – contd. From Table: Finally we get,
  • 18. CEN352, Dr. Ghulam Muhammad King Saud University 18 Partial Function Expansion Using MATLAB Example 14 The denominator polynomial can be found using MATLAB: Therefore, The solution is: Problem: Solution:
  • 19. CEN352, Dr. Ghulam Muhammad King Saud University 19 Partial Function Expansion Using MATLAB Example 15 Problem: Solution:
  • 20. CEN352, Dr. Ghulam Muhammad King Saud University 20 Partial Function Expansion Using MATLAB Example 16 Problem: Solution:
  • 21. CEN352, Dr. Ghulam Muhammad King Saud University 21 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.
  • 22. CEN352, Dr. Ghulam Muhammad King Saud University 22 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:
  • 23. CEN352, Dr. Ghulam Muhammad King Saud University 23 Example 17 – contd. Solving for A and B: Therefore, Taking inverse z-transform, we get the solution:
  • 24. CEN352, Dr. Ghulam Muhammad King Saud University 24 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 side a.
  • 25. CEN352, Dr. Ghulam Muhammad King Saud University 25 Example 18 – contd. We multiply the numerator and denominator by z2 Hnece the impulse response: Solving for A and B: Therefore,
  • 26. CEN352, Dr. Ghulam Muhammad King Saud University 26 Example 18 – contd. The input is step unit function: Corresponding z-transform: b. [Slide 24] Do the middle steps by yourself!