1. Properties and the Inverse of
Chapter 6:
Properties and the Inverse of
z-Transform

2. Properties of z-Transform
Some of the properties of the z-transform are:
Linearity
Right shift
Left shift
Final value theorem
Z-differentiation (Multiplication by n)

3. Example: Linearity
The z-transform of a unit ramp function r(nT) is Find the z-transform of the function 5r(nT). Solution Using the linearity property of z-transform,

4. Right-shifting property
Suppose that the z-transform of f(nT) is F(z).
Let y(nT) = f(nT-mT), then
assuming that f nT =0 for n<0.

5. Left-shifting property
Suppose that z-transform of f(nT) is F(z).
Let y(nT) = f(nT+mT). Then
However, if f nT =0 for n=0 to m-1, this simplifies to

6. Final value theorem
Suppose that the z-transform of f (nT ) is F(z). Then the final value of the time response is given by
Note that this theorem is valid if the poles of (1−z−1)F(z) are inside the unit circle or at z = 1.

7. Example 6.5 Given the function find the final value of g(nT).
Solution:
Using the final value theorem,

8. Multiplication by n
Let
Then
Example
Given
And
Then,
Z-differentiation

9. Inverse z-Transforms
Given the z-transform, Y(z), of a function, y(n), it is required to find the time-domain function y(n).
Here, we will study the following methods:
power series (long division).
expanding Y (z) into partial fractions and using z-transform tables to find the inverse.

10. Method 1: Power series (long division)
This method involves dividing the denominator of Y (z) into the numerator such that a power series of the form
is obtained. Notice that the values of y(n) are the coefficients in the power series.

11. Example 6.6
Find the inverse z-transform for the polynomial Solution: Dividing the denominator into the numerator gives

12. and the coefficients of the power series are y(0) = 1, y(T) = 4, y(2T) = 8, y(3T) = 8, : The required sequence is y(t) = δ(t) + 4δ(t-T) + 8δ(t-2T) + 8δ(t-3T) + … The first few samples of the time sequence y(nT) is shown below

13. Example 6.7 Find the inverse z-transform for Y(z) given by Solution:
Dividing the denominator into the numerator gives

14. and the coefficients of the power series are y(0) = 0, y(T) = 1, y(2T) = 3, y(3T) = 7, y(4T) = 15, : The required sequence is y(t) = δ(t-1) + 3δ(t-T) + 7 δ(t-2T) + 15 δ(t-3T) + … The first few samples of the time sequence y(nT) is shown below

15. The disadvantage of the power series method is that it does not give a closed form of the resulting sequence. We often need a closed-form result, and other methods should be used when this is the case.

16. Method 2: Partial fractions
A partial fraction expansion of Y (z) is found, and then tables of z-transform can be used to determine the inverse z-transform.
Looking at the z-transform tables, we see that there is usually a z term in the numerator. It is therefore more convenient to find the partial fractions of Y (z)/z and then multiply the partial fractions by z to obtain a z term in the numerator.

17. Example 6.8 Find the inverse z-transform of the function Solution
The above expression can be written as
The values of A and B can be found by equating like powers in the numerator, i.e.
We find A = − 1, B = 1, giving

18. From the z-transform tables we find that
and the coefficients of the power series are
y(0) = 0,
y(T) = 1,
y(2T) = 3,
y(3T) = 7,
y(4T) = 15, :
so that the required sequence is
y(t) = δ(t-1) + 3δ(t-T) + 7 δ(t-2T) + 15 δ(t-3T) + …
Which is the same answer as Example 6.7.

19. Example 6.9 Find the inverse z-transform of the function Answer:
The above expression can be written as

20. The values of A, B, C can be found as follows

21. Using the inverse z-transform, we find
The coefficients of the power series are
and the required sequence is

22. Example 6.11
Using the partial expansion method described above, find the inverse z-transform of Solution Rewriting the function as

23. we find that Thus

24. The inverse transform is found from the tables as and the coefficients of the power series are y(0) = 0, y(T) = 1, y(2T) = 3.3, y(3T) = 5.89, : so that the required sequence is y(t) = δ(t-T) + 3.3δ(t-2T) δ(t-3T) + …

25. Example 6.12 Find the inverse z-transform of Solution
Rewriting the function as

26. We obtain

27. We can now write Y(z) as
The inverse transform is found from the tables as
Note that for the last term, we used the multiplication by n property which is equivalent to a z-differentiation.

28. Pulse Transfer Function and Manipulation of Block Diagrams
Given the following system:
Suppose we are interested in finding the relationship between the sampled output y*(s) and the sampled input u*(s).
The continuous output y(s) is given by:
Then, sampling the output signal gives:

29. Note that if at least one of the continuous functions has been sampled, then the z-transform of the product is equal to the product of the z-transforms of each function (note that [e*(s)]* = [e*(s)], since sampling an already sampled signal has no further effect).
G(z) is the ratio of the z-transform of the sampled output and the sampled input at the sampling instants, and is called the pulse transfer function.
Notice from (6.35) that we have no information about the output y(z) between the sampling instants.

30. Example 6.17
Derive an expression for the z-transform of the output of the system.
Solution
The following expressions can be written for the system:
Which gives or

31. For example, if
Then
and the output function is given by

32. Example 6.16
The figure shows an open-loop sampled data system. Derive an expression for the z-transform of the output of the system.

33. The following expression can be written for the system:
Solution
The following expression can be written for the system: or
And
where

34. For example, if
Then, from the z-transform tables,
and the output is given by

35. Note about using Matlab
In Matlab, you can use the function c2d to convert a continuous function into discrete. For example, to convert the transfer function
using a sample period T=1, you can use the following commands:
>> G = tf([1],[1 5 6]);
>> c2d(G,1,’impulse’)
Matlab output: