Lecture 4: The z-Transform 1

The z-transform The z-transform is used in sampled data systems just as the Laplace transform is used in continuous-time systems. Given a discrete variable r(k) or r k with values r 0, r 1,..., r k−1, r k,..., the z-transform of this variable is given by Notice that the z-transform consists of an infinite series in the complex variable z, with r(kT ) are the coefficients of this power series at different sampling instants. We will look at how we can find the z-transforms of some commonly used functions. 2

Unit Step Function 3

Useful closed-form geometric series summations 4

Unit Ramp Function 5

Exponential Function 6

General Exponential Function 7

Sine Function 8

Discrete Impulse Function 9

Delayed Discrete Impulse Function 10

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The z-Transform of a Function Expressed as a Laplace Transform Given a function G(s), find G(z) which denotes the z-transform equivalent of G(s). It is important to realize that G(z) is not obtained by simply substituting z for s in G(s)! 12

Example 1 Given Determine G(z). 13

Answer: Using Inverse Laplace transform Partial fraction Inverse Laplace transform Substitute t = nT gives Finally, 14

Method 2: Laplace to z-transform table From table in Appendix A So, 15

MATLAB c2d command MATLAB function c2d can be used to convert a continuous function into discrete. For example, to convert the following transfer function with a sample period T=1: >> G = tf([1],[1 5 6]); % continuous time transfer function >> T = 1; >> Gd = c2d(G,T,’impulse’) Matlab output: Transfer function: z e z^ z Sampling time: 1 16

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) 17

Example 2: 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, 18

Right-shifting property 19

Left-shifting property 20

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 (i.e. the system is stable and so reaches final value). 21

Example 3 Given the function find the final value of g(nT). Solution: Using the final value theorem, 22

DC gain of transfer function Given the transfer function Let input u(k) be a step of magnitude u ∞, with z-transform The output is given by The final value of the output y(k) can be found using the final value theorem: Hence the DC gain of the transfer function H(z) is Again, note that when finding the DC gain of a transfer function, all poles of the transfer function must be inside the unit circle. 23

Example 4 Consider the transfer function given by Note that it is necessary to find the pole locations of this transfer function to make sure it is stable (all poles inside unit circle). The DC gain is given by Thus if this discrete system were given an input that eventually reached a constant value, the output would eventually reach twice that value. NOTE: If the denominator polynomial above were z 2 − 0.5z + 2, the DC gain would evaluate to H(1) = 0.8, but that is meaningless since the system is unstable (the roots are outside the unit circle). 24

Multiplication by n Let Then Example Given And Then, 25 Z-differentiation

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. 26

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. 27

Example 5 Find the inverse z-transform for the polynomial Solution: Dividing the denominator into the numerator gives 28

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 above. To perform long division using MATLAB, you can use the following commands: >> delta = [1 zeros(1, 4)]; % >> num = [0 1 1]; >> den = [1 -3 4]; >> yk = filter(num, den, delta) 29

Example 6 Find the inverse z-transform for Y(z) given by Solution: Dividing the denominator into the numerator gives 30

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) + 7δ(t-2T) + 15δ(t-3T) + … The first few samples of the time sequence y(nT) is shown below 31

The disadvantage of the power series (long division) method is that it does not give a closed form of the resulting sequence. When a closed-form result is required, other methods should be used. 32

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. 33

Example 7 34 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 similar powers in the numerator, i.e. We find A = − 1, B = 1, giving

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) + 7δ(t-2T) + 15δ(t-3T) + … Which is the same answer as in Example 6. Note that MATLAB has the command residue to perform partial fraction >> [r, p, k]=residue(Num, Den) >> [r, p, k]=residue([1 0],[1 -3 2]) 35

Example 8 Find the inverse z-transform of the function Answer: The above expression can be written as 36

The values of A, B, C can be found as follows 37

Using the inverse z-transform, we find The coefficients of the power series are and the required sequence is 38

Example 9 Using the partial expansion method described above, find the inverse z-transform of Solution Rewriting the function as 39

we find that Thus 40

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) + … 41

Example 10 Find the inverse z-transform of Solution Rewriting the function as 42

We obtain 43

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. 44