1 Applications of z-transforms

1.1 Transfer (or system) function

Consider a first order linear constant coefficient difference equation

$y_n+a{y}_{n-1}=b{x}_{n}n=0,1,2,\dots$ (1)

where $\left\{x_n\right\}$ is a given sequence.

Assume an initial condition $y_{-1}$ is given.

Task!

Take the z-transform of (1), insert the initial condition and obtain $Y\left(z\right)$ in terms of $X\left(z\right)$ .

Answer

The form of (2) shows us clearly that $Y\left(z\right)$ is made up of two components, $Y_1\left(z\right)$ and $Y_2\left(z\right)$ say, where

1. $Y_1\left(z\right)=\frac{bX\left(z\right)}{1+a{z}^{-1}}$ which depends on the input $X\left(z\right)$
2. $Y_2\left(z\right)=\frac{-a{y}_{-1}}{1+a{z}^{-1}}$ which depends on the initial condition $y_{-1}$ .

Clearly, from (2), if $y_{-1}=0$ (zero initial condition) then

$Y\left(z\right)=Y_1\left(z\right)$

and hence the term zero-state response is sometimes used for $Y_1\left(z\right)$ .

Similarly if $\left\{x_n\right\}$ and hence $X\left(z\right)=0$ (zero input)

$Y\left(z\right)=Y_2\left(z\right)$

and hence the term zero-input response can be used for $Y_2\left(z\right)$ .

In engineering the difference equation (1) is regarded as modelling a system or more specifically a linear discrete time-invariant system. The terms linear and time-invariant arise because the difference equation (1) is linear and has constant coefficients i.e. the coefficients do not involve the index $n$ . The term ‘discrete’ is used because sequences of numbers, not continuous quantities, are involved. As noted above, the given sequence $\left\{x_n\right\}$ is considered to be the input sequence and $\left\{y_n\right\}$ , the solution to (1), is regarded as the output sequence.

Figure 8

A more precise block diagram representation of a system can be easily drawn since only two operations are involved:

1. Multiplying the terms of a sequence by a constant.

2. Shifting to the right, or delaying, the terms of the sequence.

A system which consists of a single multiplier is denoted as shown by a triangular symbol:

Figure 9

As we have seen earlier in this workbook a system which consists of only a single delay unit is represented symbolically as follows

Figure 10

The system represented by the difference equation (1) consists of two multipliers and one delay unit. Because (1) can be written

$y_n=b{x}_n-a{y}_{n-1}$

a symbolic representation of (1) is as shown in Figure 11.

Figure 11

The circle symbol denotes an adder or summation unit whose output is the sum of the two (or more) sequences that are input to it.

We will now concentrate upon the zero state response of the system i.e. we will assume that the initial condition $y_{-1}$ is zero.

Thus, using (2),

$Y\left(z\right)=\frac{bX\left(z\right)}{1+a{z}^{-1}}$

so

$\frac{Y\left(z\right)}{X\left(z\right)}=\frac{b}{1+a{z}^{-1}}$ (3)

The quantity $\frac{Y\left(z\right)}{X\left(z\right)}$ , the ratio of the output z-transform to the input z-transform, is called the transfer function of the discrete system. It is often denoted by $H\left(z\right)$ .

Task!

1. Write down the transfer function $H\left(z\right)$ of the system represented by (1)
   1. using negative powers of $z$
   2. using positive powers of $z$ .
2. Write down the inverse z-transform of $H\left(z\right)$ .

Answer

We can represent any discrete system as follows

Figure 12

From the definition of the transfer function it follows that

$Y\left(z\right)=X\left(z\right)H\left(z\right)$ (at zero initial conditions).

The corresponding relation between $\left\{y_n\right\},\left\{x_n\right\}$ and the inverse z-transform $\left\{h_n\right\}$ of the transfer function will be discussed later; it is called a convolution summation .

The significance of $\left\{h_n\right\}$ is readily obtained.

Suppose $\left\{x_n\right\}=\left\{\begin{array}{ccc}1\hfill & \hfill \hfill & n=0\hfill \\ 0\hfill & \hfill \hfill & n=1,2,3,\dots \hfill \end{array}\right.$

i.e. $\left\{x_n\right\}$ is the unit impulse sequence that is normally denoted by ${\delta }_n$ . Hence, in this case,

$X\left(z\right)=Z\left\{{\delta }_n\right\}=1$ so $Y\left(z\right)=H\left(z\right)$ and $\left\{y_n\right\}=\left\{h_n\right\}$

In words:  $\left\{h_n\right\}$ is the response or output of a system where the input is the unit impulse sequence $\left\{{\delta }_n\right\}$ . Hence $\left\{h_n\right\}$ is called the unit impulse response of the system.

It follows from the previous Task that for the first order system (1)

$H\left(z\right)=\frac{b}{1+a{z}^{-1}}=\frac{bz}{z+a}$ is the transfer function and

$\left\{h_n\right\}=\left\{b{\left(-a\right)}^n\right\}$ is the unit impulse response sequence.

Task!

Write down the transfer function of

1. a single multiplier unit
2. a single delay unit.

Answer

Task!

Obtain the transfer function of the system.

$y_n+a_1{y}_{n-1}=b_0{x}_n+b_1{x}_{n-1}n=0,1,2,\dots$

where $\left\{x_n\right\}$ is a known sequence with $x_n=0$ for $n=-1,-2,\dots$ .

[Remember that the transfer function is only defined at zero initial condition i.e. assume $y_{-1}=0$ also.]

Answer

1.2 Second order systems

Consider the system whose difference equation is

$y_n+a_1{y}_{n-1}+a_2{y}_{n-2}=b{x}_{n}n=0,1,2,\dots$ (4)

where the input sequence $x_n=0,n=-1,-2,\dots$

In exactly the same way as for first order systems it is easy to show that the system response has a z-transform with two components.

Task!

Take the z-transform of (4), assuming given initial values $y_{-1},y_{-2}$ . Show that $Y\left(z\right)$ has two components. Obtain the transfer function of the system (4).

Answer

Example

Obtain (i) the unit impulse response (ii) the unit step response of the system specified by the second order difference equation

$y_n-\frac{3}{4}{y}_{n-1}+\frac{1}{8}{y}_{n-2}=x_n$ (5)

Note that both these responses refer to the case of zero initial conditions. Hence it is convenient to first obtain the transfer function $H\left(z\right)$ of the system and then use the relation $Y\left(z\right)=X\left(z\right)H\left(z\right)$ in each case.

We write down the transfer function of (5), using positive powers of $z$ . Taking the z-transform of (5) at zero initial conditions we obtain

We now complete the problem for inputs 1. $x_n={\delta }_n$ 2. $x_n=u_n$ , the unit step sequence, using partial fractions.

$H\left(z\right)=\frac{z^2}{\left(z-\frac{1}{2}\right)\left(z-\frac{1}{4}\right)}=\frac{2z}{z-\frac{1}{2}}-\frac{z}{z-\frac{1}{4}}$

1. With $x_n={\delta }_n$ so $X\left(z\right)=1$ the response is, as we saw earlier,

   $Y\left(z\right)=H\left(z\right)$

   so $y_n=h_n$

   where $h_n={\mathbb{Z}}^{-1}H\left(z\right)=2\times {\left(\frac{1}{2}\right)}^n-{\left(\frac{1}{4}\right)}^{n}n=0,1,2,\dots$

2. The z-transform of the unit step is $\frac{z}{z-1}$ so the unit step response has z-transform $$\begin{array}{rcll}Y\left(z\right)& =& \frac{z^2}{\left(z-\frac{1}{2}\right)\left(z-\frac{1}{4}\right)}\frac{z}{\left(z-1\right)}& \\ & & & \\ & =& -\frac{2z}{z-\frac{1}{2}}+\frac{\frac{1}{3}z}{z-\frac{1}{4}}+\frac{\frac{8}{3}z}{z-1}& \end{array}$$

   Hence, taking inverse z-transforms, the unit step response of the system is

   $y_n=\left(-2\right)\times {\left(\frac{1}{2}\right)}^n+\frac{1}{3}\times {\left(\frac{1}{4}\right)}^n+\frac{8}{3}n=0,1,2,\dots$

   Notice carefully the form of this unit step response - the first two terms decrease as $n$ increases and are called transients . Thus

   $y_n\to \frac{8}{3}\text{ as}n\to \infty$

   and the term $\frac{8}{3}$ is referred to as the steady state part of the unit step response.

1.3 Combinations of systems

The concept of transfer function enables us to readily analyse combinations of discrete systems.

Series combination

Suppose we have two systems $S_1$ and $S_2$ with transfer functions $H_1\left(z\right)$ , $H_2\left(z\right)$ in series with each other. i.e. the output from $S_1$ is the input to $S_2$ .

Figure 13

Clearly, at zero initial conditions,

so the ratio of the final output transform to the input transform is

$\frac{Y\left(z\right)}{X\left(z\right)}=H_2\left(z\right)H_1\left(z\right)$ (6)

i.e. the series system shown above is equivalent to a single system with transfer function $H_2\left(z\right)H_1\left(z\right)$

Figure 14

Task!

Obtain

1. the transfer function
2. the governing difference equation of the system obtained by connecting two first order systems $S_1$ and $S_2$ in series. The governing equations are:

   $S_1:y_n-a{y}_{n-1}=b{x}_n$

   $S_2:y_n-c{y}_{n-1}=d{x}_n$

1. Begin by finding the transfer function of $S_1$ and $S_2$ and then use (6):

   Answer

   $S_1$ : $Y\left(z\right)-a{z}^{-1}Y\left(z\right)=bX\left(z\right)$ so $H_1\left(z\right)=\frac{b}{1-a{z}^{-1}}$

   $S_2$ : $H_2\left(z\right)=\frac{d}{1-c{z}^{-1}}$

   so the series arrangement has transfer function

   $$\begin{array}{rcll}H\left(z\right)& =& \frac{bd}{\left(1-a{z}^{-1}\right)\left(1-c{z}^{-1}\right)}& \\ & & & \\ & =& \frac{bd}{1-\left(a+c\right)z^{-1}+ac{z}^{-2}}& \end{array}$$

   If $X\left(z\right)$ and $Y\left(z\right)$ are the input and output transforms for the series arrangement, then

   $Y\left(z\right)=H\left(z\right)X\left(z\right)=\frac{bdX\left(z\right)}{1-\left(a+c\right)z^{-1}+ac{z}^{-2}}$

2. By transfering the denominator from the right-hand side to the left-hand side and taking inverse z-transforms obtain the required difference equation of the series arrangement:

   Answer

   We have

   $Y\left(z\right)\left(1-\left(a+c\right)z^{-1}+ac{z}^{-2}\right)=bdX\left(z\right)$

   $Y\left(z\right)-\left(a+c\right)z^{-1}Y\left(z\right)+ac{z}^{-2}Y\left(z\right)=bdX\left(z\right)$

   from which, using the right shift theorem,

   $y_n-\left(a+c\right)y_{n-1}+ac{y}_{n-2}=bd{x}_n$ .

   which is the required difference equation.

   You can see that the two first order systems in series have an equivalent second order system. Feedback combination

   Figure 15

   

   For the above negative feedback arrangement of two discrete systems with transfer functions $H_1\left(z\right),H_2\left(z\right)$ we have, at zero initial conditions,

   $Y\left(z\right)=W\left(z\right)H_1\left(z\right)$ where $W\left(z\right)=X\left(z\right)-H_2\left(z\right)Y\left(z\right)$

Task!

Eliminate $W\left(z\right)$ and hence obtain the transfer function of the feedback system.

Answer