Microcontroller based applied digital control (D. Ibrahim, 2006)

136 SAMPLED DATA SYSTEMS AND THE Z-TRANSFORM

Zero-order

y(t)

hold

Figure 6.10 Ideal sampler and zero-order hold for Example 6.1

Zero-order

y(t)

hold

Figure 6.11 Solution for Example 6.1

Example 6.1

Figure 6.10 shows an ideal sampler followed by a zero-order hold. Assuming the input signal r(t) is as shown in the figure, show the waveforms after the sampler and also after the zero-order hold.

Solution

The signals after the ideal sampler and the zero-order hold are shown in Figure 6.11.

6.2 THE z-TRANSFORM

Equation (6.7) defines an infinite series with powers of e−snT . The z-transform is defined so that

the z-transform of the function r(t) is Z [r(t)] = R(z) which, from (6.7), is given by

Notice that the z-transform consists of an infinite series in the complex variable z, and

R(z) = r(0) + r(T )z−1 + r(2T )z−2 + r(3T )z−3 + . . . ,

i.e. the r(nT ) are the coefficients of this power series at different sampling instants.

The z-transformation is used in sampled data systems just as the Laplace transformation is used in continuous-time systems. The response of a sampled data system can be determined easily by finding the z-transform of the output and then calculating the inverse z-transform,

6.2.3 Exponential Function

6.2.4 General Exponential Function

6.2.5 Sine Function

z sin ωT

R(z) = . z2 − 2z cos ωT + 1

6.2.6 Cosine Function

140 SAMPLED DATA SYSTEMS AND THE Z-TRANSFORM

6.2.7 Discrete Impulse Function

Consider the discrete impulse function defined as

6.2.8 Delayed Discrete Impulse Function

The delayed discrete impulse function is defined as

6.2.9 Tables of z-Transforms

A table of z-transforms for the commonly used functions is given in Table 6.1 (a bigger table is given in Appendix A). As with the Laplace transforms, we are interested in the output response y(t) of a system and we must find the inverse z-transform to obtain y(t) from Y (z).

6.2.10The z-Transform of a Function Expressed as a Laplace Transform

It is important to realize that although we denote the z-transform equivalent of G(s) by G(z), G(z) is not obtained by simply substituting z for s in G(s). We can use one of the following methods to find the z-transform of a function expressed in Laplace transform format:

Given G(s), calculate the time response g(t) by finding the inverse Laplace transform of G(s). Then find the z-transform either from the first principles, or by looking at the z-transform tables.

Given G(s), find the z-tranform G(z) by looking at the tables which give the Laplace transforms and their equivalent z-transforms (e.g. Table 6.1).

Given the Laplace transform G(s), express it in the form G(s) = N (s)/D(s) and then use the following formula to find the z-transform G(z):

where D = ∂ D/∂ s and the xn , n = 1, 2, . . . , p, are the roots of the equation D(s) = 0. Some examples are given below.

Example 6.2

Let

1

G(s) = . s2 + 5s + 6

Determine G(z) by the methods described above.

Solution

Method 1: By finding the inverse Laplace transform. We can express G(s) as a sum of its partial fractions:

= (1 + e−2T z−1 + e−4T z−2 + . . .) − (1 + e−3T z−1 + e−6T z−2 + . . .)

zz

=z − e−2T − z − e−3T

142 SAMPLED DATA SYSTEMS AND THE Z-TRANSFORM

or

z(e−2T − e−3T )

G(z) = (z − e−2T )(z − e−3T ) .

Method 2: By using the z-transform transform tables for the partial product. From Table 6.1, the z-transform of 1/(s + a) is z/(z − e−aT ). Therefore the z-transform of (6.16) is

or

z(e−2T − e−3T )

(z − e−2T )(z − e−3T ) .

Method 3: By using the z-transform tables for G(s). From Table 6.1, the z-transform of

Method 4: By using equation (6.15). Comparing our expression

G(s) =

1

s2 + 5s + 6

with (6.15), we have N (s) = 1, D(s) = s2 + 5s + 6 and D (s) = 2s + 5, and the roots of D(s) = 0 are x1 = −2 and x2 = −3. Using (6.15),

2

G(z) =

N (xn ) 1

n=1 D (xn ) 1 − exn T z−1