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136 SAMPLED DATA SYSTEMS AND THE Z-TRANSFORM
Zero-order
y(t)
hold
r(t) |
r*(t) |
Figure 6.10 Ideal sampler and zero-order hold for Example 6.1
Zero-order
y(t)
hold
r(t) |
r*(t) |
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
Z = esT ; |
(6.10) |
the z-transform of the function r(t) is Z [r(t)] = R(z) which, from (6.7), is given by
∞ |
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R(z) |
(6.11) |
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= |
r(nT )z−n . |
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n |
= |
0 |
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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,
THE z-TRANSFORM |
137 |
r(nT)
1
t
0 T 2T 3T 4T
Figure 6.12 Unit step function
just like the Laplace transform techniques used in continuous-time systems. We will now look at how we can find the z-transforms of some commonly used functions.
6.2.1 Unit Step Function
Consider a unit step function as shown in Figure 6.12, defined as
r(nT ) = |
1, |
n 0. |
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0, |
n |
< |
0, |
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From (6.11), |
≥ |
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∞ |
∞ |
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R(z) = r(nT )z−n = z−n = 1 + z−1 + z−2 + z−3 + z−4 + . . . |
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= |
0 |
= |
0 |
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n |
n |
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or |
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R(z) = |
z |
, |
for |z| > 1. |
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z |
− |
1 |
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6.2.2 Unit Ramp Function
Consider a unit ramp function as shown in Figure 6.13, defined by
0, |
n < 0, |
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r(nT ) = nT, |
n |
≥ |
0. |
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From (6.11), |
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∞ |
∞ |
|||||
R(z) = |
r(nT )z−n = |
= |
nTz−n = Tz−1 + 2Tz−2 + 3Tz−3 + 4Tz−4 + . . . |
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= |
0 |
0 |
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n |
n |
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r(nT)
4
3
2
1
t 0 T 2T 3T 4T
Figure 6.13 Unit ramp function
138 |
SAMPLED DATA SYSTEMS AND THE Z-TRANSFORM |
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r(nT) |
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4 |
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3 |
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2 |
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1 |
t |
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0 T |
2T 3T 4T |
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Figure 6.14 |
Exponential function |
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or |
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R(z) = |
T z |
for |z| > 1. |
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, |
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(z |
− |
1)2 |
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6.2.3 Exponential Function
Consider the exponential function shown in Figure 6.14, defined as |
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r(nT ) = |
e−anT , |
n < |
0. |
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0, |
n |
0, |
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From (6.11), |
≥ |
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∞ |
∞ |
e−anT z−n = 1 + e−aT z−1 + e−2aT z−2 + e−3aT z−3 + . . . |
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R(z) = |
= |
r(nT )z−n = |
= |
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n |
0 |
n |
0 |
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or |
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R(z) = |
1 |
= |
z |
, |
for |z| < e−aT . |
(6.12) |
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1 |
− |
e aT z |
− |
1 |
z |
− |
e |
aT |
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− |
− |
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6.2.4 General Exponential Function
Consider the general exponential function |
pn , |
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r(n) = |
n 0. |
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0, |
n |
< |
0, |
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From (6.11), |
≥ |
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∞ |
∞ |
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R(z) = r(nT )z−n = pn z−n = 1 + pz−1 + p2z−2 + p3z−3 + . . . |
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n |
= |
0 |
n |
= |
0 |
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or |
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R(z) = |
z |
, |
for |z| < | p|. |
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z |
− |
p |
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Similarly, we can show that |
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R( p−k ) = |
z |
. |
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z |
− |
p |
− |
1 |
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THE z-TRANSFORM |
139 |
6.2.5 Sine Function
Consider the sine function, defined as |
sin nωT. |
n |
0. |
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r(nT ) = |
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0, |
n < |
0, |
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Recall that |
≥ |
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sin x |
= |
e j x |
− e− j x |
, |
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2 j |
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so that |
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r(nT ) |
= |
e jnωT |
− e− jnωT |
= |
e jnωT |
− |
e− jnωT |
. |
(6.13) |
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2 j |
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2 j |
2 j |
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But we already know from (6.12) that the z-transform of an exponential function is |
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R(e−anT ) = R(z) = |
z |
. |
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z |
− |
e |
aT |
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Therefore, substituting in (6.13) gives |
− |
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e |
jωT = 2 j z2 |
z(e jωT |
− e− jωT ) |
1 |
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R(z) = 2 j |
z e jωT |
− z |
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1 |
z |
z |
1 |
z(e jωT |
e jωT |
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or |
− |
− − |
− |
+ − |
+ |
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z sin ωT
R(z) = . z2 − 2z cos ωT + 1
6.2.6 Cosine Function
Consider the cosine function, defined as |
n |
0. |
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r(nT ) = cos nωT, |
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0, |
n < |
0, |
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Recall that |
≥ |
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cos x |
= |
e j x |
+ e− j x |
, |
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2 |
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so that |
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r(nT ) |
= |
e jnωT + e− jnωT |
= |
e jnωT |
+ |
e− jnωT |
. |
(6.14) |
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2 |
2 |
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2 |
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But we already know from (6.12) that the z-transform of an exponential function is |
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R(e−anT ) = R(z) = |
z |
. |
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z |
− |
e |
aT |
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Therefore, substituting in (6.14) gives |
− |
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+ z e |
jωT |
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R(z) = 2 z e jωT |
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1 |
z |
z |
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or |
− |
− − |
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R(z) |
= z |
2 |
z(z − cos ωT ) |
. |
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− 2z cos ωT + |
1 |
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140 SAMPLED DATA SYSTEMS AND THE Z-TRANSFORM
6.2.7 Discrete Impulse Function
Consider the discrete impulse function defined as
δ(n) = |
0, n |
= 0. |
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1, n |
0, |
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From (6.11), |
= |
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∞ |
∞ |
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R(z) = r(nT )z−n = z−n = 1. |
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= |
0 |
= |
0 |
|
n |
n |
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6.2.8 Delayed Discrete Impulse Function
The delayed discrete impulse function is defined as
δ(n − k) = |
0, n |
= |
k. |
0, |
|
1, n |
= |
k > |
|||
From (6.11), |
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∞ |
∞ |
||||
R(z) = r(nT )z−n = z−n = z−n . |
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= |
0 |
= |
0 |
||
n |
n |
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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):
p |
N (xn ) |
1 |
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G(z) = |
, |
(6.15) |
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= |
1 |
D (xn ) 1 |
− |
exn T z−1 |
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n |
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