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THE z-TRANSFORM |
145 |
Solution
Using the final value theorem,
lim g(nT ) |
lim(1 |
− |
z−1) |
0.792z |
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(z − 1)(z2 − 0.416z + 0.208) |
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n→∞ |
= z→1 |
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lim |
0.792 |
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z2 − 0.416z + 0.208 |
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= z→1 |
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= |
0.792 |
= 1. |
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1 |
0.416 |
+ |
0.208 |
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− |
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6.2.12 Inverse z-Transforms
The inverse z-transform is obtained in a similar way to the inverse Laplace transforms. Generally, the z-transforms are the ratios of polynomials in the complex variable z, with the numerator polynomial being of order no higher than the denominator. By finding the inverse z-transform we find the sequence associated with the given z-transform polynomial. As in the case of inverse Laplace transforms, we are interested in the output time response of a system. Therefore, we use an inverse transform to obtain y(t) from Y (z). There are several methods to find the inverse z-transform of a given function. The following methods will be described here:
power series (long division);
expanding Y (z) into partial fractions and using z-transform tables to find the inverse transforms;
obtaining the inverse z-transform using an inversion integral.
Given a z-transform function Y (z), we can find the coefficients of the associated sequence y(nT )at the sampling instants by using the inverse z-transform. The time function y(t) is then determined as
∞
y(t) = y(nT )δ(t − nT ).
n=0
Method 1: Power series. This method involves dividing the denominator of Y (z) into the numerator such that a power series of the form
Y (z) = y0 + y1z−1 + y2z−2 + y3z−3 + . . .
is obtained. Notice that the values of y(n) are the coefficients in the power series.
Example 6.6
Find the inverse z-transform for the polynomial
Y (z) |
z2 + z |
. |
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= z2 |
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− 3z + 4 |
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146 SAMPLED DATA SYSTEMS AND THE Z-TRANSFORM
Solution
Dividing the denominator into the numerator gives
1 + 4z−1 + 8z−2 + 8z−3
z |
2 |
− 3z + 4 |
z |
2 |
+ z |
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2 |
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z |
− 3z + 4 |
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4z − 4 |
16z−1 |
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4z |
− |
12 |
+ |
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8 |
− |
16z−1 |
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8 − 24z−1 + 32z−2 |
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8z−1 − 32z−2 |
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8z−1 − 24z−2 + 32z−3 |
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. . . |
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and the coefficients of the power series are |
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y(0) |
= 1, |
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y(T ) |
= 4, |
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y(2T ) = 8, |
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y(3T ) = 8, |
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. . . . |
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The required sequence is
y(t) = δ(t) + 4δ(t − T ) + 8δ(t − 2T ) + 8δ(t − 3T ) + . . .
Figure 6.15 shows the first few samples of the time sequence y(nT ).
Example 6.7
Find the inverse z-transform for Y (z) given by the polynomial
z
Y (z) = . z2 − 3z + 2
y(t)
8 |
|
4 |
|
1 |
time |
0 T 2T 3T
Figure 6.15 First few samples ofy(t)
THE z-TRANSFORM |
147 |
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Solution |
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Dividing the denominator into the numerator gives |
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z−1 + 3z−2 + 7z−3 + 15z−4 |
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z2 − 3z + 2 |
z |
3 |
2z− |
1 |
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z |
− |
+ |
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3 |
− |
2z−1 |
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3 − 9z−1 + 6z−2 |
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7z−1 − 6z−2 |
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7z−1 − 21z−2 + 14z−3 |
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15z−2 − 14z−3 |
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15z−2 − 45z−3 + 30z−4 |
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. . . |
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and the coefficients of the power series are |
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y(0) |
= 0 |
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y(T ) |
= 1 |
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y(2T ) = 3 y(3T ) = 7 y(4T ) = 15
. . . .
The required sequence is thus
y(t) = δ(t − T ) + 3δ(t − 2T ) + 7δ(t − 3T ) + 15δ(t − 4T ) + . . . .
Figure 6.16 shows the first few samples of the time sequence y(nT ).
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.
y(t)
15
7
3
1
time
0 |
T |
2T |
3T |
4T |
Figure 6.16 First few samples ofy(t)
148 SAMPLED DATA SYSTEMS AND THE Z-TRANSFORM
Method 2: Partial fractions. Similar to the inverse Laplace transform techniques, a partial fraction expansion of the function Y (z) can be found, and then tables of known z-transforms 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 the function Y (z)/z and then multiply the partial fractions by z to obtain a z term in the numerator.
Example 6.8
Find the inverse z-transform of the function
Y (z) =
z
(z − 1)(z − 2)
Solution
The above expression can be written as
Y (z) |
= |
1 |
= |
A |
+ |
B |
. |
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z |
(z |
− |
1)(z |
− |
2) |
z |
− |
1 |
z |
− |
2 |
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The values of A and B can be found by equating like powers in the numerator, i.e.
A(z − 2) + B(z − 1) ≡ 1.
We find A = −1, B = 1, giving |
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Y (z) |
−1 |
1 |
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z |
= z |
− |
1 |
+ z |
− |
2 |
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or |
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Y (z) = |
−z |
+ |
z |
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z 1 |
z |
− |
2 |
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− |
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From the z-transform tables we find that |
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y(nT ) = −1 + 2n |
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and the coefficients of the power series are |
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y(0) |
= 0, |
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y(T ) |
= 1, |
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y(2T ) = 3, |
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y(3T ) = 7, |
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y(4T ) = 15, |
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. . . |
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so that the required sequence is |
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y(t) = δ(t − T ) + 3δ(t − 2T ) + 7δ(t − 3T ) + 15δ(t − 4T ) + . . . . |
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Example 6.9 |
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Find the inverse z-transform of the function |
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Y (z) = |
1 |
. |
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(z |
− |
1)(z |
− |
2) |
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THE z-TRANSFORM |
149 |
Solution
The above expression can be written as
Y (z) |
= |
1 |
= |
A |
+ |
B |
+ |
C |
. |
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z |
z(z |
− |
1)(z |
− |
2) |
z |
z |
− |
1 |
z |
− |
2 |
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The values of A, B and C can be found by equating like powers in the numerator, i.e.
A(z − 1)(z − 2) + Bz(z − 2) + C z(z − 1) ≡ 1
or
A(z2 − 3z + 2) + Bz2 − 2Bz + C z2 − C z ≡ 1,
giving
A + B + C = 0,
−3A − 2B − C = 0,
2A = 1.
The values of the coefficients are found to be A = 0.5, B = −1 and C = 0.5. Thus,
Y (z) |
1 |
1 |
1 |
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z |
= 2z − z |
− |
1 + |
2(z |
− |
2) |
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or |
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Y (z) = |
1 |
− |
z |
+ |
z |
. |
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2 |
z |
− |
1 |
2(z |
− |
2) |
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Using the inverse z-transform tables, we find |
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y(nT ) |
= |
a |
− |
1 |
+ |
2n |
= |
a |
− |
1 |
+ |
2n−1 |
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2 |
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where |
a = |
10, |
n = |
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0, |
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/2, |
n |
0, |
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the coefficients of the power series are |
= |
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y(0) |
= 0 |
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y(T ) |
= 0 |
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y(2T ) = 1 |
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y(3T ) = 3 |
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y(4T ) = 7 y(5T ) = 15
. . . ,
and the required sequence is
y(t) = δ(t − 2T ) + 3δ(t − 3T ) + 7δ(t − 4T ) + 15δ(t − 5T ) + . . . .
The process of finding inverse z-transforms is aided by considering what form is taken by the roots of Y (z). It is useful to distinguish the case of distinct real roots and that of multiple order roots.