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150 SAMPLED DATA SYSTEMS AND THE Z-TRANSFORM

Case I: Distinct real roots. When Y (z) has distinct real roots in the form

Y (z) =

N (z)

,

(z − p1)(z − p2)(z − p3) . . . (z − pn )

then the partial fraction expansion can be written as

Y (z) =

A1

+

A2

+

A3

+ . . . +

An

z − p1

z − p2

z − p3

z − pn

and the coefficients Ai can easily be found as

Ai = (z − pi ) Y (z)|z= pi

for i = 1,2,3, . . . , n.

Example 6.10

Using the partial expansion method described above, find the inverse z-transform of

Y (z) =

z

.

(z

−

1)(z

−

2)

Solution

Rewriting the function as

Y (z)

=

A

+

B

,

z

z

−

1

z

−

2

we find that

A

=

(z

−

1)

1

(z

1)(z

2)

1

−

−

z

=

1

B

=

(z

−

2)

(z

1)(z

2)

2

−

−

z

=

Thus,

Y (z) =

z

+

z

z

−

1

z

−

2

and the inverse z-transform is obtained from the tables as y(nT ) = −1 + 2n ,

which is the same answer as in Example 6.7.

=−1,

=1.

Example 6.11

Using the partial expansion method described above, find the inverse z-transform of

Y (z)

=

−

z2 + z

−

.

(z

0.5)(z

−

0.8)(z

1)

THE z-TRANSFORM

151

Solution

Rewriting the function as

Y (z)

=

A

+

B

+

C

z

z

−

0.5

z

−

0.8

z

−

1

we find that

A

=

(z

−

0.5)

z + 1

=

10,

(z

1)

0.5

−

0.5)(z

−

0.8)(z

−

z

=

z

+

1

B

=

(z

−

0.8)

(z

0.5)(z

0.8)(z

1)

0.8

= −

30,

−

−

−

z

=

C

=

(z

−

1)

z

+

1

=

20.

(z

0.5)(z

0.8)(z

1

−

−

−

1) z

=

Thus,

Y (z) =

10z

−

30z

+

20z

z

−

0.5

z

−

0.8

z

−

1

The inverse transform is found from the tables as

y(nT ) = 10(0.5)n − 30(0.8)n + 20

The coefficients of the power series are

y(0)

=

0

y(T )

=

1

y(2T ) = 3.3

y(3T ) = 5.89

. . .

and the required sequence is

y(t) = δ(t − T ) + 3.3δ(t − 2T ) + 5.89δ(t − 3T ) + . . . .

Case II: Multiple order roots. When Y (z) has multiple order roots of the form

Y (z) =

N (z)

,

(z − p1)(z − p1)2(z − p1)3 . . . (z − p1)r

then the partial fraction expansion can be written as

Y (z) =

λ1

+

λ2

+

λ3

+

. . . +

λr

z − p1

(z − p2)2

(z − p1)3

(z − p1)r

and the coefficients λi can easily be found as

λr−k

1

dk

.

(6.29)

= k! dz

[(z

−

pi ) (X(z)/z)]

z

=

pi

152

SAMPLED DATA SYSTEMS AND THE Z-TRANSFORM

Example 6.12

Using (6.29), find the inverse z-transform of

Y (z)

=

+

z2 + 3z − 2

−

2)2

.

(z

5)(z

−

0.8)(z

Solution

Rewriting the function as

Y (z)

z2 + 3z − 2

A

B

C

D

E

z

= z(z

+

5)(z

−

0.8)(z

−

2)2

= z

+ z

+

5

+ z

−

0.8

+

(z

−

2)

+ (z

−

2)2

we obtain

A

=

z

z(z

z2 + 3z − 2

2)

2

0

=

5

(

−2

4

=

0.125,

+

5)(z

−

0.8)(z

−

z

×

−

0.8)

×

=

B

=

(z

+

5)

z2 + 3z − 2

=

8

=

0.0056,

z(z

+

5)(z

−

0.8)(z

−

2)

z

=−

5

−

5

×

(

−

5.8)

×

49

C

=

(z

−

0.8)

z

2

+ 3z − 2

=

1.04

=

0.16,

z(z

2)

2

0.8

0.8

5.8

1.14

+

5)(z

−

0.8)(z

−

z

×

×

=

E

=

(z

−

2)2

z2 + 3z − 2

=

8

=

0.48,

z(z

+

5)(z

−

0.8)(z

−

2)

z

=

2

2

×

7

×

1.2

D

=

d

z2 + 3z − 2

dz

z(z

5)(z

0.8)

2

+

−

z

=

+ −

+

−

2

2

+

0.29.

=

3

4.2z

2

4z)

2

+ 8.4z −

4)

= −

z(z

5)(z

0.8(2z

3)

(z

−

3z

− 2)(3z

z

+

2

=

We can now write Y (z) as

Y (z) = 0.125 +

0.0056z

+

0.016z

−

0.29z

+

0.48z

z

+

5

z

−

0.8

(z

−

2)

(z

−

2)2

The inverse transform is found from the tables as

y(nT ) = 0.125a + 0.0056(−5)n + 0.016(0.8)n − 0.29(2)n + 0.24n(2)n ,

where

a =

0,

n =

0.

1,

n

0,

=

Method 3: Inversion formula method. The inverse z-transform can be obtained using the inversion integral, defined by

y(nT ) =

2π j

Y (z)zn−1dz.

(6.30)

1

r

PULSE TRANSFER FUNCTION AND MANIPULATION OF BLOCK DIAGRAMS

153

Using the theorem of residues, the above integral can be evaluated via the expression

y(nT )

=

at

[residues of Y (z)zn−1].

(6.31)

poles of

[Y (z)zk−1]

If the function has a simple pole at z = a, then the residue is evaluated as

[residue]|z=a

= [(z − a)Y (z)zn−1].

(6.32)

Example 6.13

Using the inversion formula method, find the inverse z-transform of

Y (z) =

z

.

(z

−

1)(z

−

2)

Solution

Using (6.31) and (6.32):

y(nT )

zn

zn

1

2n

= z

1 + z

2 = −

+

−

2 z

=

−

1 z

=

which is the same answer as in Example

6.9.

Example 6.14

Using the inversion formula method, find the inverse z-transform of

Y (z) =

z

.

(z

−

1)(z

−

2)(z

−

3)

Solution

Using (6.31) and (6.32),

y(nT )

zn

zn

zn

1

2n

3n

.

= (z

2)(z

3)

1

+ (z

1)(z

3)

2

+

(z

1)(z

2)

3

=

2

−

+ 2

−

−

z

=

−

−

z

=

−

−

z

=

6.3PULSE TRANSFER FUNCTION AND MANIPULATION OF BLOCK DIAGRAMS

The pulse transfer function is the ratio of the z-transform of the sampled output and the input at the sampling instants.

Suppose we wish to sample a system with output response given by

y(s) = e*(s)G(s).

(6.33)

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