Time-Domain Analysis and z Transform - Introduction to Digital Signal Processing and Filter Design - page 91

x=[1 zeros(1,19];
y=filter(b,a,x)

The output is

y = columns 1-7:

1 . 0000 0 . 4000 0 . 2100 0 . 1040 0 . 0521 0 . 0260 0 . 0130

columns 8-14:

0 . 0065 0 . 0033 0 . 0016 0 . 0008 0 . 0004 0 . 0002 0 . 0001

columns 15-20:

0 . 0001 0 . 0000 0 . 0000 0 . 0000 0 . 0000 0 . 0000

Example 2.30

Now we consider the use of the function residuez when the transfer function
has multiple poles. Let us choose G(z) from (2.44) and (2.45) and also reduce
it to a rational function in ascending powers of z − 1 as shown in (2.80):

z( 2 z 2 − 11 z + 12 )

G(z) =

(z − 1 )(z − 2 ) 3

−

2 z

− z

3 z

−

3 z

G(z) =

− 2 ) 3 + − 2 ) 2 + − 2 ) +

(2.79)

(z (z (z (z − 1 )

2 z 3 − 11 z 2

+ 12 z

=

z 4 − 7 z 3 + 18 z 2 − 20 z + 8

2 z − 1 11 z − 2 12 z − 3

− +

=

(2.80)

1 − 7 z − 1 + 18 z − 2 − 20 z − 3 + 8 z − 4

The program used to obtain the partial fraction expansion is

b=[0 2 -11 12];
a=[1 -7 18 -20 8];
[r,p,k]=residuez(b,a)

and the following is the output data we get:

r = 3 . 0000 + 0 . 0000 i

0 . 5000 − 0 . 0000 i
− 0 . 5000
−

3 . 0000