The z-transform - Continuous and Discrete Time Signals and Systems - page 598

or alternatively as

Im{ z }

2

− 1 )(1 − z 2 z − 1 ) (1 − z m z − 1 )

(1 − z 1 z

H ( z ) = z m − n

1

p n z − 1 ) .

(13.39b)

(1 − p 1 z − 1 )(1 − p 2 z − 1 ) (1 −

Re{ z }
12 −2 −1

−1
−2

Example 13.13
Determine the poles and zeros of the following LTID systems:

(a)

Im{ z }

z
z 2 − 3 z + 2 ;

H 1 ( z ) =

(i)

1

0.5

1
( z − 0 . 1)( z − 0 . 5)( z + 0 . 2) ;

(ii) H 2 ( z ) =

Re{ z }

−1 −0.5 0.5 1

−0.5
−1

z 2 (2 z − 1 . 5)
( z + 0 . 4)( z − 0 . 5) 2 ;

(iii) H 3 ( z ) =

(b)

z 2 + 0 . 7 z + 1 . 6
( z 2 − 1 . 2 z + 1)( z + 0 . 3) .

(iv) H 4 ( z ) =

Im{ z }

1

0.5

Solution
(i) H 1 ( z ) =

z
z 2 − 3 z + 2

z
( z − 1)( z − 2) .

Re{ z }

=

−1 −0.5 0.5 1

−0.5
−1 There is one zero, at z = 0, and two poles, at z = 1 and 2.

H 2 ( z ) = 1

(c)

( z − 0 . 1)( z − 0 . 5)( z + 0 . 2) .
There is no zero, but there are three poles, at z

(ii)

Im{ z }

= 0 . 1 , 0 . 5, and − 0 . 2.

1

H 3 ( z ) = z 2 (2 z − 1 . 5)

( z + 0 . 4)( z − 0 . 5) 2 .
There are three zeros, at z

(iii) 0.5

Re{ z }

= 0, 0, and 0.75. There are three poles, at z =− 0 . 4, −1 −0.5 0.5 1

−0.5
−1

0.5, and 0.5.

H 4 ( z ) = ( z − 0 . 5)( z + 1 . 2)
(( z − 0 . 6) 2 + 0 . 8 2 )( z + 0 . 3)
= ( z − 0 . 5)( z + 1 . 2)

(iv)

(d)

Fig. 13.6. Pole and zero plots
for transfer functions in Example
13.13. Plot (a) corresponds to
part (i) of Example 13.13; plot
(b) corresponds to part (ii); plot
(c) corresponds to part (iii); and
plot (d) corresponds to part
(iv). Also note that plot (c)
includes double zeros at z

( z − 0 . 6 + j0 . 8)( z − 0 . 6 − j0 . 8)( z + 0 . 3) .
There are two zeros, at z = 0 . 5 and − 1 . 2 . There are three poles, at z = 0 . 6 −

j0 . 8 , 0 . 6 + j0 . 8, and − 0 . 3.

The poles and zeros of the above four systems are shown in Fig. 13.6. In the
plot, marks the position of a pole and • marks the position of a zero.

= 0

and double poles at z = 0.5.

13.6.2 Determination of impulse response

The impulse response h [ k ] of an LTID system can be obtained by calculating
the inverse z-transform of the transfer function H ( z ). Example 13.14 explains
the steps involved in determining the impulse response.