The z-transform - Continuous and Discrete Time Signals and Systems - page 612

used to plot the poles and zeros in the complex z-plane. In terms of Eq. (13.48),
the syntaxes for these functions are given by

>> [Z,P,K] = tf2zp(B,A); % Calculate poles and zeros

>> zplane(Z,P); % plot poles and zeros,

where the input variables B and A are defined as follows:

A=[ a n a n − 1 ... a 0 ] and B=[ b m b m − 1 ... b 0 ] .

They are obtained from the transfer function given in Eq. (13.48). The vector
Z contains the location of the zeros, vector P contains the location of the poles,
while K returns a scalar providing the gain of the numerator.

Example 13.21
For the z-transfer function

2 z (3 z + 17)
( z − 1)( z 2 − 6 z + 25) ,

H ( z ) =

compute the poles and zeros and give a sketch of their locations in the complex
z-plane.

Solution
The M ATLAB code to determine the location of zeros and poles is listed below.
The explanation follows each instruction in the form of comments.

>> B = [0, 6, 34, 0]; % Coefficients of the numerator N(z)
>> A = [1, -7, 31, -25]; % Coefficients of the denominator D(z)
>> [Z,P,K] = tf2zp(B,A) % Calculate poles and zeros
>> zplane(Z,P) % plot poles and zeros

The returned values are given by

Z = [0, -5.6667],
P = [3.0000+4.0000j 3.0000-4.0000j 1.0000] andK=6.

The transfer function H ( z ) can therefore be expressed as follows:

z ( z + 5 . 6667)
( z − (3 + j4))( z − (3 − j4))( z − 1)

H ( z ) = 6

− 1 )
(1 − (3 + j4) z − 1 )(1 − (3 − j4) z − 1 )(1 − z − 1 ) .

− 1 (1 + 5 . 6667 z

z

= 6

The pole-zero plot for H ( z ) is shown in Fig. 13.11.