Pole-Zero Analysis Next  |  Prev  |  Up  |  Top  |  Index  |  JOS Index  |  JOS Pubs  |  JOS Home  |  Search

Pole-Zero Analysis

Since our example transfer function

$\displaystyle H(z) = \frac{1 + g_1 z^{-M_1}}{1 + g_2 z^{-M_2}}

(from Eq. (3.4)) is a ratio of polynomials in $ z$, and since every polynomial can be characterized by its roots plus a scale factor, we may characterize any transfer function by its numerator roots (called the zeros of the filter), its denominator roots (filter poles), and a constant gain factor:

$\displaystyle H(z) = g \frac{(1-q_1z^{-1})(1-q_2z^{-1})\cdots(1-q_{M_1}z^{-1})}{(1-p_1z^{-1})(1-p_2z^{-1})\cdots(1-p_{M_2}z^{-1})}

The poles and zeros for this simple example are easy to work out by hand. The zeros are located in the $ z$ plane at

$\displaystyle q_k = - g_1^{\frac{1}{M_1}} e^{j2\pi\frac{k}{M_1}}, \quad

where we assume $ g_1>0$, and the poles are similarly given by

$\displaystyle p_k = - g_2^{\frac{1}{M_2}} e^{j2\pi\frac{k}{M_2}}, \quad

Figure 3.8 gives the pole-zero diagram of the specific example filter $ y(n) = x(n) + 0.5^3 x(n-3) - 0.9^5 y(n-5)$. There are three zeros, marked by `O' in the figure, and five poles, marked by `X'. Because of the simple form of digital comb filters, the zeros (roots of $ z^3+0.5^3$) are located at 0.5 times the three cube roots of -1 ( $ -e^{2k\pi/3}, k=0,1,2$), and similarly the poles (roots of $ z^5+0.9^5$) are located at 0.9 times the five 5th roots of -1 ( $ -e^{k2\pi/5}, k=0,\dots,4$). (Technically, there are also two more zeros at $ z=0$.) The matlab code for producing this figure is simply

[zeros, poles, gain] = tf2zp(B,A); % Matlab or Octave
zplane(zeros,poles); % Matlab Signal Processing Tool Box 
                     % or Octave Forge
where B and A are as given in Fig.3.7. The pole-zero plot utility zplane is contained in the Matlab Signal Processing Tool Box, and in the Octave Forge collection. A similar plot is produced by
sys = tf2sys(B,A,1); 
where these functions are both in the Matlab Control Tool Box and in Octave. (Octave includes its own control-systems tool-box functions in the base Octave distribution.)

Figure 3.8: Pole-Zero diagram of the example filter $ y(n) = x(n) + 0.5^3 x(n-3) - 0.9^5 y(n-5)$.

Next  |  Prev  |  Up  |  Top  |  Index  |  JOS Index  |  JOS Pubs  |  JOS Home  |  Search

[How to cite this work] [Order a printed hardcopy]

``Introduction to Digital Filters with Audio Applications'', by Julius O. Smith III, (August 2006 Edition).
Copyright © 2007-02-02 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
CCRMA  [Automatic-links disclaimer]