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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
k=0,2,\dots,M_1-1
$

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
k=0,2,\dots,M_2-1.
$

Figure 3.12 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 Toolbox 
                     % or Octave Forge
where B and A are as given in Fig.3.11. The pole-zero plot utility zplane is contained in the Matlab Signal Processing Toolbox, and in the Octave Forge collection. A similar plot is produced by
sys = tf2sys(B,A,1); 
pzmap(sys);
where these functions are both in the Matlab Control Toolbox and in Octave. (Octave includes its own control-systems tool-box functions in the base Octave distribution.)

Figure 3.12: Pole-Zero diagram of the example filter $ y(n) = x(n) + 0.5^3 x(n-3) - 0.9^5 y(n-5)$.
\includegraphics[width=\textwidth]{eps/epz}


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written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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