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Graphical Phase Response Calculation

The phase response is almost as easy to evaluate graphically as is the amplitude response:

\begin{eqnarray*}
\Theta(\omega) &\isdef & \angle \left\{g \frac{(1-q_1e^{-j\omega T})(1-q_2e^{-j\omega T})\cdots(1-q_Me^{-j\omega T})}{(1-p_1e^{-j\omega T})(1-p_2e^{-j\omega T})\cdots(1-p_Ne^{-j\omega T})}\right\}\\
&=& \angle{g}+\angle e^{j(N-M)\omega T}\frac{(e^{j\omega T}-q_1)(e^{j\omega T}-q_2)\cdots(e^{j\omega T}-q_M)}{(e^{j\omega T}-p_1)(e^{j\omega T}-p_2)\cdots(e^{j\omega T}-p_N)}\\
&=& \angle{g}+(N-M)\omega T\\
& & + \angle(e^{j\omega T}-q_1)+\angle(e^{j\omega T}-q_2)+\cdots+\angle(e^{j\omega T}-q_M)\\
& & - \angle(e^{j\omega T}-p_1)-\angle(e^{j\omega T}-p_2)-\cdots-\angle(e^{j\omega T}-p_N)
\protect
\end{eqnarray*}

If $ g$ is real, then $ \angle g$ is either 0 or $ \pi $ . Terms of the form $ e^{j\omega T}-z$ can be interpreted as a vector drawn from the point $ z$ to the point $ e^{j\omega T}$ in the complex plane. The angle of $ e^{j\omega T}-z$ is the angle of the constructed vector (where a vector pointing horizontally to the right has an angle of 0). Therefore, the phase response at frequency $ f$ Hz is again obtained by drawing lines from all the poles and zeros to the point $ e^{j2\pi f T}$ , as shown in Fig.8.4. The angles of the lines from the zeros are added, and the angles of the lines from the poles are subtracted. Thus, at the frequency $ \omega$ the phase response of the two-pole two-zero filter in the figure is $ \Theta(\omega) =
\theta_1+\theta_2-\theta_3-\theta_4$ .

Figure 8.4: Measurement of phase response from a pole-zero diagram.
\includegraphics{eps/kfig2p15}

Note that an additional phase of $ (N - M)2\pi fT$ radians appears when the number of poles is not equal to the number of zeros. This factor comes from writing the transfer function as

$\displaystyle H(z) = gz^{(N-M)}\frac{(z-q_1)(z-q_2)\cdots(z-q_M)}{(z-p_1)(z-p_2)\cdots(z-p_N)}
$

and may be thought of as arising from $ N - M$ additional zeros at $ z=0$ when $ N > M$ , or $ M - N$ poles at $ z=0$ when $ M>N$ . Strictly speaking, every digital filter has an equal number of poles and zeros when those at $ z=0$ and $ z=\infty$ are counted. It is customary, however, when discussing the number of poles and zeros a filter has, to neglect these, since they correspond to pure delay and do not affect the amplitude response. Figure 8.5 gives the phase response for this two-pole two-zero example.

Figure 8.5: Phase response obtained from Fig.8.4 for positive frequencies. The point of the phase response corresponding to the arrows in that figure is marked by a heavy dot. For real filters, the phase response is odd ( $ \Theta (-\omega ) = -\Theta (\omega )$ ), so the curve shown here may be reflected through 0 and negated to obtain the plot for negative frequencies.
\includegraphics{eps/kfig2p16}


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``Introduction to Digital Filters with Audio Applications'', by Julius O. Smith III, (September 2007 Edition).
Copyright © 2014-03-23 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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