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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\ome... ...gle(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.
\begin{figure}\input fig/kfig2p15.pstex_t \end{figure}

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.
\begin{figure}\input fig/kfig2p16.pstex_t \end{figure}

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Graphical Computation of Amplitude Response from Poles and Zeros