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 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 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 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

additional zeros at

when

, or

poles at

when

. Strictly speaking, every digital filter has an equal number of poles and zeros when those at

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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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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Introduction to Digital Filters
Pole-Zero Analysis
Graphical Phase Response Calculation