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Graphical Computation of Amplitude Response from Poles and Zeros

Now consider what happens when we take the factored form of the general transfer function, Eq.$ \,$(8.2), and set $ z$ to $ e^{j\omega T}$ to get the frequency response in factored form:

$\displaystyle H(e^{j\omega T}) = g\frac{(1-q_1e^{-j\omega T})(1-q_2e^{-j\omega ...
...ga T})}{(1-p_1e^{-j\omega T})(1-p_2e^{-j\omega T})\cdots(1-p_Ne^{-j\omega T})}
$

As usual for the frequency response, we prefer the polar form for this expression. Consider first the amplitude response $ G(\omega) \isdeftext
\left\vert H(e^{j\omega T})\right\vert$.
$\displaystyle G(\omega)$ $\displaystyle =$ $\displaystyle \left\vert g\right\vert\frac{\left\vert 1-q_1e^{-j\omega T}\right...
... 1-p_2e^{-j\omega T}\right\vert\cdots\left\vert 1-p_Ne^{-j\omega T}\right\vert}$  
  $\displaystyle =$ $\displaystyle \left\vert g\right\vert
\frac{\left\vert e^{-jM\omega T}\right\ve...
...vert e^{j\omega T}-p_2\right\vert\cdots\left\vert e^{j\omega T}-p_N\right\vert}$  
  $\displaystyle =$ $\displaystyle \left\vert g\right\vert
\frac{\left\vert e^{j\omega T}-q_1\right\...
...\omega T}-p_2\right\vert\cdots\left\vert e^{j\omega T}-p_N\right\vert}
\protect$ (9.3)

In the complex plane, the number $ z = x + jy$ is plotted at the coordinates $ (x, y)$ [84]. The difference of two vectors $ u = x_1 + jy_1$ and $ v = x_2 + jy_2$ is $ u - v = (x_1 - x_2) + j(y_1
- y_2)$, as shown in Fig.8.1. Translating the origin of the vector $ u-v$ to the tip of $ v$ shows that $ u-v$ is an arrow drawn from the tip of $ v$ to the tip of $ u$. The length of a vector is unaffected by translation away from the origin. However, the angle of a translated vector must be measured relative to a translated copy of the real axis. Thus the term $ e^{j\omega T} - q_i$ may be drawn as an arrow from the $ i$th zero to the point $ e^{j\omega T}$ on the unit circle, and $ e^{j\omega T} - p_i$ is an arrow from the $ i$th pole. Therefore, each term in Eq.$ \,$(8.3) is the length of a vector drawn from a pole or zero to a single point on the unit circle, as shown in Fig.8.2 for two poles and two zeros. In summary:

$\textstyle \parbox{0.8\textwidth}{%
The frequency response magnitude (amplitude...
... by the product of lengths of
vectors drawn from the poles to $e^{j\omega T}$.}$

Figure 8.1: Treatment of complex numbers as vectors in a plane.
\begin{figure}\input fig/kfig2p12.pstex_t
\end{figure}

Figure 8.2: Measurement of amplitude response from a pole-zero diagram. A pole is represented in the complex plane by `X'; a zero, by `O'.
\begin{figure}\input fig/kfig2p13.pstex_t
\end{figure}

For example, the dc gain is obtained by multiplying the lengths of the lines drawn from all poles and zeros to the point $ z = 1$. The filter gain at half the sampling rate is the product of the lengths of these lines when drawn to the point $ z = -1$. For an arbitrary frequency $ f$ Hz, we draw arrows from the poles and zeros to the point $ z =
e^{j2\pi fT}$. Thus, at the frequency where the arrows in Fig.8.2 join, (which is slightly less than one-eighth the sampling rate) the gain of this two-pole two-zero filter is $ G(\omega)
= (d_1d_2)/(d_3d_4)$. Figure 8.3 gives the complete amplitude response for the poles and zeros shown in Fig.8.2. Before looking at that, it is a good exercise to try sketching it by inspection of the pole-zero diagram. It is usually easy to sketch a qualitatively accurate amplitude-response directly from the poles and zeros (to within a scale factor).

Figure 8.3: Amplitude response obtained by traversing the entire upper semicircle in Fig.8.2 with the point $ e^{j\omega T}$. The point of the amplitude obtained in that figure is marked by a heavy dot. For real filters, precisely the same curve is obtained if the lower half of the unit circle is traversed, since $ G(\omega ) = G( - \omega )$. Thus, plotting the response over positive frequencies only is sufficient for real filters.
\begin{figure}\input fig/kfig2p14b.pstex_t
\end{figure}


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Next: Graphical Phase Response Calculation

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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