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Figure B.3: Signal flow graph for the general one-pole filter
$ y(n) = b_0 x(n) - a_1 y(n - 1).$
\begin{figure}\input fig/kfig2p20.pstex_t

Fig.B.3 gives the signal flow graph for the general one-pole filter. The road to the frequency response goes as follows:
Figure B.4: Family of frequency responses of the one-pole filter
$ y(n) = x(n) - a_1 y(n - 1)$
for various real values of $ a_1$. (a) Amplitude response. (b) Phase response.
\begin{figure}\input fig/kfig2p21.pstex_t
Difference equation: & $y(n) = b_0 x(n) - a_1 y(n-1)...
...$H(e^{j\omega T}) = \displaystyle\frac{b_0}{1+a_1e^{-j\omega T}}$
The one-pole filter has a transfer function (hence frequency response) which is the reciprocal of that of a one-zero. The analysis is thus quite analogous. The frequency response in polar form is given by
G(\omega) &=& \frac{\vert b_0\vert}{\sqrt{[1 + a_1 \cos(\omega...
... + a_1 \cos(\omega T)}\right], & b_0<0 \\
\end{array} \right..
A plot of the frequency response in polar form for $ b_0 = 1$ and various values of $ a_1$ is given in Fig.B.4. The filter has a pole at $ z = -a_1$, in the $ z$ plane (and a zero at $ z$ = 0). Notice that the one-pole exhibits either a lowpass or a highpass frequency response, like the one-zero. The lowpass character occurs when the pole is near the point $ z = 1$ (dc), which happens when $ a_1$ approaches $ -1$. Conversely, the highpass nature occurs when $ a_1$ is positive. The one-pole filter section can achieve much more drastic differences between the gain at high frequencies and the gain at low frequencies than can the one-zero filter. This difference is achieved in the one-pole by gain boost in the passband rather than attenuation in the stopband; thus it is usually desirable when using a one-pole filter to set $ b_0$ to a small value, such as $ 1 -
\left\vert a_1\right\vert$, so that the peak gain is 1 or so. When the peak gain is 1, the filter is unlikely to overflow.B.1 Finally, note that the one-pole filter is stable if and only if $ \left\vert a_1\right\vert < 1$.
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