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Figure B.1: Signal flow graph for the general one-zero filter
$ y(n) = b_0x(n) + b_1 x(n - 1)$.
\begin{figure}\input fig/kfig2p17.pstex_t

Figure B.1 gives the signal flow graph for the general one-zero filter. The frequency response for the one-zero filter may be found by the following steps:

Difference equation: & $y(n) = b_0x(n) + b_1x(n - 1)$...
...requency response: & $H(e^{j\omega T}) = b_0 + b_1e^{-j\omega T}$

By factoring out $ e^{-j\omega T/2}$ from the frequency response, to balance the exponents of $ e$, we can get this closer to polar form as follows:

H(e^{j\omega T}) &=& b_0 + b_1 e^{-j\omega T}\\
&=& (b_0 - ... T/2)\\
&=& (b_0 - b_1) + e^{-j\pi f T} 2b_1\cos(\pi f T)

Figure B.2: Family of frequency responses of the one-zero filter
$ y(n) = x(n) + b_1 x(n - 1)$
for various values of $ b_1$. (a) Amplitude response. (b) Phase response.
\begin{figure}\input fig/kfig2p19.pstex_t

We now apply the general equations given in Chapter 7 for filter gain $ G(\omega)$ and filter phase $ \Theta(\omega)$ as a function of frequency:

H(e^{j\omega T}) &=& b_0 + b_1e^{-j\omega T}\\
&=& b_0 + b_1...
...left[\frac{-b_1 \sin(\omega T)}{b_0 + b_1 \cos(\omega T)}\right]

A plot of $ G(\omega)$ and $ \Theta(\omega)$ for $ b_0 = 1$ and various real values of $ b_1$, is given in Fig.B.2. The filter has a zero at $ z = -b_1/b_0 = -b_1$ in the $ z$ plane, which is always on the real axis. When a point on the unit circle comes close to the zero of the transfer function the filter gain at that frequency is low. Notice that one real zero can basically make either a highpass ( $ b_1/b_0 < 0$) or a lowpass filter ( $ b_1/b_0 > 0$). For the phase response calculation using the graphical method, it is necessary to include the pole at $ z=0$.

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