One-Zero

Figure 10.1: Signal flow graph for the general one-zero filter

Figure 10.1 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:

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:

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

Figure 10.2: Signal flow graph for the general one-zero filter written as

This is the frequency response of the filter in Fig.10.1.

Figure 10.3: Family of frequency responses of the one-zero filter

The term $b_0 - b_1$ may be interpreted as an order 0 filter section (``volume knob'') with gain $(b_0 - b_1)$. This gain control is in parallel (i.e., summed) with a first-order section $b_1(1 + z^{-1})$ having an amplitude response that varies sinusoidally from 2 to 0 as frequency goes from 0 to half the sampling rate (which is the behavior of our simplest lowpass filter example analyzed in Chapter 1). Figure 10.2 illustrates this interpretation of the general one-zero filter. Thus, the general one-zero filter can be interpreted as a digital volume knob in parallel with the series combination of another gain control and the simplest lowpass filter.

Representing the general one-zero filter as a volume control in parallel with a scaled two-point average is useful for visualizing the possible range of frequency responses. For completeness, however, 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:

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

A plot of $G(\omega)$ and $\Theta(\omega)$ for $b_0 = 1$ and various values of $b_1$, is given in Fig.10.3. 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 $z=0$.