Stretching the unit circle with rational functions

In section 8.3.4 we saw a simple way to turn a low-pass filter into a band-pass one. It is tempting to apply the same method to turn our Butterworth low-pass filter into a higher-quality band-pass filter; but to try to preserve the higher quality of the Butterworth filter we have to be more careful in the design of the transformation we use. In this section we will prepare the way to making the Butterworth band-pass filter by introducing the idea of rational transformations of the complex plane which preserve the unit circle.

This discussion is adapted from [], pp. 201-206 (I'm grateful to Julius Smith for this pointer). There the tansformation is carried out in continuous time, but here we have adapted the method to operate in discrete time, in order to make teh discussion self-contained.

The idea is to start with any filter with a transfer function as before:

$$H(Z) = { { (1 - {Q_1}{Z^{-1}}) \cdots (1 - {Q_j}{Z^{-1}}) } \over { (1 - {P_1}{Z^{-1}}) \cdots (1 - {P_k}{Z^{-1}}) } }$$

whose frequency response (the gain at a frequency $\omega$) is given by:

$$\vert H(\cos(\omega) + i \sin(\omega)) \vert$$

Now suppose we can find a rational function, $R(Z)$, which distorts the unit circle in some desirable way. For $R$ to be a rational function means that it can be written as a quotient of two polynomials (so, for example, the transfer function $H$ is a rational function). That $R$ sends points on the unit circle to other points on the unit circle is just the condition that $\vert R(Z)\vert = 1$ whenever $Z=1$. It can easily be checked that any function of the form,

$$R(Z) = \phi { {A_n}{Z^n} + {A_{n-1}}{Z^{n-1}} + \cdots +... ...0}{Z^n} + \overline{A_1}{Z^{n-1}} + \cdots + \overline{A_n} }$$

where the leading factor $\phi$ has unit magnitude (i.e., $\vert\phi\vert = 1$), and as before we use $\overline{A}$ to denote the complex conjugate of a complex number $A$. The same reasoning as in section 8.2.2 confirms that $\vert R(Z)\vert = 1$ whenever $Z=1$.

Once we have a suitable rational function $R$, we simply fabricate a new rational function,

$$J(Z) = H(R(Z))$$

The gain of the new filter $J$ at the frequency $\omega$ is then equal to

$$\vert J(\cos(\omega) + i \sin(\omega)) \vert = \vert H(\cos(\phi) + i \sin(\phi)) \vert$$

where we choose $\phi$ so that:

$$\cos(\phi) + i \sin(\phi) = R(\cos(\omega) + i \sin(\omega))$$

For example, suppose we start with a one-zero, one-pole low-pass filter:

$$H(Z) = { -1 - {Z^{-1}} } \over { g - {Z^{-1}} }$$

and apply the function

$$R(Z) = -{Z^2} = - { 1 \cdot {Z^2} + 0 \cdot Z + 0 } \over { 0 \cdot {Z^2} + 0 \cdot Z + 1 }$$

Geometrically, this choice of $R$ stretches the unit circle uniformly to twice its circumference and wraps it around itself twice. The points $1$ and $-1$ are both sent to the point $-1$, and the points $i$ and $-i$ are sent to the point $1$. The resulting transfer function is

$$J(Z) = {{ -1 - {Z^{-2}} } \over { g - {Z^{-2}} }} = ... ...1}}) } \over { (sqrt{g} - {Z^{-1}})(-sqrt{g} - {Z^{-1}}) }}$$

The pole-zero plots of $H$ and $J$ are shown in Figure 8.19. From a low-pass filter we ended up with a band-pass filter. The points $i$ and $-i$ which $R$ sends to $1$ (where the original filter's gain is highest) become points of highest gain for the new filter.

Figure 8.19: (a). A pole-zero plot for a one-pole, one-zero low-pass filter; (b). a plot for the resulting filter after the transformation $R(Z) = {Z^2}$. THe result is a band-pass filter with center frequency $\pi /2$.