Page 88 - Analog and Digital Filter Design

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Poles and Zeroes

Frequency Response and the S-Plane
The transfer function F(w) is frequency dependent. However, this response is
complex because both the amplitude and the phase depend on frequency. For
this reason, the frequency response is described in terms of s, rather than fre-
quency, w. Hence, the transfer function becomes F(s), where S = CF * jw.

The transfer function becomes an infinite value and is referred to as a “pole”
when the denominator becomes zero. The value of s that makes the denomina-
tor zero could be a real value or a complex value, depending on the transfer
function. Thus poles occur at certain values of s = ofjo. In some cases (when
jw = 0), the value of s = o alone. Invariably, the value of CF is negative. If the
value of jo is not zero, there are a pair of values for s and these are s = o+jo
and s = 6- jw.

The transfer function becomes zero when the numerator becomes zero. The
value of s needed to make the numerator equal zero is referred to as the “zero”
location. Responses like Bessel, Butterworth, and Chebyshev have a numerator
value of 1, hence there are no zeroes. These filters are referred to as “All-Pole’’
filters. Responses like Inverse Chebyshev and Cauer have numerators that
depend on powers of s, and hence have zeroes. Invariably, the numerator zero
locations occur at values of s = 0 + jw. That is, the zero is on the (imaginary)
frequency axis.

Before looking at specific responses, whether they are Butterworth, Chebyshev,
Inverse Chebyshev, or others, I will give a brief outline of how poles, zeroes, and
the transfer function are related. This section assumes knowledge of complex
numbers; the S-plane (showing pole-zero diagrams) is introduced without expla-
nation but is described in more detail later.

Let’s start by considering the transfer function of a filter. Say, for example, a
simple second-order filter is formed from a series inductor followed by a shunt
capacitor. The transfer function of this filter can be expressed algebraically as:
k
T(s) = where S = 05 jo.
as’+bs+c
The values of k, u, and b determine the shape of the transfer function. If a
Butterworth response is required, k = 1, a = 1, c = 1, and b = a. If w = 1. .s =
jw = jl. Considering the other terms in the equation, s’ =j = -1 and bs = jfi.
So the transfer function becomes:

1
T(w) = -0.7071L-90”
-I+ jfi+l- jfi
The output is 0.7071, or -3dB, and the output phase is 90” behind the
input.