Page 96 - Analog and Digital Filter Design

Poles and Zeroes 93

R
V,=V,-
R + l/sC

Divide by R, top and bottom

I
L<, = V,
1 + li sCR

V,, sCR
=
Alternatively, - ___
K l+sCR
r:, S
By dividing top and bottom by CR, this becomes, - ~
=
I/CR+s'

Here the pole occurs at S = -l/CR, the same as for the lowpass filter, but now
there is an S in the numerator. The equation is equal to zero when S = 0, so
engineers refer to this as having a zero at the origin. The zero placed at infinity
in the lowpass filter example has moved to the origin in the highpass case. This
zero at S = 0 is intuitively correct since there is no output at zero frequency
or DC.

The transfer function is similar to the lowpass filter case, except for the S that
has appeared in the numerator. Multiplying a frequency domain equation by S
means that the time domain equation must be differentiated. This is expressed
mathematically as:

s.F(s)-f(O) = m, wheref(0) is the initial time domain condition.
di

However, analysis of a step response requires division by S, so the equation for
a highpass transfer function with a step input is simplified to:

In the Laplace Transform a/(s + b), b = UCR, and a = 1. so in the time domain:

The filter responds immediately to the fast rising edge of the step input,
giving a step output followed by an exponential decay. This is shown in
Figure 3.9.