Page 157 - Analog and Digital Filter Design

1 54 Analog and Digital Filter Design

Equations to convert from the minimum inductor lowpass model to the
minimum inductor highpass filter are given by:

Note that the inductor and capacitor values in this circuit are given by the recip-
rocal of the inductor and capacitor values, respectively, in the normalized
lowpass. Previously the capacitor values were determined by the reciprocal of
the lowpass inductor values. The reason for the change is that now the position
of capacitors in the lowpass model coincides with the position of capacitors in
the highpass model. The same is true for inductors.

Active Highpass Filters

Active filters use pole and zero locations from the frequency response’s transfer
function. Tables of pole and zero values were given in Chapter 3. The opera-
tional amplifier (op-amp), the “active” part of the circuit, buffers one stage from
the next so there is no interaction. Each stage can therefore be designed to
provide the frequency response of one pair of complex poles, or a single real
pole, or sometimes both. When all the stages are connected in series the overall
response is that which is desired.

A lowpass to highpass translation is required to find the highpass normalized
pole and zero locations. Normalized lowpass response pole and zero locations
are used as a starting point in the following formulae:

For a real pole at O, the imaginary component is zero (w = 0 in the above equa-
tion). Simplifying the equation gives oHP = l/~, which means that the highpass
pole is located at the reciprocal of the pole location in the lowpass prototype.
Similarly, for a zero on the (imaginary) frequency axis, the real component is
zero, so CT= 0 in the above equation. Simplifying the equation gives oLHp = l/wL,
which means that the highpass zero is located at the reciprocal of the zero loca-
tion in the lowpass prototype.

So, what does the S-plane diagram look like now? In Chapter 4 an example of
a fourth-order lowpass filter was given. This had a Butterworth response, with
poles on a unit circle at -0.9239 f j0.3827 and -0.3827 k j0.9239. Since the poles