Butterworth.html

Supplement to my lectures on Linear Systems

R. Brigola, Georg-Simon-Ohm-Hochschule Nürnberg

Solution for the Exercise on the Design a Specified Butterworth Lowpass Filter

As an application example for a LTI-System design a Butterworth Filter (S. Butterworth, 1930):

At first, calculate the necessary order of the filter and the cutoff angular frequency according to its specification

as we have done in classroom.

( Reference for example: R. Brigola Fourieranalysis, Distributionen und Anwendungen, Vieweg, pg. 208 )

Specification:

To be concrete, calculate H for K=1, passband edge at 3 kHz, stopband edge at 5 kHz, minimal passband gain 0.9, maximal stopband gain 0.1.

Then, calculate the n zeroes , ... , with negative real parts of

, i.e .

Give the resulting frequency characteristics H of the Butterworth filter of order n and angular cutoff frequency (bandwidth).

Plot the frequency characteristics, the phase shift and delay of the filter.

H is given by

Solution:

At first calculate the necessary order of the filter and its cutoff frequency from the given specification

> restart:

> K:=1.0;

> H1:=0.9;

> H2:=0.1;

> w1:=3000*(2*Pi);

> w2:=5000*(2*Pi);

> ord:=evalf(ln((K^2/H1^2-1)/(K^2/H2^2-1))/(2*ln(w1/w2)));

> n:=floor(ord)+1; # Order of the necessary Butterworth Filter

> Omega:=sqrt(w1*exp(-ln(K^2/H1^2-1)/(2*n))*w2*exp(-ln(K^2/H2^2-1)/(2*n)));

> evalf(Omega/(2*Pi)); # Cutoff frequency in Hz

>

H can be represented with the Butterworth polynomial in the form

The zeroes of Q with negative real parts are easily calculated by hand and give the zeroes of P

> zero:=array(1..n);

The zeroes of P are known to be ( you can compare them with the Maple solutions of Q(z)=0. Extract the zeroes with negative real part )

> for k from 1 to n do

> zero[k]:=evalc(exp(I*((2*k-1)*Pi/(2*n)+Pi/2)));

> end do;

Therefore the polynomial P is given by

> P:=unapply(simplify(evalf(expand(product((s-zero[m]),m=1..n)))),s);

>

Now the frequency characteristics of the Butterworth lowpass is

> H:=unapply(K/P(I*omega/Omega),omega);

>

> plot(abs(H(t*Omega)),t=0..2,thickness=3,labels=["cutoff frequency scaled to 1",gain]); # plot the filter gain

> plot(abs(H(t*(2*Pi))),t=0..9000,colour=blue,thickness=3,labels=["frequency in Hz",gain]); # plot the filter gain

The gain can also be represented as

> H_abs:=unapply(K/sqrt(1+(w/evalf(Omega))^(2*n)),w);

> plot(H_abs(t*Omega),t=0..2,thickness=3,labels=["cutoff frequency scaled to 1",gain]);

The frequency response can also be represented as a curve in the complex plane as follows.

Observe, that this is a qualitative graphics only unless you compute complex values for certain frequencies,

to see where you are on the curve for these.

> with(plots):complexplot(H(w),w=0..20*Omega,thickness=3);

We zoom the curve for high frequencies

> complexplot(H(w),w=15*Omega..30*Omega,thickness=3);

This shows approximately, that the curve near the origin crosses again the imaginary axis and approximates the origin from the left lower quadrant.

A mathematically well-educated engineer can see from that behavier of the curve - it begins for w=0 on the positive real axis,

goes in the mathematically negative direction to the origin, while crossing exactly as many quadrants of the complex plane as the filter order n (here n=6) indicates.

These facts tell that engineer, that the filter is stable (no resonance frequencies exist, a bounded input generates also a bounded output, it has the so-called BIBO-stability)

Remark for your future: A filter with rational frequency response is unstable, when the denominator polynomial has coefficients less or equal to zero !

Filters with frequency responses as above with respect to their start point for w=0, their orientation and number of quadrant crossings are stable.

Phase shift of the filter

> assume(w,real);

> phi:=w->-sum(argument(I*w-Omega*zero[m]),m=1..n);