Question

Create a 6th order Butterworth filter. Then change capacitor values and notice how the design specifications change. Create a plot where there is an ideal high performance filter (exact capacitor values) compared to a filter where low tolerance capacitors are used (the value of the capacitor drifts).

Answer 1

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I have done the design and plot in Matlab. Note that a 6th order Butterworth filter is formed by a cascade of 3 2nd order filters given by:

***** Matlab Code ******

R1 = 1000;                  % Resistance values in circuit
R2 = R1;

C1 = 20e-6;                 % Capacitance values in circuit
C2 = C1;

C1_minus = 15e-6;
C2_minus = C1_minus;

C1_plus = 25e-6;
C2_plus = C1_plus;

h = tf([1],[R1*R2*C1*C2 C2*(R1+R2) 1]);         % 2nd order transfer function
h_filter = h*h*h;                               % 6th order transfer function
[mag,phase,wout] = bode(h_filter);
mag = squeeze(mag);
wout = squeeze(wout);
mag = 20*log10(mag);
semilogx(wout,mag)
grid;
hold;

h_minus = tf([1],[R1*R2*C1_minus*C2_minus C2_minus*(R1+R2) 1]);     % filter with reduced capacitor values
h_minus_filter = h_minus*h_minus*h_minus;
[mag,phase,wout] = bode(h_minus_filter);
mag = squeeze(mag);
wout = squeeze(wout);
mag = 20*log10(mag);
semilogx(wout,mag)

h_plus = tf([1],[R1*R2*C1_plus*C2_plus C2_plus*(R1+R2) 1]);         % filter with increased capacitor values
h_plus_filter = h_plus*h_plus*h_plus;
[mag,phase,wout] = bode(h_plus_filter);
mag = squeeze(mag);
wout = squeeze(wout);
mag = 20*log10(mag);
semilogx(wout,mag)

legend('Ideal response','Capacitor values increased','Capacitor values decreased')
xlabel('Frequency(rad per sec)')
ylabel('Magnitude (dB)')

******* End of code ********

Output:

The differences in the filter plots can be clearly seen with changing capacitor values. However, note that the capacitor values have been changed by as much as 20% in the code, which may not happen practically. Thus, the change in the responses seen here is vast.