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The normalized analog, low pass, Butterworth filter

As a lead-in to digital filter design we look at a simple analog low pass filter - an RC filter, and its frequency response.

Example: Find the transfer function, Ha(s), impulse response, ha(t), and frequency response, Ha(jΩ), of the following system.

696_Normalized analog low pass.png

Solution This is a voltage divider. The transfer function is given by

1604_Normalized analog low pass1.png

Taking the inverse Laplace transform gives the impulse response,

1435_Normalized analog low pass2.png

The frequency response is

406_Normalized analog low pass3.png

The cut-off frequency is Ωc = 5 rad/sec. The gain at Ω = 0 is 1.

1490_Normalized analog low pass8.png

 

The MATLAB plots of frequency response of Ha(jΩ) = 5 ( jW + 5) are shown below. We use the function fplot. The analog frequency, Omega (Ω), extends from 0 to ∞; however, the plots cover the range 0 to 6π rad/sec.

 

subplot(2, 1, 1), fplot('abs(5/(5+j*Omega))', [-6*pi, 6*pi], 'k');

xlabel ('Omega, rad/sec'), ylabel('|H(Omega)|'); grid; title ('Magnitude')

%

subplot(2, 1, 2), fplot('angle(5/(5+j*Omega))', [-6*pi, 6*pi], 'k');

xlabel ('Omega, rad/sec'), ylabel('Phase of H(Omega)'); grid; title ('Phase')

499_Normalized analog low pass4.png

 

If we adjust the values of the components R and C so that 1 RC = 1, we would have Ha(s)=    1/+ 1 which is a normalized filter with cut-off frequency Ωc = 1 rad/sec and gain of 1 at Ω = 0.

Such a normalized LP filter could be transformed to another LP filter with a different cut-off frequency of, say, Ωc = 10 rad/sec by the low pass to low pass transformation s → (s /10). The transfer function then becomes

2215_Normalized analog low pass5.png

which still has a dc gain of 1. The gain of this filter could be scaled by a multiplier, say, K, so that

2367_Normalized analog low pass6.png

which has a dc gain of K and a cut-off frequency of Ωc = 10 rad/sec.

The frequency response of the normalized filter Ha(s) = 1/(s +1) is Ha(jΩ) = 1 ( jW +1) . The

corresponding MATLAB plots are shown below using the function plot. Omega is a vector,

consequently we use "./" instead of "/" etc.

 

Omega = -6*pi: pi/256: 6*pi; H = 1./(1.+ j .*Omega);

subplot(2, 1, 1), plot(Omega, abs(H), 'k');

xlabel ('Omega, rad/sec'), ylabel('|H(Omega)|'); grid; title ('Magnitude')

subplot(2, 1, 2), plot(Omega, angle(H), 'k');

xlabel ('Omega, rad/sec'), ylabel('Phase of H(Omega)'); grid; title ('Phase')

876_Normalized analog low pass7.png

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