Question
Hello, I'm very new to DSP concepts and signal processing in MATLAB so please go easy on me. I have the following code that I've begun working on which is the BZT of an RLC filter. I want to take the DFT of my impulse response y[n] and end up with a magnitude plot Y(f). I know that I can use the fft() and plot() functions for this. I've seen examples using linspace/logspace vectors which I'll need to define before being able to plot this thing, and some such things as NFFTs (to enable a faster fft?) but I'm not really sure how to put it all together here. Any help/insight would be really appreciated. CODE: R=0.1; % resistor L=0.0016; % inductor C=0.0016; % capacitor resF=1/sqrt(L*C); % resonant frequency sampRate=44100; % sampling rate (Hz) T=1/sampRate; % sample period prewarpF=(2/T)*tan(resF*T/2); % prewarp factor a=(2/T)*R*C; b=(2/(T*prewarpF))^2; e0=a/(a+b+1); e1=(2-2*b)/(a+b+1); e2=(1+b-a)/(a+b+1); N=2048; % number of taps x=zeros(1,N); % input vector x(5) = 1; % impulse at delta(n-5) y=zeros(1,N); % output vector for n = 5 : N y(n) = e0*x(n) - e0*x(n-2) - e1*y(n-1) - e2*y(n-2); end
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Expert Answer
Prashant Kumar
PhD Expert
Answered Aug 11, 2026
Since you have a discrete-time sequence with data sampled at 44.1 kHz, your frequency vector will run from -22050 Hz to 22050 Hz for a double-sided spectrum, or just 0 to 22050 for a single-sided.
Do you have the Signal Processing Toolbox? If so, then use freqz() with the coefficients of your difference equation. From this line:
y(n) = e0*x(n) - e0*x(n-2) - e1*y(n-1) - e2*y(n-2);
You difference equation has the following form as a rational transfer function.
A = [1 e1 e2]; B = [e0 -e0];
So the frequency response is
[H,F] = freqz(B,A,[],44100);
plot(F,20*log10(abs(H)));
xlabel('Hz');
You can see from the above that the response is lowpass.
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