Joseph_deep asked . 2021-12-07

How to get from discrete time impulse response vector to plot of frequency spectrum

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

dsp , filter , bzt , fft , plot , frequency response , vector

Expert Answer

Prashant Kumar answered . 2024-12-21 17:13:51

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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