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I need to find the bandwidth the signal $x = 50 \operatorname{sinc}(125(t-10))^2$.

(Where $\operatorname{sinc}(x)=\dfrac{\sin (\pi x)}{\pi x}$ [normalized] or $\operatorname{sinc}(x)=\dfrac{\sin x}{x}$ [unnormalized])

How would I go about doing this in MATLAB?

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  • $\begingroup$ I have not requested for closer of this, but do you really want to understand the theory being finding bandwidth or do you really care about HOW to do it in MATLAB? If it is later, the question will be closed. If it is former, edit your question. $\endgroup$ – Dipan Mehta Apr 17 '12 at 8:32
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That actually would be a great interview questions. Let's derive it from scratch:

  1. $\text{sinc}(x)$: The Fourier Transform of a $\text{sinc}(x)$ is a rectangle. The normalized $\text{sinc}()$ function results in a bandwidth of 1 Hz (from -0.5Hz to 0.5Hz)
  2. $t-10$: The $(t-10)$ term is irrelevant to the bandwidth: time shift is equivalent to multiplying with $e^{-j\omega t_0}$, where $t_0$ is time shift, in the frequency domain. That changes the phase but not the amplitude or bandwidth
  3. $125*t$: that is a time compression which results in frequency stretching. Since we compress time by 125 the bandwidth increases by 125, i.e. to 125 Hz (from -62.5 Hz to 62.5 Hz)
  4. $(\cdot)^2$: Squaring is multiplication in the time domain which corresponds to convolution in the frequency domain. The convolution of two equal rectangles is a triangle. The bandwidth doubles to 250 Hz (from -125 Hz to 125Hz). In general the bandwidth of the multiplication of two arbitrary signals is the sum of the individual bandwidths. That's a simple consequence of the multiplication laws for sine and cosine: you get only sum and difference frequencies.
  5. $50*(\cdot)$: that's a simple scale factor that doesn't change the shape of phase or amplitude.

So overall we would expect the spectrum to be of triangular shape (on a double linear plot) going from -125 Hz to +125 Hz with a total bandwidth of 250 Hz.

Writing the Matlab code for this is not as simple as it looks. The code itself is trivial, however you have to pick a sample rate and a proper time axis. In order to do this you need to assess the general shape of the signals first. For example the time domain signal technically goes from $-\infty$ to $+\infty$ and is centered around $t$ = 10s. t = 0:.01:5, really won't do the trick. So in essence you have to do the theoretical analysis before you can write the code (which tends to be a recurring theme in DSP :-))

% 1000 HZ sample rate, since we now the BW is smaller than 500 Hz
fs = 1000; 
% pick a good time length. Because of the time shift we need significantly more
% than 10s. Technically the sinc() has inifite length so we need to extend
% it far enough to cover the vast part of the energy AND we need to start
% at negative times, not at t = 0;
n = 2^16; % that's about 60s. Should be plenty
t = (-n/2:n/2-1)'/fs;   % time axis in seconds, symmetric around t = 0;
x = 50*sinc(125*(t-10)).^2;
% frequency axis:
f0 = fs/n;  % frequency resoution in Hz
faxis = (-n/2:n/2-1)'*f0;   % frequency axis in Hz seconds, symmetric around f = 0;
fx = fft(x);
plot(faxis,circshift(abs(fx),n/2));
title('Spectrum of 50*sinc(125*(t-10)).^2');
xlabel('Frequency in Hz');
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  • 1
    $\begingroup$ It is perhaps worth emphasizing that these calculations are for sinc defined as $\frac{\sin(\pi x)}{\pi x}$ and not for sinc defined as $\frac{\sin(x)}{x}$ (I have no idea how MATLAB defines sinc and whether the program is correct; I am speaking only of points 1-5 above). Also, some might say colloquially that this is a low pass signal of bandwidth $125$ Hz instead of the more detailed statement that the bandwidth extends from $-125$ Hz to $125$ Hz, and if they choose to give these details, they might choose to not include the words "a total bandwidth of $250$ Hz." But YMMV. $\endgroup$ – Dilip Sarwate Jan 15 '12 at 21:53

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