Fourier transform of symbolic function

Question

soup 2022-9-10

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编辑： Paul 2022-9-10

Hi. I would like to see the magnitude and phase spectrum of a symbolic function. I tried using the 'fourier' function on a basic symbolic expression, but it didn't work.

syms t w;

x=sin(2*t);

X=fourier(x);

fplot(abs(X)) // Nothing in this figure

Doing the transform manually works:

T=10;

fX=int(x* exp(1j*w*t), t, -T, T);

fplot(abs(X)/T) // This works

Please can someone let me know how to see the phase and magnitude responses using the 'fourier' function ?

Many thanks

-s

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Answer 1

Paul 2022-9-10

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编辑：Paul 2022-9-10

在 MATLAB Online 中打开

Hi soup,

The Fourier transform of x is

syms t w;

x=sin(2*t);

fourier(x)

ans =

fplot() ignores Dirac delta functions. Because there is nothing else to plot, fplot() is empty

The "manual transform"

T = 10; % not sure why T = 10?

simplify(int(x* exp(1j*w*t), t, -T, T),100)

ans =

isn't the Fourier transform of x. Rather, it's the Fourier transform of the x multiplied by a rectangular window of length 2T.

sympref('FourierParameters',[1 1]);

simplify(fourier(x*rectangularPulse(-T,T,t)),100)

ans =

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Paul 2022-9-10

编辑：Paul 2022-9-10

As you're seeing, the Fourier transform integral doesn't converge for a periodic signal. However, we can define the Fourier transform of a periodic signal as a Dirac delta impulse train scaled by the signal's Fourier series coefficients, which is exactly what we see as the Fourer transform of a sin function. See this link, for example, for further discussion. Matlab's fourier function knows how to deal with basic periodic functions, like sin and complex exponentials.

Any plot of a signal that contains a Dirac delta will be artifcial. The concept of "magnitude" doesn't really apply to a Dirac delta. Can't really plot it, just have to make up some symbol, like a vertical arrow, to signify where the Dirac delta sits on the real line, as is done in the linked reference.

soup 2022-9-10