2
$\begingroup$

What is a nice way of finding the Fourier transform of the following signal $g(t)$? It is a cosine function with frequency $f_c$ and only appearing in the interval from $-\frac{T}{2}$ to $\frac{T}{2}$ so I think $g(t)=cos(2\pi f_c t) u(-t+\frac{T}{2}) u(t+\frac{T}{2})$. Since it is a product of three functions, then by using the convolution property I would have to convolve the Fourier transforms of the step functions first, the resulting transform function of which I would then have to convolve with the Fourier transform of the cosine function to get the Fourier transform of the product function $g(t)$. Not only does my method require two convolutions, I wasn't able to find a closed form expression for the convolution of the Fourier transforms of the two step functions. Does anyone have any idea of a better way of finding the Fourier transform of $g(t)$?

enter image description here

$\endgroup$
0
$\begingroup$

You are supposed to compute $G(f) = \int_{-T/2}^{T/2} \cos(2 \pi f_c t) e^{-2i \pi f t}dt$ directly, using $\cos(2 \pi f_c t) = \frac{e^{2i \pi f_c t}+e^{-2i \pi f_c t}}{2}$

A good alternative is to compute the Fourier transform of $h(t) = 1_{|t| < T/2}$ and use the modulation property of the Fourier transform $$FT[\cos(2 \pi f_c t) h(t)] = \frac{FT[e^{2i \pi f_c t} h(t)]+FT[e^{-2i \pi f_c t} h(t)]}{2} = \frac{H(f-f_c)+H(f+f_c)}{2}$$

Finally, what you mentioned needs the Fourier transform of distributions, because $FT[\cos(2 \pi f_c t)] = \frac{\delta(f-f_c)+\delta(f+f_c)}{2}$ where $\delta$ is the Dirac delta

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.