I have the following signal and need to derive its Fourier transform:

enter image description here

I just started learning Fourier transform and don't know how to solve this kind of question. I know though how to derive the Fourier transform for: $$cos(2\pi f_0t)$$

$$\mathcal{F}\{cos(2\pi ft)\}=\int_{-\infty}^{\infty}cos2\pi f_0*e^{-j2\pi ft}dt $$ $$ =\int_{-\infty}^{\infty}\frac{1}{2}(e^{-j2\pi(f-f_0)t}dt +\int_{-\infty}^{\infty}\frac{1}{2}(e^{-j2\pi(f+f_0)t}dt$$ $$ =\frac{1}{2}(\delta(f-f_0)+\delta(f+f_0)) $$

UPDATE: I have used the Euler's relation to rewrite the signal like this: $$g_1(t)=\frac{1}{2}Ae^je^{j2\pi ft}+\frac{1}{2}Ae^{-j}e^{-j2\pi ft}$$ How do continue from here?

  • 2
    \$\begingroup\$ All that changes is the limits of your integral, instead of doing the transform between -ve infinity and +ve infinity just go between where the signal is non-zero. This is because integrals are linear and you can break them up into sections and add them. \$\endgroup\$
    – jramsay42
    Feb 14, 2018 at 0:09
  • 5
    \$\begingroup\$ Also this is really a mathematics, how to integrate question, not an electrical engineering question. \$\endgroup\$
    – jramsay42
    Feb 14, 2018 at 0:10
  • \$\begingroup\$ You can write the Fourier transform as a sum of sines and cosines, then given the limits and an even or odd functions one part of the integral may goes away. \$\endgroup\$ Feb 14, 2018 at 0:21
  • \$\begingroup\$ If no answer this may get migrated to math.stackexchange.com \$\endgroup\$
    – user105652
    Feb 14, 2018 at 1:16

1 Answer 1


Are you familiar with the multiplication/convolution rules for FT's?

The FT of a product of functions is equal to the convolution of the FT of each function $$ F(a \times b) = F(a) * F(b)$$

The FT of an infinite cosine is straightforward.

The FT of a rectangular pulse is also a very common standard transform, worth learning and committing to heart if you haven't already.

The convolution of those two transforms is the transform you seek. As the FT of the cosine is merely two spectral lines, the convolution is easy enough to be done by inspection.


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