In this answer, Jim Clay writes:

... use the fact that $\mathcal F\{\cos(x)\} = \frac{\delta(w - 1) + \delta(w + 1)}{2}$ ...

The expression above is not too different from $\mathcal F\{{\cos(2\pi f_0t)\}=\frac{1}{2}(\delta(f-f_0)+\delta(f+f_0))}$.

I have been trying to obtain the later expression by using the standard definition of the Fourier transform $X(f)=\int_{-\infty}^{+\infty}x(t)e^{-j2\pi ft}dt$ but all I end up with is an expression so different from what's apparently the answer.

Here's my work:

\begin{align} x(t)&=\cos(2\pi f_0t)\\ \Longrightarrow \mathcal F\left\{x(t)\right\}&=\int_{-\infty}^{+\infty}\cos(2\pi f_0t)e^{-j2\pi ft}dt\\ &=\int_{-\infty}^{+\infty}\frac 12 \left(e^{-j2\pi f_0t}+e^{j2\pi f_0t}\right)e^{-j2\pi ft}dt\\ &=\frac{1}{2}\int_{-\infty}^{+\infty}\left(e^{-j2\pi f_0t}e^{-j2\pi ft}+e^{j2\pi f_0t}e^{-j2\pi ft}\right)dt\\ &=\frac{1}{2}\int_{-\infty}^{+\infty}\left(e^{-j2\pi t\left(f_0+f\right)}+e^{-j2\pi t\left(f-f_0\right)}\right)dt\\ &=\frac{1}{2}\left(\int_{-\infty}^{+\infty}\left(e^{-j2\pi t(f_0+f)}\right)dt+\int_{-\infty}^{+\infty}\left(e^{-j2\pi t(f-f_0)}\right)\right) dt \end{align}

This is where I'm stuck.

  • $\begingroup$ $F[x(t)]=1/2[ξ(ω-ω_0)+ξ(ω+ω_0)]$. $\endgroup$
    – Anthony
    Commented Aug 24, 2021 at 23:32
  • $\begingroup$ $F[x(t)]=1/2j[ξ(ω-ω_0)+ξ(ω+ω_0)]$ of sine. $\endgroup$
    – Anthony
    Commented Aug 25, 2021 at 2:09

2 Answers 2


Your work is OK except for the problem that the Fourier transform of $\cos(2\pi f_0 t)$ does not exist in the usual sense of a function of $f$, and we have to extend the notion to include what are called distributions, or impulses, or Dirac deltas, or (as we engineers are wont to do, much to the disgust of mathematicians) delta functions. Read about the conditions that must be satisfied in order for the Fourier transform $X(f)$ of the signal $x(t)$ to exist (in the usual sense) and you will see that $\cos(2\pi f_0 t)$ does not have a Fourier transform in the usual sense.

Turning to your specific question, once you understand that impulses are defined only in terms of how they behave as integrands in an integral, that is, for $a < x_0 < b$, $$\int_{a}^{b} \delta(x-x_0)g(x)\,\mathrm dx = g(x_0)$$ provided that $g(x)$ is continuous at $x_0$, then it is easier to deduce the Fourier transform of $$\cos(2\pi f_0 t) = \left.\left.\frac{1}{2}\right[e^{j2\pi f_0 t} + e^{-j2\pi f_0 t}\right]$$ by musing on the fact that $$\int_{-\infty}^\infty \delta(f-f_0)e^{j2\pi ft}\,\mathrm df = e^{j2\pi f_0t}$$ and so it must be that $\cos(2\pi f_0 t)$ is the inverse Fourier transform of $\displaystyle \left.\left.\frac{1}{2}\right[\delta(f-f_0) + \delta(f+f_0)\right]$.


Then just use a table of Fourier transform pairs to see that $\delta(t) \leftrightarrow 1$, and variable substitution ($f_1 = f+f_0$ and $f_2 = f-f_0$), to get what you need.

  • 4
    $\begingroup$ Which of course begs the question as to how the person who wrote down the table came up with the answer that is in the table. $\endgroup$ Commented Nov 8, 2013 at 13:48
  • $\begingroup$ @DilipSarwate :-) Now you're asking a much, much harder question. :-) $\endgroup$
    – Peter K.
    Commented Nov 8, 2013 at 13:48
  • 1
    $\begingroup$ See my answer for a version of the answer to the much harder question that might pass muster on this stackexchange if not on math.SE! $\endgroup$ Commented Nov 8, 2013 at 13:51
  • 1
    $\begingroup$ @DilipSarwate: you've got my +1 already. Thanks, nice answer. Agreed the math.SE dudes would be appalled. Thats OK, we're engineers. :-) $\endgroup$
    – Peter K.
    Commented Nov 8, 2013 at 13:55
  • $\begingroup$ dsp.stackexchange.com/questions/14990/… $\endgroup$
    – jomegaA
    Commented Feb 3, 2020 at 20:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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