How do I compute the fourier transform of a signal:

$$Ax(t)\cos(\omega t) $$

where both $A$ and $\omega$ are constants. I tried using the fact the function was even and using Euler's formula but I ended up with $2$ equations, both of which equal zero.

  • $\begingroup$ Do you know the convolution theorem (for Fourier transforms)? $\endgroup$
    – Fabian
    Commented Oct 22, 2018 at 16:04
  • $\begingroup$ Is $x(t)$ even? Unless it is, how can you know that the whole function is even? $\endgroup$
    – David K
    Commented Oct 22, 2018 at 16:05
  • $\begingroup$ @Fabian the definition of it yes. However, I'm not sure how to apply it in this scenario. $\endgroup$ Commented Oct 22, 2018 at 17:04

1 Answer 1


The convolution theorem reads $$ \sqrt{2\pi} \mathcal{F} ( f g) =\mathcal{F}(f) * \mathcal{F}(g) $$ for any functions $f(t)$, $g(t)$ and where $$ (F*G)(\nu) = \int d\mu F(\nu-\mu) G(\mu) = \int d\mu F(\mu) G(\nu-\mu) .$$ Apply this to the functions $$f(t) = x(t)$$ and $$g(t) = A \cos(\omega t).$$

Let us denote the Fourier transform of $f(t)$ by $F(\nu)= X(\nu)$. The Fourier transform of $g(t)$ can be explicitly obtained as $$G(\nu) = \frac{\sqrt{2\pi}A}{2}\left[ \delta(\nu-\omega) + \delta(\nu+\omega)\right]\;.$$

Performing the convolution, we obtain that $$\mathcal{F}[A x(t) \cos(\omega t)](\nu)=\frac{A}{2}[X(\nu-\omega)+X(\nu+\omega)].$$ We see that the carrier frequency $\omega$ shifts the frequencies $\nu$ of the amplitude modulation to $\nu \pm \omega$.


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