# Fourier Transform of a modulated signal?

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.

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

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 $$\nu \pm \omega$$.