Fourier Transform for $x(t)\cos(w_0t)$

Question

Fourier transform table gives $\cos(w_0t) \rightarrow \pi*[\delta(w+w_0)+\delta(w-w_0)]$, but why $x(t)\cos(w_0t) \rightarrow \ 0.5*[X(w+w_0)+X(w-w_0)]$? If $\cos(w_0t) \rightarrow \ π*[δ(w+w_0)+δ(w-w_0)]$, then $x(t)\cos(w_0t)$ Fourier transform shouldn't be the convolution of $X(w)$ and $π*[δ(w+w_0)+δ(w-w_0)]$, which is $π*[X(w+w_0)+X(w-w_0)]$, but why is $0.5*[X(w+w_0)+X(w-w_0)]$?

Answer 1

I may have misunderstood your question, but first let me give my answer to my initial understanding, since I think it will help with what you're actually asking.

Let us answer the question "why is the Fourier transform of $e^{i \omega_0 t}$ equal to $2 \pi \delta(\omega-\omega_0)$ instead of just $\delta(\omega-\omega_0)$?" After all, if you pretend that you're just writing the coefficients of $e^{i \omega_0 t}$ in a sum of $e^{i \omega t} dt$, the coefficient of $e^{i \omega_0 t}$ should be "$1/dt$" (whatever that means) and the others should be zero.

The answer lies in the normalization of the Fourier transform. There are a bunch of these in common usage. For work focusing on Fourier analysis and/or distribution theory, I like the unitary normalization:

$$F(\omega)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty e^{-i \omega t} f(t) dt \\ f(t)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty e^{i \omega t} F(\omega) d \omega.$$

For any other normalization, you multiply one of these by a constant and divide the other by that constant. In particular there is a normalization which is quite popular in differential equations:

$$F(\omega)=\int_{-\infty}^\infty e^{-i \omega t} f(t) dt \\ f(t)=\frac{1}{2 \pi} \int_{-\infty}^\infty e^{i \omega t} F(\omega) d \omega$$

which is popular because it turns differentiation in the $t$ domain into multiplication by $i \omega$ in the $\omega$ domain (rather than $c i \omega$ for some $c \neq 1$). This is the normalization you are using, and you can check by the definition of the Dirac delta that $e^{i \omega_0 t}=\frac{1}{2 \pi} \int_{-\infty}^\infty e^{i \omega t} (2 \pi \delta(\omega-\omega_0)) d \omega$. Hence the Fourier transform of $e^{i \omega_0 t}$ under this normalization is $2 \pi \delta(\omega-\omega_0)$.

So that's where the $\pi$ comes from. If you keep the normalization the same, then you are right: the FT of $x(t) \cos(\omega_0 t)$ would be $\pi(X(\omega_0)+X(-\omega_0))$. My guess would be that the normalization of interest changed somewhere in your reference(s) (or else someone just started forgetting normalization constants entirely).

Answer 2

You already know that

$$\mathcal{F}\{\cos(w_0t)\} = \pi[\delta(w+w_0)+\delta(w-w_0)]$$

Considering the convolution theorem

The FT of the product of two functions is the convolution of the FT of each one. That is $$\mathcal{F}\{x(t)y(t)\}=\frac{1}{2\pi}X(\omega)*Y(\omega)$$

Also consider the shift property of $\delta(\omega-\omega_0)$ when it convolves with some function/signal $X(\omega)$:

$$\delta(\omega\pm\omega0)*X(\omega)=X(\omega\pm\omega_0)$$

Based on the above three, the appropriate scale is $0.5$. It seems you are ignoring the correct scaling ($\frac{1}{2\pi}$) when you use the convolution theorem. Note that this scale exists in the $\omega$ domain. in the $f$ domain the scale is $1$:

$$\mathcal{F}\{x(t)y(t)\}=X(f)*Y(f)$$