Find spectrum of $f(t-t_0)\sin(\omega_0(t-t_0))$

Question

If functions $f(t)$ and $F(j\omega)$ form a Fourier Transform pair, how do I find the spectrum of the function $$f(t-t_0)\sin(\omega_0(t-t_0))?$$

Answer 1

Hint: Let $g(t) = f(t)\sin(\omega_0 t) = \frac{1}{2i}( \; f(t)\exp(\omega_0 t) - f(t)\exp(-\omega_0 t) \; ) =: \frac{1}{2i}(\; g_1(t) - g_2(t) \;)$.

The Fourier transforms of $g_1$, $g_2$, and hence $g$ (and finally $g(t-t_0)$) follow from the rules outlined here.

Answer 2

Let $F(\omega) = \mathcal{L}(f(t)).$ Using the relations $$\mathcal{L}(f(t-t_0)) = F(\omega)e^{-j\omega t_0},\quad \mathcal{L}(f(t)e^{j\omega_0 t}) = F(\omega - \omega_0)$$ we obtain: $$\mathcal{L}(f(t-t_0)sin(\omega_0(t-t_0))) = e^{-j\omega t_0}\mathcal{L}(f(t)sin\:\omega_0t) = e^{-j\omega t_0}\mathcal{L}\left(f(t)\dfrac{e^{j\omega_0t}-e^{-j\omega_0t}}{2j}\right) = \dfrac{e^{-j\omega t_0}}{2j}(F(\omega-\omega_0) - F(\omega+\omega_0)).$$