I need to find the output $y(t)$ given the input $x(t)=\cos(t)$, and impulse response $h(t)=u(t)$ using the Fourier Transform.
I know that convolution in time domain is multiplication in the frequency domain, but since $\cos(t)$ has a Fourier Transform which involves the delta function, I am unable to figure out how to perform the multiplication in the frequency domain.
The answer in the time domain is $y(t)=\sin(t)$. I don't know how to get this. If someone could show the working, would be highly appreciated!
From an engineering perspective, if you assume that
$$\delta(s-a) \cdot 0 = 0$$
and thus
$$\delta(s-a)\cdot\delta(s-b) = 0 \quad a \ne b$$
then
$$\begin{align*}y(t) &= h(t) \ast x(t)\\ \\ &= u(t)\ast \cos(t)\\ \\ &= \mathscr{F}^{-1}\left\{\mathscr{F}\left\{u(t)\right\}\cdot\mathscr{F}\left\{\cos(t)\right\}\right\}\\ \\ &= \mathscr{F}^{-1}\left\{\dfrac{1}{2}\left[\dfrac{1}{i\pi s}+\delta(s)\right]\cdot\dfrac{1}{2}\left[\delta\left(s+\dfrac{1}{2\pi}\right)+\delta\left(s-\dfrac{1}{2\pi}\right)\right]\right\}\\ \\ &= \mathscr{F}^{-1}\left\{\dfrac{1}{2\pi i s}\cdot\dfrac{1}{2}\left[\delta\left(s+\dfrac{1}{2\pi}\right)+\delta\left(s-\dfrac{1}{2\pi}\right)\right]\right\}\\ \\ &= \int_{-\infty}^\infty \dfrac{e^{2\pi i st}}{2\pi i s}\cdot\dfrac{1}{2}\left[\delta\left(s+\dfrac{1}{2\pi}\right)+\delta\left(s-\dfrac{1}{2\pi}\right)\right] ds\\ \\ &= \dfrac{1}{2}\left[\dfrac{e^{2\pi i \frac{-1}{2\pi}t}}{2 \pi i \frac{-1}{2\pi}}+ \dfrac{e^{2\pi i \frac{1}{2\pi}t}}{2 \pi i \frac{1}{2\pi}}\right] \\ \\ &= \dfrac{e^{it}-e^{-it}}{2i}\\ \\ y(t) &= \sin(t) \\ \end{align*}$$
Technically, instead of using a hand-waving $\delta(s-a)\delta(s-b) = 0 $ for $a\ne b$, I suppose I should have also done the two integrals with products of delta functions:
$$ \int_{-\infty}^{\infty} \dfrac{e^{2\pi i st}}{4} \delta(s)\delta\left(s+\dfrac{1}{2\pi}\right)ds $$
$$ \int_{-\infty}^{\infty} \dfrac{e^{2\pi i st}}{4} \delta(s)\delta\left(s-\dfrac{1}{2\pi}\right)ds $$
However, using the properties of the delta function, these will come out to be 0.