what is the convolution of $sin(bx)$ and $e^{-a|x|}$?

Question

my textbook says that $F[f*g] = F[f] \cdot F[g]$ but what is $F[sinbx]$? It doesn't exist right? So how should I solve it?

$F$ is fourier transform

Answer 1

Let $f(t)=\sin(bx)$ and $g(t)=e^{-a|x|}$ with Fourier transform $$F(\xi)=i\pi\left[\delta(\xi+b)-\delta(\xi-b)\right] \quad\text{and}\quad G(\xi)=\frac{2 a}{a^2 + \xi^2} $$ Thus for $(f*g)(t)=h(t)$ the Fourier transform $\mathcal F\left\{h(t)\right\}=H(\xi)$ is $$\begin{align} H(\xi)=F(\xi)G(\xi)&=i\pi\left[\delta(\xi+b)-\delta(\xi-b)\right]G(\xi)\\ &=i\pi\left[G(-b)\delta(\xi+b)-G(b)\delta(\xi-b)\right]\tag A\\ &=i\pi\left[\delta(\xi+b)-\delta(\xi-b)\right]\frac{2 a}{a^2 + b^2}\tag B\\ &=F(\xi)\frac{2 a}{a^2 + b^2} \end{align} $$ where we use the property of Dirac delta function $ \alpha (t) \delta (t-t_0) = \alpha (t_0) \delta (t-t_0)$ in (A) and $G(-b)=G(b)=\frac{2 a}{a^2 + b^2}$ in (B).

So we have that $H(\xi)=\frac{2 a}{a^2 + b^2}F(\xi)$ and then $ h(t)=\frac{2 a}{a^2 + b^2}f(t) $ that is $$ (f*g)(t)=\frac{2 a}{a^2 + b^2}\sin(bt) $$

Answer 2

I do not see a reason to use the Fourier transform in this case, in particular, as you point out, the sine function is not square integrable so one should be careful with the convolution rule (even though the calculation works in this case, in the sense of tempered distributions).

By definition (I leave it to you to figure out what is going on in each step), $$ \begin{aligned} \bigl(\sin(b\cdot)*e^{-a|\cdot|})(t) &= \int_{-\infty}^{+\infty}\sin(b(t-\tau))e^{-a|\tau|}\,d\tau\\ &=\int_{-\infty}^{+\infty}\bigl[\sin(bt)\cos(b\tau)-\cos(bt)\sin(b\tau)\bigr]e^{-a|\tau|}\,d\tau\\ &=2\sin(bt)\int_0^{+\infty}\cos(b\tau)e^{-a\tau}\,d\tau\\ &=\frac{2a}{a^2+b^2}\sin(bt). \end{aligned} $$

Answer 3

Intuitively, the Fourier spectrum of a sinusoid is a line spectrum, i.e. all energy condensed at two frequencies, $\pm b$. Multiply that by the spectrum of the other function and you keep the same two lines just with a different amplitude, so that the convolved signal is also a sinusoid.

This can be confirmed by the following reasoning: compute the integrals of $\sin(bt)$ and $\cos(bt)$ with the other signal, hopefully giving two finite values $S$ and $C$. Then by shifting the sinusoid by a phase $bu$ (the integrand is $\sin(b(u-t)\exp(-a|t|$)), the convolved signal will be the combination $S\cos(bu)+C\sin(bu)$, another sinusoid.

Note that due to the signal parities, $S=0$.