Request
I am very new to this so please bear with me. I cannot duplicate the answer in the book although I do get very close. This tells me that my method is correct but I am making another kind of error — perhaps in my trigonometry or integral?
Given:
Find the convolution of $f(t)=sin(\omega t)$ and $g(t)=sin(\omega t)$.
$$h(t)=(f*g)(t)=\int_0^t f(\tau)g(t-\tau)d\tau$$
My Solution:
$$h(t)=sin(\omega t)*sin(\omega t)=\int_0^t sin(\omega \tau)\cdot sin(\omega(t-\tau)d\tau$$
Apply the trigonometric product formula identity…
$$=-\int_0^t sin(\omega \tau)\cdot sin(\omega(\tau-t)d\tau=-\frac{1}{2}\int_0^t[cos(\omega t)+cos(2\omega\tau-\omega t)]d\tau$$
$$=-\frac{1}{2}\left[\tau cos(\omega t)-\frac{sin(2\omega\tau-t\omega)}{2\omega}\right]_0^t$$
A bit messy just do a little house cleaning before applying bounds…
$$=\left[\frac{sin(2\omega\tau-t\omega)}{4\omega}-\frac{\tau cos(\omega t)}{2}-\right]_0^t=\frac{sin(\omega t)}{4\omega}-\frac{t cos(\omega t)}{2}-0+0$$
$$h(t)=\frac{sin(\omega t)}{4\omega}-\frac{t cos(\omega t)}{2}$$
Answer in Text:
$$h(t)=\frac{sin(\omega t)}{2\omega}-\frac{t cos(\omega t)}{2}$$
Question
Where did I go wrong?
Best Answer
The problem is in the evaluation of the term
$$\frac{\sin(2\omega\tau-\omega t)}{4\omega}{\Huge|}_0^t$$
Note that you get contributions for $\tau=t$ and for $\tau=0$:
$$\frac{\sin(2\omega\tau-\omega t)}{4\omega}{\Huge|}_0^t=\frac{\sin(\omega t)}{4\omega}-\frac{\sin(-\omega t)}{4\omega}=\frac{\sin(\omega t)}{2\omega}$$