Request:
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 integration?
Given:
Find the convolution of $f(t)=sin(2t)$ and $g(t)=sin(t)$.
$$h(t)=(f*g)(t)=\int_0^t f(\tau)g(t-\tau)d\tau$$
My Solution:
Begin by inserting the functions into the formula and simplifying...
$$h(t)=sin(2t)*sin(t)=\int_0^t sin(2\tau)\cdot sin(t-\tau)d\tau=-\int_0^t sin(2\tau)\cdot sin(\tau-t)d\tau$$
We further simplify by applying the trigonometric product formula identity...
$$=-\frac{1}{2}\int_0^t[cos(\tau+t)-cos(3\tau-t)]d\tau$$
$$=-\frac{1}{2}\left[sin(\tau+t)-\frac{sin(3\tau-t)}{3}\right]_0^t$$
$$=-\frac{1}{2}\left[sin(2t)-\frac{sin(2t)}{3}-sin(t)+\frac{sin(-t)}{3}\right]$$
$$=-\frac{sin(2t)}{2}+\frac{sin(2t)}{6}+\frac{sin(t)}{2}+\frac{sin(t)}{6}$$
$$h(t)=-\frac{sin(2t)}{3}+\frac{2}{3}sin(t)$$
Answer in Text:
$$h(t)=-\frac{sin(2t)}{3}+\frac{2}{3}sin(t)$$
Question:
See bottom.
I solved it as I was typing so nothing to do here. Thank you for checking my post regardless.
$$h(t)=-\frac{sin(2t)}{3}+\frac{2}{3}sin(t)$$