Convolution of cosine with exponential As part of an exercise, I'm trying to find the output of a cosine wave entering a low-pass filter by using a convolution integral.
The impulse response of the filter is $h(t) = \frac{1}{RC}\exp\left({-\frac{t}{RC}}\right)$
The solution should look as follows (if I'm not mistaken):
ouput = $\int_0^t\cos(wt)h(t-\tau) d\tau$
I was told that the output should be a solution of the form $A\cos(wt)$, but when I try doing this integral I get some fairly messy expressions... Could someone help me out with figuring out where I'm going wrong?
Thanks!
p.s. sorry for the lack of "math font", I'm fairly new to this site and can't quite figure it out....
 A: \begin{align*}
\int_0^t \cos(\omega t)h(t-\tau)\,d\tau&=\int_0^t \cos(\omega t){1\over RC}e^{-{t-\tau\over RC}}\\
&=\int_0^t \cos(\omega t){1\over RC}e^{-{t\over RC}}e^{\tau\over RC}\,d\tau\\
&=\cos(\omega t){1\over RC}e^{-{t\over RC}}\int_0^t e^{\tau\over RC}\,d\tau\\
&=\cos(\omega t){1\over RC}e^{-{t\over RC}} \left[RCe^{\tau\over RC}\right]_{\tau=0}^{\tau=t}\\
&=\cos(\omega t){1\over RC}e^{-{t\over RC}} RC\left[e^{t\over RC}-1\right]\\
&=\cos(\omega t)(1-e^{-{t\over RC}})
\end{align*}
so you get something close to your expectations, except that instead of a constant amplitude you get the decaying exponential above.
A: Shouldn't it be $\cos\left(\omega\tau\right)$?
Since this is a convolution, defined as $$\left(f*g\right)\left(t\right)=\intop_{-\infty}^{+\infty}f\left(\tau\right)g\left(t-\tau\right)d\tau $$
https://en.wikipedia.org/wiki/Convolution
A: The convolution is set up incorrectly. The independent variable of the cosine should be a tau, not a t. Furthermore the lower limit should be negative infinity. If this integral is performed, you will see that the answer is smaller than the input cosine by a factor 1/sqrt(1+(wRC)^2), and phase-shifted by the phase -atan(wRC/sqrt(1+(wRC)^2)).
Sincere apologies for my lack of equation formatting, but I see that this is the top Google result for "sinusoid convolved with decaying exponential," and currently a student might get the wrong result.
