I am a little bit unsure if I've set up the following problem correctly:
Consider the signal
$$f(t) = e^{-t}(\sin(5t) + \sin(3t) + \sin(t) + \sin(40t)) \quad 0 \leq t \leq \pi$$
Filter this signal with the filter:
$$h(t) = Ae^{- \alpha t} \quad t \geq 0$$ $$h(t) = 0 \quad t < 0$$
for $0 \leq t \leq \pi$. Try various values of $A = \alpha$ (starting with $A = \alpha = 10)$. Compare the filtered signal with the original signal.
Now, I have tried to set up this as follows (for $A = \alpha = 10$):
$$(f \ast h)(t) = \int_{0}^{\pi} 10 e^{-10(t - \tau) - \tau}(\sin(5 \tau) + \sin(3 \tau) + \sin(\tau) + \sin(40 \tau)) d \tau$$
$$= \int_{0}^{\pi} 10 e^{-10t + 9 \tau}(\sin(5 \tau) + \sin(3 \tau) + \sin(\tau) + \sin(40 \tau)) d \tau$$
Unfortunately I don't have MatLab available right now, so I tried running this through WolframAlpha, but was unable to get any computation made. To simplify it, I chose $t = 1$, and ran it through WolframAlpha again. But this yielded the result $5.95 \cdot 10^7$ which seems way too high.
So my question is - have I set up this problem in the wrong way? Any help will be greatly appreciated!
Check your assumptions for $f(t)$ and $h(t)$: the bounds on $t$ seem artificial, and in the case of $h$, arbitrary.
Note also that the convolution of two functions $f(t)$ and $h(t)$ in the context of an inverse Laplace transform of the product of their transforms $\hat{f}(s)$ and $\hat{h}(s)$, respectively, is
$$(f*h)(t) = \int_0^t d \tau \: f(\tau) h(t - \tau) $$
If I remove the bounds on $f(t)$ and $h(t)$, I get
$$ (f*h)(t) = A \frac{e^{-a t}+e^{-t} ((a-1) \sin (t)-\cos (t))}{a^2-2 a+2}+A \frac{3 e^{-a t}+e^{-t} ((a-1) \sin (3 t)-3 \cos (3 t))}{a^2-2 a+10}+A \frac{5 e^{-a t}+e^{-t} ((a-1) \sin (5 t)-5 \cos (5 t))}{a^2-2 a+26}+A \frac{40 e^{-a t}+e^{-t} ((a-1) \sin (40 t)-40 \cos (40 t))}{a^2-2 a+1601} $$