# Filter signal through convolution

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!

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Why don't you use fourier analysis? You correctly tagged it as such. – akkkk Jan 21 '13 at 12:14
"which seems way too high" Too high for what? And why don't you try to solve those integrals instead of asking Wolfram or Matlab? – leonbloy Jan 21 '13 at 12:20
@leonbloy: Well, if you graph $f(t)$, the values for $f$ are nowhere near that magnitude - they stay below $5$ for the entire interval $0 \leq t \leq \pi$. So I doubt we are supposed to have such a drastic increase in magnitude. Also - I could of course do this by hand, but it would be very tedious work and take a long time. That is why I thought I'd just do it on a computer. – Kristian Jan 21 '13 at 12:27
@akkkk: Not sure if I understand what you are trying to say here. In the examples given in my book, convolution is used when generating output from a filter. – Kristian Jan 21 '13 at 12:29
@Kristian: Oh, then perhaps you haven't reached this yet, but convolution in the time domain is multiplication in the Fourier domain. So just take the Fourier transform of both, multiply, and detransform. – akkkk Jan 21 '13 at 12:48

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

$$(f*h)(t) = A \frac{e^{-a t}+e^{-t} ((a-1) \sin (t)-\cos (t))}{a^2-2 a+2}+A \frac{3 e^{-a t}+e^{-t} ((a-1) \sin (3 t)-3 \cos (3 t))}{a^2-2 a+10}+A \frac{5 e^{-a t}+e^{-t} ((a-1) \sin (5 t)-5 \cos (5 t))}{a^2-2 a+26}+A \frac{40 e^{-a t}+e^{-t} ((a-1) \sin (40 t)-40 \cos (40 t))}{a^2-2 a+1601}$$

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Thanks a lot! This became a lot clearer now. Appreciate it greatly! – Kristian Jan 21 '13 at 12:54
Actually, the value of the integral stated by the OP does depend on $t$, but, as you correctly state, the upper limit should be $t$, not $\pi$. – Dilip Sarwate Jan 21 '13 at 12:56
Oh yes, I see, thanks for pointing out. I'll edit accordingly. – Ron Gordon Jan 21 '13 at 12:57