Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
up vote 1 down vote accepted

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

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.