I'm finding in trouble trying to resolve this exercise. I have to calculate the convolution of two signals:

$$y(t)=e^{-kt}u(t)*\frac{\sin\left(\frac{\pi t}{10}\right)}{(\pi t)} $$

where $u(t)$ is Heavside function

well I applied the formula that says that the convolution of this two signal is equal to


where $X(f)$ is the fourier transform of the first signal and $W(f)$ is the fourier transform of second signal

well fourier transform of $e^{-kt}u(t)$ is $X(f)=\frac{1}{k+j2\pi f}$. I have to make second signal as equals as possible to $\operatorname{sinc}\left(\frac{\pi t}{10}\right)$ so I do this operation: $\frac{\sin\left(\frac{\pi t}{10}\right)}{\left(\frac{\pi t}{10}\right)}{\left(\frac{1}{10}\right)}$. this is equal ${\left(\frac{1}{10}\right)}\operatorname{sinc}\left(\frac{\pi t}{10}\right)$

right or not?


If something is not clear please advice me

  • 3
    $\begingroup$ The convolution is not equal to the product of Fourier transforms. (It can't be, since convolution is a function of time, while the Fourier transform is a function of frequency.) Rather, the convolution is the inverse Fourier transform of the product of Fourier transforms. I haven't checked your FT's, but assuming they're correct, compute their product, then compute the inverse FT of the product. That is the convolution. $\endgroup$ – William DeMeo Mar 28 '12 at 21:26
  • $\begingroup$ I made some corrections, perhaps conceptually relevant: you are not trying to compute "the convolution of $y(t)$" (you don't compute the convolution of one signal), rather $y(t)$ is the convolution of $x(t)$ and $w(t)$ $\endgroup$ – leonbloy Jul 4 '12 at 13:09
  • $\begingroup$ @Mazzy see the edit. $\endgroup$ – Seyhmus Güngören Sep 9 '12 at 9:42

I think this question should be solved in the frequency domain. Otherwise it seems difficult (at least to me).

$$e^{-kt}u(t)\rightarrow^F\frac{1}{k+jw}$$ and $$\frac{w_b}{\pi}sinc(wbt)\rightarrow^F rect(\frac{w}{2w_b})$$ and this in your case $w_b=\frac{\pi}{10}$

As a result we have $$Y(w)=X(w)W(w)=\frac{1}{k+jw}rect(\frac{w}{2w_b})$$ where $rect(\frac{w}{2w_b})=1$ if $|w|<wb$, else $0$. Finally what we need to do is to calculate the inverse fourier transform that is $$y(t)=F^{-1}(Y(w))=\frac{1}{2\pi}\int_{-w_b}^{w_b}\frac{1}{k+jw}e^{jwt}dw.$$ I think this integral should not be a big trouble.

EDIT: We can make a change of variables as $z=k+jw$ and $dz=jdw$. From here we have

$$\frac{1}{2\pi}\int_{-w_b}^{w_b}\frac{1}{k+jw}e^{jwt}dw=\frac{-j}{2\pi}\int_{-w_b}^{w_b}\frac{1}{z}e^{(z-k)t}dz=\frac{-j}{2\pi}\int_{-w_b}^{w_b}\frac{1}{z}e^{zt}dz\cdot\int_{-w_b}^{w_b}\frac{1}{z}e^{-kt}dz$$ from http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions $$\int\frac{ e^{tz}}{z}dz=\ln|z|+\sum_{n=1}^{\infty}\frac{(tz)^n}{nn!}$$ and $$\int\frac{1}{z}e^{-kt}dz=\ln|z|e^{-kt}$$ Accordingly we have $$\frac{-j}{2\pi}\left[\ln|z|e^{-kt}\left(\ln|z|+\sum_{n=1}^{\infty}\frac{(tz)^n}{nn!}\right)\right]_{-w_b}^{w_b}$$ which is $$\frac{-j}{2\pi}\left(\ln w_be^{-kt}\left(\sum_{n=1}^{\infty}\frac{(tw_b)^n}{nn!}-\sum_{n=1}^{\infty}\frac{(-tw_b)^n}{nn!}\right)\right)$$

  • $\begingroup$ "I think this integral should not be a big trouble." Could you outline an approach that could be used to evaluate the integral or provide a reference to a table of Fourier transforms where this inverse transform can be found? $\endgroup$ – Dilip Sarwate Sep 9 '12 at 2:24
  • $\begingroup$ @DilipSarwate please see the edit. $\endgroup$ – Seyhmus Güngören Sep 9 '12 at 9:39

I think it is better to leave the Fourier transform idea, and use brute force instead.

  1. Write down the integral.

  2. Use the definition of the Heaviside function.

  3. Use a change of variable.

  4. Use Euler formula for $\sin$.

  5. End up with exponential integrals.

  • $\begingroup$ I don't think there's anything wrong with this answer. Hopefully it won't get marked down again...+1 $\endgroup$ – daniel Sep 10 '12 at 1:32
  • $\begingroup$ @daniel Thanks, perhaps the change of variables should be removed though. $\endgroup$ – AD. Sep 10 '12 at 17:17

I think William DeMeo's comment is a good answer. You are simply trying to use the table and get your function into a form that can be used.

The second function--call it $f_2$-- is currently $\frac{\sin (\pi t/10)}{\pi t}.$ Multiply it by 10, so you have $$f_3 = 10 \cdot f_2 = 10 \frac{\sin(\pi t/10)}{\pi t}= \frac{\sin(\pi t/10)}{(1/10)\pi t}= \text{sinc}(\pi t/10).$$

Letting $f_1 = e^{-kt}u(t),$ we know that the product of the transforms $F_1$ and $F_3$ pairs with the convolution of the inverses of the functions in the time domain.

So find the Fourier transforms of $f_1,f_3$, multiply those, and find the inverse transform of the product. That will be the convolution $c_1 =f_1*f_3.$ Since $AF(\omega)\leftrightarrow Af(t), c_2 = \frac{c_1}{10}$ is what you want.

  • $\begingroup$ what is the difference of your post then mine except for implicit expressions. $\endgroup$ – Seyhmus Güngören Sep 9 '12 at 9:44
  • $\begingroup$ there is no problem except that you had to refer if there was already a similar post. Your post says Find Fourier transforms of $f_1$ and $f_3$ while there is another post which provides them already (my post before the edit). $\endgroup$ – Seyhmus Güngören Sep 10 '12 at 10:37

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