Convolution - Fourier Transform

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Homework Statement


An LTI system has an impulse response h(t) = e-|t|
and input of x(t) = ejΩt

Homework Equations


Find y(t) the system output using convolution
Find the dominant frequency and maximum value of y(t)
Ω = 2rad/s

The Attempt at a Solution


I have tried using the Fourier transform to get y(t) but when you try to find X(Ω), I get infinity
as X(Ω) = ∫ x(t) * e-jΩtdt = ∫ e jΩt*e-jΩtdt = ∫ 1 dt = t between inf and -inf
h(t) I could find as 2/(1+Ω2)

Any ideas on how to solve this?
 
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Answers and Replies

  • #2
Charles Link
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You need to write the transform as ## X(\omega)=\int x(t) e^{-i \omega t} dt ##. The result is ## X(\omega)=2 \pi \, \delta(\omega-\Omega) ##. You need to read about delta functions. If you google it, you should find some useful formulas like the one I just gave you.
 
  • #3
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You need to write the transform as ## X(\omega)=\int x(t) e^{-i \omega t} dt ##. The result is ## X(\omega)=2 \pi \, \delta(\omega-\Omega) ##. You need to read about delta functions. If you google it, you should find some useful formulas like the one I just gave you.
So then using that you would get that Y(w) = 4π/(1+w2) * δ(w-2), but then how would you get that back into the time domain?
 
  • #4
Charles Link
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## H(\omega)=\int\limits_{0}^{+\infty} h(t) e^{-i \omega t} dt ##, since ## h(t)=0 ## for ## t<0 ##. (Perhaps they didn't tell you this (## h(t)=0 ## for ## t<0 ##) in the problem statement, but it is clear that that's what they want). Try recomputing ## H(\omega) ##.(You have it incorrect). ## \\ ## Now ## Y(\omega)=H(\omega)X(\omega) ##. (That part you have correct.) ## \\ ## Use an inverse transform to get ## y(t)=\frac{1}{2 \pi} \int\limits_{-\infty}^{+\infty} Y(\omega) e^{i \omega t} \, d \omega ##. ## \\ ## Wait until you process everything to put in the value for ## \Omega ##.
 
  • #5
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## H(\omega)=\int\limits_{0}^{+\infty} h(t) e^{-i \omega t} dt ##, since ## h(t)=0 ## for ## t<0 ##. Try recomputing ## H(\omega) ##.(You have it incorrect). ## \\ ## Now ## Y(\omega)=H(\omega)X(\omega) ##. (That part you have correct.) ## \\ ## Use an inverse transform to get ## y(t)=\frac{1}{2 \pi} \int\limits_{-\infty}^{+\infty} Y(\omega) e^{i \omega t} \, d \omega ##. ## \\ ## Wait until you process everything to put in the value for ## \Omega ##.
Sorry the reason why I got what i did for H(w), was because I forget to put in an absolute around the t. I've changed in now
 
  • #6
Charles Link
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Sorry the reason why I got what i did for H(w), was because I forget to put in an absolute around the t. I've changed in now
An ## h(t) ## with an absolute value would be unphysical. That is saying it responds before the impulse. These functions always begin at ## t=0 ##. If they gave you the problem in such a fashion, it is unphysical.
 
  • #7
Charles Link
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Incidentally, if ## H(t) ## is the response to a unit step function, ## h(t)=\frac{dH(t)}{dt} ## is the response to the delta function. The problem they gave you, if I'm not mistaken, is basically a series R-C circuit (along with an input voltage source) with the output voltage being measured across the capacitor, and the product ## RC=1 ##, because of the form ## h(t)=e^{-t} ##. ## \\ ## Let the input voltage source be a unit step function, and/or a delta function, and you can compute the ## h(t) ## that they give you., i.e. for the R-C circuit, with ## RC=1 ##, when ## V_{in} (t)=\delta(t) ##, you get ## V_C(t)=e^{-t} ## for ## t>0 ## . Try and see. I believe I have this result correct. ## \\ ## And additional item: If you are familiar with ac circuit analysis, try comparing the result that you get from the linear response theory with that from computing the ac impedances of the two components and considering it as a voltage divider problem-i.e. the ac voltage across the capacitor is readily computed.
 
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