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Fourier transform: signal with filter

  1. Nov 21, 2016 #1
    Hi Guys,

    I'm having trouble with the following:

    A finite-time signal is the result of a filter G(t) applied to a signal. The filter is simply “on” (1) for t ∈ [0,T] and off (“0”) otherwise. If x(t) is the signal, and x(ω),its Fourier transform, compute the Fourier transform of the filtered signal. Next, take a simple sine for x(t), x(t) = sin(ω0t), and compute the Fourier transform for the finite-time signal. Write the result, it must involve the filter, and integrations should stretch [−∞,∞]

    I don't really know what to do exactly, with the first problem.

    I can try calculating the Fourier transform of the filter:

    G(ω)= ∫0T e-iωtdt = -1/(iω)⋅(e-iωT-1)

    The Fourier transform of the signal is: x(ω)

    The convolution theorum says that the convolution of two functions is the product of the Fourier-transformed functions. Which makes: G(ω)x(ω).

    But I have the idea that this isn't right. Could one of you guys assist me?

    Peter
     
  2. jcsd
  3. Nov 21, 2016 #2

    BvU

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    The convolution theorem also works the other way around: the Fourier transform of the product of a step function and some other function is the convolution of their Fourier transforms.

    By the way: do use the template, don't erase it. It helps you order to your thinking and us to help you better
     
  4. Nov 21, 2016 #3
    Ok. If I understand you correctly, you mean:

    Fourier{x(t)g(t)} = 1/2π ⋅ X(ω)⊗G(ω) ==>

    Writing -1/(iω)⋅(e-iωT-1) to -1/iω⋅e-iωT/2(e-iωT/2-eiωT/2) = T⋅e-iωT/2⋅sinc(ωT/2)

    Fourier{x(t)g(t)}=1/2π⋅∫-∞ X(w-w')⋅T⋅sinc(ω'T/2) dω' ??

    Sorry, I will do that next time, thanks!
     
  5. Nov 21, 2016 #4

    BvU

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    Looks reasonable (all the contributions are there -- didn't check the gory details. Most of the time I use a table like this)
    I take it you mean ##\ x(\omega-\omega_0) \ ## ?

    Now for the second part ...
     
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