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Using Fourier Properties or standard Integral?

  1. May 3, 2010 #1
    Hi, sometimes when I'm trying to work out the Fourier Transform of a signal, I get different answers depending on whether I use Fourier Transform properites (such as rect(t) will go to sinc(f) etc) or whether I use the FT integral. Here's an example where I'm not sure which 1 is the correct answer:

    The signal is: [tex]g(t)=-Arect(\frac{t+T}{T})+Arect(\frac{t}{T})-Arect(\frac{t-T}{T})[/tex]

    Using FT properties:
    [tex]
    G(f)=-ATsinc(Tf)e^{j2\pi fT}+ATsinc(fT)-ATsinc(Tf)e^{-j2\pi fT}[/tex]

    Therefore:

    [tex]G(f)=-2ATsinc(Tf)cos(2\pi fT)+ATsinc(Tf)[/tex]

    Using the FT Integral:
    [tex]G(f)=-A\int_{\frac{-3T}{2}}^{\frac{-T}{2}}e^{-j2\pi ft}dt+A\int_{\frac{-T}{2}}^{\frac{T}{2}}e^{-j2\pi ft}dt-A\int_{\frac{T}{2}}^{\frac{3T}{2}}e^{-j2\pi ft}dt[/tex]

    Therefore:

    [tex]G(f)=-A[\frac{e^{j\pi fT}-e^{j\pi f3T}}{-j2\pi f}]+\frac{A}{\pi f}[\frac{e^{j\pi fT}-e^{-j\pi fT}}{2j}]-A[\frac{e^{-j\pi f3T}-e^{-j\pi fT}}{-j2\pi f}][/tex]

    Therefore:

    [tex]G(f)=2TAsinc(fT) -3TAsinc(3fT)[/tex]
     
    Last edited: May 3, 2010
  2. jcsd
  3. May 3, 2010 #2

    uart

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    I think that both answers are equivalent. After a bit of fiddling around with the usual trig identities I was able to show that :

    [tex] \sin(x) - \sin(3x) = -2 \cos(2x) \sin(x)[/tex]

    If you use that in your final equation then it reduces to the earlier one that you got "Using FT properties"
     
  4. May 3, 2010 #3
    Hey uart thanks for the reply.
    I've tried subbing in different values for A, T and f but I get different answers from the 2 equations, so they're not the same unfortunately.
     
  5. May 4, 2010 #4

    uart

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    Yeah I'm pretty sure I can show (algebraically) that they're identical. No time right now but I'll have a look at numerical values later.

    BTW Are you sure you're using the correct "sinc" definition : sinc(x) = sin(pi x)/(pi x) is the correct one to use in this case.
     
  6. May 4, 2010 #5

    uart

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    OK I'm calling "calculator error" on you frenzal_dude. :) Double check that you're using radians and also double check you have the correct sinc definition.

    I just checked and I get them both algebraically and numerically identical!
     
    Last edited: May 4, 2010
  7. May 4, 2010 #6

    uart

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    Here's the algebraic solution.

    Starting with your last resut (I took out the common factor of "fT" as it's obviously the same in both solutions) :

    [tex]G(f)=2sinc(fT) -3sinc(3fT) = sinc(fT) + sinc(fT) - 3 sinc(3fT)[/tex]

    [tex] = sinc(fT) + \frac{1}{\pi f T} \left( \sin(\pi f T) - \sin(3 \pi f T) \right)[/tex]

    Now sub in the trig identity I posted previosly and you get :

    [tex]G(f)=sinc(fT) -2 sinc(fT) \cos(2 \pi f T)[/tex]

    Which is identical to your "using FT properties" result.
     
  8. May 4, 2010 #7
    uart, thankyou so much for your help!
    I'm relieved to know I'm on the right track now.
     
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