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Fourier transform of a time integral

  1. May 21, 2012 #1
    Question:

    Derive the relationship

    [tex]\int^t_{- \infty} f(\tau) d \tau \Leftrightarrow \frac{F(\omega)}{j \omega} + \pi F(0) \delta (\omega)[/tex]

    (where [itex]\Leftrightarrow[/itex] means "Fourier transforms into").

    Attempt:

    I have already proved the relationship

    [tex]\frac{dg(t)}{dt} \Leftrightarrow j \omega G( \omega)[/tex]

    so define [itex]h(t) = \frac{dg(t)}{dt}[/itex]. Then we have

    [tex]\int^t_{- \infty} h(\tau) d \tau = \int^t_{- \infty} \frac{dg}{d \tau} d \tau = [g(\tau)]^t_{- \infty} = g(t) - g(-\infty)[/tex]

    so [itex]g(t) = \int ^t _{- \infty} h(\tau) d \tau + g(- \infty)[/itex]

    Using the Fourier differentiation relationship above we get

    [tex]\mathcal{F}[h(t)] = j \omega \mathcal{F}[g(t)] = j \omega \mathcal{F} \left [ \int ^t _{- \infty} h(\tau) d \tau + g(- \infty) \right ] = j \omega \mathcal{F} \left [ \int ^t _{- \infty} h(\tau) d \tau \right ] + j \omega \mathcal{F}[g(- \infty)] [/tex]
    [tex]= j \omega \mathcal{F} \left [ \int ^t _{- \infty} h(\tau) d \tau \right ] + j \omega 2 \pi g(-\infty) \delta(\omega)[/tex]

    so

    [tex]\mathcal{F} \left [ \int ^t _{- \infty} h(\tau) d \tau \right ] = \frac{\mathcal{F}[h(t)]}{j \omega} - 2 \pi g(-\infty) \delta(\omega)[/tex]

    I'm not sure if i've done something wrong here or if somehow this is equivalent to the correct relationship? Could someone help please? :smile:

    Thanks!
    Jon.
     
  2. jcsd
  3. May 21, 2012 #2

    vela

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    Suggestion: Try writing
    $$\int^t_{-\infty} f(\tau)\,d \tau = \int_{-\infty}^\infty f(\tau)u(t-\tau)\,d\tau$$ and taking the Fourier transform of that.
     
  4. May 22, 2012 #3
    Thanks Vela, I can see what you mean there. That's the convolution integral so the Fourier transform will just be [itex]\mathcal{F}[f(t)] \mathcal{F}[u(t)][/itex], but then that has lead me to another problem because I can't seem to work out the Fourier transform of the step function :tongue:

    I tried the direct method but get a nonsense answer, then I tried using the derivative property and get

    [tex]\mathcal{F}[u(t)] = \frac{1}{\omega j}\mathcal{F}[du/dt] = \frac{1}{\omega j}\mathcal{F}[\delta (t)] = \frac{1}{\omega j}[/tex]
    I cheated and googled it, but don't really understand the explanations as to why this doesn't work. Most explanations seem to be along the lines of:

    "In general, any two function f(t) and g(t)=f(t)+c with a constant difference c have the same derivative, df(t)/dt, and therefore they have the same transform according the above method. This problem is obviously caused by the fact that the constant difference c is lost in the derivative operation. To recover this constant difference in time domain, a delta function needs to be added in frequency domain."

    So the condition for use of the equation [itex]\mathcal{F}[\frac{dg(t)}{dt}] = j \omega \mathcal{F}[g(t)][/itex] is that g(0)=0? I just can't really see why this is the case from the proof... if anything, things seem to go awry for [itex]\omega = 0[/itex].

    Thanks :smile:
     
    Last edited: May 22, 2012
  5. May 22, 2012 #4

    vela

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    F[g'(t)] = jωG(ω) always works. If two functions f(t) and g(t) have the same derivative, it will turn out that jωF(ω) will equal jωG(ω). The problem is going the other way. If jωF(ω)=jωG(ω), you can't recover the difference between f(t) and g(t). The information was lost when taking the derivatives.

    One way to solve the problem is to calculate the inverse Fourier transform of 1/jω using contour integration. You should find that
    $$\frac{1}{2\pi}\int_{-\infty}^\infty \frac{1}{j\omega} e^{j\omega t}\,d\omega = \begin{cases}+1/2 & t>0 \\ -1/2 & t<0\end{cases}$$ which differs from the unit step by a constant. You can then deduce what the Fourier transform of u(t) must be.

    Another way is to calculate the Fourier transform of u(t) with the use of a convergence factor to get around the problem you probably ran into in your earlier attempt:
    $$F[u(t)] = \lim_{\epsilon \to 0^+} \int_{-\infty}^\infty u(t)e^{-\epsilon t}e^{-j\omega t}\,dt$$
     
  6. May 22, 2012 #5

    vela

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    By the way, I think I got the convention you're using for the Fourier transforms right, but if not, there could be differences in signs and factors of ##2\pi##. I'll let you sort that out.
     
  7. May 23, 2012 #6
    Thanks for your help. I haven't done much complex analysis so need to study contour integration. It's like opening one can of worms after another! :eek:

    P.S. I don't see how using a convergence factor helps. Do you not just get

    $$F[u(t)] = \lim_{\epsilon \to 0^+} \int_{-\infty}^\infty u(t)e^{-\epsilon t}e^{-j\omega t}\,dt = \lim_{\epsilon \to 0^+} \int_{0}^\infty e^{-(\epsilon + j \omega)t}dt = \lim_{\epsilon \to 0^+} \frac{1}{\epsilon + j \omega} = \frac{1}{j \omega}$$

    the same as before? It's not important though, I will accept it for now and learn the contour integration method eventually :smile:

    Thanks!
     
  8. May 23, 2012 #7

    vela

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    It would seem that way, but you have to be careful when you have j in the denominator. Evaluate it using this:
    $$F[u(t)] = \lim_{\epsilon \to 0^+} \frac{1}{\epsilon + j \omega}\cdot\frac{\epsilon - j \omega}{\epsilon - j \omega}$$ and the fact that
    $$\delta(\omega) = \lim_{\epsilon\to 0^+} \frac{1}{\pi}\frac{\epsilon}{\epsilon^2+\omega^2} $$ This is one of various representations of the Dirac delta function.
     
  9. May 24, 2012 #8
    Oh wow, that's interesting! Thanks :-)
     
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