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    Hello.

    Quick question about a F.T. problem that is bugging me. See attached, it's part b) that I am a little unsure of

    I know that the Fourier transform of exp(-kt) = 1 / (k + jw)

    Then the F.T. of derivative of f(t) - which is (-k)*exp(-kt) will simply be the above multiplied by -k: (-k) / (k + jw)

    But now using the derivative property of fourier transforms:

    d/dt[f(t)] <==> jw[F(w)]

    the transform of the derivative of f(t) should be jw multiplied by the transform of the original function, or in other words: (jw) / (k + jw)

    These two answers are supposed to be equal but they're not! I cant re-arrange one into the other, and when I put dummy values in for k and w, I am left with a result of -1/1 when I take one away from the other

    Can anyone spot where I went wrong please?

    Thanks.
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    late night bump
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    Can't be bothered to look up the definitions, but an obvious problem here is that the FT of exp(-kt) doesn't strictly exist (since \exp(-kt)e^{ikt} is unbounded as x \to -\infty).

    I think this is one of the cases where you don' need to worry about convergence if you (or your examiners) choose the problem carefully, but things can go wrong once you step outside that.
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    I'm doing an Engineering course so we sort of take for granted much of the convergence and other Maths stuff (i.e. we dont bother with them!)

    so presuming t is within an acceptable range....

    do you think both solutions to the FT of the derivative are correct?
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    (Original post by DFranklin)
    Can't be bothered to look up the definitions, but an obvious problem here is that the FT of exp(-kt) doesn't strictly exist (since \exp(-kt)e^{ikt} is unbounded as x \to -\infty).
    I don't think that's a problem though...? The function given has a cutoff at 0:
    \displaystyle f(t) = \begin{cases} e^{-k t} & t \ge 0 \\ 0 & t &lt; 0 \end{cases}
    so \displaystyle \int_{-\infty}^{\infty} f(t) e^{-i \omega t} \, dt certainly exists (for k > 0).

    The problem, I think, is that f'(0) is undefined. Using integration by parts on \displaystyle \int_{a}^{b} f'(t) e^{-i \omega t} \, dt = f(b) e^{-i \omega b} - f(a) e^{-i \omega a} + i \omega \int_{a}^{b} f(t) e^{-i \omega t} \, dt is certainly fine when 0 \not \in [a, b], but to calculate the Fourier transform you need to integrate over (-\infty, \infty) - and we run into a problem near 0. So really what you should do is skirt around the problem by writing \displaystyle \int_{-\infty}^{\infty} f'(t) e^{-i \omega t} \, dt = \int_{-\infty}^{0} f'(t) e^{-i \omega t} \, dt + \int_{0}^{\infty} f'(t) e^{-i \omega t} \, dt and using the IBP formula above on each half separately, taking limits appropriately. This seems to produce an answer which agrees with the direct calculation \displaystyle \int_{0}^{\infty} -k e^{-kt} e^{-i \omega t} \, dt = - \frac{k}{k + i \omega} = -1 + \frac{i \omega}{k + i \omega}.

    (Informally, we may regard f'(t) = -k f(t) + \delta(t) where \delta is the Dirac delta function.)
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    there is that mystery -1 that came up for me, unfortunately I cant see where it came from in your calculations.

    Thanks for the post though, very informative.
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    It comes from taking the limit \displaystyle \int_{a}^{b} f'(t) e^{-i \omega t} \, dt with (a, b) \to (0, \infty).

    Note that if we take f'(t) = -k f(t) + \delta(t), then \displaystyle \int_{-\infty}^{\infty} f'(t) e^{-i \omega t} \, dt = \frac{i \omega}{k + i \omega}, exactly as expected, whereas if we take f'(t) = -k f(t) we end up with the aforementioned awkwardness with a mysterious -1 appearing.
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    [QUOTE=Zhen Lin;28878280]I don't think that's a problem though...? The function given has a cutoff at 0:
    \displaystyle f(t) = \begin{cases} e^{-k t} & t \ge 0 \\ 0 & t &lt; 0 \end{cases}[/qupte]I assumed the OP would have described f correctly and didn't look at the JPG...

    For the given f, as you say, f'(0) is going to be a delta function.
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    Thank you so much Zhen.

    I should've remembered that having that instantaneous jump from 0 to 1 in the function would mean that you'd get a sigma(t) appearing at t=0 in f'(t). I'm happy now
 
 
 
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