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Derivative of Convolution

  1. May 13, 2010 #1
    Derivative of a Convolution

    1. The problem statement, all variables and given/known data

    How do I find the derivative of a convolution, meaning [tex]\frac{d}{dt}(f \ast g)(t)[/tex]?

    2. Relevant equations
    [tex](f \ast g)(t)=\int^{}_{} f(t-\tau)g(\tau)d\tau[/tex]


    3. The attempt at a solution
    I want to use the fundamental theorem of calculus, but I can't just directly substitute t for [tex]\tau[/tex], because that would make[tex]f(t-\tau)g(\tau)=f(0)g(t)[/tex], which doesn't make sense. How would I correctly apply the fundamental theorem of calculus in this case?
     
    Last edited: May 13, 2010
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  3. May 13, 2010 #2

    jbunniii

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    If you know basic Fourier transform theory, you can answer the question as follows:

    Define

    [tex]h(t) = \frac{d}{dt}(f*g)(t)[/tex]

    Then

    [tex]\begin{align*}
    \hat{h}(\omega) &= \mathcal{F}\{h(t)\} \\
    &= \mathcal{F}\{\frac{d}{dt}(f*g)(t)\} \\
    &= i\omega \mathcal{F}\{(f*g)(t)\} \\
    &= i\omega \hat{f}(\omega)\hat{g}(\omega)
    \end{align*}
    [/tex]

    You can now associate the [itex]i \omega[/itex] with either [tex]\hat{f}[/tex] or [tex]\hat{g}[/tex] and invert the Fourier transform to obtain

    [tex]\frac{d}{dt}(f*g)(t) = \left(\left(\frac{d}{dt}f\right) * g\right)(t) = \left(f * \left(\frac{d}{dt}g\right)\right)(t)[/tex]

    Now the above is not rigorous, because I don't specify under what conditions the operations are valid. Also, it doesn't answer how to solve the problem without resorting to Fourier transforms. But now you know what answer to work toward (which you could equally well have found by looking in some table, so I don't think I'm giving the game away).
     
    Last edited: May 13, 2010
  4. May 13, 2010 #3
    I'm sorry, but unfortunately I don't know about Fourier transforms, but I do know about Laplace transforms. In fact, it is in the context of solving differential equations using Laplace transforms that I came up with this question. Is there any way that I can use the fundamental theorem of calculus to do this? Intuitively, it seems like this problem is just a simple case of differentiating an integral.
     
  5. May 13, 2010 #4

    jbunniii

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    You can solve it very similarly by using Laplace transforms. I'll leave the details to you.

    I don't see how that would work, because you are dealing with a definite integral (from [itex]-\infty[/itex] to [itex]\infty[/itex]), not an indefinite one.

    But notice that the variable of integration is [itex]\tau[/itex], not [itex]t[/itex]. Therefore, I claim that you can simply slide the differentiation operator inside the integral as follows:

    [tex]\frac{d}{dt}\int_{-\infty}^{\infty}f(t-\tau) g(\tau) d\tau = \int_{-\infty}^{\infty} \frac{d}{dt} f(t-\tau) g(\tau) d\tau[/tex]

    See if you can justify why (and under what conditions) that is valid.
     
  6. May 13, 2010 #5
    Apparently, we are using different definitions of the covolution. In the definition I am using, the integral goes from 0 to t, not negative infinity to infinity.
     
  7. May 13, 2010 #6

    jbunniii

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    That makes the implicit assumption that both [itex]f(t)[/itex] and [itex]g(t)[/itex] are zero for [itex]t < 0[/itex]. But that's OK. If they are both zero for negative [itex]t[/itex] then you can equally well set the integration limits to [itex]-\infty[/itex] to [itex]\infty[/itex] without changing the result. Do you see why?
     
  8. May 13, 2010 #7
    No, I don't. I see why setting the lower limit to negative infinity rather than zero would make no difference if f and g are zero for t<0. But how can you just arbitrarily change the upper limit from t, a variable, to infinity?
     
  9. May 13, 2010 #8

    gabbagabbahey

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    I see...in that case, you might want to make use of the Heaviside theta function:

    [tex]H(x)=\left\{\begin{array}{lr}0, & x<0 \\ 1, & x>0\end{array}\right.[/tex]

    by realizing that

    [tex]\int_0^t f(t-\tau)g(\tau)d\tau=\int_0^\infty H(t-\tau)f(t-\tau)g(\tau)d\tau[/tex]
     
  10. May 13, 2010 #9
    Why can't the fundamental theorem of calculus just be used directly? If we just leave the limits as 0 and t, then we have an indefinite integral. In other words, the problem is to take the derivative with respect to t of an integral the upper limit of which is t, precisely the case that the fundamental theorem of calculus is supposed to deal with. So why doesn't it work in this case?
     
  11. May 13, 2010 #10

    jbunniii

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    Because [itex]f(t) = 0[/itex] for [itex]t < 0[/itex] if and only if [itex]f(t - \tau) = 0[/itex] for [itex]\tau > t[/itex]. Therefore I can use any upper endpoint for the integration as long as it is greater than or equal to [itex]t[/itex]. If I want one upper endpoint that satisfies this condition no matter what [itex]t[/itex] is, then that endpoint has to be [itex]\infty[/itex].
     
  12. May 13, 2010 #11

    jbunniii

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    Good question. The answer is that the INTEGRAND also depends on [itex]t[/itex], so the FTC doesn't apply, at least not directly.
     
  13. May 13, 2010 #12
    I see what you mean. So in essence we are dealing with an expression of the form [tex]\frac{\partial}{\partial t}\int^{t}_{0}F(t, \tau) \partial \tau[/tex]. Is there any formula for such a beast?
     
  14. May 13, 2010 #13

    jbunniii

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    Not in general, as far as I know.
     
  15. May 13, 2010 #14
    This seems to be a far more interesting topic than the original question I asked. I think I'll start a non-homework thread about it.
     
  16. May 13, 2010 #15

    gabbagabbahey

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    Sure, you can derive it yourself using the Heaviside theta function...
     
  17. May 13, 2010 #16
    Last edited by a moderator: Apr 25, 2017
  18. May 13, 2010 #17
    How would I use the Heaviside theta function in this more general case?
     
  19. May 13, 2010 #18

    gabbagabbahey

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    Same way as before...

    [tex]\int_0^t F(t,\tau)d\tau=\int_0^\infty H(t-\tau)F(t,\tau)d\tau[/tex]

    Then take the derivative...
     
  20. May 13, 2010 #19

    jbunniii

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    But H isn't differentiable at 0, so you either need distribution theory to proceed along these lines, or you hand-wave something about a delta function and cross your fingers that the answer turns out valid. I guess it depends on whether you are a mathematician or a physicist.:wink:

    [Edit]: I'm sure it turns out fine, and can probably be made rigorous by choosing a sequence of differentiable functions [itex]H_n[/itex] that "almost" have the desired property and which converge to [itex]H[/itex], then making an appropriate limiting argument.
     
  21. May 13, 2010 #20
    Maybe I'm missing something obvious, but why is this true?
     
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