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Derivative of p-fold convolution

  1. Jul 8, 2014 #1

    What is the derivative of a p-fold convolution?

    \frac{\partial}{\partial Y(\omega) } \underbrace{Y(\omega) * \dots * Y(\omega)}_{p-\text{times}}

    EDIT: I have two contradicting approaches - I guess both are wrong ;-)

    As a simple case, take the 2-fold convolution. FIRST approach:

    \frac{\partial}{\partial Y(\omega)} Y(\omega)*Y(\omega) = \frac{\partial}{\partial Y(\omega)} \int_{-\infty}^{\infty} Y(\tau) Y(\omega-\tau) d\tau = \int_{-\infty}^{\infty} Y(\tau) \underbrace{\frac{\partial Y(\omega-\tau)}{\partial Y(\omega)}}_{\delta(\tau)} d\tau = \int_{-\infty}^{\infty} Y(\tau)\delta(\tau) d\tau = 1

    Here I think the derivative is wrong but intuitively it makes sense at least: Differentiation is the opposite of smoothing, smoothing is convolution, so taking away the convolution is taking away smoothing.

    SECOND approach:

    \frac{\partial}{\partial Y(\omega)} Y(\omega)*Y(\omega) = \mathcal{F}\left\{ \mathcal{F}^{-1}\left\{ \frac{\partial}{\partial Y(\omega)} Y(\omega)*Y(\omega) \right\} \right\} = \mathcal{F}\left\{ -j y(t) y^2(t) \right\}= \mathcal{F}\left\{ -j y^3(t) \right\} = -jY(\omega)*Y(\omega)*Y(\omega)

    Here is the point where I am stuck - I am sure I can't apply the derivative theorem here because it's not a derivative by [tex]\omega[/tex]. But how to do this? Anyways, this result does the opposite from above and is also against my inuition, so I think it's wrong ...
    Last edited: Jul 8, 2014
  2. jcsd
  3. Jul 9, 2014 #2
    Ok I think I got it. Is this correct? (I was not able to do simple differentiation)

    \frac{\partial}{\partial Y(\omega)} Y(\omega)*Y(\omega) =
    \frac{\partial}{\partial Y(\omega)} \int_{-\infty}^{\infty} Y(\tau) Y(\omega-\tau) d\tau = \\
    \int_{-\infty}^{\infty} Y(\tau) \frac{\partial Y(\omega-\tau)}{\partial Y(\omega)} d\tau =
    \int_{-\infty}^{\infty} Y(\tau) \frac{\partial \omega}{\partial \omega} \frac{\partial Y(\omega-\tau)}{\partial Y(\omega)} d\tau =
    \int_{-\infty}^{\infty} Y(\tau) \frac{\partial \omega}{\partial Y(\omega)} \frac{\partial Y(\omega-\tau)}{\partial \omega} d\tau = \\
    \int_{-\infty}^{\infty} Y(\tau) \left(\frac{\partial Y(\omega)}{\partial \omega}\right)^{-1} \frac{\partial Y(\omega-\tau)}{\partial (\omega-\tau)} \frac{\partial (\omega-\tau)}{\partial \omega} d\tau =\\
    \int_{-\infty}^{\infty} Y(\tau) \frac{1}{Y'(\omega)} Y'(\omega-\tau) d\tau =
    \frac{Y'(\omega) * Y(\omega)}{Y'(\omega)}

    In any case, now I am struggling with something different where I do not know if there is even a solution to it!

    Suppose the same setup as above. However, instead of having just a convolution, I have a function which operates on [itex]Y(\omega)[/itex]:

    \frac{\partial}{\partial Y(\omega)} F(Y(\omega))

    However, I do not have the analytical form of F. However, I know the analytical form of its Fourier transform f.

    Can I write the equation above in terms of the Fourier transform [itex]y(t)[/itex] and [itex]f[/itex] ?

    I can do it but I still have the factor of [itex]1/Y'(\omega)[/itex]. For the example above:

    \frac{1}{Y'(\omega)} j \mathcal{F}\left( t \cdot y^2(t) \right)

    Oh, wait, I didn't realize that I can also trivially expand [itex]1/Y'(\omega)[/itex]:

    -\frac{\mathcal{F}\left\{ t \cdot y^2(t) \right\}}{\omega \mathcal{F}\left\{(t) \right\}}

    Is this correct?
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