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What is the derivative of a p-fold convolution?

[tex]

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

[/tex]

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

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

[tex]

\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

[/tex]

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:

[tex]

\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)

[/tex]

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 ...

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# Derivative of p-fold convolution

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