Derivative of p-fold convolution

[divB]

Hi,

What is the derivative of a p-fold convolution?

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

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

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

<br /> \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<br />

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:

<br /> \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)<br />

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 \omega. 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 ...

 

[divB]

Ok I think I got it. Is this correct? (I was not able to do simple differentiation)

<br /> \frac{\partial}{\partial Y(\omega)} Y(\omega)*Y(\omega) =<br /> \frac{\partial}{\partial Y(\omega)} \int_{-\infty}^{\infty} Y(\tau) Y(\omega-\tau) d\tau = \\<br /> \int_{-\infty}^{\infty} Y(\tau) \frac{\partial Y(\omega-\tau)}{\partial Y(\omega)} d\tau =<br /> \int_{-\infty}^{\infty} Y(\tau) \frac{\partial \omega}{\partial \omega} \frac{\partial Y(\omega-\tau)}{\partial Y(\omega)} d\tau =<br /> \int_{-\infty}^{\infty} Y(\tau) \frac{\partial \omega}{\partial Y(\omega)} \frac{\partial Y(\omega-\tau)}{\partial \omega} d\tau = \\<br /> \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 =\\<br /> \int_{-\infty}^{\infty} Y(\tau) \frac{1}{Y&#039;(\omega)} Y&#039;(\omega-\tau) d\tau =<br /> \frac{Y&#039;(\omega) * Y(\omega)}{Y&#039;(\omega)}<br />


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 Y(\omega):


<br /> \frac{\partial}{\partial Y(\omega)} F(Y(\omega))<br />

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 y(t) and f ?

I can do it but I still have the factor of 1/Y&#039;(\omega). For the example above:

<br /> \frac{1}{Y&#039;(\omega)} j \mathcal{F}\left( t \cdot y^2(t) \right)<br />

Oh, wait, I didn't realize that I can also trivially expand 1/Y&#039;(\omega):

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

Is this correct?