- #1
divB
- 87
- 0
Hi,
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 ...
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 ...
Last edited: