Partial derivative of convolution integral

cdsi385
Messages
1
Reaction score
0
Does anyone know how to take the partial derivative of a convolution integral where the derivative is taken with respect to one of the functions of the convolution integral?

In the following example, the best I can come up with is:

\frac{\partial}{\partial g(t)}\int L(t-\tau)g(t)\,d\tau=\int L(t-\tau)\,d\tau

Is this correct, or does it even make sense?

To put this in context, what I usually do (successfully) is perform the convolution integral in a simulation (without the partial differentiation) where L(t) is the impulse response function of a system and g(t) is the velocity of my system which is calculated on the fly during the simulation.

What I'm trying to do now is make a new simulation which relies on this partial derivative which I'm trying to express analytically before simulating it. If what I've expressed above is correct then all I need to simulate is: \int L(t-\tau)\,d\tau

Thanks in advance...
cdsi385
 
Physics news on Phys.org
To be a convolution, it should have g(τ) inside the integral, not g(t).
 

Similar threads

Replies
2
Views
3K
Replies
1
Views
3K
Replies
4
Views
3K
Replies
1
Views
1K
Replies
4
Views
3K
Replies
2
Views
2K
Back
Top