cdsi385
Oct17-11, 10:39 AM
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
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