Partial derivative of convolution integral

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SUMMARY

The discussion focuses on taking the partial derivative of a convolution integral with respect to one of the functions involved, specifically the function g(t). The user presents an attempt to derive the expression, stating that the partial derivative results in the integral of L(t-τ) with respect to τ. However, another participant points out that for the convolution to be valid, g(τ) should be used instead of g(t). This indicates a fundamental misunderstanding of the convolution operation in the context of the simulation being developed.

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cdsi385
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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
 
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To be a convolution, it should have g(τ) inside the integral, not g(t).
 

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