Partial Derivative of Convolution

Click For Summary
SUMMARY

The discussion focuses on calculating the partial derivative of a convolution, specifically the expression \(\frac{\partial}{\partial r}(x(t) * y(t, r))\). The user seeks clarification on how to differentiate when only one function, \(y(t, r)\), depends on the variable \(r\). The correct approach is established as \(\frac{\partial}{\partial r} (x * y)(t,r) = \int_{-\infty}^{\infty} x(\tau)\frac{\partial}{\partial r}y(t - \tau, r)d\tau = (x*\frac{\partial y}{\partial r})(t,r)\), confirming that the differentiation can be applied directly to \(y\) while treating \(x\) as a constant with respect to \(r\.

PREREQUISITES
  • Understanding of convolution operations in signal processing
  • Familiarity with partial derivatives in multivariable calculus
  • Knowledge of integral calculus and its applications
  • Basic understanding of functions of multiple variables
NEXT STEPS
  • Study the properties of convolution in signal processing
  • Learn about differentiation under the integral sign
  • Explore applications of partial derivatives in machine learning
  • Investigate the implications of convolution in neural networks
USEFUL FOR

Mathematicians, signal processing engineers, and data scientists interested in advanced calculus and its applications in convolutional operations.

jstop
Messages
2
Reaction score
1
TL;DR
Calculating the partial derivative of a convolution
Hello, I am trying to calculate the partial derivative of a convolution. This is the expression:
##\frac{\partial}{\partial r}(x(t) * y(t, r))##​

Only y in the convolution depends on r. I know this identity below for taking the derivative of a convolution with both of the functions only depending on t:
##\frac{d}{dt}(x(t)*y(t) = (\frac{dx(t)}{dt}*y(t)) = (x(t)*\frac{dy(t)}{dt})##​

I'm not sure how this changes when taking a partial derivative with only one of the functions depending on the variable that the partial derivative is being taken with respect to. Any help/resources would be appreciated! (This is my first post here, please let me know if I need to improve my post formatting/structure).
 
Physics news on Phys.org
Your convolution is (x * y)(t,r)=\int_{-\infty}^{\infty} x(\tau)y(t - \tau, r)\,d\tau, is it not? What happens if you differentiate that with respect to r?
 
  • Like
Likes   Reactions: Twigg
Ah thanks, I think I see what you mean:
##\frac{\partial}{\partial r} (x * y)(t,r) = \int_{-\infty}^{\infty} x(\tau)\frac{\partial}{\partial r}y(t - \tau, r)d\tau = (x*\frac{\partial y}{\partial r})(t,r)##​
 
  • Like
Likes   Reactions: Twigg

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 34 ·
2
Replies
34
Views
3K