Convolution of a box function with itself Ask Question Asked 10 years, 9 months ago Modified 5 months ago

Transforms such as Fourier transform or Laplace transform, takes a product of two functions to the convolution of the integral transforms, and vice versa. This is called the Convolution …

Convolution of two independent uniform Random Variables Ask Question Asked 8 years, 10 months ago Modified 8 years, 10 months ago

Since the Fourier Transform of the product of two functions is the same as the convolution of their Fourier Transforms, and the Fourier Transform is an isometry on $L^2$, all we need find is an …

Convolution with sinc pulses What we want to do to reconstruct the signal is a convolution between the samples and scaled and shifted versions of sinc. This technique is known as …

But in general, convolution of functions is almost a ring (there's no exact identity element). The linear space of compactly supported distributions forms an actual ring under convolution, and …

Explore related questions integration probability-distributions normal-distribution convolution See similar questions with these tags.

I'm wondering if the 2D laplacian kernel 0 1 0 1 -4 1 0 1 0 is also a separable kernel. How can I find that out?

1 The purpose of this answer is to show how a direct application of convolution may lead to the desired result. I take the following results from Cohn, Measure Theory. Definition of …

Apr 16, 2016 · You should end up with a new gaussian : take the Fourier tranform of the convolution to get the product of two new gaussians (as the Fourier transform of a gaussian is …