Sep 6, 2015 · 3 The definition of convolution is known as the integral of the product of two functions (f ∗ g)(t)∫∞ − ∞f(t − τ)g(τ)dτ But what does the product of the functions give? Why are is it being …

I am currently learning about the concept of convolution between two functions in my university course. The course notes are vague about what convolution is, so I was wondering if anyone could giv...

Closed 9 years ago. Why do we define the convolution? Why is convolution useful? What is the purpose of the geometry of convolution of two functions in plane? Can we draw the convolution of two …

EDIT: You define convolution integral in [0, t] [0, t] for bounded signals. The integral limits depend on where your signal is non-zero. If you have two signals as you suggested f(t) =eat f (t) = e a t and g(t) …

Feb 27, 2024 · Definition of Convolution of functions of two variables Ask Question Asked 1 year, 10 months ago Modified 1 year, 10 months ago

Mar 14, 2015 · I doubt you can define convolution without some group structure on the manifold.

A shift-invariant linear operator T T is completely determined by its impulse response T(δ) = f T (δ) = f (where δ δ is the Dirac delta function). You can show that for any function g g, T(g) = f ∗ g T (g) = f ∗ …

Jul 2, 2024 · Particularly the case where one of the functions is zonal and we can define convolution as $ (f*g) (x) = \int_M g (d (x,y))f (y)d\mu (y)$. I know there is a result like this for spherical harmonics …

Oct 7, 2025 · Woronowicz's paper (Quantum deformation of Lorentz group) has some information about the convolution, but the actual definition is slightly different. Question 2.

Jun 20, 2018 · When I read the notes, a convolution is defined as: $(f*g)(x) =\\int_{-\\infty}^{+\\infty} f(\\tau)g(x-\\tau)\\rm{d}\\tau.$ What is the difference if we define a ...