Convolution with sign function

[PhysicsGirl90]

Hello all,

I am having some trouble calculating a convolution. For the question, and my attempt at a solution, please take a look at the pic. Can someone please point me in the right direction?

 

[vela]

I didn't check your work closely. If it's correct, you probably want to now look at two cases separately: t<0 and t>0.

The other suggestion I'd have is to try using $\tau-t$ as the argument of sgn instead of the exponential. It might make the integral easier to analyze.

 

[PhysicsGirl90]

Ok, I will try it and see what i get

 

[rude man]

Tried it that way, got 2exp(-t) -1

 

[paranormal]

Sure, I'd be happy to help. Convolution with the sign function can be a bit tricky, but here are some steps you can follow to calculate it:

1. Start by writing out the definition of convolution: (f * g)(t) = ∫f(τ)g(t-τ)dτ, where * represents convolution, f and g are functions, and τ is a dummy variable for integration.

2. In this case, our functions are the sign function, sgn(x), and another function, h(x). So the convolution will be (sgn * h)(t) = ∫sgn(τ)h(t-τ)dτ.

3. Next, we need to consider the limits of integration. Since the sign function is only defined for positive and negative numbers, we can split the integral into two parts: one for positive τ and one for negative τ. This gives us (sgn * h)(t) = ∫sgn(τ)h(t-τ)dτ = ∫[sgn(τ)h(t-τ)]dτ + ∫[sgn(τ)h(t-τ)]dτ.

4. Now, we can use the definition of the sign function to simplify the integrands. For positive τ, sgn(τ) = 1, and for negative τ, sgn(τ) = -1. So our integral becomes (sgn * h)(t) = ∫h(t-τ)dτ - ∫h(t-τ)dτ.

5. Finally, we can use a change of variables to simplify the integrals. Let u = t-τ, so du = -dτ. This gives us (sgn * h)(t) = ∫h(u)du - ∫h(u)du = 0.

So the convolution of the sign function with any other function will always be 0. I hope this helps! Let me know if you have any further questions or need clarification. Good luck!