Help me understand this convolution function

[Highlander-SP]

if i have a a function g(t) and a Heaveside function = h(t) , the convolution g(t)*h(t) = ∫g(σ).h(t-σ) dσ

. why h(t-σ) is 1 inside the integral?

. in same way, if h(t) was a delta Dirac function, why it's is 1 also?

tks

 

[neoaspilet11]

convultion of the heaviside function

Hello,

The Dirac delta function has a unique mathematical property. Geometrically it is viewed as an impulse or spike in y - axis on very short interval.
One of its interesting property is that

∫f(x)δ(x-a) = f(a).

The above expression means that if a function if convoluted over a delta function, the result is the function itself.

To answer your questions, you have to clarify first what is your f(x). Generally, the convoluted function - the function that has been shifted to (x -a) is not unity or 1! It happens only in speacial cases like if it is a Heavide or Delta function.

If you send to me your f(x) and g(x) and you want me to to find the result of the convolution of these two functions I'll be happy to help you. Just email me at neoaspilet11@yahoo.com.

Added after 7 minutes:

and Heaviside function has also a very interesting and very important property and that is


h(t - a) = 0, t < a and
= 1, t=>a

This explains why you get unity in the heaviside function inside the integral. I am asking you before what is your g(t) or you first function because you werent sure if it is defined and convergent from t = a to infinity. Because you will use it in the convolution

 

[Highlander-SP]

convolution of heaviside function

Hy Neo,

In my example, if h(t) is a Heaveside function, why its integral of h(t-σ) is 1?

rgds,

 

[neoaspilet11]

function convoluted with heaviside

Hello Highlander,
Yes it should be 1. becuase heaviside or Unit step function is ZERO from negative infinity to delta < t and ONE from delta <= t to positive infinity. delta is you variable of integration.

Thanks

 

[Highlander-SP]

unity of convolution function

but what's the explanation for that? How to desenvolve it?

the convolution is: ∫ g(σ).h(t-σ) dσ ; the h(t-σ) is a Heaviside function shifted in σ and g(t) is any function. How to resolve it step by step?


tks

 

[neoaspilet11]

convolution of the heaviside function

h(t-δ) is a Unit Step function
∫g(δ) h(t-δ)dδ = ∫g(δ) * 0 *dδ (limits negative infinity to δ = t) +
∫g(δ) *(1)* dδ (limits δ = t to positive infinity)

I think this might help.