# Convolution integral

1. Mar 5, 2012

### zezima1

Determine the confolution of f with itself where f is:

f(t) = 1 for ltl<1 and 0 everywhere else

Then deduce that:

-∞ sin2ω/ω2 dω = ∏

Fouriertransform of f gives:

f(ω) = 2/√(2∏) sin(ω)/ω

and using the convolution theorem gives:

f*f = 4/√(2∏) sin2(ω)/ω2

But I'm clueless of what to do from this point. Should I evaluate the convolution integral and equate that to the above?

In that case we have:

f*f = ∫-∞ f(τ)f(t-τ) dτ

But how do I evaluate that? I can see that τ must be between -1 and 1. Thus t must be between -2 and 2? But what does that make the integral look like?

Last edited: Mar 5, 2012
2. Mar 5, 2012

### Dick

You are integrating a step function times a shifted step function where t is the size of the shift. The product is going to be 1 on some interval and 0 off that interval. Draw a picture and try working out say f*f(0), f*f(1), f*f(2)... until it becomes clear what function of t that is. Though for your given problem you only need the value of f*f(0).

And the convolution theorem actually tells you that you take the inverse fourier transform of f(ω)^2 to get the convolution f*f.

Last edited: Mar 5, 2012