Without Prefix: How to Evaluate the Convolution of f with Itself?

In summary, the conversation discusses determining the convolution of a function f with itself, where f is defined as 1 for ltl<1 and 0 everywhere else. It also introduces the Fourier transform of f and the convolution theorem. To solve the problem, one needs to evaluate the convolution integral f*f and take the inverse Fourier transform of f(ω)^2. The result of this process is 4/√(2∏) sin2(ω)/ω2.
  • #1
zezima1
123
0
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:
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  • #2
zezima1 said:
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?

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.
 
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What is a convolution integral?

A convolution integral is a mathematical operation that combines two functions to produce a third function. It is used to describe how the shape of one function is altered when it is passed through another function.

What is the purpose of a convolution integral?

The purpose of a convolution integral is to help understand how systems respond to input signals. It is commonly used in signal processing, image processing, and other areas of science and engineering.

What is the difference between a convolution integral and a regular integral?

A regular integral calculates the area under a curve, while a convolution integral calculates the product of two functions integrated over a specific range. Additionally, a regular integral is a scalar value, while a convolution integral produces a new function as the result.

How is a convolution integral calculated?

A convolution integral is calculated by first flipping one of the functions, then sliding it across the other function, multiplying the two functions at each point, and finally integrating the product over the entire range of the function.

What are some real-world applications of convolution integrals?

Convolution integrals have many applications in science and engineering, such as in signal and image processing, electrical engineering, and physics. They are also used in fields such as economics, finance, and statistics to model and analyze complex systems.

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