Convolutions (differential Equations)

[Dusty912]

Homework Statement

Compute the convolution f*g for the given function f and g.
f(t)=cost g(t)=U₂(t)

Homework Equations

f*g=∫f(t-u)g(u)du

The Attempt at a Solution

So I pretty much only know how to plug in the functions into the integral for convolutions. Not really sure how to evaluate it with the U₂(t)

f*g=∫cos(t-u)U₂(u)du

so do I evaluate thi integral directly or are there other methods to evaluate this.?

 

[Mark44]

Homework Statement

Compute the convolution f*g for the given function f and g.
f(t)=cost g(t)=U₂(t)

Homework Equations

f*g=∫f(t-u)g(u)du

The Attempt at a Solution

So I pretty much only know how to plug in the functions into the integral for convolutions. Not really sure how to evaluate it with the U₂(t)

f*g=∫cos(t-u)U₂(u)du

so do I evaluate thi integral directly or are there other methods to evaluate this.?

What does U₂(t) mean in this context? I think I know, but I want to be sure that you do, as well.

 

[Dusty912]

well it is a heaviside function. The U₂(t) has a discontinuity at 2. 2 is when the function gets "turned on" so to speak.

 

[Dusty912]

sorry for the late reply

 

[Mark44]

well it is a heaviside function. The U₂(t) has a discontinuity at 2. 2 is when the function gets "turned on" so to speak.

What effect does this have on your integral as well as on the limits of integration?

 

[Dusty912]

Well it causes for the integral to be evaluated from 2 to t. Because the function would be 0 when t is less than 2

 

[Mark44]

Well it causes for the integral to be evaluated from 2 to t. Because the function would be 0 when t is less than 2

OK, so what does your convolution integral look like then?

 

[Dusty912]

∫^t₂cos(t-u)U₂(u)du

 

[Mark44]

∫^t₂cos(t-u)U₂(u)du

But what is U₂(u) on that interval? What's the value of any Heaviside function?

 

[Dusty912]

1?

 

[Mark44]

1?

You're not sure?

 

[Dusty912]

no I'm sure

 

[Mark44]

So, if U₂(t) = 1 for t ≥ 2, what does the integral now look like? This is what I was asking you in post #7.

U₂ should NOT appear in the integral.