# Convolutions (differential Equations)

## 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
Mentor

## 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.

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.

Mark44
Mentor
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?

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
Mentor
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?

t2cos(t-u)U2(u)du

Mark44
Mentor
t2cos(t-u)U2(u)du
But what is U2(u) on that interval? What's the value of any Heaviside function?

Last edited:
1?

Mark44
Mentor
1?
You're not sure?

no I'm sure

Mark44
Mentor
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.