# 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.?

## 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.

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

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

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

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1?

1?
You're not sure?

no I'm sure

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.