# Convolutions (differential Equations)

1. May 22, 2016

### Dusty912

1. The problem statement, all variables and given/known data
Compute the convolution f*g for the given function f and g.
f(t)=cost g(t)=U2(t)

2. Relevant equations
f*g=∫f(t-u)g(u)du

3. 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.?

2. May 22, 2016

### Staff: Mentor

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

3. May 22, 2016

### 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.

4. May 22, 2016

### Dusty912

sorry for the late reply

5. May 22, 2016

### Staff: Mentor

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

6. May 22, 2016

### 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

7. May 22, 2016

### Staff: Mentor

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

8. May 22, 2016

### Dusty912

t2cos(t-u)U2(u)du

9. May 23, 2016

### Staff: Mentor

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

Last edited: May 23, 2016
10. May 23, 2016

### Dusty912

1?

11. May 23, 2016

### Staff: Mentor

You're not sure?

12. May 23, 2016

### Dusty912

no I'm sure

13. May 23, 2016

### Staff: 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.