Question

convolution of y(t)=exp(-3t)u(t)* u(t+3) is??

Answer 1

having a problem to eliminate the step function from the integral can anyone help?

Answer 2

So you already put this into the form of an integral? If so, what did you have so far?

Answer 3

L[exp(-3t)u(t)]=1/(s+3),
L{u(t+3)}=exp(3s)/s
so the out put is {1/(s+3)}*{exp(3s)/s}=exp(3s)/s(s+3)
now i'm stuck what to do next?

Answer 4

should i take inverse laplace now?@AccessDenied

Answer 5

Oh, yes. The convolution equals the inverse Laplace of the product of their Laplace transforms.

Answer 6

hold on i'm doing it

Answer 7

=1/3.exp(3s){1/s -1/(s+3)}
=1/3{1-exp(-3t)}u(t+3)
it is correct @AccessDenied

Answer 8

i mean
inverse laplace[1/3.exp(3s){1/s -1/(s+3)}]=1/3{1-exp(-3t)}u(t+3)

Answer 9

is it okay @AccessDenied

Answer 10

Inverse Laplace of 1/3 exp(3s) (1/s - 1/(s+3)) = 1/3 inv.laplace(exp(3s) 1/s) - 1/3 inv.laplace(exp(3s) 1/(s+3))
Property: inverse laplace of e^(-cs) F(s) = u(t-c) f(t-c)
So inverse laplace of exp(3s) 1/s) = u(t+3) *1
inverse laplace of exp(3s) 1/(s+3) = u(t+3) * e^(-3(t+3))
Is what I was getting