Question

Convolution theorem
laplace { integral from 0 to t
(e^(-tau)cos(tau) d(tau)}

Answer 1

f(tau) and g(t-tau)
how should I manipulate e^(-tau) so that I could extract the function of g?

Answer 2

still doesn't sound right
f(tau) and g(t-tau)
f*g= integral 0-->t f(tau)g(t-tau)dtau
if f(tau)=cos(tau)
and g(t-tau)=e^(-tau) how should i manipulate the function in order to get the funtion of g

Answer 3

My first thought was something like this:
\( e^{- \tau} = e^{t - \tau - t} = e^{-t} e^{t - \tau} \)
Adding and subtracting t.

Answer 4

You get an extra e^(-t), which can be pulled outside of the integral because it does not depend on tau. From that though, I think there is a theorem to work with the e^(-t) multiplied to it.

Answer 5

for this problem, am suppose to transform it back to f*g and take the laplace and get F(s)G(s) though

Answer 6

and according to the solution manual
1/s Laplace[e^(-t) cost} is the next step
am not sure where they pull out the 1/s

Answer 7

Strange. I was looking at a table and found a specific identity that gives exactly that. But it did not explain how it was derived in terms of convolutions. Unless it is just calling one function 1?
\( \mathcal{L} \left[ f(v) \ dv \right) = \dfrac{F(s)}{s} \)
The way I was going about it does not seem to lead there.

Answer 8

\( \displaystyle \mathcal{L} \left[ \int_{0}^{t} f(v) \ dv \right] = \dfrac{F(s)}{s} \)
Fixed that...

Answer 9

I suppose...that's the only way I could explain the 1/s

Answer 10

what exactly is the convolution? lol

Answer 11

So, the g(t - tau) = 1, and f(tau) = e^(-tau) cos (tau). I've honestly never seen that one before. o.O

Answer 12

hmmm, it still hard to picture where that 1/s came from if g(t - tau) = 1, and f(tau) = e^(-tau) cos (tau was the case. I'll just flow with it!

Answer 13

wait... Laplace{1} =1/s... ahh

Answer 14

thanks again!

Answer 15

Yup, glad to help! I learned something here too.. lol
I cannot seem to get the same answer with my original route, either. So I guess that was it.