Question

evalute using convolution theorem
1/(s+1)(s+9)^2

Answer 1

To be honest I've never even come across the convolution theorem. Sorry I can't be of more help! I'll tag a few people of higher level who may be able to assist.
@agent0smith @primeralph @dan815

Answer 2

thanks for ur coperation miss sarahusher>>>...........

Answer 3

@TuringTest

Answer 4

I suppose the theorem we are supposed to use here is\[\mathcal L^{-1}\{F(s)G(s)\}=(f*g)(t)\]

Answer 5

so let \[F(s)=\frac1{s+1}\implies f(t)=e^{-t}\]by form 2 on this page http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx
and\[G(s)=\frac1{(s+9)^2}\implies g(t)=e^{-9t}t\]by form 23
so now we need to use the theorem

Answer 6

\[\large(f*g)(t)=\int_0^tf(t-\tau)g(\tau)d\tau=\int_0^te^{-(t-\tau)}\cdot e^{-9\tau}\tau d\tau\\\large\mathcal L^{-1}\left\{{1\over(s+1)(s+9)^2}\right\}=\int_0^te^{-t-8\tau}\tau d\tau\]which looks doable by integration by parts

Answer 7

Turing 1stly thanks alot for solving my question..............r u a person or a comunity????

Answer 8

I'm just a student living in Mexico :)

Answer 9

WHATS UR NAME???????