Question

I'm having trouble with evaluating another inverse laplace transform. Posted in a moment in the comments section below.

Answer 1

\[L^{-1} \left\{ \vphantom{} \frac{1}{s^2+3s} \right\}\]

Answer 2

I'm not sure what I should do from here, but I think I'm going to fiddle with the denominator for a moment before I say "I don't know" and ask for help again.

Answer 3

It's a rather ugly solution, but I'm thinking of...well, no, nevermind, I can't do that. Assume that the Heaviside/Unit Step Function cannot be used and you can't "shift" anything. How should this be approached, then?

Answer 4

@perl

Answer 5

use partial fractions

Answer 6

1 / ( s^2 + 3s) = 1 / ( s ( 3+s) ) , now find partial fraction form

Answer 7

i would like to use the draw button, but its not working :/ so i have to type things horizontally

Answer 8

1 / ( s^2 + 3s) = 1 / ( s ( 3+s) ) = A / S + B / ( s + 3)

Answer 9

so solve A, B for
1 / ( s ( 3+s) ) = A / S + B / ( s + 3)

Answer 10

Awesome! I just needed that initial idea, thank you.

Answer 11

after you find A, B, then you can use your Laplace table (known transforms)

Answer 12

hey we can use another property of inverse laplace here
\[L^{-1}\frac{ f(s)}{ s }=\int\limits_{0}^{t}F(t).dt\\so~we~Have\\f(s)=\frac{ 1 }{ s(s+3) }\\now, ~L^{-1}\frac{ 1 }{ s(s+3) }=\int\limits_{0}^{t}L^{-1}\frac{ 1 }{ s+3 }.dt\\=\int\limits_{0}^{t}e^{-3t}.dt\]
this makes it easiler @perl @Mendicant_Bias

Answer 13

Yeah, wait a minute, when I use partial fraction decomposition, I get a contradiction; can it still be done with PFD?

Answer 14

\[(A+B)s + 3 = 1\]This relationship must be true for all values of s, set s equal to zero.

\[1 \neq 3\]

Answer 15

And @sidsiddhartha , did you just use convolution, or what? I'm totally unaware of that 1/s property and can't seem to find it anywhere else.

Answer 16

no i did'nt use convolution
if u multiply a function in S-Domain with s then it means-diffrentiation
and if u multiply with 1/S then it means integration

Answer 17

take a small example

Answer 18

\[L^{-1}\frac{ s }{ s^2+1 }=?\]

Answer 19

so here we can see \[\frac{ 1 }{ s^2+1 }~is~multiplied~with~s\]

Answer 20

so first take inverse laplace of it