Inverse laplace
((s+1)/(s^2))(e^-s)

Question

Inverse laplace
((s+1)/(s^2))(e^-s)

anonymous · Answer

Please do help

tkhunny · Answer

Have you considered $\dfrac{s+1}{s^{2}} = \dfrac{1}{s}+\dfrac{1}{s^{2}}$

anonymous · Answer

not yet
is that so?

tkhunny · Answer

I think addition of fractions still works, doesn't it?

anonymous · Answer

its a inverse laplace transform

tkhunny · Answer

Yes.  It's okay to use algebra to simplify your task.

tkhunny · Answer

$\mathscr{L}^{-1}\left(\dfrac{s+1}{s^{2}}\cdot e^{-s}\right) = \mathscr{L}^{-1}\left(\dfrac{e^{-s}}{s}\right) + \mathscr{L}^{-1}\left(\dfrac{e^{-s}}{s^{2}}\right)$ and one or both of those may look more familiar.

anonymous · Answer

is that the final answer?

tkhunny · Answer

No.  There is no inverse, yet.  Just algebra.

You don't recognize those and being sufficiently common to include in a table?

anonymous · Answer

no

anonymous · Answer

how about $$\frac{6 (s+1) }{ s^4 }$$

anonymous · Answer

its also an inverse laplace

tkhunny · Answer

How did you get "s"?  Normally, it moves to 't'.

It is very important that you know $\mathscr{L}^{-1}\left(\dfrac{e^{-s}}{s}\right) = \Phi(t-1)$

tkhunny · Answer

t wouldn't hurt also to know that $\mathscr{L}^{-1}\left(\dfrac{e^{-s}}{s^{2}}\right) = (t-1)\cdot \Phi(t-1)$

Have you seen those things?

anonymous · Answer

nope

tkhunny · Answer

Okay, I'm a little stunned by that.  It's usually Page 1 in Laplace Transform world.  Please look up the Heaviside Step Function and its Laplace Transform.

Really, why would you be given this material if you are not also introduced to the tools to achieve or to understand the solution?  Something seriously wrong.

anonymous · Answer

yes really I really dont understand this thing thats whay I beg for help 
thanks for helping.

tkhunny · Answer

Well, we won't get far without Heaviside.  Let's try the easy one.

$\mathscr{L}^{-1}\left(\dfrac{s+1}{s^{4}}\right) = \mathscr{L}^{-1}\left(\dfrac{1}{s^{3}}\right) + \mathscr{L}^{-1}\left(\dfrac{1}{s^{4}}\right)$

Now what?

anonymous · Answer

54

tkhunny · Answer

No.  Not even close.  Why are you in this material if you have not actually been introduced to it?  Please go have a very sincere chat with your academic advisor.  You will spend a very frustrated academic year if you continue on this path.