# Inverse Laplace transform of ex(-s)/(s+3)^3

1. Dec 15, 2008

### rootX

matlab:
>> ilaplace(exp(-1*s)/(s+3)^3)

ans =

1/2*heaviside(t-1)*(t-1)^2*exp(-3*t+3)

But I think there should be an additional exp(3) multiplied by all.

=exp(-1*s)/(s+3)^3
=exp(-1*(s+3)+3)/(s+3)^3
Taking out exp(-3t). In above step I did (s+3)+3 because exp(-3t) says all variables should be shifted by 3 ..
=[exp(-1*(s)+3)/(s)^3]*exp(-3t)
Now I take care anything but Heaviside
=[exp(-1*(s))]*exp(-3t)*[t^2/2]*exp(3)
now I take care of Heaviside
=exp(-3(t-1))*[(t-1)^2/2]*exp(3)

I might have made some error while explaining what I was doing - just ignore it. But, the thing I am concerned about is the one in red. It gives me exp(3) in the end which matlab does not.

Thanks

2. Dec 15, 2008

### gabbagabbahey

Re: ilaplace(exp(-1*s)/(s+3)^3)

This makes no sense. Are you trying to apply a frequency-shift or a time-shift? The $e^{-s}$ should correspond to a time shift. The rule for such is:

$$u(t-a)f(t-a)=\mathcal{L}^{-1}[e^{-as}F(s)]$$

where $f(t)=\mathcal{L}^{-1}[F(s)]$ and $u(t-a)$ is the Heaviside function.

In this case, you would take $a=3$ and $$F(s)=\frac{1}{(s+3)^3}$$. What does that make $f(t)$? And so $f(t-a)$ is___?

3. Dec 15, 2008

### rootX

Re: ilaplace(exp(-1*s)/(s+3)^3)

I was using the formula:

ILaplace{F(s-3)} = ILaplace{F(s)}e^3t
Yes exp(-as) corresponds to Heaviside but shouldn't it also be exp(-a(s-3)) if I want to take out e^3t (or use the above formula)?

For F(s-3) e^3t makes it F(s) or f(t) .. (?)

4. Dec 15, 2008

### gabbagabbahey

Re: ilaplace(exp(-1*s)/(s+3)^3)

Okay, so you are doing a frequency shift first and then a time-shift.

So for the frequency shift you have:

$$\mathcal{L}^{-1}[\frac{e^{-s}}{(s+3)^3}]=e^{-3t}\mathcal{L}^{-1}[\frac{e^{-(s-3)}}{s^3}]$$

But when you apply the time-shift, you seem to be shifting the $e^{-3t}$. Why would you do that? The $e^{-3t}$ is just a factor multiplying the stuff you are taking the inverse transform of; it shouldn't be affected.

Last edited: Dec 15, 2008