Inverse Laplace Transform of 5/(s+1)exp(-2s): Solution & Explanation

[daf1]

I want to find the following inverse Laplace transform:

[5/(s+1)]exp(-2s)

I was thinking of using the second shift theorem to no avail.

I have tried the question and my answer is :
5(exp-t)u(t-2)

This does not seem to be a plausible answer since I modified the second
shift theorem.

{If only the question was to inverse [5/(s+2)]exp(-2s),
then my answer would have been 5(exp-2t)u(t-2) (which hopefully is right).
However the question is different.}

 

[EugP]

I just want to make sure I understand the equation. Is it:

$$\frac{5e^{-2s}}{s+1}$$ ?

If it is, I think this is simpler than you're making it.

$$\frac{5e^{-2s}}{s+1} = \frac{K}{s+1}$$

Multiply both sides by $$s+1$$ and evaluate at $$s = -1$$:

$$5e^{-2(-1)} = K$$

$$5e^{2} = K$$

Now plug it back in:

$$\frac{5e^{2}}{s+1}$$

Now the inverse laplace would be:

$$L^{-1} = 5e^{2}e^{-t} u(t)$$

i.e.

$$L^{-1} = 5e^{2-t} u(t)$$

I hope this helped.

 

[tacman]

The correct answer to this inverse Laplace transform is 5(exp(-t+2))u(t-2). This can be obtained by using the first shift theorem, which states that:

L{f(t-a)u(t-a)} = exp(-as)F(s)

where F(s) is the Laplace transform of f(t). In this case, f(t) = 5/(s+1), so the Laplace transform is F(s) = 5/(s+1).

Now, using the first shift theorem, we can write the inverse Laplace transform as:

L^-1{5/(s+1)exp(-2s)} = exp(2t)L^-1{5/(s+1)} = 5(exp(2t))u(t)

Note that we have used the property that L^-1{exp(as)F(s)} = f(t-a)u(t-a).

Substituting a = -2, we get the final answer as 5(exp(-t+2))u(t-2).

In summary, we have used the first shift theorem to obtain the correct inverse Laplace transform. The second shift theorem is not applicable in this case as the function inside the exponential term is not of the form f(t-a).