- #1

ColdFusion85

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## Homework Statement

We are to use the convolution theorem to compute the inverse Laplace transform of the function

[tex]L=\frac{1}{s^2 + 16}e^{-2s}[/tex]

## Homework Equations

Using a table, I find that [tex]L^{-1}[\frac{1}{s^2 + 16}] = \frac{1}{4}sin(4t)[/tex]

and

[tex]L^{-1}[e^{-2s}] = \delta(t-2)[/tex]

## The Attempt at a Solution

Using the convolution theorem, [tex]L^{-1}[\frac{1}{s^2 + 16}e^{-2s}] = \int_{0}^{t} \frac{1}{4}sin(4\tau)\delta(\tau-2)d\tau[/tex]

My question is, how do I evaluate that integral? I know that when you have the delta function, e.g. [tex]\delta(t-2)[/tex] times a function f(x), and the limits are negative infinity to infinity, it is just f(2). But what if the limits are 0 to t?