# Laplace transform and region of convergence

1. Oct 30, 2014

### redundant6939

Find the LT and specify ROC of:
x(t) = e-at, 0 ≤ t ≤ T
= 0, elsewhere
where a > 0

Attempt:
X(s) = - 1/(s+a)*e-(s+a) integrated from 0 to T
=> -1/(s+a)[e-(s+a) + 1]
Converges to X(s) = 1/(s+a) , a ⊂ R, if Re{s} > -a for 0≤t≤T
Elsewhere ROC is empty (LT doesn't exist).

Is this correct?

2. Nov 1, 2014

### CopyOfA

The Laplace transform of $f\left(t\right)$ is:
$$\mathscr{L}\left\{f\left(t\right)\right\} = F\left(s\right) = \int\limits_{0}^{\infty} e^{-st}f\left(t\right) dt$$
Hence, the Laplace transform of your function is:
$$X\left(s\right) = \int\limits_{0}^{T} e^{-at}e^{-st}dt + \int\limits_{T}^{\infty} 0 \cdot e^{-st}dt= \int\limits_{0}^{T} e^{-\left(a+s\right)t}dt = -\dfrac{1}{s+a}e^{-\left(s+a\right)t}\bigg|_{0}^{T}$$
$$X\left(s\right) = \dfrac{1}{s+a}\left(1-e^{-\left(s+a\right)T}\right)$$
The ROC is as you stated: $\Re\left\{s\right\}>-a$