# Laplace transform

I want to find the Laplace transform of the following signal but I don't know what to do with the absolute value.

$$x(t)=e^{-|t|}\; u(t+1)$$

The first thing it came to my mind is to split in negative and positive sides and then find each one and add them. The problem is that I checked back to the solutions and it not the same.

Any ideas?

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Isn't $|-t|=|t|$ for all $t$? – Babak S. Jun 25 '12 at 20:02
@BabakSorouh I fixed it. Thanks. – jeanleauveux22 Jun 25 '12 at 20:04
What LT are you considering? Two sided? – Peter Tamaroff Jun 25 '12 at 20:15
@PeterTamaroff Yes two sided. – jeanleauveux22 Jun 25 '12 at 20:17

## 2 Answers

Since you're considering the two sided Laplace Transform you need to evaluate

$$\mathcal L(s)=\int_{-\infty}^\infty e^{-|t|} \theta(t+1) e^{- st}dt$$

Since $\theta(t+1)=0$ for $t<-1$ you integral becomes

$$\mathcal L(s)=\int_{-1}^\infty e^{-|t|} e^{- st}dt$$

Now we consider that for $(-1,0)$, $-|t|=t$, and for $(1,\infty)$, $-|t|=-t$, so that

$$\mathcal L(s)=\int_{-1}^0 e^{t} e^{- st}dt+\int_{1}^\infty e^{-t} e^{- st}dt$$

Can you take it from there?

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Yes I understand. What's different with the right sided LT? – jeanleauveux22 Jun 25 '12 at 20:31
@jeanleauveux22 The "right" sided is usually called "one sided" and is $\mathcal L(s) =\int_0^\infty \theta(t)f(t) e^{-st}dt$ where $\theta$ is the Heaviside step function. – Peter Tamaroff Jun 26 '12 at 0:15

$$e^{-|t|} = e^{t}u(-t)+e^{-t}u(t)$$ $$x(t)=e^{-|t|} u(t+1)=e^{t}[u(t+1)-u(t)]+e^{-t}u(t)$$ Two Side Laplace Transform : $$X(S)= \int_{-\infty}^{\infty} x(t)e^{-st}dt$$ $$X(S)=\int_{-1}^0 e^t e^{-st}dt + \int_0^{\infty} e^{-t} e^{-st}dt$$ $$X(S)=\int_{-1}^0 e^{-(s-1)t}dt + \int_0^{\infty} e^{-(s+1)t}dt$$ $$X(S)= \frac{-1} {s-1}e^{-(s-1)t} |_{-1}^0 + \frac{-1} {s+1}e^{-(s+1)t}|_0^{\infty}$$ $$X(S)=\frac{-1+e^{s-1}} {s-1} +\frac{1} {s+1}$$

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