- #1
jaus tail
- 615
- 48
Homework Statement
If laplace of [e^-(at)] u(t) is 1/(s+a) and ROC is s > -a
Find laplace and ROC of -e^(-at) u(-t)
Homework Equations
Laplace is integral over minus infinity to plus infinity of f(t) e^(-st) dt
The Attempt at a Solution
Well i integrated f(t) over the limits with e(-st)u(-t) dt
the u(-t) changes limits to from (-inf to +inf) to (-inf to zero.)
So i integrated it and got Laplace as 1/(s+a)
Now ROC is region for which laplace goes positive that is more than zero
So in case of (1/(s+a)) s+a must be > 0 for F(s) to be positive.
For that s > (-a)
The book says that laplace is right but ROC is s < (-a)
Book didnt integrate. It went through other formula.
like L(f(t)) is F(s)
then laplace of f(-t) is F(-s)
so Laplace of e^(at)u(-t) becomes 1/(-s + a) ROC is -s + a > 0 so -s > -a
Now a becomes -a
so Laplace of e ^ (-at) u(-t) becomes 1/(-s -a) = -1 /(s + a) ROC is s + a > 0 so s > -a
Multiply both side f(t) and F(s) by -1
Laplace of -1 * e ^ (at) u(-t) becomes 1 / (s + a) Here it said that ROC remains same as power remains same. And ROC is -s > a
But shouldn't ROC for (1/(s+a) ) be s + a > 0, so s > -a
I'm confused as to why is ROC different. What does underlined part mean?