Laplace and ROC of function(- [e^(-at)]u(-t))

[jaus tail]

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?

 

[I like Serena]

Hi jaus tail!

Simply calculating the Laplacian gives us:
$$\mathcal L[-e^{-at} u(-t)] = \int_{-\infty}^\infty e^{-st}\cdot-e^{-at} u(-t)dt
= \int_{+\infty}^{-\infty} e^{s\tau}\cdot e^{a\tau} u(\tau)d\tau
=\int_{+\infty}^{0} e^{(s+a)\tau}d\tau
=\frac{1}{s+a}e^{(s+a)\tau} \Big|_{+\infty}^0
=\frac{1}{s+a}
$$
with ROC $(s+a)<0\Rightarrow s<-a$.

 

[jaus tail]

Thanks for the reply.
But isn't ROC values for which denominator of laplace is positive. Like in the below exam from screen show of book.

Here ROC is s > -a, means s + a is more than 0. Laplace of function is 1 / (s + a), so it means denominator of laplace must be positive for ROC, right?

But then same book as other example as:

Why is -a to right of Y axis? here ROC is s < -a, means s + a < 0, whereas above it was s + a > 0.

 

[I like Serena]

Thanks for the reply.
But isn't ROC values for which denominator of laplace is positive. Like in the below exam from screen show of book.

Not quite.
ROC is the Region Of Convergence.
That's where the Laplacian converges.
In our case it converges iff:
$$\lim_{\tau\to+\infty} e^{(s+a)\tau} < \infty$$
And that's only if $(s+a)<0$.
All other rules about ROC derive from that and in case of doubt we have to go back to when the integral converges exactly.

In both your examples we have $e^{-(s+a)t}$ inside the integral.
After integration that will turn out unchanged due to the nature of the exponential function.
In the first example the corresponding infinite boundary is $+\infty$ meaning that we require that $(s+a) > 0$ to ensure the integral converges.
And in the second example the infinite boundary is $-\infty$ meaning that we require that $(s+a) < 0$.

 

[jaus tail]

Oh...so by converges it means the value must never go infinite. Like an energy signal? The value must always be bounded. Right?

Another doubt: In second ROC picture why have they put -a, in right side of Y axis? Shouldn't -a be on left side?

 

[I like Serena]

Oh...so by converges it means the value must never go infinite. Like an energy signal? The value must always be bounded. Right?

Another doubt: In second ROC picture why have they put -a, in right side of Y axis? Shouldn't -a be on left side?

Right!
$a$ could be either positive or negative - we don't know. So it's an arbitrary choice to put $-a$ left or right of the y-axis

 

[jaus tail]

Thanks for the help.