# Homework Help: Laplace and ROC of function(- [e^(-at)]u(-t))

1. Nov 15, 2016

### jaus tail

1. The problem statement, all variables and given/known data
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)

2. Relevant equations
Laplace is integral over minus infinity to plus infinity of f(t) e^(-st) dt

3. 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 shouldnt 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?

2. Nov 15, 2016

### 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 $(s+a)<0\Rightarrow s<-a$.

3. Nov 16, 2016

### jaus tail

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.

#### Attached Files:

File size:
13.3 KB
Views:
80
4. Nov 16, 2016

### I like Serena

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 $(s+a) < 0$.

5. Nov 16, 2016

### 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?

6. Nov 16, 2016

### I like Serena

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

Last edited: Nov 16, 2016
7. Nov 18, 2016

### jaus tail

Thanks for the help.