Another Laplace Transform problem, need region of convergence help

[Color_of_Cyan]

Homework Statement

Find L[x(t)], where $$ x(t) = tu(t) + 3e^{-1}u(-t) $$

Also determine the region of convergence

Homework Equations

Laplace properties, Laplace table:

L[te^-at = 1/(s+a)²

L[u(t)] = 1/s

L[t] = 1/s²

The Attempt at a Solution

I don't really know what to do with this as my table doesn't give the product of these two.

Do you just combine them like this?:

tu(t) ---> (1/s)(1/s²)

3e^-3u(-t) ---> wait...

I don't even know if my table says anything about a transform for u(-t), could it be -1/s?

I was going to do this and then add them to each other. Also need help with RoC

 

[NascentOxygen]

tu(t) ---> (1/s)(1/s2)

No. You make use of this rule:

L t.f(t) = -F'(s)

so if you have a function f(t) whose Laplace transform you know to be F(s)
then multiplying that function by t results in a Laplace transform which can be calculated by you as the derivative of F(s) multiplied by -1

F'(s) means the derivative of F(s). So you need to know how to differentiate F(s) with respect to s.

 

[NascentOxygen]

Check that e^-1
If it's really e^-1 then that's just a constant, it's a number.

I don't know anything about u(-t)

Paging rude man ☎ ...

 

[Color_of_Cyan]

Yeah that was actually -t and I read it too fast.

Find L[x(t)], where $$ x(t) = tu(t) + 3e^{-t}u(-t) $$

Also determine the region of convergence

No. You make use of this rule:

L t.f(t) = -F'(s)

so if you have a function f(t) whose Laplace transform you know to be F(s)
then multiplying that function by t results in a Laplace transform which can be calculated by you as the derivative of F(s) multiplied by -1

F'(s) means the derivative of F(s). So you need to know how to differentiate F(s) with respect to s.

So since I'm converting it to s-domain does that mean I integrate it instead (so it becomes F(s) now)?

 

[NascentOxygen]

So since I'm converting it to s-domain does that mean I integrate it instead (so it becomes F(s) now)?

No. You start with F(s) and by differentiating F(s) you're calculating F'(s) which is what you need.

 

[NascentOxygen]

BTW, wolfram alpha is happy to give a result (but should we believe it?) for L u(-t)

http://m.wolframalpha.com/input/?i=Laplace Transform UnitStep(-t)&x=0&y=0

It is probably intended that you work out the transform yourself some way, though, not merely quote wolfram alpha.

 

[Color_of_Cyan]

Ok for that the derivative is tδ(t) + u(t)? And that's it at least for that right? Still not sure what RoC is

 

[NascentOxygen]

We're looking at only the t.u(t) term right now.

What is the Laplace Transform of u(t)?

 

[rude man]

This is a math problem more than an engineering problem. The usual application of the Laplace transform is to solve a linear differential equation with constant coefficients and with given initial conditions.

This problem on the other hand is purely math and probably purely useless, but here goes:

The mathematically correct Laplace transform is L{f(t)} = integral from -∞ to +∞ of f(t)exp(-st)dt.
In the real problems to which I referred, the transform is integral from 0 to +∞ of the same integrand.
Thus, taking your expression, the fact that it includes a u(-t) forces you to integrate from -∞ rather than zero.
In other words, and we've gone thru all this before, pick you limits to accord with the u function's argument.

As to convergence, look at the given time function. The region of convergence is simply the region where the time function does not blow up to ∞. The "region" is the region in the complex s plane, with x-axis = σ = Re{s} and y-axis = Im{s} = jw.

Example:
f(t) = exp(-at) u(t)
F(s) = integral fro 0 to infinity of exp(-at)exp(-st)dt
= integral from 0 to infinity of exp[-(s+a)t]dt
= (-1/(s+a)[exp(-(s+a)t evaluated from t=0 to infinity.

Now, you can see that, since 0<t<∞, the expression (s+a) must be positive or you get infinity for evaluating the integral between its limits.
But s = σ + jw
So σ must be > -a
and the region of convergence is the region to the right of σ = -a in the s plane, sine there σ> -a.

The u(t) term is handled similarly.

Look at the attached for more info.