Laplace transform of tu(t) + 3e^-1u(-t) and region of convergence

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Color_of_Cyan
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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
 

Discussion

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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.
 
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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 ☎ ...[/size]
 
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
NascentOxygen said:
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)?
 
Color_of_Cyan said:
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.
 
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
 
We're looking at only the t.u(t) term right now.

What is the Laplace Transform of u(t)?
 
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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.
 

Attachments