Region of Convergence of Laplace Transform

cunnins4

Hi folks, could someone help me out with this one. I can't see why the transform converges for all values of s? If Re{s}=-a would that not make the denominator zero, and then the transform wouldn't converge?

Michael Collins

Hi. You are quite correct that the above answer in undefined at s = -a + 0 j. But there is actually a valid transform at that point - if you put s = -a into the function and then integrate that, you should see it comes out to be T. (This only works because the integral limit is finite - thanks to the two unit step functions cancelling out. In general there would what's referred to as an "essential singularity" at s = -a)

Now the trick is that your final answer will also give T for s = -a if you take the limit of it as s -> -a. So in that sense, your answer is valid for all s in the complex plane.

Fremen

Not sure I understand how you got from line two to line three there. What happend to the U(t) terms, and why are you changing limits?

Sorry if I missed something obvious there.

Edit #2: Asked a dumb question in my previous edit

Michael Collins

Fremen wrote: »

Not sure I understand how you got from line two to line three there. What happend to the U(t) terms, and why are you changing limits?

Are you familiar with the Laplace Transform Fremen? The U(t) is a unit step function which is 0 for t < 0 and 1 for t >= 0. So you can adjust the infinite limits appropriately, since integrating through a range where the integrand is 0 isn't going to change the final result.

Edit: surely integral (U(t) - U(T-t)) from zero to T = 0?
Feck what am I missing?

It's U(t) - U(t-T) though (not T- t in the second term)! Which is 1 for 0 <= t < T, and 0 everywhere else. Hence that's the only range you have to worry about...

Fremen

Michael Collins wrote: »

Are you familiar with the Laplace Transform Fremen? The U(t) is a unit step function which is 0 for t < 0 and 1 for t >= 0. So you can adjust the infinite limits appropriately, since integrating through a range where the integrand is 0 isn't going to change the final result.



It's U(t) - U(t-T) though (not T- t in the second term)! Which is 1 for 0 <= t < T, and 0 everywhere else. Hence that's the only range you have to worry about...

Ah, fair enough. Was never too strong on that stuff.

cunnins4

Cheers Mick, that's not the first time you've helped me out!

I'm still unclear about the Laplace Transform though. I can perform the transform no problemo-we spent a lot of time on it last year, but unfortunately we haven't been shown the application of it really, just "Find the laplace transform of ........ *insert function here*" and this year it's "... and its region of convergence". Sometimes lecturers can be so frustrating! So much of our stuff just comes to an answer with no practical meaning (that we're made aware of!) and as student engineers that's what we're after!

Michael Collins

Do you see applications for differential equations? Laplace Transforms allow you to solve differential equations by reducing the problem to one of simple algebra (plus the added computation of finding the Transform and Inverse Transform, of course).

So, for example take the case of damped oscillations - e.g. a mass-spring system. A general example of a differential equation for this system is:

my'' + cy' +ky = 0

where y denotes the distance the spring travels about its equilibrim point
where y' denotes the derivative with respect to time t, and y'' the 2nd derivative.
m is the mass tied onto the end of the spring,
c is the damping constant - this tries to slow the motion down,
k is Hooks constant.

So as a particular example take:

y'' + y' +9y = 0, inital conditions are y(0) = 0.16, y'(0) = 0

You can now solve this using the Laplace Transform, quite easily actually (do you know how to do this?). And it takes less work than solving the diff. equation by standard ODE methods.

Son Goku

cunnins4 wrote: »

Cheers Mick, that's not the first time you've helped me out!

I'm still unclear about the Laplace Transform though. I can perform the transform no problemo-we spent a lot of time on it last year, but unfortunately we haven't been shown the application of it really, just "Find the laplace transform of ........ *insert function here*" and this year it's "... and its region of convergence". Sometimes lecturers can be so frustrating! So much of our stuff just comes to an answer with no practical meaning (that we're made aware of!) and as student engineers that's what we're after!

Well, if you understand vectors, I'm sure you are used to getting their components in a certain basis. For instance you're given a vector in Cartesian coordinates and then you get it in spherical coordinates.

You can do the same with functions. Take the polynomial:
(x^3) + 3*(x^2) + 6*(x) + 7
= (x^3) + 3*(x^2) + 6*(x^1) + 7*(x^0)

If I use powers of x as "basis functions" I could write this as a vector:
(7, 6, 3, 1, 0, 0, 0, 0,..........................)
Let's just call this the function vector.

Another example:
Sin(x) = x - (1/3!)*(x^3) + (1/5!)*(x^5) -..................
So Sin(x) would be:
(0, 1, -1/3!, 1/5!,..................)

However I could use different functions as my basis functions. For example e^(-sx).

What the Laplace transform does is obtain the components of the function vector in the e^(-st) basis. Several differential equations are easier to solve in this basis, as has been explained above.

cunnins4

Yeah I know that they're a useful for solving ODEs but we've only scratched the surface so far. Sometimes our lecturer gets a bit caught up in the theory and although it's supposed to be a module geared towards engineering problems and analysis, we don't see many examples like what you've posted above, although we have seen a couple like that. I'm gonna drop him an email and ask for some more practical examples and the implications of things like the ROC etc...

Sorry for my little rant in my second post, I was just getting frustrated with my study is all!

Thanks for the help though man, it's much appreciated!