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can this be a proper Laplace transform?

Started by lucy October 3, 2005
sin(s)/(s^2+4),

can it be a proper Laplace transform?

why?

-----------------

I honestly claim this is not a HW problem. Thanks a lot!

lucy wrote:
> sin(s)/(s^2+4), > > can it be a proper Laplace transform? > > why? > > ----------------- > > I honestly claim this is not a HW problem. Thanks a lot! >
Is it for a test?
I honestly claim that this is not a test problem either. thanks

"lucy" <losemind@yahoo.com> wrote in message
news:1128385887.712472.76740@z14g2000cwz.googlegroups.com...
> I honestly claim that this is not a test problem either. thanks >
Have you taken the comp.dsp oath first? McC
>>>>> "lucy" == lucy <losemind@yahoo.com> writes:
lucy> sin(s)/(s^2+4), lucy> can it be a proper Laplace transform? What do you mean by proper Laplace transform? Ray
in article 1128377002.944857.263010@z14g2000cwz.googlegroups.com, lucy at
losemind@yahoo.com wrote on 10/03/2005 18:03:

> sin(s)/(s^2+4), > > can it be a proper Laplace transform? > > why? > > ----------------- > > I honestly claim this is not a HW problem. Thanks a lot!
well since you honestly say it is not, the "constructive" way to deal with this is to find an function x(t) that has LT of X(s) = sin(s)/(s^2+4). to try to do that, try substituting s = j*w ("w" is "omega") and see if X(jw) = sin(j*w)/((j*w)^2 + 4) = j*sinh(w)/(4 - w^2) = j/2 * (exp(w) - exp(-w))/(4 - w^2) now try to find if 1/(4 - w^2) is the "proper Fourier Transform" of something. or if 1/(s^2+4) is the L.T. of something. then ask, what does multiplying X(jw) by exp(w) do? -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
lucy wrote:
> sin(s)/(s^2+4), > > can it be a proper Laplace transform?
I don't think so. The Laplace Inversion formula doesn't work for it.
> > why?
>>>>> "dave" == dave <david_lawrence_petry@yahoo.com> writes:
dave> lucy wrote: >> sin(s)/(s^2+4), >> >> can it be a proper Laplace transform? dave> I don't think so. The Laplace Inversion formula doesn't dave> work for it. What fails? I took a quick stab at it by replacing sin(s) with its power series. Then we have terms of the form s^(2*k+1)/(s^2+4), for which we know the inverse Laplace transform. I think the resulting infinite series converges. But I did leave out all the terms having to do with the initial values, so perhaps it doesn't really converge. Ray
Raymond Toy wrote:
> >>>>> "dave" == dave <david_lawrence_petry@yahoo.com> writes: > > dave> lucy wrote: > >> sin(s)/(s^2+4), > >> > >> can it be a proper Laplace transform? > > dave> I don't think so. The Laplace Inversion formula doesn't > dave> work for it. > > What fails? > > I took a quick stab at it by replacing sin(s) with its power series. > Then we have terms of the form s^(2*k+1)/(s^2+4), for which we know > the inverse Laplace transform.
Really? What is the inverse Laplace transform of s^5/(s^2+4) ? "Proper" (I'm not sure what that means) Laplace transforms usually tend to 0 as s goes to infinity.
> I think the resulting infinite series > converges. But I did leave out all the terms having to do with the > initial values, so perhaps it doesn't really converge.
I suspect you left out the key parts.
In article <1128377002.944857.263010@z14g2000cwz.googlegroups.com>,
lucy <losemind@yahoo.com> wrote:
>sin(s)/(s^2+4),
>can it be a proper Laplace transform?
>why?
The Laplace transform F(s) of an exponentially bounded function f(t) (say with |f(t)| <= K exp(Bt)) converges for Re(s) > B with |F(s)| <= K/(Re s - B). But |sin(s)| ~ exp(|Im(s)|)/2 as |Im(s)| -> infty. So sin(s)/(s^2+4) is not the Laplace transform of an exponentially bounded function. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada