> ... To obtain a spectral
> analysis using Laplace, you essentially have to multiply a sine wave by
> a step (heaviside) function. The FT range is from -inf to inf.

The continuous FT, yes. Most work here involves the DFT, which always
"sees" the data through a window of finite extent in time.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

Reply by Fred Stevens●August 3, 20052005-08-03

Jerry Avins wrote:

> Raghavendra Mahuli wrote:
> > Hi,
> > Just to answer ur question regarding when Laplace works and when FT fai=

ls to

> > work?
> > FT works only for those signals where dirichlets condition is satisfied=

.=2E.

> > i.e. the signal should have finite number of maxima and minima.... and =

it

> > should have a finite number of discontiuities...
>
> A finite number of finite discontinuities.
> =AF=AF=AF=AF=AF=AF
> Jerry
> --
> Engineering is the art of making what you want from things you can get.
> =AF

=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF
Fourier analysis gives steady state behaviour of a dynamical system,
because one is only using the jw part of s. The Laplace transform uses
s=3Dsigma+jw where the sigma part gives transient information. Thus the
Laplace transform gives a more detailed picture of the system behaviour
because it contains both the transient and steady state parts.
fred.

Reply by Ikaro●August 3, 20052005-08-03

Raghavendra Mahuli wrote:

> Hi,
> Just to answer ur question regarding when Laplace works and when FT fails to
> work?
> FT works only for those signals where dirichlets condition is satisfied...
> i.e. the signal should have finite number of maxima and minima.... and it
> should have a finite number of discontiuities...

A sine wave has infinite number of maxima and minima....
Also, the Laplace transform is used mainly for signals with transient
components. If you look at the definition you will see that the laplace
transform starts from t=0 (and *not* -inf). To obtain a spectral
analysis using Laplace, you essentially have to multiply a sine wave by
a step (heaviside) function. The FT range is from -inf to inf.
-Ikaro

Reply by Jerry Avins●August 3, 20052005-08-03

Raghavendra Mahuli wrote:

> Hi,
> Just to answer ur question regarding when Laplace works and when FT fails to
> work?
> FT works only for those signals where dirichlets condition is satisfied...
> i.e. the signal should have finite number of maxima and minima.... and it
> should have a finite number of discontiuities...

A finite number of finite discontinuities.
������
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

Reply by ●August 3, 20052005-08-03

Hi,
Just to answer ur question regarding when Laplace works and when FT fails to
work?
FT works only for those signals where dirichlets condition is satisfied...
i.e. the signal should have finite number of maxima and minima.... and it
should have a finite number of discontiuities...
Take for example a growing exponential.. e^qt... it does not have a FT...
but it will have a laplace transform for a particular condition on which the
transform converges....
the condition is |a| > |q|... where 'a' is the real part of 's'... so if u
want to anlayzse a signal does not satisfy dirichlet's condition, laplace
is the only option.
In this sense, we say laplace transform is a more general case and FT a more
specialised case...
In the discrete domain also the same argument holds good with Z-transform
being the equivalent of Laplace transform...
regards,
raghavendra

Reply by robert bristow-johnson●July 26, 20052005-07-26

> I'll bet the LT works where the FT doesn't.
> The reason would be that, as you said, there exists some function
> where the LT converges and the FT does not.

i dunno, Rune, the LT has more trouble with x(t) = 1 than the FT has.
r b-j

Reply by Rune Allnor●July 23, 20052005-07-23

Firebird skrev:

> I understand the mathematical differences between the two - e.g. -
>
> a) LT is more general b/c it is a function of a complex variable 's',
> whereas FT is a function of an imaginary variable (real part = 0)

Agreed

> b) LT converges for a larger range of functions

I don't know, you could well be right.

> But when to use which? I have used LT for solving differential eqns and
> thus for anything that is derived from diff eqs, and FT for frequency
> analysis. But can it be the other way round? LT used for frequency
> analysis

Yep. With s = sigma + jw, set sigma = 0 and you have yor frequency
analysis.

> and FT for diff eqs?

Sure. That's done all the time. Fourier himself developed what we
know as the "Fourier series" while studing the differential equation
governing heat transfer.

> When to use which?

I would say that the LT is used when time-domain responses are
important, while the FT is used for frequency-domain analysis.
I have noted that the FT is used for signals with constant-mean
signals, while the LT can be used for, say, ramp signals.

> Why is FT preferred for frequency analysis?

Because it is specialized for the case sigma = 0, as you mentioned
above. It gets to the point with far less work than the LT. I don't
think there exists such a thing as the Fast Laplace Transform.
The Fast Fourier Transform, FFT, is used all over the place.

> Why not just have one transform that works for
> everything? The FT and LT look so similar - it certainly should be possible
> to combine the two into a more general transform (maybe LT would be enough)
> that works for everything. If this is not possible, why not?

There are two different uses. It's like cars. Some drive a general-
purpose vehicle, like an SUV, where you can move people, move cargo,
drive on the road, drive off the road. It does everything, but none
of it particularly well. Then you have the specialized vehicles,
like the F1 cars, that does one thing, go fast, extremely well, but
is utterly useless for anything else one might consider a car to do.
Perhaps not a very good comparision, but as far as I can see, the LT
is the general-purpose tool that does everything but requires lots of
work, while the FT is a more specialized tool that is more efficient
with what it does, not least because of the FFT algorithm.

> Are there any cases where only one of them works?

I don't know. If it is, I'll bet the LT works where the FT doesn't.
The reason would be that, as you said, there exists some function
where the LT converges and the FT does not.
Rune

Reply by ●July 23, 20052005-07-23

I understand the mathematical differences between the two - e.g. -
a) LT is more general b/c it is a function of a complex variable 's',
whereas FT is a function of an imaginary variable (real part = 0)
b) LT converges for a larger range of functions
But when to use which? I have used LT for solving differential eqns and
thus for anything that is derived from diff eqs, and FT for frequency
analysis. But can it be the other way round? LT used for frequency
analysis and FT for diff eqs? When to use which? Why is FT preferred for
frequency analysis? Why not just have one transform that works for
everything? The FT and LT look so similar - it certainly should be possible
to combine the two into a more general transform (maybe LT would be enough)
that works for everything. If this is not possible, why not? Are there any
cases where only one of them works?
Thanks.