      # Cas Question Laplace, Z-transf, Fourier

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### #1 Griott

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Posted 08 August 2004 - 09:40 PM

CAS question Laplace, Fourier, Z-Transf

Hello, I have been asked a question which I think you could resolve. I was asked if the CP300 is able to do Laplace transformation, z-trasnformation and fourier expansion.

For the Laplace transformation I read in this forum that it was possible manually with a few steps. I took as a reference the Hp49, which handles this calculation with a single function:

LAP(1) -> 1/x
LAP(x) -> 1/x^2
LAP(e^x) -> 1/(x-1)

I then tried to follow the steps with the CP300 and the results confused me:

S(e^(-s*t) * 1, t, 0, oo) -> Undefined
S(e^(-s*t) * t, t, 0, oo) -> Undefined
S(e^(-s*t) * e^t, t, 0, oo) -> Undefined

Then, as somewhere in this forum it says that the AFX was able to do these calculations I tried with it:

S(e^(-s*t) * 1, t, 0, oo) -> 1/s - (e^(-oo*s)/(s))
S(e^(-s*t) * t, t, 0, oo) -> 1/s^2 - (oo*e^(-oo*s))/(s)
S(e^(-s*t) * e^t, t, 0, oo) -> 1/(s-1) - (oo*e^(-oo*s))/(s-1)

The things that confused me are the following:
1. If it is supposed that the CP300's CAS is better than that of AFX, then why it gives an undefined result while the AFX gives a defined result?. Is this a bug on the Cp300's CAS.
2. What is the long second term involved in the AFX's answer, the one that differs from te HP49 answer (which I assumed as the correct one)? Can we tell the AFX's answer is also correct?

I used S as the integral symbol and oo as infinity.

Additionaly I was asked for the z-trasnformation, for which I have no clear idea of the steps to accomplish it.

For the fourier expansion I know that the ECON addin application comeswith a functionality of taking sampled microphone/audio data and making the expansion from a selected range, for which it can generate the corresponding function. Besides that, I dont know if a program can be made to do this task.

Hope you could help with this questions. Thanks a lot

CrimsonCasio: I deleted the duplicate topic in the FDF, i assume it was just a mistake.

### #2 Overlord

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Posted 08 August 2004 - 10:18 PM

after a search about laplace transformation :

the laplace transform (not sure about translation) of the function f(t)=1
is p -> 1/p
but the integral is convergent only if p > 0 (if p<0 the laplace transform doesn't exist). that explains the undefined result : the cp doesn't know if your s if positive or negative, so it can't show an answer

it explains the long answer of the AFX : is s positive, e^(-inf*s)/s = 0
if s negative, e^(-inf*s)/s = -inf and the expression 1/s - e^(-inf*s)/s = +inf
the AFX shows a tricked answer.

for f(t)=e^at, the integral converges only if s > a so it's more complicated

you can't do a laplace transformation only with the integral, without the conditions s>0, s>a, ...)

to implement a laplace function, it must be asserted that the conditions are always respected, and when you enter a "s" it refers to any number

i can't help you more for the moment, i know nothing about laplace, fourier and z-trans ### #3 Daruosh

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Posted 09 August 2004 - 11:58 AM

Nice topic.
How can we consider above conditions?. I tried S(e^(-abs(s)*t) * 1, t, 0, oo) and got right answer, but for other expresions these conditions are not as simple as this. I tried yo use with (|) operator, but I got undefined answer: (S(e^(-s*t) * e^s, t, 0, oo)|s>1).

Does anyone know how we can use | operator in integral command????

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