Problem with Laplace transform
01-18-2016, 11:58 AM
Post: #1
 Natoe Junior Member Posts: 5 Joined: Sep 2015
Problem with Laplace transform
Hi !

When i want to do a Laplace transform like :

Laplace(u(t),t,p) i want to have for the result : 1/p but didn't have this ...

And when i do ILaplace(1/p) i have for result : Dirac/p ...

What is the problem ? Ans how can i fix it ?

Thanks and have a Nice day !
01-18-2016, 02:13 PM
Post: #2
 JMB Member Posts: 95 Joined: Jan 2016
RE: Problem with Laplace transform
Hi

Both the "laplace" and "ilaplace" functions have three parameters: expression, variable of the expression, and variable of the result. The last two variables are optional, but it looks like if you don't specify them, the calculator uses the variable x as the default variable.

With firmware 2015 6 17 (8151), you have the following results:

laplace(1): 1/x

laplace(1,t,p): 1/p

ilaplace(1/x): 1

ilaplace(1/p): Dirac(x)/p ---> wrong

ilaplace(1/p,p,t): 1

So, it seems that sometimes the two last variables are necessary, in order to get correct answers.

I hope this heps.
01-18-2016, 03:09 PM
Post: #3
 parisse Senior Member Posts: 1,013 Joined: Dec 2013
RE: Problem with Laplace transform
The default variable is x, therefore nothing is wrong.
01-18-2016, 04:52 PM
Post: #4
 JMB Member Posts: 95 Joined: Jan 2016
RE: Problem with Laplace transform
You're right.

The case ilaplace(1/p) = Dirac(x)/p is correct because p is interpreted by the HP Prime as a constant.
01-18-2016, 10:51 PM (This post was last modified: 01-18-2016 10:52 PM by Natoe.)
Post: #5
 Natoe Junior Member Posts: 5 Joined: Sep 2015
RE: Problem with Laplace transform
Thanks ! , but for the U(t) like in this exemple :

http://image.noelshack.com/fichiers/2016...-titre.png

U(t) on Hp prime = "1" because U(t) = 1/P and when i did : "ilaplace(1/p,p,t)" the result is "1"

So i did on the hp prime :"Laplace((1+(t-2)².1(t-2),t,p) for the exercice and the result is wrong or not the same ...

01-19-2016, 08:21 AM
Post: #6
 JMB Member Posts: 95 Joined: Jan 2016
RE: Problem with Laplace transform
I guess that by U(t) you mean the step function. In that case you need to use the HP Prime function Heaviside().

The step function at t=0 is Heaviside(t), and the step function at other values, say t=2, is Heaviside(t-2).

Using this function you get:

laplace(Heaviside(t),t,p) = 1/p

laplace(Heaviside(t-2),t,p) = e^(-2*p)/p

laplace(Heaviside(t)+(t-2)^2*Heavisede(t-2),t,p) = (p^2+2*e^(-2*p)) / p^3
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