Simplify the result of inverse laplace transform of a simple second order transfunction

[Ken]

I've just started learning Sympy. I wrote a few lines of code to perform a inverse laplace transform on a simple 2nd order transfunction:

H(s) = 1 / ((s+p1) * (s+p2)).

The result I got from Sympy is

(e^(p1*t) - e^(p2*t))*e^-t*(p1+p2) / (p1 - p2)

Is there a way to simplify this result to the one like in Maxima (e^-t*p1 + e^-t*p2) / (p1-p2) ?

[Christophe Bal]

Hello.

Have you triez to expand 1nd then to factorize the formula ?

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[Ken]

Hi thanks for your reply, but I am not sure what exactly do you mean by expand "then to factorize".

I found that if I set the transfer function to be H = k / ( (s+p1) * (s+p2) ), then the inverse laplace transform becomes:

k*e^(-p1*t) / (p1-p2) + k*e^(-p2*t) / (p1-p2)

which is what I prefer. It is weird to me that adding a constant 'k' change the form that Sympy chooses to show the result.

[Christophe Bal]

Can you give your code ?

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[Ken]

Please see the following (if you replace the '1' in hs with 'k', the resulted form will be different).

from sympy import *

init_printing()

var('k p1 p2 t', real=True, positive=True)
var('s')

hs = 1 / ( (s + p1) * (s + p2) )

ht = inverse_laplace_transform(hs, s, t)

[Ken]

It looks like it returns the same form with either '1' or 'k'...

I don't see I did anything different before.