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)
On Friday, January 1, 2016 at 6:58:43 PM UTC-5, Christophe Bal wrote:
>
> Can you give your code ?
> Le 1 janv. 2016 22:44, "Ken" <[email protected] <javascript:>> a écrit :
>
>> 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.
>>
>> On Friday, January 1, 2016 at 2:46:19 AM UTC-5, Christophe Bal wrote:
>>>
>>> Hello.
>>>
>>> Have you triez to expand 1nd then to factorize the formula ?
>>> Le 1 janv. 2016 03:42, "Ken" <[email protected]> a écrit :
>>>
>>>> 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) ?
>>>>
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