Can you give your code ?
Le 1 janv. 2016 22:44, "Ken" <[email protected]> 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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