It looks like it returns the same form with either '1' or 'k'...
I don't see I did anything different before.
On Friday, January 1, 2016 at 8:28:01 PM UTC-5, Ken wrote:
>
> 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]> 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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