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Odd Output From Inverse Laplace Transform

조회 수: 2 (최근 30일)
Alex
Alex 2012년 2월 14일
댓글: sach van 2019년 4월 10일
Hello All,
I'm trying to find the inverse laplace transform of the function:
(((s^3)+0.2*(s^2)+(0.8*s)+1)^-1)
OR equivalent
1/(((s^3)+0.2*(s^2)+(0.8*s)+1)
Here is my code:
syms step3S;
step3S=(((s^3)+0.2*(s^2)+(0.8*s)+1)^-1)*(1/s);
step3Y=ilaplace(step3S,s,t)
Here is the output:
1 - sum((4*exp(r15*t) + r15*exp(r15*t) + 5*r15^2*exp(r15*t))/(15*r15^2 + 2*r15 + 4), r15 in RootOf(s15^3 + s15^2/5 + (4*s15)/5 + 1, s15))
Why doesn't this have a clean execution on the output in terms of sin, cosine, and exponentials? I did this with several functions but MatLab balked at this one for some reason. I tried it in Mathematica and it spit right out but I don't understand what I'm doing wrong in Matlab.
If someone could propose a method that will output this inverse laplace transform clean, I would greatly appreciate it!
Thank you , - Alex
  댓글 수: 3
Andrew Newell
Andrew Newell 2012년 3월 6일
The first line should be
syms s t
sach van
sach van 2019년 4월 10일
Use vpa(ilaplace()) to output correctly

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답변 (1개)

Walter Roberson
Walter Roberson 2012년 3월 6일
Do you want the numeric form, or do you want the analytic form?
By default MuPad converts all floating point values in to rational values, and then solves using those rational values. Pure rational coefficients are the signal to MuPAD to find an analytic solution. The analytic solution happens to have RootOf(), corresponding to the 3 roots of the cubic in the denominator of the original expression.

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