# Why am i getting an unknown variable while taking the inverse Laplace of a 4th order system?

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Preeti Prajapati on 13 Jul 2015
Commented: Star Strider on 19 Jul 2015
I am calculating the inverse Laplace of a 4th order transfer function with the following code:
syms s t
B=((0.8236*s^2+0.4877*s+1.315)/(s^4+1.385*s^3+1.978*s^2+1.317*s));
y=ilaplace(B,s,t)
Here is the ilaplace i get:
y =
1315/1317 - (3682969*sum((r3*exp(r3*t))/(2*(1500*r3^2 + 1385*r3 + 989)), r3 in RootOf(s3^3 + (277*s3^2)/200 + (989*s3)/500 + 1317/1000, s3)))/6585 - (1315000*sum((r3^2*exp(r3*t))/(2*(1500*r3^2 + 1385*r3 + 989)), r3 in RootOf(s3^3 + (277*s3^2)/200 + (989*s3)/500 + 1317/1000, s3)))/1317 - (19587691*sum(exp(r3*t)/(2*(1500*r3^2 + 1385*r3 + 989)), r3 in RootOf(s3^3 + (277*s3^2)/200 + (989*s3)/500 + 1317/1000, s3)))/13170
After that i am calculating integral squared error with the following code:
h=(1/s);
Y=ilaplace(h,s,t);
Ise=int((Y-y)^2,t,0,100)
and getting :
Ise =
((5260*exp(100*r3))/1734489 + (1729225*exp(200*r3))/3468978 - 193305/385442)/r3 + 400/1734489
Now, the problem is that i am getting this unknown variable 'r3' due to which i am unable to calculate a numerical value of integral squared error when i am using :ISE=double(Ise)
Can anyone please let me know how to get a numerical value of integral squared error. Thank you in advance.
Star Strider on 19 Jul 2015
Yes.
I also posted the inverse Laplace transform of the partial fraction decomposition.