Obviously I'm doing something wrong here
Why am i getting such a strange answer for an inverse laplace??
syms stf = 10/(s^4 + 6*s^3 + 8*s^2 + 10*s)>> ilaplace(tf)ans =1 - 8*symsum(exp(root(s3^3 + 6*s3^2 + 8*s3 + 10, s3, k)*t)/(3*root(s3^3 + 6*s3^2 + 8*s3 + 10, s3, k)^2 + 12*root(s3^3 + 6*s3^2 + 8*s3 + 10, s3, k) + 8), k, 1, 3) - 6*symsum((root(s3^3 + 6*s3^2 + 8*s3 + 10, s3, k)*exp(root(s3^3 + 6*s3^2 + 8*s3 + 10, s3, k)*t))/(12*root(s3^3 + 6*s3^2 + 8*s3 + 10, s3, k) + 3*root(s3^3 + 6*s3^2 + 8*s3 + 10, s3, k)^2 + 8), k, 1, 3) - symsum((exp(root(s3^3 + 6*s3^2 + 8*s3 + 10, s3, k)*t)*root(s3^3 + 6*s3^2 + 8*s3 + 10, s3, k)^2)/(3*root(s3^3 + 6*s3^2 + 8*s3 + 10, s3, k)^2 + 12*root(s3^3 + 6*s3^2 + 8*s3 + 10, s3, k) + 8), k, 1, 3)
The answer is still in terms of S???
It did not even do the inverse? Its only a 4th order function? I don't get it
This TF happens to be the the step response of a system.
My goal was to transfer it back to the time domain so i could plot it in time.
I wanted to compare the manual plot i made to the internal step function of matlab
step(tf_closed_loop)
Obviously they should of been the same
but i can't get a reasonable answer from ilapalce???
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