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I am working with a system(closed-loop feedback control) that has a transfer function like this

F = 20*(s+5)/(s^4+13.5s^3+41s^2+80*s+100)

this one is in the s-domain but for the response analysis, I needed it to be on the time-domain.

I used python sympy to get the inverse Laplace transform of this

import sympy as sp
s, t = sp.symbols('s, t', real=True)
F = 20*(s+5)/(s**4+13.5*s**3+41*s**2+80*s+100)
ilt = sp.inverse_laplace_transform(F, s, t)

the thing that surprised me was that answer contained some imaginary part as well:-

Here is the image containing the resultant answer from sympy

so my quest is about the existence, correctness, and if it is possible to have an imaginary part then what could be the significance of such a system.

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  • $\begingroup$ How did you arrive at that transfer function? $\endgroup$ – Chuck Sep 30 '20 at 20:49
  • $\begingroup$ The system was given with specified Gc(s), Gp(s) so with simple equations relating output and input $\endgroup$ – ShivamPAI21 Sep 30 '20 at 21:44
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We can solve this problem by using MATLAB and specifically by applying the partial fraction expansion by means of the function residue.

The MATLAB code below

[r, p] = residue([20 100], [1 13.5 41 80 100]);

syms s;
F1 = r(1) / (s - p(1));
F2 = r(2) / (s - p(2));
F3 = r(3) / (s - p(3));
F4 = r(4) / (s - p(4));

F34 = (2*real(r(3))*s - 2*real(r(3))*real(p(3)) - 2*imag(r(3))*imag(p(3))) / ...
      (s^2 - 2*real(p(3))*s + abs(p(3)));

F = F1 + F2 + F34;
f = ilaplace(F);

disp('F(s) = ');
pretty(vpa(F1 + F2 + F3 + F4, 5));
disp(' = ');
pretty(vpa(F, 5));

disp('f(t) = ');
pretty(vpa(f, 5));

outputs the following result:

F(s) = 
  1.0121       0.13579     - 0.57396 - 0.74149i    - 0.57396 + 0.74149i
---------- + ---------- + --------------------- + ---------------------
s + 2.2394   s + 10.138   s + 0.56117 - 2.0223i   s + 0.56117 + 2.0223i

 = 
  1.0121     (1.1479 s - 2.3549) 1.0     0.13579
---------- - ----------------------- + ----------
s + 2.2394     2                       s + 10.138
              s  + 1.1223 s + 2.0987

f(t) = 
exp(-2.2394 t) 1.0121 + exp(-10.138 t) 0.13579 - exp(-0.56117 t)

   (cos(1.3356 t) - sin(1.3356 t) 1.9561) 1.1479

The key is that the fractions F3 and F4 containing imaginary numbers can be conveniently grouped up into F34, which is instead a fraction with a second-order denominator containing only real numbers that in turn yields the well-known oscillatory behavior with an exponential profile.

In general, polynomials with real coefficients may exhibit only up to complex and conjugate roots, which can be always handled following the procedure above that basically lets imaginary numbers fade away into fractions with second-order denominators.

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