How to solve transfer function inside summation?

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I tried this method:
kr=600;
kp=9;
s=tf('s');
Gc_r= (0.866*s - n*50*pi)/((s^2) + 6*s + (n*100*pi)^2 );
sol= subs(Gc_r, n, 1:2:9);
f=sum(sol)
r_term=f*kr
Gc=kp+r_term
But, it is giving this error : Incorrect dimensions for raising a matrix to a power. Check that the matrix is square and the power is a scalar. To
perform elementwise matrix powers, use '.^'.
I corrected above mentioned error but it didn't work.
error : Undefined operator '.^' for input arguments of type 'tf'.
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I also tried this method:
kr=600;
kp=9;
syms s n
Gc_r= (0.866*s - n*50*pi)/((s^2) + 6*s + (n*100*pi)^2 );
sol= subs(Gc_r, n, 1:2:9);
f=sum(sol)
r_term=f*kr
Gc=kp+r_term
It didn't simplified answer either. returned in the form of sum of products.

Accepted Answer

Walter Roberson
Walter Roberson on 4 Nov 2020
Symbolic variables do not mix with tf. You cannot use subs() with a tf
You need to use your second approach. Then
[N, D] = numdem(Gc);
collect(N,s) / D
or just
collect(Gc, s)
which will collect the denominator too when the first approach leaves it factored.
  4 Comments
Walter Roberson
Walter Roberson on 4 Nov 2020
kr=600;
kp=9;
s=tf('s');
n = 1;
f = (0.866*s - n*50*pi)/((s^2) + 6*s + (n*100*pi).^2 );
for n = 3:2:9
f = f + (0.866*s - n*50*pi)/((s^2) + 6*s + (n*100*pi).^2 );
end
r_term = f*kr;
Gc = kp+r_term;
Gc
Gc = 9 s^10 + 2868 s^9 + 1.443e08 s^8 + 3.731e10 s^7 + 7.432e14 s^6 + 1.467e17 s^5 + 1.408e21 s^4 + 1.895e23 s^3 + 8.176e26 s^2 + 5.702e28 s + 6.099e31 -------------------------------------------------------------------------------------------------------------------------------------------------- s^10 + 30 s^9 + 1.629e07 s^8 + 3.908e08 s^7 + 8.551e13 s^6 + 1.539e15 s^5 + 1.661e20 s^4 + 1.994e21 s^3 + 1.003e26 s^2 + 6.019e26 s + 8.363e30 Continuous-time transfer function.
Paul
Paul on 5 Nov 2020
Though it may not matter for this particular example, you should consider using ss objects instead of tf objects for this type of computation.

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