Hi Daam,
To get the stability margins at 'e', I believe this is correct:
Le = getLoopTransfer(CL,'e',-1); % open-loop function at e
[margins] = allmargin(Le);
To get the stability margins at 'u_c', I believe this is correct.
L_P_input = getLoopTransfer(CL,'u_c',-1);
[margins_input] = allmargin(L_P_input); % Margins at the plant input
Note that I removed the loop break at 'y2' which is 'a_n'. You could compute the margins at u_c with the loop broken at 'a_n', but the real system has the loop closed at a_n.
Similary, at 'q'
L_P_out_q = getLoopTransfer(CL,'y1',-1);
[margins_q] = allmargin(L_P_out_q); % Margins at the output of q
If in doubt, it's always easy to verify the gain margins (and not so difficult to verify the phase margins). For example, rebuild your system with the nominal value of Kq multiplied by a gain margin from margins_q.GainMargin. The closed loop system should have poles (or one pole at the origin) on the imaginary axis at the corresponding margins_q.GMFrequency.
"I found that breaking the loop at both a_n and Kp*a_n and only at a_n gives the same transfer function when evaluating q."
Breaking the loop at a_n eliminates the feedback loops through Kp and Ki. So additionally breaking the loop at the output of Kp doesn't change anything. In either case, the loop transfer at q
L_q = getLoopTransfer(CL,'y1',-1,{'a_n'});
should be L_q(s) = -Kq*G21(s).
"Futhermore, I also saw that the OL transfer functions for u_c and q are exactly the same (when I break the outer loop at a_n)."
When you open the loop at a_n, the two outer loops are eliminated when computing the open loop transfer at either u_c or q. The lone remaining loop contains both q and u_c, so the open loop tranfer fucntion at either should be L(s) = -Kq*G21(s).
If you save G, Kp, Ki, Kq in a .mat file and upload that file using the paperclip icon on the Insert menu, then we can verify all of the above, which at this point is based only on my view of the diagram.
"Lastly, is it enough to find the stability margins at the highlighted locations? I reckoned it wouldnt be necessary to also do it at a_n since the error is already on this signal."
As discussed above, opening the loop at a_n is actually breaking two feedback loops and the resulting open loop transfer fucntion is not the same as it would be at e. One apprach is to find the stability margins at the control inputs and feedback outputs, i.e., u_c, q, and a_n. However, you can, of course, compute stabilty margins at any point in the system should you feel there is a need to do so.
