Transfer Function Sensitivity & Complete Derivations

조회 수: 5 (최근 30일)
Richard
Richard 2023년 11월 5일
답변: Sam Chak 2023년 11월 6일
I have a transfer function block which looks like the following in SIMULINK:
Ignore the temporary error messages as 'a', 'k1', & 'k2' are not defined. When I find the CLTF of this system or T(s) I get this expression:
I am unsure how to derive these steps (if they are even correct!) in MATLAB.
My second problem is that I cannot determine the following sensitivity paramaters, definitely not in MATLAB and unsure about my own derivations:
=
&
(chain rule!)
I have = , however I did not expect the denominator powers and I remain unsure. For I simply could not figure this problem out myself.
I would appreciate any help on this subject matter, thank you!

답변 (1개)

Sam Chak
Sam Chak 2023년 11월 6일
I followed the formulas you provided and arrived at these solutions.
syms s a k1 k2
Gp = 1/(s^2 + a*s);
Gc = k1 + 1;
H = k2;
G = Gc*Gp;
% Closed-loop transfer function
Gcl = G/(1 + G*H);
T = simplify(Gcl)
T = 
% sensitivity of T to changes in Gp
STG = 1/(1 + (Gc*Gp)*H);
STG = simplify(STG, 'steps', 100)
STG = 
% sensitivity of Gp to changes in a
dGda = diff(Gp, a);
dGda = simplify(dGda, 'steps', 100);
SGa = dGda*a/G;
SGa = simplify(SGa, 'steps', 100)
SGa = 
% sensitivity of T to changes in a
STa = STG*SGa;
STa = simplify(STa, 'steps', 100)
STa = 

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