Hi @Davide
If you examine the transfer function 
from control input to θ, you will observe that it is a 4th-order function, with the coefficients of 
, 
, and 
all being negative. Although the proportional action of the PID controller can manipulate the 
term and the derivative action can influence the 
term, the sign of the 
term remains negative. According to the Routh–Hurwitz stability theorem, the PID-controlled system will still be unstable.
from control input to θ, you will observe that it is a 4th-order function, with the coefficients of , , and all being negative. Although the proportional action of the PID controller can manipulate the term and the derivative action can influence the term, the sign of the term remains negative. According to the Routh–Hurwitz stability theorem, the PID-controlled system will still be unstable.A 4th-order θ compensator can stabilize the 
transfer function, resulting in the inner closed-loop becoming an 8th-order system. However, placing an 8th-order system in series with a 2nd-order unstable system 
(from θ to α) results in a 10th-order transfer function. In other words, you will need to design a 10th-order α compensator. Consequently, the outer closed-loop will become a substantial 20th-order system. The stabilization gains in a 20th-order system will be ridiculously high, as you need to stabilize the inverted pendulum at a very fast rate. If your computation can handle such precision in calculations, theoretically, it will work.
transfer function, resulting in the inner closed-loop becoming an 8th-order system. However, placing an 8th-order system in series with a 2nd-order unstable system (from θ to α) results in a 10th-order transfer function. In other words, you will need to design a 10th-order α compensator. Consequently, the outer closed-loop will become a substantial 20th-order system. The stabilization gains in a 20th-order system will be ridiculously high, as you need to stabilize the inverted pendulum at a very fast rate. If your computation can handle such precision in calculations, theoretically, it will work.A practical approach is to use a parallel output-feedback PD controller configuration, where the PD gains are obtained from the LQR design method. Please see the block diagram below.
% Motor
Rm = 8.4; % Resistance
kt = 0.042; % Current-torque (N-m/A)
km = 0.042; % Back-emf constant (V-s/rad)
% Rotary Arm
mr = 0.095; % Mass (kg)
r = 0.085; % Total length (m)
Jr = mr*r^2/3; % Moment of inertia about pivot (kg-m^2)
% Pendulum Link
mp = 0.024; % Mass (kg)
Lp = 0.129; % Total length (m)
l = Lp/2; % Pendulum center of mass (m)
Jp = mp*Lp^2/3; % Moment of inertia about pivot (kg-m^2)
Jt = 1.314528060e-08;
br = 4.7078e-05; % From identification
bp = 4e-4;
g = 9.81; % Gravity constant
K = 0.0061; % Stiffness coefficient as for least squares
a_31= -Jp*K/Jt;
a_32= mp^2*l^2*r*g/Jt;
a_33= -(Jp*(km^2/Rm + br))/Jt;
a_34= -(mp*l*r*bp)/Jt;
a_41= -K*l*mp*r/Jt;
a_42= (mp*g*l*Jr)/Jt;
a_43= -(mp*l*r*(km^2/Rm +br))/Jt;
a_44= -Jr*bp/Jt;
% State space description
A = [0 0 1 0;
0 0 0 1;
a_31 a_32 a_33 a_34;
a_41 a_42 a_43 a_44];
B = [0;
0;
Jp*km/Rm;
mp*r*l*km/Rm]/Jt;
C = [1 0 0 0;
0 1 0 0];
D = [0;
0];
sys = ss(A, B, C, D);
G = tf(sys); % G(1) for input-to-theta G(2) for input-to-alpha
G01 = G(1) % input-to-theta TF
G12 = minreal(G(2)/G(1)) % theta-to-alpha TF
%% LQR design method
K = lqr(A, B, eye(4), 1)
% check closed-loop eigenvalues
ecl = eig(A - B*K)
