Typically, you extract coefficients from a controller obtained from another function, such aspidtuneorgetBlockValue. For this example, create a 2-DOF PID controller that has random coefficients.

rng('default');% for reproducibilityC2 = pid2(rand,rand,rand,rand,rand,rand);

Extract the PID coefficients, filter time constant, and setpoint weights.

[Kp,Ki,Kd,Tf,b,c] = piddata2(C2);

Create a 2-DOF PID controller in standard form.

C2 = pidstd2(2,3,4,10,0.5,0.5)

C2 = 1 1 s u = Kp * [(b*r-y) + ---- * --- * (r-y) + Td * ------------ * (c*r-y)] Ti s (Td/N)*s+1 with Kp = 2, Ti = 3, Td = 4, N = 10, b = 0.5, c = 0.5 Continuous-time 2-DOF PIDF controller in standard form

Compute the coefficients of an equivalent parallel-form PID controller.

[Kp,Ki,Kd,Tf,b,c] = piddata2(C2);

Check some of the coefficients to confirm that they are different from the standard-form coefficients.

Ki

Ki = 0.6667

Kd

Kd = 8

Extract coefficients from a two-input, one-output dynamic system that represents a valid 2-DOF parallel-form PID controller.

The following A, B, C, and D matrices form a discrete-time state-space model that represents a 2-DOF PID controller.

A = [1,0;0.09975,0.995]; B = [0.00625,-0.00625;0.1245,-0.1241]; C = [0,4]; D = [2.875,-5.75]; sys = ss(A,B,C,D,0.1)

sys = A = x1 x2 x1 1 0 x2 0.09975 0.995 B = u1 u2 x1 0.00625 -0.00625 x2 0.1245 -0.1241 C = x1 x2 y1 0 4 D = u1 u2 y1 2.875 -5.75 Sample time: 0.1 seconds Discrete-time state-space model.

Extract the PID gains, filter time constant, and setpoint weights of the model.

[Kp,Ki,Kd,Tf,b,c,Ts] = piddata2(sys);

For a discrete-time system,piddata2calculates the coefficient values using the defaultForwardEulerdiscrete integrator formula for bothIFormulaandDFormula.

Typically, you obtain an array of controllers by usingpidtuneon an array of plant models. For this example, create an 2-by-3 array of 2-DOF PI controllers with random values ofKp,Ki, andb.

rng('default'); C2 = pid2(rand(2,3),rand(2,3),0,0,rand(2,3),0);

Extract all the coefficients from the array.

[Kp,Ki,Kd,Tf,b,c] = piddata2(C2);

Each of the outputs is itself a 2-by-3 array. For example, examineKi.

Ki

Ki =2×30.2785 0.9575 0.1576 0.5469 0.9649 0.9706

Extract only the coefficients of entry (2,1) in the array.

[Kp21,Ki21,Kd21,Tf21,b21,c21] = piddata2(C2,2,1);

Each of these outputs is a scalar.

Ki21

Ki21 = 0.5469