# Plot step response for control signal (u1 u2 ...) in a MIMO system with feedback

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Cesar Adolfo Cruz Vargaya on 21 Nov 2022
Answered: Sam Chak on 26 Nov 2022
0) Especify aditional variables and its values (use space as separator)
otherVars = "K1 K2 B1 B2 m1 m2"; otherVars = split(otherVars);
valueVars = "1 1 1 1 1 1"; valueVars = split(valueVars);
if length(otherVars) == length(valueVars)
arrayfun(@eval,strcat(otherVars,'=',valueVars));
else
warning("Number of variables not equal to its values")
return
end
K1 = 1
K2 = 1
B1 = 1
B2 = 1
m1 = 1
m2 = 1
1) Define the state space matrices
% STATE SPACE REPRESENTATION
AA = [0 0 1 0;0 0 0 1;-K1/m1 K1/m1 -B1/m1 B1/m1;K1/m2 -(K1+K2)/m2 -B1/m2 -(B1+B2)/m2];
BB = [0 0;0 0;1/m1 0;0 1/m2];
CC = [1 0 0 0;0 1 0 0];
DD = [0 0;0 0];
1.1) Define the value of for every
T = [4 4]; I = eye(size(DD,1)); s = tf('s');
2) Obtain the plant transfer function
Gp = tf(ss(AA,BB,CC,DD))
Gp = From input 1 to output... s^2 + 2 s + 2 1: ----------------------------- s^4 + 3 s^3 + 6 s^2 + 4 s + 1 -s + 1 2: ----------------------------- s^4 + 3 s^3 + 6 s^2 + 4 s + 1 From input 2 to output... s + 1 1: ----------------------------- s^4 + 3 s^3 + 6 s^2 + 4 s + 1 s^2 + s + 1 2: ----------------------------- s^4 + 3 s^3 + 6 s^2 + 4 s + 1 Continuous-time transfer function.
Gp.TimeUnit = 'seconds';
Gp.InputName = 'u';
Gp.OutputName = 'y';
3) Obtain the open loop transfer function
Go = (1/s)*diag(1./T)
Go = From input 1 to output... 0.25 1: ---- s 2: 0 From input 2 to output... 1: 0 0.25 2: ---- s Continuous-time transfer function.
4) Obtain the full process transfer function
G = feedback(Go,I)
G = From input 1 to output... 0.25 1: -------- s + 0.25 2: 0 From input 2 to output... 1: 0 0.25 2: -------- s + 0.25 Continuous-time transfer function.
5) Obtain the PID controller block transfer function
Gc = Gp\Go; Gc = minreal(Gc)
Gc = From input 1 to output... 0.25 s^2 + 0.25 s + 0.25 u(1): ------------------------ s - 2.378e-15 0.25 s - 0.25 u(2): ------------- s - 1.513e-15 From input 2 to output... -0.25 s - 0.25 u(1): -------------- s - 1.644e-15 0.25 s^2 + 0.5 s + 0.5 u(2): ---------------------- s + 4.121e-15 Continuous-time transfer function.
Gc.TimeUnit = 'seconds';
Gc.InputName = 'e';
Gc.OutputName = 'u';
6) Plot the step response from input r to output y
timeLimit = 40;
Sum1 = sumblk('e=r-y',size(T,2));
Cnt = connect(Gc,Gp,Sum1,'r','y','u');
step(Cnt,timeLimit,'r'), grid on 7) Plot the response from input r to output u (problem here)
Tp = getIOTransfer(Cnt,'r','u');
step(Tp,timeLimit,'r'), grid on
Error using DynamicSystem/step
Cannot simulate the time response of improper (non-causal) models.
I want to obtain the plots for U as this one, is the same problem but the data is obtained through a for loop This is the original diagram block that I'm trying to simulate without using for loops Amey Waghmare on 25 Nov 2022
As per my understanding, you want to plot the response of the system from ‘r’ to ‘u’ but receive the error Cannot simulate the time response of improper (non-causal) models’.
In order to obtain the step response of a system, the system should be ‘causal’ or ‘proper’. A proper system is a one having at least as many poles as zeros or number of poles is greater than number of zeros. To check if a system is proper, you can use the following command;
isproper(sys)
In the code, the Transfer function ‘Tp’ is improper, specifically the transfer functions from r1 to u1 and r2 to u2 are non-causal, with number of poles being 12 and number of zeros being 13 and hence it is not possible to obtain the step response.
##### 2 CommentsShow 1 older commentHide 1 older comment
Paul on 26 Nov 2022
Hi Cesar,
Tp is not the controller block. That is, Tp is not equal to Gc. Tp is the closed-loop transfer function from r to u; Gc is just the transfer function from e to u. These are not the same thing.
Correct. connect does not execute minreal AFAIK. But I don't see how that's relevant here.
Can you clarify this statement: "if a T.F. is improper then it's graph shouldn't be able to tend to a value."
The design approach used in the Question could be slightly modified so that we can generate the step response from r to us. But those won't look like either set of plots that have been shown.

Sam Chak on 26 Nov 2022
The original plant is a coupled 2nd-order MIMO system. If the desired closed-loop transfer function is a 1st-order system, this suggests there is loss of information, and in your case, the velocity becomes unobservable.
By redesigning your desired , see example below, you should be able to make proper.
Moreover, the settling time of the closed-loop transfer function is almost the same as the time in the previous .
otherVars = "K1 K2 B1 B2 m1 m2"; otherVars = split(otherVars);
valueVars = "1 1 1 1 1 1"; valueVars = split(valueVars);
if length(otherVars) == length(valueVars)
arrayfun(@eval,strcat(otherVars,'=',valueVars));
else
warning("Number of variables not equal to its values")
return
end
K1 = 1
K2 = 1
B1 = 1
B2 = 1
m1 = 1
m2 = 1
AA = [0 0 1 0;0 0 0 1;-K1/m1 K1/m1 -B1/m1 B1/m1;K1/m2 -(K1+K2)/m2 -B1/m2 -(B1+B2)/m2];
BB = [0 0;0 0;1/m1 0;0 1/m2];
CC = [1 0 0 0;0 1 0 0];
DD = [0 0;0 0];
T = [4 4];
I = eye(size(DD, 1));
s = tf('s');
Gp = tf(ss(AA, BB, CC, DD))
Gp = From input 1 to output... s^2 + 2 s + 2 1: ----------------------------- s^4 + 3 s^3 + 6 s^2 + 4 s + 1 -s + 1 2: ----------------------------- s^4 + 3 s^3 + 6 s^2 + 4 s + 1 From input 2 to output... s + 1 1: ----------------------------- s^4 + 3 s^3 + 6 s^2 + 4 s + 1 s^2 + s + 1 2: ----------------------------- s^4 + 3 s^3 + 6 s^2 + 4 s + 1 Continuous-time transfer function.
Gp.TimeUnit = 'seconds';
Gp.InputName = 'u';
Gp.OutputName = 'y';
w = 5/13;
Go = tf(w^2, [1 2*w 0])
Go = 0.1479 -------------- s^2 + 0.7692 s Continuous-time transfer function.
Gc = Gp\Go;
Gc = minreal(Gc)
Gc = From input "y(1)" to output... 0.1479 s^2 + 0.1479 s + 0.1479 u(1): ------------------------------ s^2 + 0.7692 s 0.1479 s - 0.1479 u(2): ----------------- s^2 + 0.7692 s From input "y(2)" to output... -0.1479 s - 0.1479 u(1): ------------------ s^2 + 0.7692 s 0.1479 s^2 + 0.2959 s + 0.2959 u(2): ------------------------------ s^2 + 0.7692 s Continuous-time transfer function.
Gc.TimeUnit = 'seconds';
Gc.InputName = 'e';
Gc.OutputName = 'u';
timeLimit = 40;
Sum1 = sumblk('e = r - y', size(T, 2));
Cnt = connect(Gc, Gp, Sum1, 'r', 'y', 'u');
step(Cnt, timeLimit), grid on Tp = getIOTransfer(Cnt, 'r', 'u');
Tp = minreal(tf(Tp))
Tp = From input "r(1)" to output... 0.1479 s^14 + 1.377 s^13 + 6.816 s^12 + 22.15 s^11 + 51.71 s^10 + 90.14 s^9 + 119.6 s^8 + 121.7 s^7 + 94.97 s^6 + 56.18 s^5 + 24.7 s^4 + 7.804 s^3 + 1.673 s^2 + 0.2175 s + 0.01295 u(1): --------------------------------------------------------------------------------------------------------------------------------------------------------------- s^14 + 9.077 s^13 + 43.31 s^12 + 134.2 s^11 + 294.1 s^10 + 472.1 s^9 + 562.6 s^8 + 500.1 s^7 + 331.8 s^6 + 163.4 s^5 + 58.89 s^4 + 15.09 s^3 + 2.606 s^2 + 0.2719 s + 0.01295 0.1479 s^13 + 1.081 s^12 + 4.21 s^11 + 10.05 s^10 + 15.3 s^9 + 13.08 s^8 + 1.061 s^7 - 11.99 s^6 - 15.83 s^5 - 10.96 s^4 - 4.689 s^3 - 1.25 s^2 - 0.1916 s - 0.01295 u(2): --------------------------------------------------------------------------------------------------------------------------------------------------------------- s^14 + 9.077 s^13 + 43.31 s^12 + 134.2 s^11 + 294.1 s^10 + 472.1 s^9 + 562.6 s^8 + 500.1 s^7 + 331.8 s^6 + 163.4 s^5 + 58.89 s^4 + 15.09 s^3 + 2.606 s^2 + 0.2719 s + 0.01295 From input "r(2)" to output... -0.1479 s^13 - 1.377 s^12 - 6.668 s^11 - 20.92 s^10 - 46.27 s^9 - 74.65 s^8 - 88.8 s^7 - 77.87 s^6 - 50.04 s^5 - 23.25 s^4 - 7.599 s^3 - 1.66 s^2 - 0.2175 s - 0.01295 u(1): --------------------------------------------------------------------------------------------------------------------------------------------------------------- s^14 + 9.077 s^13 + 43.31 s^12 + 134.2 s^11 + 294.1 s^10 + 472.1 s^9 + 562.6 s^8 + 500.1 s^7 + 331.8 s^6 + 163.4 s^5 + 58.89 s^4 + 15.09 s^3 + 2.606 s^2 + 0.2719 s + 0.01295 0.1479 s^14 + 1.525 s^13 + 8.193 s^12 + 28.82 s^11 + 72.63 s^10 + 136.4 s^9 + 194.2 s^8 + 210.5 s^7 + 172.8 s^6 + 106.2 s^5 + 47.95 s^4 + 15.4 s^3 + 3.332 s^2 + 0.4351 s + 0.0259 u(2): --------------------------------------------------------------------------------------------------------------------------------------------------------------- s^14 + 9.077 s^13 + 43.31 s^12 + 134.2 s^11 + 294.1 s^10 + 472.1 s^9 + 562.6 s^8 + 500.1 s^7 + 331.8 s^6 + 163.4 s^5 + 58.89 s^4 + 15.09 s^3 + 2.606 s^2 + 0.2719 s + 0.01295 Continuous-time transfer function.
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