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

22 次查看（过去 30 天）
Cesar Adolfo Cruz Vargaya
回答： Sam Chak 2022-11-26
Ran in:
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

请先登录，再回答此问题。

采纳的回答

Amey Waghmare
Amey Waghmare 2022-11-25
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 个评论
Cesar Adolfo Cruz Vargaya
Cesar Adolfo Cruz Vargaya 2022-11-25
编辑：Cesar Adolfo Cruz Vargaya 2022-11-26
Thanks for the answer, as you've stated, it's true that the function Tp is improper. After running step by step I've noticed that my Tp function, which is the controller block, has more poles and zeros than it has to have. In fact, I should've obtained the same T.F. as specified in the fifth step, where Gc is my controller. I tried to use minreal, but that only works when doing manually. It appears that the connect command doesn't take that in account. Additionally, if I recall correctly, if a T.F. is improper then it's graph shouldn't be able to tend to a value. But as shown in the last picture, the response of u1 and u2 is stable and tends to a value.
Pd: I've checked my teacher code, and in the plot code, he changed the ccontroller values,here is the right image with the original values, and as I've stated, both of them (u1 and u2) tend to a value
Paul
Paul 2022-11-26
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.

请先登录，再进行评论。

更多回答（1 个）

Sam Chak
Sam Chak 2022-11-26
Ran in:
Hi @Cesar Adolfo Cruz Vargaya
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.
step(Tp, timeLimit, 'r'), grid on

类别

在 Help Center 和 File Exchange 中查找有关 Classical Control Design 的更多信息

标签

产品


版本

R2022b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by