lsim() vs step() : are different responses expected?
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Are different responses expected if running lsim() and step(), both with the same unit step inputs?
I have a state-space system representing a closed-loop system, that gives the expected output with a unit step input to lsim() -- but not with a unit step input to step().
Even if the system itself is incorrectly built, i'd have expected equivalent responses from step() and lsim().
Background:
The system is attached in sys.mat, and is shown at the end of the post as well. The .m file that plots these is also attached.
The states are: x = [x1, x2, xi, x1^, x2^, xi^], where ^ are estimated states. x1,x2 are pos/vel, and xi is the integral action state.
x1, x1^ , and x2, x2^ , and xi^ are all as expected.
But xi is not as expected. For some reason, with the same ref step input of 1,
step() gives an unexpected xi=0, and
lsim() gives the expected xi .
Questions:
What's the difference between step() and lsim() -- if the same system and same ref inputs are used -- that would cause outputs to differ?
Are step() and lsim() both inputting the ref to u? If they are both setting u = 1, then shouldn't sys.B provide inputs to both xi and xi^ level even when using step(), since B is [0;0;1, 0;0;1]?
For step(): see 2nd subplot, blue line = 0.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1353908/image.png)
For lsim(): both plots seem as expected.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1353913/image.png)
The details are perhaps not important: regardless of whether the system is correctly built or not, i'd have expected equivalent responses from step() and lsim()
But in case it matters, in the state space sys, the two relevant rows are xi, and xi^:
xi_dot = -C*x + 1*u
xi^_dot = -C*x^ + 1*u
with u = reference step input, and -C*x = y (theta) and -C*x^ = y^ (theta^)
Here's the system. It's also in the attached .mat file.
sysCl_lqgi =
A =
theta angVel xi theta^ angVel^ xi^
theta 0 1 0 0 0 0
angVel -6.317e+04 -160 0 -4.468e+07 -6.376e+04 1.596e+10
xi -1 0 0 0 0 0
theta^ 4.212e+04 0 0 -4.212e+04 1 0
angVel^ 4.398e+08 0 0 -4.845e+08 -6.392e+04 1.596e+10
xi^ 0 0 0 -1 0 0
B =
u F
theta 0 0
angVel 0 1e+05
xi 1 0
theta^ 0 0
angVel^ 0 0
xi^ 1 0
C =
theta angVel xi theta^ angVel^ xi^
theta 1 0 0 0 0 0
D =
u F
theta 0 0
Continuous-time state-space model.
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採用された回答
Paul
2023 年 4 月 18 日
編集済み: Paul
2023 年 4 月 18 日
I was able to recreate your strange result on R2022a.
After some investigation, I think I know what's happening.
load sys.mat
sys = sysCl_lqgi;
The six states of this system are:
sys.StateName.'
After examining the state space matrices and based on the state names, it looks like you started with a 2-state model, augmented it with an integrator, then designed an observer for the augmented plant (including the integrator) and then closed the loop through the control applied to the estimated states, where the external reference signal forms the error that feeds xidot and xihatdot. However, the third column of sys.A is all zeros, so even if xi is excited by the reference input, xi will have no effect on the output, which is theta. Same thing with xi being excited by the disturbance input. Consequently, xi can be eliminated from the system with no impact on the input/output response. This elimination is called structural minimal realization sminreal.
msys = sminreal(sys);
msys.StateName.'
Note that sminreal eliminates the third state.
I'm 99% certain (can't be 100% because I ran into a function that was implemented in p-code and so I couldn't examine it) that deep in the bowels of the code that computes the step response, it does a structural minimal realization (after converting the input sys to its discrete time approximation) of the system. But, in the outputs returned to the user, the eliminated state(s) is filled with zeros, so as to keep the dimensions of the state vector output of step() aligned with the definitions of sys. I don't have 2023a installed locally and so can't see what changed in 2023a.
You can test this by changing the C matrix to
sys.c = [0 0 1 0 0 0];
In so doing, that third state can't be eliminated because it's needed for the output
msys = sminreal(sys);
msys.StateName.'
You'll see that if you run your code with this sys that you get the step response plots you expect for the state vector.
5 件のコメント
Paul
2023 年 4 月 21 日
I think a better approach would be
sys = ss(a,b,[c;eye(size(a))],[d;zeros(size(a,1),size(b,2))]);
[y,tout,x] = step(sys);
which preserves the original outputs specified by the user as the first elements in y.
Paul
2023 年 4 月 22 日
I wonder why the update in 2023a is not listed as bug fix or as a change in behavior in the Release Notes
その他の回答 (1 件)
Paul
2023 年 4 月 12 日
The code and the system in the .mat file does not recreate the step plot. There is also an undefined variable.
load sys.mat
% step()
sys = sysCl_lqgi;
[y,tout,x] = step(sys);
figure
subplot(2, 1, 1);
plot(tout, x(:, 1, 1), 'b', 'LineWidth', 1.5);
hold on
plot(tout, x(:, 4, 1), 'k:', 'LineWidth', 2.5)
grid on;
title('x1', 'FontSize', 14)
legend({'$x_1$', '$\hat{x_1}$'},'Interpreter','latex', 'FontSize', 16);
xlim([0, 0.02]);
subplot(2, 1, 2);
plot(tout, x(:, 3, 1), 'b', 'LineWidth', 1.5);
hold on
plot(tout, x(:, 6, 1), 'k:', 'LineWidth', 2.5);
grid on;
title('xi', 'FontSize', 14)
legend({'$x_i$', '$\hat{x}_i$'},'Interpreter','latex', 'FontSize', 16);
xlim([0, 0.02]);
xlabel('Time [s]', 'FontSize', 14)
sgtitle('STEP(): Ref response');
% lsim()
step inputs a step simultaneously to both inputs. So the second column of u should also be ones, not zeros.
u = [ones(size(t)) zeros(size(t))];
[y, tout, x] = lsim(sys, u, t, zeros(6,1));
figure;
subplot(2, 1, 1);
plot(t, x(:, 1, 1), 'b', 'LineWidth', 1.5);
hold on;
plot(t, x(:, 4, 1), 'k:', 'LineWidth', 2.5);
grid on;
title('x1')
legend({'$x_1$', '$\hat{x_1}$'},'Interpreter','latex', 'FontSize', 16);
xlim([0, 0.02]);
subplot(2, 1, 2);
plot(t, x(:, 3, 1), 'b', 'LineWidth', 1.5);
hold on;
plot(t, x(:, 6, 1), 'k:', 'LineWidth', 2.5);
grid on;
title('xi', 'FontSize', 14);
legend({'$x_i$', '$\hat{x}_i$'},'Interpreter','latex', 'FontSize', 16);
xlim([0, 0.02]);
xlabel('Time [s]', 'FontSize', 14)
sgtitle('LSIM(): Ref response', 'FontSize', 14)
16 件のコメント
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