Control System Analysis Techniques course error

In the Time domain section of the course when I use the step function to plot the grpah it keeps saying i am wrong even though it is the exact same code and graph. Those anyone have anny solutions
below is the code
s = tf('s');
quadcopter_ol = 0.04133/(s + 0.01479);
controller = zpk(-5,0,100);
Task 1
step(quadcopter_ol)

2 个评论

Is it look for step(controller) ?
s = tf('s');
quadcopter_ol = 0.04133/(s + 0.01479);
controller = zpk(-5,0,100);
step(controller)
No it’s you’re meant to use the step function on the quadcopter, even the solution in the course does the same thing I do and the graphs are the same but it keeps saying I’m wrong

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回答(1 个)

If this topic is related to control systems, you are probably required to implement the controller on the open-loop quadcopter to form a closed-loop control system with the desired characteristics, such as the settling time.
s = tf('s');
% Open-loop system
quadcopter_ol = 0.04133/(s + 0.01479)
quadcopter_ol = 0.04133 ----------- s + 0.01479 Continuous-time transfer function.
% Controller
controller = zpk(-5,0,100)
controller = 100 (s+5) --------- s Continuous-time zero/pole/gain model.
% Closed-loop system
closedLoop = feedback(controller*quadcopter_ol, 1)
closedLoop = 4.133 (s+5) ---------------------- (s^2 + 4.148s + 20.66) Continuous-time zero/pole/gain model.
% Step response of the closed-loop system
step(closedLoop), grid on
Warning: Graphics acceleration hardware is unavailable. Graphics quality and performance might be diminished. See MATLAB System Requirements.

2 个评论

Thank you for your reply but the task specifically askes you to use the step function on quadcopter_ol, and what you did is the immediate task after completing what I said.
My issue is that I used the step function correctly but it still says I failed
I’m sorry to hear that. Have you resolved the issue? If not, please contact technical support for assistance.

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2026-5-1,20:48

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