Control System Analysis Techniques course error

Question

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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)

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Walter Roberson 2026-5-2，0:00

在 MATLAB Online 中打开

Is it look for step(controller) ?

s = tf('s');

quadcopter_ol = 0.04133/(s + 0.01479);

controller = zpk(-5,0,100);

step(controller)

Dunbarin 2026-5-3，10:06

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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Answer 1

Dunbarin 2026-5-12，8:57

2 个投票

I reached out to technical support and they were able to give me a work around to the problem,I will paste it below

Use stepplot function instead of step function as follows:

output = stepplot(<variable_name>); output.Responses.Name = '<variable_name>';

Similarly, the following function mapping + output.Responses.Name workaround can be used to get around the issue in other parts of the course:

bode -> bodeplot nyquist -> nyquistplot pzmap -> pzplot

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Sam Chak 2026-5-12，9:35

在 MATLAB Online 中打开

Hi @Dunbarin

Thanks for the update. Virtually no difference between "step(sys)" and "stepplot(sys)", but the grading system can only accept one answer. The "stepplot(sys, plotoptions)" command can be used if targeted customization of the step plot is required, but "step(sys, timeDuration)" is generally good enough for initial visualization of the step response.

s = tf('s');

% Open-loop system

quadcopter_ol = 0.04133/(s + 0.01479)

quadcopter_ol = 0.04133 ----------- s + 0.01479 Continuous-time transfer function.

% method 1

figure(1)

step(quadcopter_ol)

% method 2

figure(2)

stepplot(quadcopter_ol)

Agasthya 2026-5-21，5:01

It seems to work for me. Thankyou

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Answer 2

Sam Chak 2026-5-4，13:04

在 MATLAB Online 中打开

0 个投票

Hi @Dunbarin

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.

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Andre 2026-5-6，19:50

Is there anything yet on this issue? I'm stuck on it.

Elijah 2026-5-7，23:32

I am trying to do the same but it isn't working and I am stuck on it.

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