I couldn't do the 2nd question in my project homework

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Emre Komur 2022-6-15

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评论： Sam Chak 2022-6-17

MCH3008 Lab Project2.pdf

I need urgent help for question 2.

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Answer 1

Sam Chak 2022-6-15

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You should at least provide the model of the harmonic drive in order to work on Q2.

If you do it properly, then it is possible to guide you along the way.

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Sam Chak 2022-6-15

编辑：Sam Chak 2022-6-15

在 MATLAB Online 中打开

@Emre Komur

Please check if you obtain the same model from your derivation of the Laplace transforms.

A = [0 0 1 0; 0 0 0 1; -100/10 100/10 -1/10 0; 100/2 -100/2 0 -0.5/2];

B = [0; 0; 0; 1/2];

C = [1 0 0 0]; % only gives theta_l as output

D = 0;

sys = ss(A, B, C, D);

Gp = tf(sys)

Gp = 5 ---------------------------------------------- s^4 + 0.35 s^3 + 60.02 s^2 + 7.5 s + 1.557e-13 Continuous-time transfer function.

n = order(Gp) % system order

n = 4

z = zero(Gp) % zeros of the TF

z = 0×1 empty double column vector

p = pole(Gp) % poles of the TF

p =

-0.1125 + 7.7449i -0.1125 - 7.7449i -0.1250 + 0.0000i -0.0000 + 0.0000i

impulse(Gp)

The impulse response of the system can be used to determine the stability of harmonic drive system. Comment based on the observation.

1.d requires you to plot the step responses of the angular positions and velocities of the link and motor. So, the output matrix needs to modified.

Cc = eye(n);

Dd = zeros(4, 1);

sys = ss(A, B, Cc, Dd);

step(sys, 30)

Sam Chak 2022-6-15

编辑：Sam Chak 2022-6-15

在 MATLAB Online 中打开

This is a 4th-order system. It is still possible to achieve the response with the percentage overshoot is less than 5%. However, to achieve a peak time of 1 second, the motor will have to exert a pretty aggressive torque and it may destabilize the system. A high gain compensator is possible.

I didn't use the root locus, but the pidtune instead. You can use the root locus to show that you can design a similar gain value of Gc as shown below:

A = [0 0 1 0; 0 0 0 1; -100/10 100/10 -1/10 0; 100/2 -100/2 0 -0.5/2];

B = [0; 0; 0; 1/2];

C = [1 0 0 0]; % only gives theta_l as output

D = 0;

sys = ss(A, B, C, D);

Gp = tf(sys)

Gp = 5 ---------------------------------------------- s^4 + 0.35 s^3 + 60.02 s^2 + 7.5 s + 1.557e-13 Continuous-time transfer function.

Gc = pidtune(Gp, 'PD', 0.0592)

Gc = Kp = 0.0982 P-only controller.

Gcl = feedback(Gc*Gp, 1);

step(Gcl, 120)

S = stepinfo(Gcl)

S = struct with fields:

RiseTime: 23.1876 TransientTime: 66.2475 SettlingTime: 66.2475 SettlingMin: 0.9138 SettlingMax: 1.0498 Overshoot: 4.9791 Undershoot: 0 Peak: 1.0498 PeakTime: 47.8965

Emre Komur 2022-6-15

Thank you for your time

God bless you

Sam Chak 2022-6-17

You are welcome, @Emre Komur. If my demonstration of using pidtune() in the Comment has helped in your project, consider accepting ✔ and voting 👍 the Answer as a closure to this. Thanks!

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