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Unit 4 The Performance of Second Order System

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Open Loop & Close Loop Open Loop: Close Loop:

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The Performance of Second Order System

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The Response of Second Order System

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Homework 1 1. Steady State Error : 2. Overshoot : 0.163 3. The Peak Time : 1.8138 s 4. The Rise Time : 0.9 s 5. The Setting Time : 4 s [Hint] : max

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The Response of

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Homework2 : The effect of damping ratio

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P Controller This type of control action is formally known as proportional control (Gain) Homework3 : K=1, K=4, K=8, K=12, K=36 Please explain the effect of P controller to the second order system

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Solution of Homework3

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PD Controller

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The Performance of P Controller : The error signal is positive, the torque is positive and rising rapidly. The large overshoot and oscillations in the output because lack of damping. : The error signal is negative, the torque is negative and slow down causes the direction of the output to reverse and undershoot. : The torque is again positive, thus tending to reduce the undershoot, the error amplitude is reduced with each oscillations.

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The contributing factors to the high overshoot The positive correcting torque in the interval is too large ( ) Decrease the amount of positive torque The retarding torque in the interval is inadequate ( ) Increase the retarding torque

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The Effect of PD Controller : is negative; this will reduce the original torque due to alone. : both and is negative; the negative retarding torque will be greater than that with only P controller. : and have opposite signs. Thus the negative torque that originally contributes to the undershoot is reduced also.

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Homework 4

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Solution of PD Controller clear; x1=0;x2=0;dt=0.01;r=1;step=2000; kp=36;kd=6;pe=r-x1; for k=1:step t(k)=k*dt; e=r-x1; de=(e-pe)/dt; u=kp*e+kd*de; x1=x2*dt+x1; x2=(u-4*x2)*dt+x2; pos(k)=x1;vel(k)=x2;pe=e; end

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PI Controller

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HW5 : The Effect of PI Controller Adds a zero at to the forward-path T.F. Adds a pole at to the forward-path T.F. This means that the steady-state error of the original system is improved by one order. a=2,b=8,k=1

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Program of PID Controller clear; x1=0;x2=0;dt=0.01;r=1;step=2000; kp=1;kd=6;ki=0.1;pe=r-x1;ie=(r-x1)*dt; for k=1:step t(k)=k*dt; e=r-x1; de=(e-pe)/dt; ie=ie+e*dt; u=kp*e+kd*de+ki*ie; x1=x2*dt+x1; x2=(u-2*x2-8*x1)*dt+x2; pos(k)=x1;vel(k)=x2;pe=e; end

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PID Controller

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Homework6, PID,

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Homework7 : Ziegler-Nichols Tuning Step 1 : Let until the occur of critical stable Step 2 : Optimal Parameter Tuning

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Homework8: Pendulum System

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Feedback Controller Design State Feedback Controller

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