Unit 4 The Performance of Second Order System

Open Loop & Close Loop Open Loop: Close Loop:

The Performance of Second Order System

The Response of Second Order System

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

The Response of

Homework2 : The effect of damping ratio

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

Solution of Homework3

PD Controller

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.

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

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.

Homework 4

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

PI Controller

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

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

PID Controller

Homework6, PID,

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

Homework8: Pendulum System

Feedback Controller Design State Feedback Controller