Cs REPORT
Complex Engineering Problem
SUBMITTED TO: Dr. Naveed Ishtiaq
SUBMITTED BY: Asfandiyar # 261-FET/BSEE/F17
International Islamic university
Islamabad
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Cs REPORT
Complex Engineering Problem
SUBMITTED TO: Dr. Naveed Ishtiaq
SUBMITTED BY: Asfandiyar # 261-FET/BSEE/F 17
International Islamic university
Islamabad
Introduction:
Proportional-integral - derivative (PID) controllers are the most widespread controllers used in industrial automation operations because of its straightforward structure and good efficiency in a wide variety of operating conditions. The design of such a controller involves three parameters to be identified: proportional gain, integral time constant and time constant derivatives. The PID controller parameters Kp, Ki , and Kd (or Kp, Ti, and Td) can be modified to create different response curves from a given system.
Transfer function:
𝑺^ (^ 𝑱^ 𝑿^ 𝒔^ 𝑿^ 𝒃)^ 𝑿^ 𝜽(𝒔)^ =^ 𝒌^ 𝑿^ i(s) (𝒗^ −^ 𝒌^ ). (s).^ 𝜽(𝒔)^ =^ 𝒊(𝒔)^.^ (𝑳^ 𝑿^ 𝒔^ +^ 𝑹) The open loop transfer function is 𝜽 𝑽 = 𝒌 ( (^) 𝑱 𝑿 𝒔 𝑿 𝒃) (^) 𝑿 (𝑳 𝑿 𝒔 + 𝑹) (^) + 𝒌𝟐 ⁄
Design Requisites:
First, the motor with an input voltage of 1volt can only rotate at 0.1 rad / sec. The steady state error of motor speed will be below 1 percent. The motor will accelerate to its steady-state speed as soon as it switches on, the settling time should be under 2 seconds. We would like an overflow of less than 5 percent because a speed faster than the reference will harm the equipment. Mat Lab Code: clear all clc b=0.1; k=0.01; J=0.01; R=1; L=0.5; A=k; B=[(JL) ((JR)+(Lb)) ((bR) + k^2)]; step(A,B,0:0.01:5) title('OL Response')
sys_c3 = feedback(DP_motor,1); Kp = 70; E = pid(Kp); sys_c4 = feedback(EP_motor,1); Kp = 100; F = pid(Kp); sys_c5 = feedback(F*P_motor,1); step(sys_cl,sys_c3,sys_c4,sys_c5,'g--',t) grid on title('Response with Proportional Control') legend('pd=10', 'pd=50','pd=70','pd=100') Fig. 2. Closed loop response with different values of kp kp Rise time(sec) Maximum OS (%) Peak amplitude Steady State error 10 0.491 - 0.5 0. 50 0.159 12.8 0.94 0. 70 0.126 18.4 1.04 0. 100 0.099 24.9 1.14 0. The proportional controller kp reduces the raise time for kp=10 from (0.49sec) to (0.099sec) for kp=100 and will never eliminate the steady-state error from (0.5) for kp=10 to (0.909) for kp=100.
So, by increasing the gain kp, the response becomes more rapid. However the maximum overshoot will increase from (12 per cent to 25 per cent).
Proportional Integral Control (PI):
The settling period is still too long, we have to raise the gain ki with kp=100,. Mat Lab Code: J = 0.01; b = 0.1; K = 0.01; R = 1; L = 0.5; s = tf('s'); P_motor = K/((Js+b)(Ls+R)+K^2); Ki = 10; C = pid(Kp,Ki); sys_cl = feedback(CP_motor,1); t = 0:0.01:2; Ki = 50; D = pid(Kp,Ki); sys_c3 = feedback(DP_motor,1); Ki = 70; E = pid(Kp,Ki); sys_c4 = feedback(EP_motor,1); Ki = 100; F = pid(Kp,Ki); sys_c5 = feedback(F*P_motor,1); step(sys_cl,sys_c3,sys_c4,sys_c5,'g--',t) grid on title('Response with Integral Control') legend('pi=10', 'pi=50','pi=70','pi=100') ki settling time(sec) Rise time (sec) Maximum OS% Peak amplitude Steady state error 50 - 0.107 17.9 1.18 0. 70 1.69 0.106 19.6 1.2 0. 100 1.03 0.104 22.2 1.22 0. 200 0.774 0.0989 30.5 1.3 0 The settling time becomes 0.774 sec for ki=200, and the steady-state error becomes very small and eliminated.
Fig. 4. closed loop response with different values of kd. If we use a pid controller with: kp=100,ki=200 and kd=10. All our design objectives are fulfilled and the response could very well look like below. Fig. 5. Closed loop response with PID control
References:
- http://ctms.engin.umich.edu/CTMS/index.php?example=Introduction§ion=Con trolPID
- https://www.mathworks.com/help/control/ref/step.html
- https://www.mathworks.com/help/control/ref/feedback.html
- http://ctms.engin.umich.edu/CTMS/index.php?aux=Extras_step Template:
- https://www.researchgate.net/publication/310416585_DC_MOTOR_SPEED_CONT ROL_USING_PID_CONTROLLER