2. 1 MATLAB AND CONTROLS PRESENTED BY:- AGILESWARI K. RAMASAMY DR. FARRUKH HAFIZ NAGI

3. 2 Controls & MATLAB INRTRODUCTION  Control system consists of subsystems and processes assembled for the purpose of controlling the outputs of the processes.  DC MOTOR  Physical Modeling of a DC Motor  Physical Modeling of a DC Motor in STATE SPACE  Designing the full-state feedback controller  Bode Plot  PID CONTROLLER  LTIVIEW

4. 3 A common actuator in control systems is the DC motor. It directly provides rotary motion and, coupled with wheels or drums and cables, can provide transitional motion. Physical Modeling of a DC Motor

5. 4 The motor torque, T, is related to the armature current, i, by a constant factor Kt. The back emf, e, is related to the rotational velocity by the following equations: Kt (armature constant) = Ke (motor constant). Physical Modeling of a DC Motor rotational speed is the output voltage is the input

6. 5 Newton's law combined with Kirchoff's law : The modeling equations in Laplace Transforms Physical Modeling of a DC Motor

7. 6  Open-loop transfer function  The value of the constants :- moment of inertia of the rotor (J) = 0.01 kg.m^2/s^2 damping ratio of the mechanical system (b) = 0.1 Nms electromotive force constant (K=Ke=Kt) = 0.01 Nm/Amp electric resistance (R) = 1 ohm electric inductance (L) = 0.5 H rotational speed is the output voltage is the input Physical Modeling of a DC Motor

8. 7 Motor speed design criteria based on step input  Settling time less than 2 seconds  Overshoot less than 5%  Steady-state error less than 1% Physical Modeling of a DC Motor

9. 8 MATLAB representation of Open loop transfer function is as follows:  Create a new m-filem-file  Enter the following commands in m-file:  J=0.01 %Defining constants  b=0.1  K=0.01  R=1  L=0.5  num=K %Defining the numerator  den=[(J*L) ((J*R)+(L*b)) ((b*R)+K^2)] %Defining denominator  step(num,den,0:0.1:3) %Obtaining the step response step  title('Step Response for the Open Loop System') Physical Modeling of a DC Motor

10. 9 Alternative MATLAB representation  J=0.01; b=0.1;  K=0.01;  R=1;  L=0.5;  num=K  den=[(J*L) ((J*R)+(L*b)) ((b*R)+K^2)]  T=tf(num,den)%Defining transfer function  step(T,0:0.1:3) %Obtaining step response step  title('Step Response for the Open Loop System') %Title of the figure Physical Modeling of a DC Motor

11. 10 Physical Modeling of a DC Motor  When 1 volt is applied to the system, the motor can only achieve a maximum speed of 0.1 rad/sec, ten times smaller than our desired speed.  It takes the motor 3 seconds to reach its steady-state speed; this does not satisfy our 2 seconds settling time criterion. Overshoot less than 5% Settling time less than 2 seconds Steady-state error less than 1%

12. 11 Physical Modeling of a DC Motor in STATE SPACE DC Motor Differential Equation is represented in state-space by choosing rotational speed and electric current as the state variables  voltage as an input  rotational speed as output State space representation DC Motor Differential Equation

13. 12 Physical Modeling of a DC Motor in STATE SPACE MATLAB representation of Open loop transfer function using the state-space equations.  Create a new m-filem-file  Enter the following commands in m-file:  J=0.01 ;  b=0.1;  K=0.01;  R=1;L=0.5;  A=[-b/J K/J ; -K/L -R/L] %Defining the matrix A,B,C,D  B=[0 ; 1/L]  C=[1 0]  D=0  step(A, B, C, D) % Obtaining the step response

14. 13 Designing the full-state feedback controller The schematic of full state feedback system is

15. 14 Designing the full-state feedback controller The characteristic polynomial for this closed- loop system is the determinant of (sI-(A-BK)). The matrices A and B*K are both 2x2 matrices, there should be 2 poles for the system. The two poles will be placed at -5 + i and -5-i (note that this corresponds to a zeta = 0.98 which gives 0.1% overshoot and a sigma = 5 which leads to a 1 sec settling time). MATLAB will find the controller matrix,K using these two poles. Settling time less than 2 seconds Overshoot less than 5% Steady-state error less than 1%

16. 15  J=0.01;  b=0.1;  K=0.01;  R=1;L=0.5;  A=[-b/J K/J ; -K/L R/L];  B=[0 ; 1/L];  C=[1 0];  D=0;  p1 = -5 + i% Pole 1  p2 = -5 - i% Pole 2  K = place(A,B,[p1 p2]) % Obtaining the value for the controller  t=0:0.01:3;%Defining the time step  step(A-B*K,B,C,D,1,t)%Obtaining the step response step MATLAB representation of full state feedback system. Designing the full-state feedback controller

17. 16 Step response with K controller Designing the full-state feedback controller

18. 17 BODE PLOT The main idea of frequency-based design is to use the Bode plot of the open-loop transfer function to estimate the closed-loop response. JJ=0.01; bb=0.1; KK=0.01; RR=1;L=0.5; nnum=K; dden=[(J*L) ((J*R)+(L*b)) ((b*R)+K^2)];  b bode(num,den)%Obtaining the bode plot MATLAB representation to draw bode plot.

19. 18  A controller will be designed to satisfy the design requirements. The closed loop DC motor with the controller Overshoot less than 5% Settling time less than 2 seconds Steady-state error less than 1% PID CONTROLLER

20. 19 PID CONTROLLER Transfer function of PID controller: K p =porportional gain K D =derivative gain K i =intergral gain

21. 20 PID CONTROLLER  J=0.01;  b=0.1;  K=0.01;  R=1;L=0.5;  num=K;  den=[(J*L) ((J*R)+(L*b)) ((b*R)+K^2)];  Kp=100 %Defining the Proportional constant  numa=Kp*num %Obtaining the new numerator  dena=den %Obtaining the denominator  [numac,denac]=cloop(numa,dena) %Obtaining the num and den of the %closed loop system  t=0:0.01:5;  step(numac,denac,t) step  title('Step response with Proportion Control') MATLAB representation for proportional controller.

22. 21 PID CONTROLLER  J=0.01;  b=0.1;  K=0.01;  R=1;L=0.5;  num=K;  den=[(J*L) ((J*R)+(L*b)) ((b*R)+K^2)];  Kp=100 %Defining the Proportional constant  numa=Kp*num%Obtaining the new numerator  dena=den %Obtaining the denominator  T=tf(numa,dena)%Obtaining the new OL transfer function  G=feedback(T,1)%Obtaining the new CL transfer function  t=0:0.01:5;  step(G,t) step  title('Step response with Proportion Control') Alternative MATLAB representation for proportional controller

23. 22 PID CONTROLLER Step response with Proportional Control

24. 23 PID CONTROLLER Derivative control – reduce overshoot Integral control – reduce steady state  J=0.01; b=0.1; K=0.01;  R=1; L=0.5; num=K;  den=[(J*L) ((J*R)+(L*b)) ((b*R)+K^2)];  Kp=100; %Defining proportional contstant  Ki=1; Kd=1; %Defining integral and derivative constant  numc=[Kd, Kp, Ki]; %Obtaining num and den of the controller  denc=[1 0];  numa=conv(num,numc) %Obtaining the num and den of the whole % system  dena=conv(den,denc)  [numac,denac]=cloop(numa,dena) %Obtaining the num and den of the closed %loop system  step(numac,denac)  title('PID Control with small Ki and Kd')

25. 24 PID CONTROLLER Derivative control – reduce overshoot Integral control – reduce steady state  J=0.01; b=0.1; K=0.01;  R=1; L=0.5; num=K;  den=[(J*L) ((J*R)+(L*b)) ((b*R)+K^2)];  Kp=100; %Defining proportional constant  Ki=1; Kd=1; %Defining integral and derivative constant  numc=[Kd, Kp, Ki]; %Obtaining num and den of the controller  denc=[1 0];  C=Tf(numc,denc)%Transfer function of the controller  P=tf(num,den) %Transfer function of the Plant  G=series(C,P) % Transfer function of the plant and controller  T=feedback(G,1)%Closed Loop transfer function  step(T)  title('PID Control with small Ki and Kd')

26. 25 PID CONTROLLER Step response of the system with small Ki and Kd for the PID controller

27. 26 PID CONTROLLER Increase K i to reduce settling time  Ki=200

28. 27 PID CONTROLLER Increase K d to reduce overshoot K d =10 K i =200 K p =100

29. 28 LTIVIEW  LTIVIEW (GUI Tools)- convenient way to obtain time and frequency response plots of LTI transfer function.  Define the transfer function in the command window  Type LTIVIEW  Import the function  Right click on the plot and then select the type of responses.

30. 29 THE END