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1 Matlab and Simulink for Control Automatica I (Laboratorio) 1/78 Matlab and Simulink CACSD 2/78

2 Matlab and Simulink for Control Matlab introduction Simulink introduction Control Issues Recall Matlab design Example Simulink design Example 3/78 Part I Introduction 4/78

3 Introduction What is Matlab and Simulink? What is Matlab High-Performance language for technical computing Integrates computation, visualisation and programming Matlab = MATrix LABoratory Features family of add-on, application-specific toolboxes 5/78 Introduction What is Matlab and Simulink? What are Matlab components? Development Environment The Matlab Mathematical Function Library The Matlab language Graphics The Matlab Application Program Interface 6/78

4 Introduction What is Matlab and Simulink? What is Simulink? Software Package for modelling, simulating and analysing dynamic systems Supports linear & Non-linear systems Supports continuous or discrete time systems Supports multirate systems Allows you to model real-life situations Allows for a top-down and bottom-up approach 7/78 Introduction What is Matlab and Simulink? How Simulink Works? 1. Create a block diagram (model) 2. Simulate the system represented by a block diagram 8/78

5 Matlab Environment Introduction Matlab&Simulink Getting Started 9/78 The Matlab Language Introduction Matlab&Simulink Getting Started Dürer s Matrix 1 A=[ ; ; ; ]; 2 sum(a) %ans = sum(a ) %ans = sum( diag(a)) %ans = 34 Operators 1 100: 7:50 % sum(a(1:4,4)) % ans = / 78

6 Introduction Matlab&Simulink Getting Started The Matlab API You can use C or FORTRAN Pipes on UNIX, COM on Windows You can call Matlab routines from C/FORTRAN programs and vice versa You can call Java from Matlab 11 / 78 Introduction Matlab&Simulink Getting Started Simulink Environment 12 / 78

7 Starting Simulink Introduction Matlab&Simulink Getting Started Just type in Matlab 1 simulink 13 / 78 Part II Matlab Background 14 / 78

8 Laplace Transform Matlab Background Transfer Functions and Laplace Transforms Definition The Laplace Transform is an integral transform perhaps second only to the Fourier transform in its utility in solving physical problems. The Laplace transform is defined by: L [f (t)] (s) 0 f (t)e st dt Source: [1, Abramowitz and Stegun 1972] 15 / 78 Laplace Transform Matlab Background Transfer Functions and Laplace Transforms Several Laplace Transforms and properties f L [f (t)] (s) range t t n n! e at 1 s a s s > 0 s > 0 s 2 n Z > 0 s n+1 s > a ] L t [f (n) (t) (s) =s n L t [f (t)] s n 1 f (0) s n 2 f (0)... f (n 1) (0) (1) This property can be used to transform differential equations into algebraic ones. This is called Heaviside calculus 16 / 78

9 Matlab Background Transfer Functions and Laplace Transforms Heaviside Calculus Example Let us apply the Laplace transform to the following equation: f (t)+a 1 f (t)+a 0 f (t) =0 which should give us: { s 2 L t [f (t)] (s) sf (0) f (0) } + +a 1 {sl t [f (t)] (s) f (0)} + +a 0 L t [f (t)] (s) =0 which can be rearranged to: L t [f (t)] (s) = sf (0) + f (0) + a 1 f (0) s 2 + a 1 s + a 0 17 / 78 Matlab Background Transfer Functions and Laplace Transforms Transfer Functions For Matlab modelling we need Transfer Functions To find the Transfer Function of a given system we need to take the Laplace transform of the system modelling equations (2) & (3) System modelling equations F = m v + bv (2) y = v (3) Laplace Transform: F (s) =msv (s)+bv (s) Y (s) =V (s) 18 / 78

10 Matlab Background Transfer Functions and Laplace Transforms Transfer Functions cntd. Assuming that our output is velocity we can substitute it from equation (5) Transfer Function Laplace Transform: Transfer Function: F (s) =msv (s)+bv (s) (4) Y (s) =V (s) (5) Y (s) F (s) = 1 ms + b (6) 19 / 78 Matlab Background Matlab Functions Matlab Functions Transfer Function I What is tf? Specifies a SISO transfer function for model h(s) =n(s)/d(s) 1 h = tf(num, den ) What are num & den? row vectors listing the coefficients of the polynomials n(s) and d(s) ordered in descending powers of s Source: Matlab Help 20 / 78

11 Matlab Background Matlab Functions Matlab Functions Transfer Function II tf Example T (s) = 2s 3 s+1 h=tf([2-3], [1 1]) T (s) = 2s+1 4s 2 +s+1 h=tf([2 1], [4 1 1]) 21 / 78 Matlab Background Matlab Functions Feedback I Matlab Functions Matlab code 1 sys = feedback( forward, backward ); Source: [2, Angermann et al. 2004] 22 / 78

12 Matlab Background Matlab Functions Feedback II Matlab Functions obtains a closed-loop transfer function directly from the open-loop transfer function no need to compute by hand Example Forward = 1 st i (7) Backward = V (8) T (s) = C(s) 1 R(s) = st i 1+V 1 st i feedback(tf(1,[ti 0]), tf(v, 1)) (9) Matlab Functions Step Response 23 / 78 Matlab Background Matlab Functions system=tf ([2 1],[4 1 1]); t =0:0.1:50; step(100 system) axis ([ ]) 24 / 78

13 Matlab Background Steady-State Error Definition Steady-State Error Figure 1: Unity Feedback System Steady-State Error The difference between the input and output of a system in the limit as time goes to infinity 25 / 78 Steady-State Error Matlab Background Steady-State Error Steady-State Error e( ) = lim s 0 sr(s) 1+G(s) (10) e( ) = lim s 0 sr(s) 1 T (s) (11) 26 / 78

14 Matlab Background Proportional, Integral and Derivative Controllers Feedback controller How does it work I? Figure 2: System controller e represents the tracking error e difference between desired input (R) an actual output (Y) e is sent to controller which computes: derivative of e integral of e u controller output is equal to / 78 Matlab Background Proportional, Integral and Derivative Controllers Feedback controller How does it work II? u controller output is equal to: K p (proportional gain) times the magnitude of the error + Ki (integral gain) times the integral of the error + K d (derivative gain) times the derivative of the error Controller s Output u = K p e + K i edt + K d de dt Controller s Transfer Function K p + K i s + K ds = K ds 2 + K p s + K i s 28 / 78

15 Matlab Background Characteristics of PID Controllers Proportional, Integral and Derivative Controllers Proportional Controller K p reduces the rise time reduces but never eliminates steady-state error Integral Controller K i eliminates steady-state error worsens transient response Derivative Controller K d increases the stability of the system reduces overshoot improves transient response 29 / 78 Example Problem Matlab Background Proportional, Integral and Derivative Controllers Figure 3: Mass spring and damper problem Modelling Equation mẍ + bẋ + kx = F (12) 30 / 78

16 Matlab Background Proportional, Integral and Derivative Controllers Example Problem Laplace & Transfer Functions mẍ + bẋ + kx = F ms 2 X (s)+bsx (s)+kx (s) =F (s) (13) X (s) F (s) = 1 ms 2 + bs + k (14) 31 / 78 Matlab Background Proportional, Integral and Derivative Controllers Matlab System Response Assumptions Let: m =1[kg], b = 10[Ns/m], k = 20[N/m] Matlab code 1 %{ Set up variables%} 2 m=1; b=10; k=20; 3 %{ Calculate response%} 4 num=1; 5 den=[m, b, k ] ; 6 plant=tf(num, den ) ; 7 step(plant) 32 / 78

17 Matlab Background Proportional, Integral and Derivative Controllers Matlab System Response Figure 4: Amplitude Displacement 33 / 78 Matlab Background Proportional, Integral and Derivative Controllers Problems The steady-state error is equal to 0.95 equation (11) Therisetimeisabout1second The settling time is about 1.5 seconds The PID controller should influence (reduce) all those parameters 34 / 78

18 Matlab Background Proportional, Integral and Derivative Controllers Controllers Characteristics Type Rise time Overshoot Settling time S-S Error K p decrease increase small change decrease K i decrease increase increase eliminate K d small change decrease decrease small change These correlations may not be exactly accurate, because K p, K i,andk d are dependent on each other. In fact, changing one of these variables can change the effect of the other two. 35 / 78 Matlab Background Proportional, Integral and Derivative Controllers Proportional Controller P Transfer Function Matlab code X (s) F (s) = K p s 2 + bs +(k + K p ) 1 %{ Set up proportional gain%} 2 Kp=300; 3 %{ Calculate controller%} 4 sys ctl=feedback(kp plant, 1); 5 %{ Plot results%} 6 t=0:0.01:2; 7 step(sys ctl, t) 36 / 78

19 Matlab Background Proportional Controller Plot Proportional, Integral and Derivative Controllers Figure 5: Improved rise time & steady-state error 37 / 78 Matlab Background Proportional Derivative Controller Proportional, Integral and Derivative Controllers PD Transfer Function Matlab code X (s) F (s) = K d s + K p s 2 +(b + K d )s +(k + K p ) 1 %{ Set up proportional and derivative gain%} 2 Kp=300; Kd=10; 3 %{ Calculate controller%} 4 contr=tf ([Kd, Kp],1); 5 sys ctl=feedback(contr plant, 1); 6 %{ Plot results%} 7 t=0:0.01:2; 8 step(sys ctl, t) 38 / 78

20 Matlab Background Proportional Derivative Controller Plot Proportional, Integral and Derivative Controllers Figure 6: Reduced over-shoot and settling time 39 / 78 Matlab Background Proportional Integral Controller Proportional, Integral and Derivative Controllers PI Transfer Function X (s) F (s) = K p s + K i s 3 + bs 2 +(k + K p )s + K i Matlab code 1 %{ Set up proportional and integral gain%} 2 Kp=30; Ki=70; 3 %{ Calculate controller%} 4 contr=tf ([Kp, Ki],[1, 0]); 5 sys ctl=feedback(contr plant, 1); 6 %{ Plot results%} 7 t=0:0.01:2; 8 step(sys ctl, t) 40 / 78

21 Matlab Background Proportional Integral Controller Plot Proportional, Integral and Derivative Controllers Figure 7: Eliminated steady-state error, decreased over-shoot 41 / 78 Matlab Background Proportional, Integral and Derivative Controllers Proportional Integral Derivative Controller PID Transfer Function X (s) F (s) = K d s 2 + K p s + K i s 3 +(b + K d )s 2 +(k + K p )s + K i Matlab code 1 %{ Set up proportional and integral gain%} 2 Kp=350; Ki=300; Kd=50; 3 %{ Calculate controller%} 4 contr=tf ([Kd, Kp, Ki],[1, 0]); 5 sys ctl=feedback(contr plant, 1); 6 %{ Plot results%} 7 t=0:0.01:2; 8 step(sys ctl, t) 42 / 78

22 Matlab Background Proportional, Integral and Derivative Controllers Proportional Integral Derivative Controller Plot Figure 8: Eliminated steady-state error, decreased over-shoot 43 / 78 Part III Matlab Cruise Control System 44 / 78

23 Matlab Cruise Control System Design Criteria How does Cruise Control for Poor work? Figure 9: Forces taking part in car s movement Based on Carnegie Mellon University Library Control Tutorials for Matlab and Simulink 45 / 78 Matlab Cruise Control System Design Criteria Building the Model Using Newton s law we derive Where: m = 1200[kg], b = 50[ Ns m ], F = 500[N] F = m v + bv (15) y = v (16) 46 / 78

24 Matlab Cruise Control System Design Criteria Design Criteria ForthegivendataV max = 10[m/s] = 36[km/h] The car should accelerate to V max within 6[s] 10% tolerance on the initial velocity 2% of a steady-state error 47 / 78 Matlab Cruise Control System Matlab Representation Transfer Function System Equations: F = m v + bv y = v Laplace Transform: F (s) =msv (s)+bv (s) (17) Y (s) =V (s) (18) Transfer Function: Y (s) F (s) = 1 ms + b (19) 48 / 78

25 Matlab Cruise Control System Matlab Representation Matlab Representation Now in Matlab we need to type Matlab code 1 m=1200; 2 b=50; 3 num = [ 1 ] ; 4 den=[m,b ] ; 5 cruise=tf(num, den ) ; 6 step = (500 cruise ); 49 / 78 Results Matlab Cruise Control System Matlab Representation Figure 10: Car velocity diagram mind the design criteria 50 / 78

26 Matlab Cruise Control System Matlab Representation Design criteria revisited Our model needs over 100[s] to reach the steady-state The design criteria mentioned 5 seconds 51 / 78 Matlab Cruise Control System PID Controller Feedback controller To adjust the car speed within the limits of specification We need the feedback controller Figure 11: System controller 52 / 78

27 Matlab Cruise Control System Decreasing the rise time PID Controller Proportional Controller Y (s) R(s) = K p ms +(b + K p ) (20) Matlab code 1 Kp=100; m=1200; b=50; 2 num = [ 1 ] ; den =[m, b ] ; 3 cruise=tf(num, den ) ; 4 sys ctl=feedback(kp cruise, 1); 5 t=0:0.1:20; 6 step(10 sys cl,t) 7 axis ([ ]) 53 / 78 Matlab Cruise Control System Under- and Overcontrol PID Controller Figure 12: K p =100 Figure 13: K p = / 78

28 Matlab Cruise Control System PID Controller Making Rise Time Reasonable Proportional Integral Controller Y (s) R(s) = K p s + K i ms 2 +(b + K p )s + K i (21) Matlab code 1 Kp=800; Ki=40; m=1200; b=50; 2 num = [ 1 ] ; den =[m, b ] ; 3 cruise=tf(num, den ) ; 4 contr=tf ([Kp Ki],[1 0]) 5 sys ctl=feedback(contr cruise, 1); 6 t=0:0.1:20; 7 step(10 sys cl,t) 8 axis ([ ]) 55 / 78 Results Matlab Cruise Control System PID Controller Figure 14: Car velocity diagram meeting the design criteria 56 / 78

29 Part IV Simulink Cruise Control System 57 / 78 Simulink Cruise Control System Building the Model How does Cruise Control for Poor work? Figure 15: Forces taking part in car s movement Based on Carnegie Mellon University Library Control Tutorials for Matlab and Simulink 58 / 78

30 Physical Description Simulink Cruise Control System Building the Model Summing up all the forces acting on the mass Forces acting on the mass F = m dv dt + bv (22) Where: m=1200[kg], b=50[ Nsec m ], F=500[N] 59 / 78 Simulink Cruise Control System Physical Description cntd. Building the Model Integrating the acceleration to obtain the velocity Integral of acceleration a = dv dt dv dt = v (23) 60 / 78

31 Simulink Cruise Control System Building the Model in Simulink Building the Model Figure 16: Integrator block from Continuous block library 61 / 78 Simulink Cruise Control System Building the Model in Simulink Building the Model Obtaining acceleration Acceleration a = dv dt = F bv m (24) 62 / 78

32 Simulink Cruise Control System Building the Model Building the Model in Simulink Figure 17: Gain block from Math operators block library 63 / 78 Simulink Cruise Control System Building the Model Elements used in Simulink Model Friction (Gain block) Subtract (from Math Operators) Input (Step block from Sources) Output (Scope from Sinks) 64 / 78

33 Complete Model Simulink Cruise Control System Building the Model 65 / 78 Simulink Cruise Control System Building the Model Mapping Physical Equation to Simulink Model 66 / 78

34 Simulink Cruise Control System Setting up the Variables Simulating the Model Now it is time to use our input values in Simulink... F=500[N] In Step block set: Step time = 0 and Final value = and adjust simulation parameters... Simulation Configuration Parameters... Stop time = and set up variables in Matlab 1 m=1200; 2 b=50; 67 / 78 Running Simulation Simulink Cruise Control System Simulating the Model Choose Simulation Start Double-click on the Scope block / 78

35 Simulink Cruise Control System Extracting Model into Matlab Simulating the Model Replace the Step and Scope Blocks with In and Out Connection Blocks 69 / 78 Simulink Cruise Control System Verifying Extracted Model Simulating the Model We can convert extracted model into set of linear equations into transfer function Matlab code 1 [A, B, C, D]=linmod( cc2mtlb ); 2 [ num, den ]= ss2tf(a, B, C, D); 3 step(500 tf(num, den ) ) ; 70 / 78

36 Simulink Cruise Control System Simulating the Model Matlab vs Simulink Figure 18: Matlab Figure 19: Simulink 71 / 78 Simulink Cruise Control System Implementing the PI Control The open-loop system In Matlab section we have designed a PI Controller Kp = 800 Ki =40 We will do the same in Simulik First we need to contain our previous system in a Sybsystem block Choose a Subsystem block from the Ports&Subsystems Library Copy in the model we used with Matlab 72 / 78

37 Subsystem Simulink Cruise Control System Implementing the PI Control Figure 20: Subsystem Block and its Contents 73 / 78 PI Controller I Simulink Cruise Control System Implementing the PI Control Figure 21: Step: final value=10, time=0 74 / 78

38 PI Controller II Simulink Cruise Control System Implementing the PI Control Figure 22: We use Transfer Fcn block from Continuous-Time Linear Systems Library 75 / 78 Results Simulink Cruise Control System Implementing the PI Control Runnig simulation with time set to 15[s] 76 / 78

39 References Course basic references 77/78 Textbooks Digital Control of Dynamic Systems (3rd Edition) by Gene F. Franklin, J. David Powell, Michael L. Workman Publisher: Prentice Hall; 3 edition (December 29, 1997) ISBN: Lecture slides Computer Lab Exercises 78/78

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