# MATHEMATICAL MODELING OF CONTROL SYSTEMS

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1 1 MATHEMATICAL MODELING OF CONTROL SYSTEMS Sep-14 Dr. Mohammed Morsy

2 Outline Introduction Transfer function and impulse response function Laplace Transform Review Automatic control systems Signal Flow Graph Modeling In state-space Transformation of mathematical models with Matlab Linearization of nonlinear mathematical models Sep-14 2

3 Introduction A mathematical model of a dynamic system is defined as a set of equations that represents the dynamics of the system accurately The dynamics of many systems may be described in terms of differential equations obtained from physical laws governing a particular system In obtaining a mathematical model, we must make a compromise between the simplicity of the model and the accuracy of the results of the analysis In general, in solving a new problem, it is desirable to build a simplified model so that we can get a general feeling for the solution Sep-14 3

4 Introduction Linear Systems A system is called linear if the principle of superposition applies. The principle of superposition states that the response produced by the simultaneous application of two different forcing functions is the sum of the two individual responses. Hence, for the linear system, the response to several inputs can be calculated by treating one input at a time and adding the results. Linear Time-Invariant (LTI) Systems and Linear Time- Varying Systems A differential equation is linear if the coefficients are constants or functions only of the independent variable Systems that are represented by differential equations whose coefficients are functions of time are called linear timevarying systems. Sep-14 4

5 System Model Differential equation Transfer function Frequency characteristic Study time-domain response Linear system study frequency-domain response Transfer function Differential Laplace equation transform Frequency Fourier characteristic transform Sep-14 5

6 Transfer Function and Impulse Response Transfer Function: The transfer function of an LTI system is defined as the ratio of the Laplace transform of the output (response function) to the Laplace transform of the input (driving function) under the assumption that all initial conditions are zero. Sep-14 6

7 Transfer Function The applicability of the transfer function concept is limited to LTI systems The transfer function is a property of a system itself, independent of the magnitude and nature of the input or driving function. If the transfer function of a system is known, the output or response can be studied for various forms of inputs If the transfer function of a system is unknown, it may be established experimentally by introducing known inputs and studying the output of the system. Sep-14 7

8 Transfer Function (2) For an LTI system the transfer function G(s) is Multiplication in the complex domain is equivalent to convolution in the time domain Sep-14 8

9 Impulse Response Consider the output (response) of a linear time-invariant system to a unit-impulse input when the initial conditions are zero. Since the Laplace transform of the unit-impulse function is unity, the Laplace transform of the system output The impulse-response function g(t) is thus the response of an LTI system to a unit-impulse input when the initial conditions are zero It is hence possible to obtain complete information about the dynamic characteristics of the system by exciting it with an impulse input and measuring the response Sep-14 9

10 Faculty of Engineering - Alexandria University Control Systems and Components 2013 Laplace Transform Review Note: This review is not included in the textbook (Slides 10-42) Sep-14 10

11 Solving Differential Equation Example Solving linear differential equations with constant coefficients: To find the general solution (involving solving the characteristic equation) To find a particular solution of the complete equation (involving constructing the family of a function) Sep-14 11

12 Laplace Transform Difficult Time-domain ODE problems Laplace Transform (LT) s-domain algebra problems Easy Solutions of timedomain problems Inverse LT Solutions of algebra problems Sep-14 12

13 Laplace Transform The Laplace transform of a function f(t) is defined as F( s) f ( t) 0 f ( t) e st dt Laplace, Pierre-Simon where variable. s j is a complex Sep-14 13

14 Examples Step signal: f(t)=a st 0 0 F( s) f ( t) e dt Ae st dt A e s st 0 A s Exponential signal f(t)= Fs () at st e e dt 0 s e at 1 ( as) t a e 0 s 1 a Sep-14 14

15 Laplace Transform Table f(t) F(s) f(t) F(s) δ(t) 1 1(t) t e at 1 s 1 2 s 1 s a sin wt cos wt e at sin wt e at cos wt s s w w 2 2 s w 2 2 w 2 2 ( s a) w s a 2 2 ( s a) w Sep-14 15

16 Properties of Laplace Transform (1) Linearity [ af ( t) bf ( t)] a [ f ( t)] b [ f ( t)] (2) Differentiation df () t sf ( s ) f (0) dt Using Integration By Parts method to prove where f(0) is the initial value of f(t). n d f () t (1) s F s n s f s f f dt n n 1 n 2 ( n 1) ( ) (0) (0) (0) Sep-14 16

17 Properties of Laplace Transform (2) (3) Integration t F( s) f() d 0 s Using Integration By Parts method to prove ) t1 t2 t n o o o f () ddt dt dtn F( s) s n Sep-14 17

18 Properties of Laplace Transform (3) (4)Final-value Theorem lim t f ( t) limsf( s) s0 The final-value theorem relates the steady-state behavior of f(t) to the behavior of sf(s) in the neighborhood of s=0 (5) Initial-value Theorem lim t0 f ( t) lim s sf( s) Sep-14 18

19 Properties of Laplace Transform (4) (6)Shifting Theorem: a. shift in time (real domain) [ f( t )] e s F() s b. shift in complex domain [ e at f ( t)] (7) Real convolution (Complex multiplication) Theorem t 0 [ f ( t ) f ( ) d ] F ( s) F ( s) F( s a) Sep-14 19

20 Inverse Laplace Transform Definition:Inverse Laplace transform, denoted by is given by 1 [ Fs ( )] st f ( t) [ F( s)] F( s) e ds( t 0) 2 j 1 1 where C is a real constant Cj Cj Note: The inverse Laplace transform operation involving rational functions can be carried out using Laplace transform table and partial-fraction expansion. Sep-14 20

21 Partial Fraction Expansion Ns () b s b s b s bm F( s) ( m n) D() s s a s a s a m m1 0 1 m1 n n1 1 n1 If F(s) is broken up into components F( s) F ( s) F ( s) F ( s) 1 2 If the inverse Laplace transforms of components are readily available, then F( s) F ( s) F ( s) F ( s) f ( t) f ( t)... f ( t) 1 2 n n n n Sep-14 21

22 Poles and Zeros Poles A complex number s 0 is said to be a pole of a complex variable function F(s) if F(s 0 )=. Zeros A complex number s0 is said to be a zero of a complex variable function F(s) if F(s 0 )=0. Examples: ( s 1)( s 2) ( s 3)( s 4) 2 s s 1 poles: -3, -4; zeros: 1, -2 2s 2 poles: -1+j, -1-j; zeros: -1 Sep-14 22

23 Faculty of Engineering - Alexandria University Control Systems and Components 2013 Case 1: F(s) has simple real poles Fs () Ns () b s b s b s b D() s s a s a s a m m1 0 1 m1 n n1 1 n1 n m where p ( i 1, 2,, n) are eigenvalues of D( s) 0, and i c c cn s p s p s p Ns () ci ( s pi ) () Ds s pi Partial-Fraction Expansion n Inverse LT c e p t c e p t... c e n f() t Parameters pk give shape and numbers ck give magnitudes. n pt Sep-14 23

24 Faculty of Engineering - Alexandria University Control Systems and Components 2013 Example 1 1 Fs () ( s 1)( s 2)( s 3) Partial-Fraction Expansion c c c3 s 1 s 2 s c1 ( s 1) ( 1)( 2) ( 3) s s s 6 s1 1 c2 ( s 2) 1 ( 1)( 2)( 3) s s s 15 s ( 1)( 2) ( 3) ( 3) c s s s s 10 s Fs () 6 s 1 15 s 2 10 s 3 1 t 1 2t 1 3t f () t e e e Sep-14 24

25 Faculty of Engineering - Alexandria University Control Systems and Components 2013 Case 2: F(s) has simple complex-conjugate poles Example 2 Laplace transform Applying initial conditions Ys () s Partial-Fraction Expansion Inverse Laplace transform 2 s 5 4s5 yt s 5 2 s s ( s 2) 1 ( s 2) 1 2 ( s 2) 1 2t 2t () e cos t 3e sin s 23 ( 2) 1 t Sep-14 25

26 Faculty of Engineering - Alexandria University Control Systems and Components 2013 Case 3: F(s) has multiple-order poles N( s) N( s) Fs () l D( s) ( s p1)( s p2) ( s pnr) ( s pi ) c c b b b s p s p s p s p s p 1 nl l l1 1 l l1 1 nl ( i) ( i) Simple poles The coefficients c,, of simple poles can be calculated as Case 1; 1 cn l Multi-order poles The coefficients corresponding to the multi-order poles are determined as b l d l b ( ) ( ), 1 ()( ) l F s s pi bl,, Fs s pi s p 1 ds lm 1 s pi 1 m 1 ( ) ( ), 1 l d N s l d N s l ( ) 1 ( )! s pi ( ) b ( 1)! s pi m ( ) ds D s l ds D s sp sp Sep i i

27 Faculty of Engineering - Alexandria University Control Systems and Components 2013 Example 3 Laplace transform: s Y( s) s y(0) sy( 0) y(0) 3 s Y( s) 3 sy(0) 3 y(0) 3 sy( s) 3y( 0) Y( s) Applying initial conditions: Partial-Fraction Expansion Ys () Ys () 1 s 1 ss ( 1) 3 s= -1 is a 3 rd order pole c1 b3 b2 b1 3 2 s ( s 1) ( s 1) s 1 Sep-14 27

28 Faculty of Engineering - Alexandria University Control Systems and Components 2013 Determining coefficients: b 1 [ ( s 1) ] 1 ss ( 1) s1 c 1 s ss ( 1) s s1 ( 1) 1 ds s s s ds s s b1 (2 s ) 1 2! s1 d 1 d 1 b [ ( s 1) ] [ ( )] ( s ) Ys () s s s s 3 2 ( 1) ( 1) 1 Inverse Laplace transform: 1 y t t e te e 2 t t t ( ) Sep-14 28

29 Matlab Application 1. Laplace Transform L= laplace(f) 2. Inverse Laplace Transform F= ilaplace(l) >> syms t >> L=laplace(t) L= 1/s^2 >> L=laplace(sin(t)) L= 1/(s^2+1) >> F=ilaplace(L) F= sin(t) Sep-14 29

30 Transfer Function Input u(t) LTI system Output y(t) Consider a linear system described by differential equation y ( t) a y ( t) a y( t) b u ( t) b u ( t) bu ( t) b u( t) ( n) ( n1) ( m) ( m1) (1) n1 0 m m1 0 Assume all initial conditions are zero, we get the transfer function(tf) of the system as TF G() s L L output y() t input u() t zero initial condition m Y() s bms b s... b s b n U ( s) s a s... a s a m1 m1 1 0 n1 n1 1 0 Sep-14 30

31 Faculty of Engineering - Alexandria University Control Systems and Components 2013 Example 1. Find the transfer function of the RLC Solution: Input u(t) R 1) Writing the differential equation of the system according to physical law: L C i(t) u c (t) LCu ( t) RCu ( t) u ( t) u( t) Output 2) Assuming all initial conditions are zero and applying Laplace transform 3) Calculating the transfer function Gs () as Gs () C C C 2 LCs Uc s RCsU c s Uc s U s ( ) ( ) ( ) ( ) U () s 1 U s LCs RCs c ( ) 2 1 Sep-14 31

32 Exercise Find the transfer function of the following system : 2 d y t ( ) dy( t) 5 4 y( t) u( t) 2 dt dt Sep-14 32

33 Transfer Function of Typical Components Component ODE TF vt () it () R v( t) Ri( t) G() s V() s Is () R vt () it () L v() t di() t L dt G() s V() s Is () sl vt () it () 1 t v( t) i( ) d C C 0 Gs () V( s) 1 I() s sc Sep-14 33

34 Properties of Transfer Functions The transfer function is defined only for a linear time-invariant system, not for nonlinear system All initial conditions of the system are set to zero The transfer function is independent of the input of the system The transfer function G(s) is the Laplace transform of the unit impulse response g(t) Sep-14 34

35 Faculty of Engineering - Alexandria University Control Systems and Components 2013 How poles and zeros relate to system response Why we strive to obtain TF models? Why control engineers prefer to use TF models? How to use a TF model to analyze and design control systems? we start from the relationship between the locations of zeros and poles of TF and the output responses of a system Sep-14 35

36 Faculty of Engineering - Alexandria University Control Systems and Components 2013 Transfer function X() s s A a Time-domain impulse response x() t Ae at Position of poles and zeros j -a 0 i 0 Sep-14 36

37 Faculty of Engineering - Alexandria University Control Systems and Components 2013 Transfer function Time-domain impulse response A1s B1 X() s 2 2 at ( s a) b x( t) Ae sin( bt ) Position of poles and zeros j b -a 0 i 0 Sep-14 37

38 Faculty of Engineering - Alexandria University Control Systems and Components 2013 Transfer function X() s A s B s b Time-domain impulse response x( t) Asin( bt ) Position of poles and zeros j b 0 i 0 Sep-14 38

39 Faculty of Engineering - Alexandria University Control Systems and Components 2013 Transfer function Time-domain impulse response A X() s s a () at x t Ae Position of poles and zeros j 0 -a i Sep-14 39

40 Faculty of Engineering - Alexandria University Control Systems and Components 2013 Transfer function X() s A s B ( s a) b Time-domain impulse response at x( t) Ae sin( bt ) Position of poles and zeros j -a b 0 i 0 Sep-14 40

41 Summary of Pole Position & System Dynamics Sep-14 41

42 Faculty of Engineering - Alexandria University Control Systems and Components 2013 Characteristic equation obtained by setting the denominator polynomial of the transfer function to zero n n1 s an 1s a1s a0 0 Note: stability of linear single-input, single-output systems is completely governed by the roots of the characteristics equation. 42 Sep-14 42

43 Transfer Function (TF) models in Matlab Suppose a linear SISO system with input u(t), output y(t), the transfer function of the system is G( s) num den Y( S) U( s) b m bms n s b a m, b,..., b m 1 1, a,..., a 1 n 0 0 s m1 m1 n1 n1s TF in polynomial form >> Sys = tf(num,den) >> [num, den] = tfdata (sys)... b1s b... a s a Descending power of s Sep-14 43

44 Faculty of Engineering - Alexandria University Control Systems and Components 2013 TF in zero-pole form >> sys = zpk(z, p, k) >> [z, p,k] = tfdata (sys) Transform TS from zero-pole form into polynomial form >> [z, p, k] = tf2zp(num, den) 44 Sep-14 44

45 Review Questions What is the definition of transfer function? When defining the transfer function, what happens to initial conditions of the system? Does a nonlinear system have a transfer function? How does a transfer function of a LTI system relate to its impulse response? Define the characteristic equation of a linear system in terms of the transfer function Sep-14 45

46 Automatic Control Systems A control system may consist of a number of components To show the functions performed by each component, in control engineering, we commonly use a diagram called the block diagram A block diagram of a system is a pictorial representation of the functions performed by each component and of the flow of signals A block diagram has the advantage of indicating more realistically the signal flows of the actual system Sep-14 46

47 Block Diagrams In a block diagram all system variables are linked to each other through functional blocks The transfer functions of the components are usually entered in the corresponding blocks Blocks are connected by arrows to indicate the direction of the flow of signals Note: The dimension of the output signal from the block is the dimension of the input signal multiplied by the dimension of the transfer function in the block Element of a block diagram Sep-14 47

48 Block Diagrams(2) The advantage of the block diagram representation is the simplicity of forming the overall block diagram for the entire system by connecting the blocks of the components according to the signal flow A block diagram contains information concerning dynamic behavior, but it does not include any information on the physical construction of the system A number of different block diagrams can be drawn for a system Sep-14 48

49 Block Diagram Representation The transfer function relationship Y( s) G( s) U( s) can be graphically denoted through a block diagram. U(s) G(s) Y(s) Sep-14 49

50 Equivalent Transform of Block Diagram 1. Connection in series U(s) G 1 (s) X(s) G 2 (s) Y(s) U(s) G(s) Y(s) Gs ( )? Ys () G( s) G1( s) G2( s) Us () Sep-14 50

51 Faculty of Engineering - Alexandria University Control Systems and Components Connection in parallel U(s) G1(s) G2(s) Y1(s) Y2(s) Y(S) U(s) G(s) Y(s) Gs ( )? Ys () G( s) G1( s) G2( s) Us () Sep-14 51

52 Faculty of Engineering - Alexandria University Control Systems and Components Negative feedback R(s) _ U(s) G(s) Y(s) R(s) M(s) Y(s) H(s) Y ( s) U( s) G( s) U ( s) R( s) Y( s) H( s) Transfer function of a negative feedback system: Gs ( ) gain of the forward path M() s 1G( s) H( s) 1 gain of the loop Y( s) R( s) Y( s) H( s) G( s) Sep-14 52

53 Matlab Application Obtaining Cascaded, Parallel, and Feedback (Closed- Loop) Transfer Functions with MATLAB Suppose that there are two components G 1 (s)andg 2 (s) MATLAB has convenient commands to obtain the cascaded, parallel, and feedback (closed-loop) transfer functions. Example: Sep-14 53

54 Matlab Application (2) To obtain the transfer functions of the cascaded system, parallel system, or feedback (closedloop) system The following commands may be used [num, den] = series(num1,den1,num2,den2) [num, den] = parallel(num1,den1,num2,den2) [num, den] = feedback(num1,den1,num2,den2) Check Matlab Program 2-1 in the textbook Sep-14 54

55 Automatic Controllers An automatic controller compares the actual value of the plant output with the reference input (desired value), determines the deviation, and produces a control signal that will reduce the deviation to zero or to a small value. Sep-14 55

56 Classifications of Industrial Controllers 1. Two-position or on off controllers 2. Proportional controllers 3. Integral controllers 4. Proportional-plus-integral controllers 5. Proportional-plus-derivative controllers 6. Proportional-plus-integral-plus-derivative controllers Sep-14 56

57 Two-Position or On Off Control Action In a two-position control system, the actuating element has only two fixed positions, which are, in many cases, simply on and off, e.g. Liquid-level control system Let the output signal from the controller be u(t)and the actuating error signal be e(t) Block diagram of an on off controller Sep-14 57

58 Proportional Control Action For a controller with proportional control action, the relationship between the output of the controller u(t)and the actuating error signal e(t)is Where K p is termed the proportional gain Whatever the actual mechanism may be and whatever the form of the operating power, the proportional controller is essentially an amplifier with an adjustable gain Sep-14 58

59 Integral Control Action In a controller with integral control action, the value of the controller output u(t)is changed at a rate proportional to the actuating error signal e(t) where K i is an adjustable constant Sep-14 59

60 Other Control Actions Proportional-Plus-Integral Control Action Proportional-Plus-Derivative Control Action Proportional-Plus-Integral-Plus-Derivative Control Action Block diagram of a proportional-plusintegral-plus-derivative controller Sep-14 60

61 Closed-Loop System Subjected to a Disturbance Figure shows a closed-loop system subjected to a disturbance We can use superposition of LTI systems to obtain the output For R(s)=0 For D(s)=0 Sep-14 61

62 Closed-Loop System Subjected to a Disturbance The response to both inputs is Consider now the case where G 1 (s)h(s) >>1 and G 1 (s)g 2 (s)h(s) >>1. In this case, the closed-loop transfer function C D (s)/d(s) becomes almost zero, and the effect of the disturbance is suppressed. This is an advantage of the closed-loop system The closed-loop transfer function C R (s)/r(s) approaches 1/H(s) as the gain of G 1 (s)g 2 (s)h(s) increases This means that if G 1 (s)g 2 (s)h(s) >>1, then the closed-loop transfer function C R (s)/r(s) becomes independent of G 1 (s) and G 2 (s) and inversely proportional to H(s) Sep-14 62

63 Procedures for Drawing a Block Diagram 1. Write the equations that describe the dynamic behavior of each component 2. Take the Laplace transforms of these equations, assuming zero initial conditions 3. Represent each Laplace-transformed equation individually in block form 4. Assemble the elements into a complete block diagram. Sep-14 63

64 Example Consider the RC circuit shown in Figure The equations for this circuit are, Sep-14 64

65 Block Diagram Reduction Any number of cascaded blocks can be replaced by a single block, the transfer function of which is simply the product of the individual transfer functions Blocks can be connected in series only if the output of one block is not affected by the next following block (no feedback) A complicated block diagram involving many feedback loops can be simplified by a step-by-step rearrangement Simplification of the block diagram by rearrangements considerably reduces the labor needed for subsequent mathematical analysis Sep-14 65

66 Example Simplify this diagram Solution: By moving the summing point of the negative feedback loop containing H 2 outside the positive feedback loop containing H 1, we obtain Sep-14 66

67 Example (2) Eliminating the positive feedback loop The elimination of the loop containing H 2 /G 1 gives Finally, eliminating the feedback loop results in Sep-14 67

68 Example (3) Notice that the numerator of the closed-loop transfer function C(s)/R(s) is the product of the transfer functions of the feed-forward path. The denominator of C(s)/R(s) is equal to The positive feedback loop yields a negative term in the denominator Sep-14 68

69 Faculty of Engineering - Alexandria University Control Systems and Components 2013 Signal Flow Graph Note: This part is not included in the textbook (Slides 69-76) Sep-14 69

70 Signal Flow Graph (SFG) SFG was introduced by S.J. Mason for the cause-andeffect representation of linear systems. 1. Each signal is represented by a node. 2. Each transfer function is represented by a branch. U(s) G(s) Y(s) G(s) U(s) Y(s) R(s) _ U(s) G(s) Y(s) R(s) 1 U(s) G(s) Y(s) H(s) -H(s) Sep-14 70

71 Block Diagram and Signal-Flow Graph Ur () s - 1 I () s 1-1 sc R1 1 U () s 1 1 R 2 Uc I () s 2 sc 1 () s - 2 U () s I () s U () s I () s r R1 1 sc1 1 2 R 2 U () c s Note: A signal flow graph and a block diagram contain exactly the same information (there is no advantage to one over the other; there is only personal preference) sc Sep-14 71

72 Mason s Rule N M k N Ys ( ) 1 M () s M kk Us () k1 Total number of forward paths between output Y(s) and input U(s) Path gain of the k th forward path 1 ( all individual loop gains) ( gain products of all possible two loops that do not touch) ( gain products of all possible three loops that do not touch) k Value of for that part of the block diagram that does not touch the k th forward path Sep-14 72

73 Faculty of Engineering - Alexandria University Control Systems and Components 2013 Example 1 Find the transfer function for the following block diagram U(s) _ 1/s 1/s 1/s b1 b2 b Y(s) + a1 Solution: Forward path a2 Path gain 1 M1 1 ( b1)(1) s 1 1 M2 1 ( b2)(1) s s M3 1 ( b3)(1) s s s and the determinates are a1 a2 a s s s a3 Sep-14 81

74 Faculty of Engineering - Alexandria University Control Systems and Components 2013 Example 1 Find the transfer function for the following block diagram U(s) _ 1/s 1/s 1/s b1 b2 b Y(s) + a1 Solution: Y() s M() s U() s N k1 a2 a3 Applying Mason s rule, we find the transfer function to be M k k b s b s b s a s a s a Sep-14 82

75 Faculty of Engineering - Alexandria University Control Systems and Components 2013 Example 2 Find the transfer function for the following SFG H 4 Us () 1 Solution: Forward path H 5 H 1 Path gain M H H H 1256 M2 H4 Loop path Path gain l1 H1H5 343 l H H H H l3 H3H l4 H4H7H6H5 H 3 and the determinates are 1 l l l l ( l l ) H 7 1 HH Ys () Sep-14 83

76 Faculty of Engineering - Alexandria University Control Systems and Components 2013 Example 2 Find the transfer function for the following SFG H 4 Us () M() s 1 H 5 Y() s U() s H 1 Solution: Applying Mason s rule, we find the transfer function to be N k1 M k k H H H H H H H H H H H H H H H H H H H H H H H 2 H 3 H 7 1 Ys () Sep-14 84

77 MODELING IN STATE SPACE The modern trend in engineering systems is toward greater complexity Complex systems may have multiple inputs and multiple outputs (MIMO) and may be nonlinear and/or time varying Because of the increase in system complexity, and easy access to large scale computers, modern control theory has been developed since around 1960 This new approach is based on the concept of state Modern control theory is essentially time-domain approach Modern control theory addresses MIMO, nonlinear, time varying control problems with non-zero initial conditions in addition to classical problems Sep-14 85

78 State-Space Terminology A linear combination of n variables, x i, for i=1 to n, is given by the following sum, S: A set of variables is said to be linearly independent if none of the variables can be written as a linear combination of the others Formally, variables x i, for i=1 to n, are said to be linearly independent if their linear combination, S, equals zero only if every K i =0 and no x i =0 for all t 0 A system variable is any variable that responds to an input or initial conditions in a system Sep-14 86

79 State-Space Terminology (2) State variables are the smallest set of linearly independent system variables such that the values of the members of the set at time t 0 along with known forcing functions completely determine the value of all system variables for all t t 0 State vector is the vector whose elements are the state variables State-space is the n-dimensional space whose axes are the state variables Sep-14 87

80 State-Space Terminology (3) State equations are a set of n simultaneous, firstorder differential equations with n variables, where the n variables to be solved are the state variables Output equation is the algebraic equation that expresses the output variables of a system as linear combinations of the state variables and the inputs. Sep-14 88

81 State-Space Representation An LTI system is represented in state-space by the following equations: x(t) = state vector x(t) = derivative of the state vector with respect to time y(t) = output vector u(t) = input or control vector A = system matrix B = input matrix C = output matrix D = feedforward or direct transmission matrix Sep-14 89

82 Block diagram of the system in state-space. Sep-14 90

83 Example The external force u(t)is the input to the system, and the displacement y(t) of the mass is the output The system equation is This system is of second order. This means that the system involves two integrators. Let us define state variables x 1 (t) and x 2 (t) as Sep-14 91

84 Example (2) Then we obtain The output equation is Sep-14 92

85 Example (3) In a vector-matrix form, the state and output equations can be written as, They are in the standard form Sep-14 93

86 Example (4) We can draw the block diagram of the mechanical system as shown in Figure Sep-14 94

87 Transfer Functions and State-Space Let us consider the system whose transfer function is given by This system may be represented in state space by the following equations The Laplace transforms of the state-space equations is Sep-14 95

88 Transfer Functions and State-Space (2) To get the transfer function from the state-space equations, set initial conditions to zero (x(0)=0) or Substitute in the output equation The eigenvalues of A are identical to the poles of G(s) Sep-14 96

89 Example Obtain the transfer function for the mechanical system in the previous example from the state-space equations. Solution: Sep-14 97

90 Example (2) Note: Thus Sep-14 98

91 STATE-SPACE REPRESENTATION An n th -order differential equation may be expressed by a first-order vector-matrix differential equation If n elements of the vector are a set of state variables, then the vector-matrix differential equation is a state equation In this section we will present methods for obtaining state-space representations of continuous-time systems Sep-14 99

92 State-Space Representation of n th -Order Systems Linear Differential Equations in which the Forcing Function Does Not Involve Derivative Terms: Let us define Sep

93 State-Space Representation of n th -Order Systems (2) This equation is on the form where Sep

94 State-Space Representation of n th -Order Systems (3) The output can be given by Or where Note that the transfer function of the system is Sep

95 State-Space Representation of n th -Order Systems (4) Linear Differential Equations in which the Forcing Function Involves Derivative Terms: The main problem in defining the state variables for this case lies in the derivative terms of the input u The state variables must be expressed such that they will eliminate the derivatives of u in the state equation Sep

96 State-Space Representation of n th -Order Systems (5) Let us define the following n variables as a set of n state variables: Sep

97 State-Space Representation of n th -Order Systems (6) Sep

98 State-Space Representation of n th -Order Systems (7) With this choice of state variables the existence and uniqueness of the solution of the state equation is guaranteed With the present choice of state variables, we obtain Sep

99 State-Space Representation of n th -Order Systems (8) The state and output equations can be written as Sep

100 State-Space Representation of n th -Order Systems (9) Or Where Note that the transfer function of the system is Sep

101 LINEARIZATION OF NONLINEAR SYSTEMS A system is nonlinear if the principle of superposition does not apply Most linear systems are really linear only in limited operating ranges In practice, many systems involve nonlinear relationships among the variables For example, the output of a component may saturate for large input signals There may be a dead space that affects small signals Square-law nonlinearity may occur in some components Sep

102 Nonlinear Systems In reality, most systems are indeed nonlinear, e.g. the pendulum system, which is described by nonlinear differential equations. Example 2 d ML Mg sin ( t) 0 2 dt It is difficult to analyze nonlinear systems, however, we can linearize the nonlinear system near its equilibrium point under certain conditions. L Mg 2 d ML Mg( t) 0 (when is small) 2 dt Sep

103 Linearization of Nonlinear Systems Several typical nonlinear characteristics in control system. output output 0 input 0 input Saturation (Amplifier) Dead-zone (Motor) Sep

104 Linearization Methods (1)Weak nonlinearity neglected If the nonlinearity of the component is not within its linear working region, its effect on the system is weak and can be neglected. (2)Small perturbation/error method Assumption: In the system control process, there are just small changes around the equilibrium point in the input and output of each component. This assumption is reasonable in many practical control system: in closed-loop control system, once the deviation occurs, the control mechanism will reduce or eliminate it. Consequently, all the components can work around the equilibrium point. Sep

105 Example y Faculty of Engineering - Alexandria University Control Systems and Components 2013 y0 0 x0 x dy y f ( x) y Saturation 饱和 ( 放大器 (Amplifier) ) dx The input and output only have small n variance around the equilibrium point. x ( x x0 ),( x) 0 dy y y ( x x ) 0 0 dx x y A(x0,y0) 0 kx y=f(x) A(x0,y0) is equilibrium point. Expanding the nonlinear function y=f(x) into a Taylor series about A(x0,y0) yields ( x x0 ) ( x x 2 0 ) x 2! dx 0 x 1 d y 0 This is linearized model of the nonlinear component. Sep

106 Faculty of Engineering - Alexandria University Control Systems and Components 2013 Note:this method is only suitable for systems with weak nonlinearity. 0 0 继电特性 Relay 饱和特性 Saturation For systems with strong nonlinearity, we cannot use such linearization method. 114 Sep

107 Linear Approximation of Nonlinear Mathematical Models To obtain a linear mathematical model for a nonlinear system, we assume that the variables deviate only slightly from some operating condition Consider a system whose input is x(t)and output is y(t).the relationship between y(t)and x(t) is given by The output equation may be expanded into a Taylor series about this equilibrium point as follows Sep

108 Linear Approximation of Nonlinear Mathematical Models (2) If the variation x x is small, we may neglect the higher-order terms in x x The output equation can be rewritten as Or where, This indicates that y y is proportional to x which gives a linear mathematical model for the nonlinear system x Sep

109 Linear Approximation of Nonlinear Mathematical Models (3) The linearization technique presented here is valid in the vicinity of the operating condition If the operating conditions vary widely, however, such linearized equations are not adequate, and nonlinear equations must be dealt with Sep

110 Example Linearize the nonlinear equation: in the region 5 x 7, 10 y 12. Find the error if the linearized equation is used to calculate the value of z when x=5, y=10 Solution: Choose x, y as the average values of the given ranges Then Sep

111 Example (2) Expanding the nonlinear equation into a Taylor series about points x = x, y = y and neglecting the higherorder terms Where Sep

112 Example (3) Hence the linearized equation is or When x=5, y=10,the value of z given by the linearized equation is The exact value of z is z = xy =50 The error is thus 50-49=1 or 2% Sep

113 Example Problems and Solutions Check the textbook examples and solutions (Pages 46-60) Sheet #1 includes Chapter 2 problems. Sep

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