MATHEMATICAL MODELING OF CONTROL SYSTEMS


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1 1 MATHEMATICAL MODELING OF CONTROL SYSTEMS Sep14 Dr. Mohammed Morsy
2 Outline Introduction Transfer function and impulse response function Laplace Transform Review Automatic control systems Signal Flow Graph Modeling In statespace Transformation of mathematical models with Matlab Linearization of nonlinear mathematical models Sep14 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 Sep14 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 TimeInvariant (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. Sep14 4
5 System Model Differential equation Transfer function Frequency characteristic Study timedomain response Linear system study frequencydomain response Transfer function Differential Laplace equation transform Frequency Fourier characteristic transform Sep14 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. Sep14 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. Sep14 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 Sep14 8
9 Impulse Response Consider the output (response) of a linear timeinvariant system to a unitimpulse input when the initial conditions are zero. Since the Laplace transform of the unitimpulse function is unity, the Laplace transform of the system output The impulseresponse function g(t) is thus the response of an LTI system to a unitimpulse 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 Sep14 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 1042) Sep14 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) Sep14 11
12 Laplace Transform Difficult Timedomain ODE problems Laplace Transform (LT) sdomain algebra problems Easy Solutions of timedomain problems Inverse LT Solutions of algebra problems Sep14 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, PierreSimon where variable. s j is a complex Sep14 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 Sep14 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 Sep14 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) Sep14 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 Sep14 17
18 Properties of Laplace Transform (3) (4)Finalvalue Theorem lim t f ( t) limsf( s) s0 The finalvalue theorem relates the steadystate behavior of f(t) to the behavior of sf(s) in the neighborhood of s=0 (5) Initialvalue Theorem lim t0 f ( t) lim s sf( s) Sep14 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) Sep14 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 partialfraction expansion. Sep14 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 Sep14 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, 1j; zeros: 1 Sep14 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 PartialFraction 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 Sep14 23
24 Faculty of Engineering  Alexandria University Control Systems and Components 2013 Example 1 1 Fs () ( s 1)( s 2)( s 3) PartialFraction 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 Sep14 24
25 Faculty of Engineering  Alexandria University Control Systems and Components 2013 Case 2: F(s) has simple complexconjugate poles Example 2 Laplace transform Applying initial conditions Ys () s PartialFraction 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 Sep14 25
26 Faculty of Engineering  Alexandria University Control Systems and Components 2013 Case 3: F(s) has multipleorder 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 Multiorder poles The coefficients corresponding to the multiorder 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: PartialFraction 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 Sep14 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 ( ) Sep14 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) Sep14 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 Sep14 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 Sep14 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 Sep14 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 Sep14 33
34 Properties of Transfer Functions The transfer function is defined only for a linear timeinvariant 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) Sep14 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 Sep14 35
36 Faculty of Engineering  Alexandria University Control Systems and Components 2013 Transfer function X() s s A a Timedomain impulse response x() t Ae at Position of poles and zeros j a 0 i 0 Sep14 36
37 Faculty of Engineering  Alexandria University Control Systems and Components 2013 Transfer function Timedomain 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 Sep14 37
38 Faculty of Engineering  Alexandria University Control Systems and Components 2013 Transfer function X() s A s B s b Timedomain impulse response x( t) Asin( bt ) Position of poles and zeros j b 0 i 0 Sep14 38
39 Faculty of Engineering  Alexandria University Control Systems and Components 2013 Transfer function Timedomain impulse response A X() s s a () at x t Ae Position of poles and zeros j 0 a i Sep14 39
40 Faculty of Engineering  Alexandria University Control Systems and Components 2013 Transfer function X() s A s B ( s a) b Timedomain impulse response at x( t) Ae sin( bt ) Position of poles and zeros j a b 0 i 0 Sep14 40
41 Summary of Pole Position & System Dynamics Sep14 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 singleinput, singleoutput systems is completely governed by the roots of the characteristics equation. 42 Sep14 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 Sep14 43
44 Faculty of Engineering  Alexandria University Control Systems and Components 2013 TF in zeropole form >> sys = zpk(z, p, k) >> [z, p,k] = tfdata (sys) Transform TS from zeropole form into polynomial form >> [z, p, k] = tf2zp(num, den) 44 Sep14 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 Sep14 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 Sep14 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 Sep14 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 Sep14 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) Sep14 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 () Sep14 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 () Sep14 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) Sep14 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 (closedloop) transfer functions. Example: Sep14 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 21 in the textbook Sep14 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. Sep14 55
56 Classifications of Industrial Controllers 1. Twoposition or on off controllers 2. Proportional controllers 3. Integral controllers 4. Proportionalplusintegral controllers 5. Proportionalplusderivative controllers 6. Proportionalplusintegralplusderivative controllers Sep14 56
57 TwoPosition or On Off Control Action In a twoposition control system, the actuating element has only two fixed positions, which are, in many cases, simply on and off, e.g. Liquidlevel 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 Sep14 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 Sep14 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 Sep14 59
60 Other Control Actions ProportionalPlusIntegral Control Action ProportionalPlusDerivative Control Action ProportionalPlusIntegralPlusDerivative Control Action Block diagram of a proportionalplusintegralplusderivative controller Sep14 60
61 ClosedLoop System Subjected to a Disturbance Figure shows a closedloop system subjected to a disturbance We can use superposition of LTI systems to obtain the output For R(s)=0 For D(s)=0 Sep14 61
62 ClosedLoop 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 closedloop transfer function C D (s)/d(s) becomes almost zero, and the effect of the disturbance is suppressed. This is an advantage of the closedloop system The closedloop 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 closedloop transfer function C R (s)/r(s) becomes independent of G 1 (s) and G 2 (s) and inversely proportional to H(s) Sep14 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 Laplacetransformed equation individually in block form 4. Assemble the elements into a complete block diagram. Sep14 63
64 Example Consider the RC circuit shown in Figure The equations for this circuit are, Sep14 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 stepbystep rearrangement Simplification of the block diagram by rearrangements considerably reduces the labor needed for subsequent mathematical analysis Sep14 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 Sep14 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 Sep14 67
68 Example (3) Notice that the numerator of the closedloop transfer function C(s)/R(s) is the product of the transfer functions of the feedforward path. The denominator of C(s)/R(s) is equal to The positive feedback loop yields a negative term in the denominator Sep14 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 6976) Sep14 69
70 Signal Flow Graph (SFG) SFG was introduced by S.J. Mason for the causeandeffect 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) Sep14 70
71 Block Diagram and SignalFlow Graph Ur () s  1 I () s 11 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 Sep14 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 Sep14 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 Sep14 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 Sep14 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 () Sep14 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 () Sep14 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 timedomain approach Modern control theory addresses MIMO, nonlinear, time varying control problems with nonzero initial conditions in addition to classical problems Sep14 85
78 StateSpace 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 Sep14 86
79 StateSpace 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 Statespace is the ndimensional space whose axes are the state variables Sep14 87
80 StateSpace 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. Sep14 88
81 StateSpace Representation An LTI system is represented in statespace 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 Sep14 89
82 Block diagram of the system in statespace. Sep14 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 Sep14 91
84 Example (2) Then we obtain The output equation is Sep14 92
85 Example (3) In a vectormatrix form, the state and output equations can be written as, They are in the standard form Sep14 93
86 Example (4) We can draw the block diagram of the mechanical system as shown in Figure Sep14 94
87 Transfer Functions and StateSpace 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 statespace equations is Sep14 95
88 Transfer Functions and StateSpace (2) To get the transfer function from the statespace 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) Sep14 96
89 Example Obtain the transfer function for the mechanical system in the previous example from the statespace equations. Solution: Sep14 97
90 Example (2) Note: Thus Sep14 98
91 STATESPACE REPRESENTATION An n th order differential equation may be expressed by a firstorder vectormatrix differential equation If n elements of the vector are a set of state variables, then the vectormatrix differential equation is a state equation In this section we will present methods for obtaining statespace representations of continuoustime systems Sep14 99
92 StateSpace Representation of n th Order Systems Linear Differential Equations in which the Forcing Function Does Not Involve Derivative Terms: Let us define Sep
93 StateSpace Representation of n th Order Systems (2) This equation is on the form where Sep
94 StateSpace 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 StateSpace 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 StateSpace Representation of n th Order Systems (5) Let us define the following n variables as a set of n state variables: Sep
97 StateSpace Representation of n th Order Systems (6) Sep
98 StateSpace 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 StateSpace Representation of n th Order Systems (8) The state and output equations can be written as Sep
100 StateSpace 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 Squarelaw 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) Deadzone (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 closedloop 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 higherorder 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 5049=1 or 2% Sep
113 Example Problems and Solutions Check the textbook examples and solutions (Pages 4660) Sheet #1 includes Chapter 2 problems. Sep
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