Chapter 3 State Variable Models - Engineering

Chapter 3State Variable ModelsThe State Variables of a Dynamic SystemThe State Differential EquationSignal-Flow Graph State VariablesThe Transfer Function from the State Equation1

Introduction In the previous chapter, we used Laplace transform to obtain thetransfer function models representing linear, time-invariant, physicalsystems utilizing block diagrams to interconnect systems. In Chapter 3, we turn to an alternative method of system modelingusing time-domain methods. In Chapter 3, we will consider physical systems described by annth-order ordinary differential equations. Utilizing a set of variables known as state variables, we can obtaina set of first-order differential equations. The time-domain state variable model lends itself easily to computersolution and analysis.2

Time-Varying Control System With the ready availability of digital computers, it is convenient toconsider the time-domain formulation of the equations representingcontrol systems. The time-domain is the mathematical domain that incorporates theresponse and description of a system in terms of time t. The time-domain techniques can be utilized for nonlinear, timevarying, and multivariable systems (a system with several input andoutput signals). A time-varying control system is a system for which one or more ofthe parameters of the system may vary as a function of time. For example, the mass of a missile varies as a function of time asthe fuel is expended during flight3

Terms State: The state of a dynamic system is the smallest set of variables(called state variables) so that the knowledge of these variables at t t0, together with the knowledge of the input for t t0, determinesthe behavior of the system for any time t t0.State Variables: The state variables of a dynamic system are thevariables making up the smallest set of variables that determine thestate of the dynamic system.State Vector: If n state variables are needed to describe thebehavior of a given system, then the n state variables can beconsidered the n components of a vector x. Such vector is called astate vector.State Space: The n-dimensional space whose coordinates axesconsist of the x1 axis, x2 axis, ., xn axis, where x1, x2, ., xn are statevariables, is called a state space.State-Space Equations: In state-space analysis, we are concernedwith three types of variables that are involved in the modeling ofdynamic system: input variables, output variables, and statevariables.4

The State Variables of a Dynamic System The state of a system is a set of variables such that the knowledgeof these variables and the input functions will, with the equationsdescribing the dynamics, provide the future state and output of thesystem.For a dynamic system, the state of a system is described in terms ofa set of state variables.y1(t)u1(t)Systemu2(t)Input Signalsy2(t)Output Signals5

State Variables of a Dynamic Systemx(0) initial conditiony(t) Outputu(t) InputDynamic SystemState x(t)The state variables describe the future response of a system,given the present state, the excitation inputs,and the equations describing the dynamics6

The State Differential EquationThe state of a system is described by the set of first-order differentialequations written in terms of the state variables (x1, x2, ., xn).x1 a11 x1 a12 x2 . a1n xn b11u1 . b1m u m.x 2 a 21 x1 a 22 x2 . a 2 n xn b21u1 . b2 m u m.x n a n1 x1 a n 2 x2 . a nn xn bn1u1 . bnm u mdxx& dt x1 a11 a12 a1n d x2 a 21 a 22 a 2 n . . .dt . xn a n1 a n 2 a nnA x1 b11.b1m u1 x2 . . . bn1.bnm u m xn xBu.A : State matrix; B : input matrixC : Output matrix; D : direct transmission matrixx Ax Bu (State differential equation)y Cx Du (Output equation - output signals)7

Block Diagram of the Linear, Continuous Time Control SystemD(t) .u(t)B(t)x (t ) dtx(t)C(t)y(t) A(t).x (t ) A(t )x(t) B(t ) u (t )y (t ) C(t ) x (t ) D(t ) u (t )8

Mass Grounded, M (kg)Mechanical system described by the first-order differential equationAppied torque Ta (t ) (N - m)Linear velocity v(t ) (m/sec)Linear position x(t ) (m)d 2 x(t )dvFa (t ) M Mdtdt 21 tv(t ) Fa (t )dt M t0x (t) v (t)Fa(t)M9

Mechanical Example: Mass-Spring DamperA set of state variables sufficient to describe this system includes theposition and the velocity of the mass, therefore, we will define a set ofstate variables as (x1, x2)x1 (t ) y (t )dy (t )x2 (t ) dtd2ydyM 2 b ky u (t )dtdtdx2 bx2 kx1 u (t )Mdtdx1 x2 ;dtdx21bk x2 x1 udtmMMu(t)bWall frictionKMy(t)k : Spring constant10

.Example 1: Consider thepreviousmechanicalsystem. Assume that thesystem is linear. Theexternal force u(t) is theinput to the system, andthe displacement y(t) ofthe mass is the output.The displacement y(t) ismeasuredfromtheequilibrium position in theabsence of the externalforce. This system is asingle-input-single-outputsystem.m y b y ky uThis is a second order system. It means it involves two integrators.Let us define two variables : x1 (t ) and x2 (t ).x1 (t ) y (t ); x2 (t ) y (t ); then x1 x2.kb1x1 x2 ummmThe output equation is : y x1x2 In a vector matrix form, we have . 0 x1 . - k x 2 m1 0 x1 b 1 u (State Equation) x 2 m m x y [1 0] 1 (Output Equation) x2 The state equation and the output equation are in the standard form :.x Ax Bu; y Cx Du1 0 0 A kb , B 1 , C [1 0], D 0 m m m 11

Electrical and Mechanical CounterpartsEnergyMechanicalElectricalKineticMass / Inertia0.5 mv2 / 0.5 jω2Inductor0.5 Li2PotentialGravity: mghSpring: 0.5 kx2Capacitor0.5 Cv2DissipativeDamper / Friction0.5 Bv2ResistorRi212

Resistance, R (ohm)Appied voltage v(t )Current i (t )v(t ) Ri (t )i(t)v(t)R1i (t ) v(t )R13

Inductance, L (H)Appied voltage v(t )Current i (t )di (t )v(t ) Ldtt1i (t ) v(t )dtL t0i(t)v(t)L14

Capacitance, C (F)Appied voltage v(t )Current i (t )1 tv(t ) i (t )dtC t0i(t)v(t)Cdv(t )i (t ) Cdt15

Electrical Example: An RLC Circuitx1 vC (t ); x2 i L (t )ξ (1 / 2)Li L2 (1 / 2)Cvc2LiLx1 (t 0 ) and x2 (t 0 ) is the total initialenergy of the networkUSE KCL at the junctionic Cdvc u (t ) i Ldtdi L RiL vcdtThe output of the system is represented by : vo RiL (t )iCu(t) vCC VRLo-dx111 x2 u (t )dtCCdx2 1R x1 x2dt LLThe output signal is then : y1 (t ) vo (t ) Rx216

Example 2: Use Equations from the RLC circuit1 1 . 0 - C x x C u (t ) 1 - R 0 LL The output isy [0 R ] xWhen R 3, L 1, C 1/2, we have 2 0 - 2 x ux 1 - 3 0 y [0 3] x.17

Signal-Flow Graph ModelA signal-flow graph is a diagram consisting of nodes that areconnected by several directed branches and is a graphicalrepresentation of a set of linear relations. Signal-flow graphs areimportant for feedback systems because feedback theory is concernedwith the flow and processing of signals in system.Vf(s)R1(s)G(s)θ (s)G11(s)Y1(s)G12(s)G21(s)R2(s)G22(s)Y2(s)Read Examples : 2.8 - 2.1118

Mason’s Gain Formula for Signal Flow GraphsIn many applications, we wish to determine the relationship between aninput and output variable of the signal flow diagram. The transmittancebetween an input node and output node is the overall gain betweenthese two nodes.P 1Pk k kPk path gain of k th forward path determinant of graph 1 - (sum of all individual loop gain) (sum of gain of all possible combinations of two nontouching loops)- (sum of gain products of all possible combinations of these nontouching loops) . 1 - La Lb Lc - Ld Lc L fab,cd,e,f k cofactor of the kth forward path determinant of the graph with the loopstouching the kth forward path removed, that is, the cofactor k is obtained from by removing the loops that touch path Pk .19

Signal-Flow Graph State Models-R/L1/C1/L1/s1/sx1U(s)Rx2vo-1/CVo ( s )α G(s) U ( s) s 2 βs γ.11x2 u (t )CC.1Rx 2 x1 x2 ; vo Rx2LLVo( s )R / LCs 2R / LC ; U ( s ) 1 (R / Ls ) 1 / LCs 2s 2 (R / L )s (1 / LC )x1 ()20

Y ( s ) s m bm 1s m 1 . b1s boG (s) nU ( s ) s an 1s n 1 . a1s aoY ( s ) s ( n m ) bm 1s ( n m 1) . b1s ( n 1) bo s n G (s) U ( s)1 a n 1s 1 . a1s ( n 1) ao s nY ( s ) k Pk k G (s) U ( s) Some of the forward - path factors k PkG (s) N1 q 1 Lq 1 - sum of the feedback loop factor21

Phase Variable Format: Let us initially consider the fourth-ordertransfer function. Four state variables (x1, x2, x3, x4); Number ofintegrators equal the order of the 1/s-a1x11/sx2Y(s)bo-a0x1Y(s)b0Y (s)G ( s) 4U ( s ) s a3 s 3 a 2 s 2 a1s a0 1 a3 s 1b0 s 4 a2 s 2 a1s 3 a0 s 422

b3b1b21/s1U(s)G ( s) 1/sx3x4-a3-a2b3 s b2 s b1s 3 b0 sx2.-a0box1Y(s).s 4 a3 s 3 a2 s 2 a1s a0 2.-a11/sx1 x 2 ; x 2 x3 ; x 3 x 4b3 s 3 b2 s 2 b1s b0 11/s 41 a3 s 1 a 2 s 2 a1s 3 a0 s 4Read Examplev 3.1 of the textbookx 4 a0 x1 a1 x2 a 2 x3 a3 x4 uy (t ) b0 x1 b1 x2 b2 x3 b3 x4 x1 0 d x 2 0 dt x3 0 x4 - a01 00010001- a1 - a2 - a3 x1 x2 y (t ) Cx [b0 b1 b2 b3 ] x3 x4 x1 0 x 2 0 u (t ) x3 0 x4 1 23

Alternative Signal-Flow Graph State ModelsMotor and LoadR(s)5( s 1)ControllerG ( s) cGc ( s ) ( s 1)I(s)U(s)5( s 1)( s 5)Y(s)16( s 3)( s 2)6(s 3)1( s 2)511/sR (s)511/sI (s) 61/sU (s)-5 - 3 6.x 0 - 2 001Y (s)-20 0 - 20 x 5 r (t ) 1 - 5 -3y [1 0 0]x24

The State Variable Differential Equations1/s1-51/s1-20-2011/s-330Diagonal form or Canonical form30( s 1)Y(s)q( s) T ( s) ( s 5)( s 2)( s 3) ( s s1 )( s s 2 )( s s3 )R( s)k3k1k2Y(s) T ( s) ( s 5) ( s 2) ( s 3)R( s)k1 20, k 2 -10, and k 3 30 - 5 0 0 1 x 0 - 2 0 x 1 r (t ); y (t ) [- 20 - 10 30]x 0 0 - 3 1 .25

The State Variable Differential Equations - 3 6 0 0 x 0 - 2 - 5 x 5 r (t ) 0 0 - 5 1 Y(s)30( s 1)q(s) T (s) R( s)( s 5)( s 2)( s 3) ( s s1 )( s s 2 )( s s3 ).k3k1k2Y(s) T (s) R(s)( s 5) ( s 2) ( s 3)k1 20, k 2 -10, and k 3 30 - 5 0 0 1 x 0 - 2 0 x 1 r (t ) 0 0 - 3 1 y (t ) [- 20 - 10 30]x.26

The Transfer Function from the State EquationGiven the transfer function G(s), we may obtain the state variable equationsusing the signal-flow graph model. Recall the two basic equations.x Ax Buy is the single output andy Cxu is the single input.sX ( s ) A X ( s ) B U ( s )Take the Laplace transformY ( s ) CX ( s )(sI A ) X( s) BU ( s)Since [sI - A ] 1 Φ ( s )X (s) Φ (s) B U (s)Y (s) C Φ (s) B U (s)Y (s)G (s) C Φ (s) BU ( s)27

Exercises: E3.2 (DGD)A robot-arm drive system for one joint can be represented by the differential equation,dv(t ) k1v(t ) k 2 y (t ) k 3i (t )dtwhere v(t) velocity, y(t) position, and i(t) is the control-motor current. Put the equationsin state variable form and set up the matrix form for k1 k2 1dyv dtdv k1v(t ) k 2 y (t ) k 3i (t )dt1 y 0 d y 0 i - k1 v k 3 dt v k 2Define u i, and let k1 k 2 1 0 0 1 y x Ax Bu; A , B , x - 1 - 1 v k 3 .28

E3.3: A system can be represented by the state vector differentialequation of equation (3.16) of the textbook. Find the characteristicroots of the system (DGD).1 0A 11 .x Ax Bu λDet (λI - A) Det 1 (λ 1) -1 λ (λ 1) 1 λ λ 1 021313λ1 j; λ2 j222229

E3.7: Consider the spring and mass shown in Figure 3.3 where M 1kg, k 100 N/m, and b 20 N/m/sec. (a) Find the state vectordifferential equation. (b) Find the roots of the characteristic equation forthis system (DGD).x1 x 2.x 2 100 x1 20 x2 u01 0 x x u - 100 - 20 1 -1 λ2 λ 20λ 100Det (λI - A) Det 100 λ 20 . (λ 10 )2 0; λ1 λ2 -1030

E3.8: The manual, low-altitude hovering task above a moving land deckof a small ship is very demanding, in particular, in adverse weather andsea conditions. The hovering condition is represented by the A matrix(DGD) 0 1 0 A 0 0 1 0 - 5 - 2 λ Det(λI - A) Det 0 0(2)-1λ50 -1 λ 2 λ λ 2λ 5) 0λ1 0; λ2 -1 j 2; λ3 -1 - j 231

E3.9: See the textbook (DGD).1x1 x2 x12.x 2 x1 x2 - 1 1/2 x, y [1 - 3/2]xx 1 - 3/2 51 s s 1 (s 2) s 022 20 2z z; y [- 0.35 - 1.79]z 0 - 1/2 .32

P3.1 (DGD-ELG4152):Apply KVLdiv(t ) Ri (t ) L vcdt1vc idtC(a) Select the state variables as x1 i and x2 vc(b) The state equations are :.1R1v x1 x2LLL.1x 2 x1C. - R/L - 1/L 1/L (c) x x u 0 1/C 0 x1 33

4 Terms State: The state of a dynamic system is the smallest set of variables (called state variables) so that the knowledge of these variables at t t 0, together with the knowledge of the input for t t 0, determines the behavior of the system for any time t t 0. State Variables:The state variables of a dynamic system are