Chapter 3 MATHEMATICAL MODELING OF DYNAMIC SYSTEMS 3. System Modeling Mathematical Modeling In designing control systems we mst be able to model engineered system dynamics. The model of a dynamic system is a set of eqations (differential eqations) that represents the dynamics of the system sing physics laws. The model permits to stdy system transients and steady state performance. Model compleity As model becomes more detailed it also can become more accrate. Model accracy needed for control system design is normally simpler than the model sed for system simlation. Simpler models: -ignore some physical phenomena -approimate linearly nonlinear characteristics -se lmped parameters approimation of distribted parameters systems. For the design of a control system: -an initial simplified model is bilt for conceptal design -a more model is sed for controller design and parameters determination 3-
Linear Systems For linear systems the principle of sperposition is valid and the response to a comple inpt can be calclated by smming p the responses to its components. Linear Time Invariant (LTI ) Systems verss Linear Time Varying Systems Linear Time Invariant (LTI ) Systems = systems: - represented by lmped components- - described by linear differential eqations - parameters of the eqations are time invariant. Systems with parameters that vary in time are called linear time varying systems. Eamples: a car in motion or a rocket in flight have weight that diminishes as the fel is consmed. 3-
3. Transfer Fnction L{otpt} Transfer Fnction = G(s) = L{inpt} for zero initial conditions. Inpt (s) Transfer Fnction=G(s) Otpt y(s) G (s) = y(s) (s) The transfer fnction of a system represents the link between the inpt to the system to the otpt of the system. The transfer fnction of a system G(s) is a comple fnction that describes system dynamics in s-domains opposed t the differential eqations that describe system dynamics in time domain. The transfer fnction is independent of the inpt to the system and does not provide any information concerning the internal strctre of the system. Same transfer fnction can represent different systems. The transfer fnction permits to calclate the otpt or response for varios inpts. The transfer fnction can be calclated analytically starting from the physics eqations or can be determined eperimentally by measring the otpt to varios known inpts to the system. 3-3
Eample: Car sspension model m d Y o /dt m m k b Y o Y i k(y o -Y i ) B (dy o /dt - dy i /dt) d Y m dt o dyodt dyidt b dt k(y o Y ) = 0 i Laplace transform for zero initial conditions gives ms Y (s) bs(y (s) Y (s)) k(y (s) Y (s)) = 0 or o o ms Y (s) bsy (s) ky (s) = bsy (s) ky (s) o The transfer fnction is o i o i o i i Y o (s) Y (s) i = bs k ms bs k 3-4
Implse Response The Laplace transform of an implse fnction δ(t) is given by L { δ (t)} = The otpt of a system de to an implse inpt (s)= δ(s) = is y (s) = G(s) (s) = G(s) The implse response of a system is identical to the transfer fnction of that system. The inverse Laplace transform of the implse response G(s) L {G(s)} = g(t) The transfer fnction that contains complete information abot the dynamic characteristics of a system can be obtained by applying an implse inpt (t)= δ(t) and measring system response y(t) which in this case is identical to g(t). The transfer fnction will then be G(s)=L{g(t)} 3-5
3.3 Block Diagrams Block diagrams are a graphical representation of the system model. The blocks represent physical or fnctional components of the system. Each block has inscribed the transfer fnction of that component the relate the otpt of the component to its inpt. Block Diagrams Block diagrams consist of. blocks. smmation jnctions 3. paths 4. branching points. Block Eample: car sspension system Y i (s) bsk ms bsk Y o (s) Blocks represent physical or fnctional components in the system. In the block is inscribed the transfer fnction of that component of the system 3-6
. Smmation Jnction B A C - D=AB-C A B Σ D=AB-C -C A smming jnction reslts in the addition or sbtraction of inpt signals for a single otpt. 3. Path X(s) Signal X(s) flows along the directed path 4. Branching Point X(s) X(s) X(s) At the branching point a signal splits into two signals of the same vale. 3-7
3.4 Block Diagram of a Closed Loop System In a closed loop control system also called feedback control system the otpt variable y(s) is measred as y m (s) sbtracted from its desired vale y d (s) to calclate the error e(s)= y d (s) - y m (s) y d (s) _ e (s)= y d (s)- y m (s) G(s) y y m(s) H(s) G(s) is feed forward transfer fnction (of the controller actator and the system) G(s) = y(s) / e(s) H(s) is feedback transfer fnction (of the sensor) H(s) = y m (s) / y(s) G(s) H(s) = open loop transfer fnction G(s) H(s) = [y m (s) / y(s)][ y(s) / e(s)]= y m (s) / e(s) y d (s) / y(s) is closed loop transfer fnction obtained from the above eqations by eliminating e(s) and y m (s) y(s) = G(s) e(s) = G(s) [y d (s) - y m (s)]= G(s) [y d (s) H(s)y (s)] y(s) G(s) H(s) y(s) =G(s) y d (s) y(s)/y d (s)= G(s) / [ G(s) H(s)] This eqation gives the single block eqivalent of the above closed loop system y d (s) G(s) G(s)H(s) y(s) 3-8
The transfer fnction represents the closed loop system dynamics with comple fnctions. The otpt y(s) is given by y(s)= G(s) / [ G(s) H(s)] y d (s) and depends on the closed loop transfer fnction and the desired vale of the otpt y d (s) called also the inpt (to the closed loop system). The following positive feedback block diagram y d (s) G(s) y (s) y m(s) H(s) is eqivalent to a negative feedback one if H(s) is replaced by H(s) y d (s) - G(s) y (s) y m(s) -H(s) which is eqivalent to the block y d (s) G(s) G(s)H(s) y(s) 3-9
Closed loop Control System Advantage Ideally i.e it is reqired that y(s)/y d (s) G(s) / [ G(s) H(s)] or / [/ G(s) H(s)] which is achieved for a very high vale of G(s) G(s) G(s) represents the controller actator and the system and G(s) and the high feed forward gain can be achieved by a very high vale of the controller transfer fnction. In the stability stdy it will be seen that there is a limit to sch high vale de to stability constraints. 3-0
Error Error of a closed loop control system can be obtained from: e(s)=y d (s)-y m (s)= y d (s)-h(s) y (s)=y d (s){- H(s)y(s)/y d (s)}= y d (s) {- H(s)G(s) / [ G(s) H(s)] }= {- H(s)G(s) / [ G(s) H(s)] }y d (s)= { / [ G(s) H(s)] }y d (s) y d (s) G(s)H(s) e(s) Steady State Error Final vale theorem gives the steady state error of a system e( ) i.e. error when t By taking the limit for s 0 of e( )=lim t e(t)= lim s 0 {se(s)}= lim s 0 { / [ G(s) H(s)] }y d (s)= { / [ G(0) H(0)] }y d (0) The Components of Closed Loop Control System -The system to be controlled S (s) -The transdcer (sensor) H(s) to measre the otpts y(s) and feed them back as y m (s) -the comparator to calclate the error e(s) between the desired vale of the otpt y d (s) and otpt measrement y m (s) -the controller C(s) that ses this error signal e\(s) and generates the command c (s) to the system -the actator M(s) sch that 3-
G(s) = C(s) M(s) S(s) y d (s) _ e (s) C(s) CONTROLLER c power spply M(s) (s) ACTUATOR S(s) SYSTEM y (s) y m(s) H(s) Pertrbations Pertrbations p (s) to a control system generally act on the system G(s) and occr as an additional power inpt that adds to the power inpt (s) to the system. y d (s) _ e (s) C(s) c power spply M(s) (s) p(s) S(s) y (s) y m(s) H(s) Sperposition principle is applied for linear systems and consider each inpt independently and the otpts corresponding to each inpt alone can be added to give the total otpt. The principle of sperposition is applied as follows -the otpt of the system y (s) is calclated as a reslt of the inpt command y d (s) for p(s)=0. 3-
y G(s) (s) = G(s)H(s) y (s) d -the otpt of the system y (s) is calclated as a reslt of the p(s) for y d (s)=0. y (s)=s(s){c(s)/[c(s)m(s)s(s)h(s)]}p(s) or y C(s) (s) = p(s) G(s)H(s) - add the two otpts y(s)= y (s) y (s) (s) = y (s) y (s) G(s) = G(s)H(s) y y (s) d C(s) G(s)y (s) C(s)p(s) p(s) = d G(s)H(s) G(s)H(s) The effect of the pertrbation p(s) is cancelled for G(s)H(s)>> and M(s) >>. For G(s)H(s)>> G(s)H(s) G(s)H(s) and C(s)/[G(s)H(s)]= C(s)/[C(s)M(s)S(s)H(s)] C(s)/[C(s)M(s)S(s)H(s)] /[ M(s)S(s)H(s)] sch that for y(s) = y (s) y(s) = y (s) p(s) H(s) d M(s)S(s)H(s) For M(s) >> /[ M(s)S(s)H(s)] 0 and y(s) = y (s) y(s) = y (s) H(s) d i.e. the otpt y(s) is not affected by p(s) and G(s). H(s)= will reslt in the ideal y(s)= y d (s) 3-3
Block Diagram Redction Block diagram for actal systems can contain a large nmber of blocks and block diagram redction is reqired to redce it to a single eqivalent block sing the following two rles:. The prodct of the transfer fnctions in the feedforward direction shold remain the same. The prodct of the transfer fnctions arond the loop shold remain the same Eample: G (s) - G (s) H (s) G 3 (s) H (s) The top negative feedback loop cannot be redced nless the smming point is moved ahead G (s). - G (s) G (s) a(s) G 3 (s) H (s) We will determine the new nknown feedback transfer fnction a(s) sing the above two rles.. The first rle is actally satisfied as both block diagrams have the same prodct of the transfer fnctions in the feedforward direction G (s) G (s) G 3 (s) 3-4
. The prodct of the transfer fnctions arond the loop for the initial block diagram is G (s) [G (s) G 3 (s)/( G (s)g 3 (s) H (s)]= G (s) G (s) G 3 (s)/[( G (s)g 3 (s) H (s)] and for the modified block diagram G (s) G (s) G 3 (s)/( G (s)g (s)g 3 (s) a(s)] In order to be the same given the same nmerators the denominators have to be eqal G (s)g 3 (s) H (s)= G (s)g (s)g 3 (s) a(s) or G (s)g 3 (s) H (s)= G (s)g (s)g 3 (s) a(s) or H (s)= G (s) a(s) which gives the soltion for the nknown new feedback transfer fnction a(s) = H (s)/g (s) Previos block diagram is eqivalent to H (s)/g (s) - G (s) G (s) G 3 (s) H (s) as the switch of the smming points satisfy the two rles. Now the bottom positive feedback loop can be redced to G (s)g (s)/[-g (s)g (s)h (s)] sch that 3-5
- G (s)g (s)/[-g (s)g (s)h (s)] H (s)/g (s) G 3 (s) and final redction gives the final transfer fnction G (s)g (s) G 3 (s)/[-g (s)g (s)h (s)]/{ G (s)g (s) G 3 (s)/[- G (s)g (s)h (s) H (s)/g (s)] (See Eample 3- and A-3- to A-3-5) 3-6
Controllers Controllers are the part of the overall feedback control system that are the main focs of control engineering. desired otpt or error C(s) CONTROLLER otpt measrement controller otpt H(s) power spply M(s) ACTUATOR inpt S(s) SYSTEM otpt of the system y d (s) _ e (s) C(s) power spply c (s) M(s) (s) p(s) S(s) y (s) y m(s) H(s) Classifications of Controllers By the type of implementation: -pnematic controller -hydralic controller -analog electronic controllers -digital controllers Modern implementation is as digital controllers. By control law: On-off Proportional P-control PD-control PID-control In Control I and II only the last three are stdied. 3-7
P-control The transfer fnction is c (s) = k e(s) p where k p is the proportional gain The block diagram is e(s) k p c (s) The P-controller corresponds to an operational amplifier with an adjstable gain k p. PD Control The transfer fnction is c (s) e(s) = k p k s d = k p ( T s) d where k d is the derivative gain and T d = k d / k p The block diagram is e(s) k p k p s c (s) 3-8
PID Control The PDI controller transfer fnction of is c (s) e(s) = k p k s d k i s = k p ( T s d ) T s i where k i is the integral gain and /T i = k i / k p The block diagram is e(s) k p k p sk i /s c (s) PID controllers are freqently sed in applications. 3-9
3.5 State Space Representation State space based control approaches are developed in Rssia in 960s. This is called Modern Control Theory or State space Control Theory or Time Domain Approach (as opposed to Conventional Control Theory of Classic Control Theory or Freqency Domain Approach The major limitation of the conventional freqency domain models is the reqirement that the system be linear and Single Inpt Single Otpt (SISO). Definitions State Variables: are the minimm set of variables that niqely define the state of a dynamic system at any instant of time. States: are the vales of the set of state variables. For known crrent state vales and of the inpt the state of the system in the ftre time can be calclated. State Vector: is the vector of state variables. State space: is the hyperspace in which the state of the system takes vales. State Space Model consists of: a) State eqations for a nonlinear system 3-0
3- t) k... n... ( n f n..................... t) k... n... ( f t) k... n... ( f = = = & & & b) otpt eqations for a nonlinear system t) k... n... ( m g m y... t) k... n... ( g y t) k... n... ( g y = = = Matri state space model is defined for -state vector [n ] = (t).. (t) (t) (t) n -otpt vector [m ] = (t) y.. (t) y (t) y y(t) m
-inpt vector [k ] (t) = k (t) (t).. (t) - right hand side of state eqation f (... n f (... n f ( t) =.. f... k... (...... n n k -right hand side of otpt eqation t) t) t) k g ( t) f ( f ( = f... n... n. (...... n n k.... k... t) t) t) k the above state and otpt eqations can be written in a compact form as matri eqations & (t) = f ( t) y(t) = g( t) 3-
Linearization of Nonlinear Eqations The operation of a system for a short time dration is arond an operating point with small variations abot the eqilibrim point. In this case is sitable to model the operation by approimations of the nonlinear system sing local linear approimations. Sch a linearized LTI model is sed for controllers design. Linearizing state and otpt nonlinear eqations abot a given vale of the state vector gives Linear Time Variant eqations & (t) = A(t) (t) B(t) (t) y(t) = C(t) (t) D(t) (t) A(t) is state matri B(t) is the inpt matri C(t) is the otpt matri D(t) is the direct transmission matri. A Linear Time Invariant (LTI) system is given by & (t) = A(t) B(t) y(t) = C(t) D(t) where A B C and D are matrices of constant vales. Eample: A mass-damper-spring vertical M-B-K system is sbject to a force f(t) K B M Y(t) f(t) The free body diagram is 3-3
KY(t) BdY(t)\dt M f(t) Y(t) The eqation of motion is given by Newton second law d Y(t) dy(t) M = B KY(t) f (t) dt dt Inertia force is M d Y(t)/dt and gives the following free body diagram KY(t) M d Y(t)/dt BdY(t)\dt M f(t) Y(t) D Alembert principle gives the force balance eqation d Y(t) dy(t) M B KY(t) f (t) = 0 dt dt Let s define the following set of state variables (t) and (t) (t)=y(t) (t)=dy(t)/dt= d (t)/dt and d (t) /dt= d Y(t)/dt =(/M)[-B dy(t)/dt-ky(t)f(t)] The above second order eqation of motion can be replaced by two first order state eqations sing state variables (t) and (t) and inpt variable (t)=f(t) 3-4
d (t)/dt= (t) d (t) /dt= -(K/M) (t) - (B/M) (t) (/M) (t)] and system otpt eqation y(t) = (t) in case that the otpt is the position (or y(t) = (t) if the otpt is the velocity (t)=dy(t)/dt of the mass M) Matri form of state dynamics and otpt eqations is & (t) & (t) y(t) = or = [ 0] 0 - K/M (t) (t) & (t) = A(t) B(t) y(t) = C(t) D(t) where (t) = (t) (t) - B/M (t) (t) 0 / M (t) A = 0 - K/M - B/M 0 B = / M C = D=0 [ 0] 3-5
Relationship between Transfer Fnctions and State-Space Eqations System transfer fnction y(s) G (s) = (s) has to be obtained from the matri form of state and otpt eqations by elimination the other variable the vector (s). This reqires Laplace transform of these matri eqations for zero initial conditions s(s) = A(s) B(s) y(s) = C(s) D(s) or (si A) (s) = B(s) y(s) = C(s) D(s) where I is a by identity matri. The elimination is carried ot by solving first eqation for (s) ( s) = (si A) B (s) and replacing (s) in the second eqation y ( s) = C(sI A) B (s) D(s) or y ( s) = [C(sI A) B D] (s) For single inpt y( s) = y(s) and single otpt ( s) = (s) sch that y (s) = [C(sI A) B D](s) This eqation gives the relationship between the transfer fnction and the matrices A B C and D of the state space representation 3-6
y(s) G (s) = = [C(sI A) B D] (s) Given that the inverse of the matri si-a is given by the matri of adjoints si-a divided by the determinant si-a (si-a) - = si-a / si-a Assming D=0 G(s) =C si-a B/ si-a Sch that the poles of the transfer fnction G(s) are given by si-a =0 that is actally the eqation that permits the calclation of the vector of eigenvales λ of the sqare matri A λ I-A =0 This indicates that the characteristic eqation of the system is given by (si-a). That means that the eigenvales of A is identical to the poles of G(s) in the Laplace domain. The poles of the transfer fnction G(s) can conseqently be calclated as eigenvales of the matri A. Eample: Consider the M-B-K system analysed before. State space and otpt eqations for the system were given by & (t) (t) 0 = 0 (t) & (t) (t) - K/M - B/M / M (t) y(t) = [ 0] (t) To obtain the transfer fnction from the state space eqations see (See Eample 3-4 from Ogata tetbook) The reslt is G(s)= /( Ms bsk) The poles of G(s) are given by Ms BsK=0 Which is the determinant of the matri A A =0 3-7
State Space representation of nth-order systems of Linear Differential Eqations a) The case when the forcing fnction does not involve derivative terms b) When the inpt involves derivative terms in the forcing fnction (See Ch. 3-5 from Ogata tetbook) (See Eample 3-5 in Ogata tetbook) 3-8