Chapter 2: Mathematical Models of Systems Objectives We use quantitative mathematical models of physical systems to design and analyze control systems.
Download ReportTranscript Chapter 2: Mathematical Models of Systems Objectives We use quantitative mathematical models of physical systems to design and analyze control systems.
Chapter 2: Mathematical Models of Systems Objectives We use quantitative mathematical models of physical systems to design and analyze control systems. The dynamic behavior is generally described by ordinary differential equations. We will consider a wide range of systems, including mechanical, hydraulic, and electrical. Since most physical systems are nonlinear, we will discuss linearization approximations, which allow us to use Laplace transform methods. We will then proceed to obtain the input–output relationship for components and subsystems in the form of transfer functions. The transfer function blocks can be organized into block diagrams or signal-flow graphs to graphically depict the interconnections. Block diagrams (and signal-flow graphs) are very convenient and natural tools for designing and analyzing complicated control systems Illustrations Introduction Six Step Approach to Dynamic System Problems • Define the system and its components • Formulate the mathematical model and list the necessary assumptions • Write the differential equations describing the model • Solve the equations for the desired output variables • Examine the solutions and the assumptions • If necessary, reanalyze or redesign the system Illustrations Differential Equation of Physical Systems Ta( t) Ts( t) Ta( t) ( t) Ta( t) 0 Ts( t) s( t) a( t) = th ro ug h - v ari abl e an gu lar rate d ifferen ce = acros s-v ari abl e Illustrations Differential Equation of Physical Systems Electrical Inductance Energy or Power Describing Equation v 21 d L i dt v21 1 d F k dt E 1 2 2 L i Translational Spring 2 E 1 F 2 k Rotational Spring 21 1d T k dt P21 d I Q dt 2 E 1 T 2 k Fluid Inertia Illustrations E 1 2 2 I Q Differential Equation of Physical Systems Electrical Capacitance i d C v21 dt E F d M v 2 dt E T d J 2 dt E Q d Cf P21 dt E q d Ct T2 dt E 1 2 2 M v21 Translational Mass 1 2 2 M v 2 Rotational Mass 1 2 2 J 2 Fluid Capacitance 1 2 Thermal Capacitance Illustrations 2 Cf P21 Ct T2 Differential Equation of Physical Systems Electrical Resistance 1 1 2 v 21 P F b v21 P b v21 T b 21 P b 21 i R R v 21 Translational Damper 2 Rotational Damper 2 Fluid Resistance Q 1 P21 P T2 1 P Rf 1 Rf 2 P21 Thermal Resistance q Illustrations 1 Rt 1 Rt T2 1 Differential Equation of Physical Systems M Illustrations d 2 d y ( t) b y ( t) k y ( t) 2 dt dt r( t) Differential Equation of Physical Systems t 1 d C v ( t ) v ( t) d t R L 0 dt v( t) y( t) Illustrations 1 t K 1 e r( t ) sin 1 t 1 Differential Equation of Physical Systems Illustrations Differential Equation of Physical Systems K2 1 2 .5 y ( t) K2 e 2 t y 1( t) K2 e 2 1 0 2 2 si n 2 t 2 2 t y 2( t) K2 e 2 t 1 y ( t) y 1( t ) 0 y 2( t ) 1 0 1 2 3 4 t Illustrations 5 6 7 Linear Approximations Illustrations Linear Approximations Linear Systems - Necessary condition Principle of Superposition Property of Homogeneity Taylor Series http://www.maths.abdn.ac.uk/%7Eigc/tch/ma2001/notes/node46.html Illustrations Linear Approximations – Example 2.1 M 2 00g m g 9 .8 m s L 1 00cm 2 0 0rad 1 5 16 T0 M g L si n 0 T1 M g L si n T2 M g L co s 0 0 T0 10 5 T 1( ) T 2( ) 0 5 10 4 3 2 1 0 1 2 3 4 Illustrations St udents are encouraged to inv es t igate linear approxim at ion acc uracy f or dif f erent val 0 The Laplace Transform Historical Perspective - Heaviside’s Operators Origin of Operational Calculus (1887) Illustrations Historical Perspective - Heaviside’s Operators Origin of Operational Calculus (1887) p 1 R L p 1 p n H( t) H( t) t v = H(t) 0 Z( p ) Z( p ) i i p v i t 1d u 1 d dt Expanded in a power series R L p 1 R L p 1 L p H( t) R 1 R 2 1 R 3 1 ..... H( t) R L p L 2 L 3 p p 1 n n R 2 t2 R 3 t3 1 R t .. R L L 2 L 3 i (*) Oliver Heaviside: Sage in Solitude, Paul J. Nahin, IEEE Press 1987 . Illustrations R t 1 L 1 e R The Laplace Transform D ef inition L( f ( t) ) f ( t) e s t dt = F(s) 0 H ere the complex f requency is s j w The Laplac e Trans f orm ex ist s when 0 Illustrations f ( t) e s t dt t his means that the int egral conv erges The Laplace Transform D et ermine the Laplac e transf orm f or the f unc tions a) f1( t ) 1 F1( s ) e s t dt 0 b) F2( s ) f2( t ) 0 Illustrations e e t0 f or 1 ( s t ) e s = 1 s ( a t ) ( a t ) ( s t ) e dt = 1 s1 e [ ( s a) t ] F2( s ) 1 sa The Laplace Transform Ev aluat e t he laplace transf orm of t he deriv ativ e of a f unct ion d f ( t) d t L ( s t ) d f ( t) e dt dt 0 by t he use of where u e udv ( s t ) = dv u v v d u d f( t) and, f rom whic h s e du ( s t ) d t and v f ( t) we obt ain udv = f ( t) e ( s t ) 0 f ( t) s e ( s t ) dt 0 = -f (0+) + s 0 f ( t) e ( s t ) dt d f ( t) d t = s F(s ) - f (0+) not e that t he init ial condit ion is inc luded in the transf ormat ion L Illustrations The Laplace Transform Prac tic al Ex ample - C onsider the circuit . The KVL equation is d 4 i( t ) 2 i( t ) dt ass ume i(0+) = 5 A 0 Apply ing t he Laplac e Trans f orm, we hav e 4 i( t) 2 d i( t) e dt ( s t ) dt 0 I( s ) 5 s2 i( t) e 0 0 4 I( s ) 2 ( s I( s ) i( 0) ) 4 ( s t ) d t 2 ( s t ) d i( t) e dt dt 0 0 4 I( s ) 2 s I( s ) 1 0 0 0 t rans f orming back t o t he t ime domain, with our pres ent knowledge of Laplac e trans f orm, we may s ay t hat t ( 0 0 .01 2) 6 i( t ) 5 e ( 2 t ) 4 i( t ) 2 0 Illustrations 0 1 t 2 The Laplace Transform The Partial-Fraction Expansion (or Heaviside expansion theorem) Suppose that F(s ) s z1 The partial fraction expansion indicates that F(s) consists of ( s p1 ) ( s p2 ) a sum of terms, each of which is a factor of the denominator. The values of K1 and K2 are determined by combining the individual fractions by means of the lowest common denominator and comparing the resultant numerator or coefficients with those of the coefficients of the numerator F(s ) K1 s p1 K2 before separation in different terms. s p2 Evaluation of Ki in the manner just described requires the simultaneous solution of n equations. An alternative method is to multiply both sides of the equation by (s + pi) then setting s= - pi, the right-hand side is zero except for Ki so that Ki Illustrations ( s pi ) ( s z1 ) ( s p1 ) ( s p2 ) s = - pi The Laplace Transform Property Tim e Domain Frequenc y Domain e ( s T ) f ( t T) u ( t T) 1. Time delay a a 1 2. Time s caling 3. Frequency 4. Frequency 5. Frequenc y f ( at) s hif t ing I nt egrat ion f ( t) e Theorem ( a t ) Illustrations Theorem s Li mf ( ( t) ) F( s a) f ( t) F( s ) d s 0 f ( 0) t -> 0 7. Final-v alue F d F( s ) ds dif f erent iationt f ( t) t 6. I nit ial-v alue F( s ) Li ms ( F( s ) ) s -> inf inite Li mf ( ( t) ) Li ms ( F( s ) ) t -> inf inite s -> 0 The Laplace Transform Useful Transform Pairs Illustrations The Laplace Transform Consider the mass-spring-damper system ( Ms b ) y o Y( s ) equat ion 2.21 2 Ms b s k y( s ) s1 s2 s b y M o s 2 b s k M M 2 s 2 n n 2 2 n n 1 s 2 n n k M 2 b k M 2 n n 1 R oots R eal R eal repeat ed I maginary (conjugat es ) Illustrations C omplex (c onjugat es) 2 2 s1 n j n 1 s2 n j n 1 The Laplace Transform Illustrations The Transfer Function of Linear Systems V1( s ) R 1 I( s ) Cs V2( s ) Cs I( s ) V2( s ) V1( s ) Illustrations 1 Z 2( s ) 1 Cs R Z 1( s ) Z2( s ) 1 Cs Z 1( s ) Z 2( s ) R 1 Cs The Transfer Function of Linear Systems Example 2.2 d2 The partial f raction expansion y ields: d y( t) 4 y( t) 3 y( t) 2 r( t) 2 dt dt Initial Conditions: Y( 0) 1 d y( 0) dt 0 1 1 3 2 1 2 2 3 3 Y( s ) r( t) 1 ( s 1) ( s 3) ( s 1) ( s 3) s The Laplace transf orm y ields: Theref ore the transient response is: s 2Y(s ) s y(0) 4(s Y(s ) y(0)) 3Y(s ) Since R(s)=1/s and y (0)=1, we obtain: ( s 4) 2 Y( s ) 2 2 s 4s 3 s s 4s 3 Illustrations 2 R( s ) y( t) 3 e t 1 e 3 t 1e t 1 e 3 t 2 2 3 2 3 The steady -state response is: lim y( t) t 2 3 The Transfer Function of Linear Systems Illustrations The Transfer Function of Linear Systems Illustrations The Transfer Function of Linear Systems Illustrations The Transfer Function of Linear Systems Illustrations The Transfer Function of Linear Systems Kf if Tm K1 Kf if( t ) ia( t ) f ield cont roled m ot or - Lapalc e Transf orm Vf( s ) K1 Kf Ia If( s ) Rf Lf s If( s ) Tm( s ) TL( s ) Td ( s ) TL( s ) J s ( s ) b s ( s ) Tm( s ) 2 rearranging equat ions TL( s ) Tm( s ) Td ( s ) Tm( s ) Km If( s ) If( s ) Illustrations Vf( s ) Rf Lf s Td ( s ) 0 (s ) Km Vf( s ) s ( J s b ) Lf s Rf The Transfer Function of Linear Systems Illustrations The Transfer Function of Linear Systems Illustrations The Transfer Function of Linear Systems Illustrations The Transfer Function of Linear Systems Illustrations The Transfer Function of Linear Systems V 2( s ) 1 V 1( s ) RCs V 2( s ) V 1( s ) Illustrations RCs The Transfer Function of Linear Systems Illustrations V 2( s ) R 2 R 1 C s 1 V 1( s ) R1 V 2( s ) R 1 C 1 s 1 R 2 C 2 s 1 V 1( s ) R 1 C 2 s The Transfer Function of Linear Systems (s ) V f(s ) (s ) V a( s ) Illustrations Km s ( J s b) L f s R f Km s R a L a s ( J s b) K b K m The Transfer Function of Linear Systems (s ) Km s s 1 Vc( s ) J ( b m) m = slope of linearized t orque-speed c urv e (normally negativ e) K R R c q Vo( s ) s c 1 s q 1 Vc( s ) c Lc Rc q Lq Rq For the unloaded c as e: id 0 c q 0 .05s c 0 .5s V1 2 Illustrations Vq V3 4 Vd The Transfer Function of Linear Systems Y( s ) K X( s ) s ( Ms B) A kx K B kp kx g d g dx g ( x P) kp 2 A b k p d dP fl ow A = area of pist on Gear Rat io = n = N 1/ N2 N2 L Illustrations N1 m L n m L n m g The Transfer Function of Linear Systems V2( s ) R2 R2 V1( s ) R R1 R2 R2 R max ks 1( s ) 2( s ) V2( s ) ks error( s ) ks Illustrations V2( s ) Vbattery max The Transfer Function of Linear Systems V2( s ) Kt Kt ( s ) Kt s ( s ) constant V2( s ) ka V1( s ) s 1 R o = out put resis t ance C o = out put capac itanc e Ro Co 1s and is of t en negligible f or cont roller am plif ier Illustrations The Transfer Function of Linear Systems y ( t) xin( t) xo( t) s Xo( s ) Xin( s ) 2 s 2 b s k M M For low f requenc y os cillat ions, where n Xo j Xin j 2 k M T( s ) q(s ) 1 Ct s Q S T Ct Q S Rt q( s ) Illustrations R 1 To Te = t emperat ure dif f erenc e due to t hermal proc ess = = = = = t hermal c apacit ance f luid f low rat e = cons t ant s pecif ic heat of water t hermal resis tanc e of insulat ion rat e of heat f low of heating element The Transfer Function of Linear Systems x r conv erts radial motion to linear motion Illustrations Block Diagram Models Illustrations Block Diagram Models Illustrations Block Diagram Models Illustrations Original Diagram Equivalent Diagram Original Diagram Equivalent Diagram Block Diagram Models Illustrations Original Diagram Equivalent Diagram Original Diagram Equivalent Diagram Block Diagram Models Illustrations Original Diagram Equivalent Diagram Original Diagram Equivalent Diagram Block Diagram Models Illustrations Block Diagram Models Example 2.7 Illustrations Block Diagram Models Illustrations Example 2.7 Signal-Flow Graph Models For complex systems, the block diagram method can become difficult to complete. By using the signal-flow graph model, the reduction procedure (used in the block diagram method) is not necessary to determine the relationship between system variables. Illustrations Signal-Flow Graph Models Illustrations Y1( s ) G1 1( s ) R1( s ) G1 2( s ) R2( s ) Y2( s ) G2 1( s ) R1( s ) G2 2( s ) R2( s ) Signal-Flow Graph Models Illustrations a11 x1 a12 x2 r1 x1 a21 x1 a22 x2 r2 x2 Signal-Flow Graph Models Example 2.8 Illustrations Y( s ) G 1 G 2 G 3 G 4 1 L 3 L 4 G 5 G 6 G 7 G 8 1 L 1 L 2 R( s ) 1 L 1 L 2 L 3 L 4 L 1 L 3 L 1 L 4 L 2 L 3 L 2 L 4 Signal-Flow Graph Models Example 2.10 Illustrations Y( s ) G 1 G 2 G 3 G 4 R( s ) 1 G 2 G 3 H 2 G 3 G 4 H 1 G 1 G 2 G 3 G 4 H 3 Signal-Flow Graph Models P1 1 Illustrations Y( s ) P1 P2 2 P3 R( s ) G1 G2 G3 G4 G5 G6 P2 G1 G2 G7 G6 P3 G1 G2 G3 G4 G8 1 L1 L2 L3 L4 L5 L6 L7 L8 L5 L7 L5 L4 L3 L4 3 1 2 1 L5 1 G4 H4 Design Examples Illustrations Design Examples Speed control of an electric traction motor. Illustrations Design Examples Illustrations Design Examples Illustrations Design Examples Illustrations Design Examples Illustrations The Simulation of Systems Using MATLAB Illustrations The Simulation of Systems Using MATLAB Illustrations The Simulation of Systems Using MATLAB Illustrations The Simulation of Systems Using MATLAB Illustrations The Simulation of Systems Using MATLAB Illustrations The Simulation of Systems Using MATLAB Illustrations The Simulation of Systems Using MATLAB Illustrations The Simulation of Systems Using MATLAB Illustrations The Simulation of Systems Using MATLAB Illustrations The Simulation of Systems Using MATLAB Illustrations The Simulation of Systems Using MATLAB Illustrations The Simulation of Systems Using MATLAB Illustrations The Simulation of Systems Using MATLAB Illustrations The Simulation of Systems Using MATLAB Illustrations The Simulation of Systems Using MATLAB Illustrations The Simulation of Systems Using MATLAB Illustrations The Simulation of Systems Using MATLAB error Sys1 = sysh2 / sysg4 Illustrations The Simulation of Systems Using MATLAB Illustrations The Simulation of Systems Using MATLAB error Num4=[0.1]; Illustrations The Simulation of Systems Using MATLAB Illustrations The Simulation of Systems Using MATLAB Illustrations Sequential Design Example: Disk Drive Read System Illustrations Sequential Design Example: Disk Drive Read System Illustrations Sequential Design Example: Disk Drive Read System = Illustrations P2.11 Illustrations P2.11 1 1 L d L a s R d R a L c s R c +Vd Vq Id K2 K1 1 L q s R q Illustrations J s b s Vc Ic 1 Tm Km -Vb 1 K3 Illustrations http://www.jhu.edu/%7Esignals/sensitivity/index.htm Illustrations http://www.jhu.edu/%7Esignals/ Illustrations