Technische Informatik Systeme Leitung: Prof. Dr.-Ing. D.P.F. Möller Mathematical and Computational Modeling and Simulation Prof. Dr.-Ing. D.P.F.Möller VAK 8. Sommersemester 005
Content. Modeling Continuous-Time and Discrete-Time Systems. Mathematical Description of Continuous-Time Systems 3. Mathematical Description of Discrete-Time Systems 4. Simulation Languages of Continuous-Time and Discrete-Time Systems 5. Parameter Estimation of Dynamic Systems 6. Soft Computing in Simulation 7. Distributed Simulation 8. Virtual Reality in Simulation
. Mathematical Description of Continuous Time Dynamic Systems
Content. Mathematical Description of Continuous Time Dynamic Systems. Introduction. State Space Systems Concept.. Solution of the State Space Differential Equation.. Stability Analysis.3 Use of State Space Models.4 First Order Linear State Space Models of Time Continuous Systems.4. Systems Theory of First Order Linear State Space Systems Models.5 Second Order Linear State Space Models.5. Systems Theory of Second Order Linear State Space Models.6. Higher Order Linear State Models
. Introduction Based on the phenomenological principles relevant to a particular dynamical system, the equations which characterize the dynamic system can usually be written in the notation of ordinary differential equations or, in the notation of state space variables. Hence we may decide upon the input and output variables; this subjective decision depends on what aspects of the dynamic responses are of interest, e.g. system analysis. The input and output variables typically are certain of the physical variables, the input variable relates the effects of the system on the external environment. State space variables of the system which are unknown may be identified to be known. State space variables should have the following property: If the current state is known and the future input is known then the future state and output are uniquely determined. The basic methodological concepts behind are state space controllability, and its inverse, state space observability. Modeling and Simulation
Definition. A dynamic system. Introduction is said to be completely controllable if there exist a control signal u(t), defined over the finite interval t 0 <t<t F, which transfers the dynamic system from any initial state x(t 0 ) = x 0 to any desired final state x(t F ) = x F in the defined time interval. Definition. is true if and only if the (nxnr)-controllability matrix has full rank n. Modeling and Simulation S: = [B AB A²B² A n- B ]
. Introduction A dynamic system may be described by the state space model dx/dt = A*x(t) + B*u(t) With -3 x 0 = -.5 x + [u] A = -3 -.5 and B = 0 The dynamic system given in the example is completely controllable if B and AB are linear independent and the rank of the controllability matrix C: [B, AB] =, which means the system is completely controllable.
. Introduction Definition.-: A dynamic system is said to be completely observable within A the finite interval t 0 <t<t F, if every initial state x(t 0 ) = x 0 can be determined from the output y(t), observed over the time interval. Definition.- is true if and only if the (nxnr)-matrix O: = [C T C T A T C T ( A) n- ] has full rank n, with C T as the transform of C.
. Introduction
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. Introduction Definition.-3: A dynamic system is said to be structurally identifiable if its parameter vector Θ can be determined from the output y(t), observed over the same interval t 0 <t< t F. A dynamic system is called at the true model parameter vector Θ T parameter identifiable if there exists an input sequence {u}such that Θ and Θ T are distinguished for all Θ Θ T, system identifiable if there exists an input sequence {u}such that Θ and Θ T are distinguishable for all Θ Θ T but a finite set unidentifiable in all other cases.
. Introduction The conditions for controllability, and observability of dynamic systems, defined by Kalman, rely on the system state-space representation, which can successfully be applied both in continuous time and discrete time dynamic systems. The state variable concept of dynamic systems completely characterize the systems past, since the past input is not required to determine the future output of the dynamic system. This seemingly elementary notation of state space is of importance in the systems state space approach. In fact, this modern mathematical approach describing dynamic systems is fundamentally based upon the state space variables concept Although the state space notation has been explicitly emphasized during the last several decades due to the successive use of digital computers when modeling and simulating dynamic system.
. Introduction Unfortunately, no general prescription can be given for choosing state space variables in the sense of general guidelines, i.e., in electrical RLC circuits the charge on each capacitor and the current through each inductor in the network usually serve to define the state space of the respective network; in a mechanical system the force and mass of each body, usually serve to define the state space of a mechanical system. In other domains choosing state space variables may be much more difficult. Once state space variables are chosen the mathematical equations characterizing the state space variables, the state space equations, can be derived. Depending on the particular form of the equations used to describe the physical system the state space equations may be in one of the manyfold mathematical forms. It is possible classifying state space equations on the basis of their mathematical structure.
. Introduction Time is usually an independent variable in state space models: sometimes the time variable is considered being a discrete variable and the state space model is described by recursive equations or the independent variable time is considered to be real valued sometimes there can be additional independent variables in which case the state space model is said to be distributed; such state space models can often be described by partial differential equations if time is the only independent variable then the state space model is said to be lumped further, the state model may include random effects in which case the state space model is said to be stochastic if no such effects are included the state space model is said to be deterministic.
. Introduction State space equations considered here are assumed to be described by ordinary differential equations, in their respective notation of n-order with u(t) as control variable and y(t) as output variable u(t) = a n * y (n) (t) + a n- * y (n-)) (t) +, a 0 * y(t) Rewriting the above differential equation, which is of order n, using n first order differential equations, we find: x (t) = y (t) x (t) = y () (t). x n (t) = y (n-) (t)
. Introduction This results in n differential equations if first order x (t) = f ( x (t),, x n (t), u(t) ) x n (t) = f n ( x (t),, x n (t), u(t) ) The output variable is assumed to depend on the state space variables and the input variable u according to an algebraic equation y = g (x,, x n, u)
. Introduction From a more general point of view, the model is expressed in terms of n state space variables labeled x,, x n If a mathematical model for a dynamic system can be written in the above form, then the variables x,, x n do have state property. State equations can be assumed to serve as basis for the application of systems theory to dynamic systems modeling; the state space equation, in their first order form, is said to be in the standard or normal form State equations also serve as a very general model for the manifold of physical, electrical, mechanical, medical, biological, etc. processes, that is, it is often possible to develop state models of the above form for the manifold of physical, etc. processes; thus there is a large range of potential application of any mathematical systems theory based on the above state equations
. Introduction The previously defined differential state space model, defined in terms of the arbitrary functions f,, f n and g, is sufficiently general to include models developed for many dynamic systems. A special class of state space models, of much practical importance in applications, are those for which the functions f,, f n and g have a linear mathematical structure so that the state space equations can be written as dx /dt = a x + + a n x n + b u.. dx n /dt = a n x + + a nn x n + b n u y = c x +..., c n x n + d u where a,,a nn, b,,b n, c,, c n, d are the systems constants. State models with such special structure said to be linear state models
. State Space Concepts It should be noticed that the state space model, defined in terms of dx /dt = f ( x,, x n, u ).. dx n /dt = f n ( x,, x n, u ) y = g ( x,..., x n, u ) is a sufficient way describing dynamic systems in terms of a mathematical model. Our interest when using state space nation is both formalization of state space noticed models analysis of state space models as a way of better understanding the intrinsic systems dynamic
.. Solution of the State Space Differential Equation The time dependent behavior of a linear state space differential system is which means the dependence of the output vector y(t) form is input vector u(t), where the system matrix A and the control matrix B assumed to be constant, can be solved applying the Laplace transform hence: L [ f(t) ] = f(t) * e -st dt = L [ f(t) ] = F(s) s*x(s) - x(0) = A x(s) + B u(s) with x(0) as initial state of the dynamic system under test, and x(s) = e -st x(t)dt as the Laplace transform, and s as the Laplace operator for d/dt. Modeling and Simulation
.. Solution of the State Space Differential Equation Modeling and Simulation Solving equation s*x(s) - x(0) = A*x(s) + B*u(s) we receive s*x(s) - A*x(s) = x(0) + B*u(s) x(s)[ s A ] = x(0) + B*u(s) and with I = equal to the (n,n)-unit matrix, and after rewriting x(s)[ s I A] = x(0) + B*u(s) x(s) = x(0) / [ s I A] + B*u(s) / [ s I A] x(s) = (s I A) - x(0) +( s I A) - B*u(s) with x(s) as Laplace transform of x(t), and I, as mentioned, as a (n,n)- unit matrix.
.. Solution of the State Space Differential Equation Modeling and Simulation Assuming u(s) being known, the state vector x(t) can be calculated from u(s) as follows, while L{f(t)} = F(s) and the inversion property is written as we receive f(t) =L - { F(s)} x(t) = L - {(s I A) - } x(0) + L - {( s I A) - }B*u(s)
.. Solution of the State Space Differential Equation Modeling and Simulation Introducing the notation x(t) = L - {(s I A) - } x(0) = e At and the respective correspondency from the Laplace transform we may rewrite the equation above as follows Φ(t) = L - {(s I A) - } x(0) = e At in order to calculate the solution for the state space differential equation
.. Solution of the State Space Differential Equation We finally receive Φ(t) = e At in order to calculate the solution for the state space differential equation, where Φ(t) is called fundamental-, transistion-, or state space transient matrix, which with the state for any future time step of a dynamic system can be calculated, if the initial state is known.
.. Solution of the State Space Differential Equation Modeling and Simulation For the state variables we may introduce the well known concept of solving ODEs by the partial and the homogeneous solution, which is x(t) = x h (t) + x p (t) For the homogeneous part we receive the expression d x h (t) /dt = A * x h (t)
.. Solution of the State Space Differential Equation Modeling and Simulation With the solution x h (t) = e At x h (0) = Φ(t) x h (0) while e At can be expressed in terms as Taylor series expansion at t=0 which results in e A*t = ( A*t ) / n! = I + A*t + (½) ( A²*t² ) +...
.. Solution of the State Space Differential Equation Modeling and Simulation For the particular solution of the state space which is x p (t) = Φ(t) * g(t) we may use Langrange s method of variation of the constants, which yields in general terms y = a * b + a * b which results in the solution which has to satisfy the state equation as d/dt { Φ(t) * g(t)} = Φ (t) * g(t) + Φ(t) * g (t)
.. Solution of the State Space Differential Equation The derivation of Φ(t) can be expressed using the well known equation e A*t = Φ(t) which results in dφ/dt = d/dt *e At = A * e At = A* Φ(t) Using this equation, the particular equation may be rewritten as follows d/dt { Φ(t) * g(t)} = Φ (t) * g(t) + Φ(t) * g (t) x p (t) = A* Φ(t) * g(t) + Φ(t) * g (t) = A* Φ(t) * g(t) + B * u(t) or g (t) = Φ - (t) * B * u(t) or g(t) = Φ - (τ) * B * u(τ) dτ Modeling and Simulation
.. Solution of the State Space Differential Equation Modeling and Simulation The solution of the particular as well as for the homogeneous part is now well known as x(t) = x h (t) + x p (t) x(t) = Φ(t) x(0) g(t) + Φ(t) * Φ - (τ) * B * u(τ) dτ
.. Stability Analysis An important class of state responses of the state space equations are those which are periodic or oscillatory. If for some initial state and some input function the solution of the state space equations satisfy x i (t+t) = x i (t), i=,.., n for all t>0, where T>0, then the state response is said to be periodic. The smallest such value of T is said to be the periodic of the oscillation. A special class of periodic state responses are those which are constant functions, i.e. for some initial state and some input function the solution of the state equations satisfy x i (t) = x i (0), i=,.., n
.. Stability Analysis For all t>0. Such constant state responses are said to define an equilibrium state, denoted by x i (0),..., x n (0) An important property of a class of responses of the state equations in the property of stability. An equilibrium state is stable, roughly, if for any initial state close to the equilibrium state the state response tends, in the limit as time increases, to the equilibrium state.
.. Stability Analysis Stability analysis of linear state space systems of type dx/dt = A*x is possible by calculating the Eigenvalues λ of the system matrix A. A System S is said to be stable if the Eigenvalues λ of the system matrix A are negativ.
.. Stability Analysis Example: Inspecting a linear system whether it satisfy the stability criteria, which means the system has to have only one equilibrium state x s 0 x s = = x s 0 with the system matrix -k 0 A = -k -k Eigenvalues of system matrix A are calculated with polynomial det(a-λi)= 0 -k -λ 0 det = (-k -λ)(-k -λ) =0 k -k -λ The Eigenvalues λ = -k and λ =-k are negative which satisfies the stability criteria. The solution of the equations tends to 0 for t which means that b the equilibrium state is said to be asymptotic stable state.
.3 Use of State Models The most direct use of state space models is as basis for the analysis of the dynamic responses of the dynamic system. This requests analysis of response properties of the state space equations of the dynamic system A related but usually more difficult use of state space models as basis for the modification of the dynamic system in order to achieve some desired objective. This process is usually referred to as synthesis or design of the dynamic system, often used in controller design. State space models are useful in developing insight into some aspects of a dynamical system. This insight may be qualitative in nature or it may involve quantitative information about the dynamic responses of the dynamic system under test. State space models are useful in obtaining both types of information.
.4 First Order Linear State Space Models The time dependent behavior of a linear state space differential system is with u as input variable, y as output variable, and x is a state variable. The state model is defined by the constants a, b, c. A block diagram interconnection is as follows, where the symbol /s is used to denote the integration operator.
.4. System Theory of First Order Linear State Space Models Due to the simplicity of state equations it is easy matter to determine an explicit representation for the solution response as a function of initial state and the input function. Suppose that U(s) denotes the Laplace transform of u(t), Y(s) denotes the Laplace transform of y(t) and X(s) denotes the Laplace transform of x(t). In general terms the Laplace transformation converts or transforms a real function of a real variable into a complex function of a complex variable. In particular let f(t) be a real function defined for t > 0; then the Laplace transform of the function is defined by L [ f(t) ] = f(t) * e -st dt and is often written as L [f(t)] = F(s). The integral is defined for complex values of s, F(s) is a complex function and the methods of complex analysis are applicable. Laplace transform is defined for a wide class of functions, so long as the complex values s are restricted to a region in the complex plane for which the indefinite integral in the definition converges. Modeling and Simulation
.4. System Theory of First Order Linear State Space Models Modeling and Simulation For the systems state equations we obtain, with the Laplace transform with s as Laplace operator of d/dt s*x(s) - x(0) = a*x(s) + b*u(s) or sx -x(0) = a X + b U and Y = c X where x(0) denotes the initial state.
.4. System Theory of First Order Linear State Space Models We may rewrite this equation as sx - a X = x(0) + b U X (s - a) = - x(0) + b U X = [ x(0) ]/[ (s - a) ] + [ b U ] / [(s - a) ] and Y = c * {[ x(0) ]/[ (s - a) ] + [ b U ] / [(s - a) ]} Y = [ c*x(0) ] / [ ( s-a ) ] + [ c*b*u ] / [ (s-a) ]
.4. Systems Theory of First Order Linear State Space Models Let a be an arbitrary real number and let f(t), t > 0, have a Laplace transform F(s). Then we find the expressions and L[ e -at * f(t) ] = F (s+a) L[ e at * f(t) ] = F (s-a) Let f(t), t > 0, have the Laplace transform F(s) and let g(t), t > 0, have the Laplace transform G(s). Then the convolution integral has the Laplace transform f(τ) g(t-τ) dτ L [ f(τ) g(t-τ) dτ ] = F(s) * G(s)
.4. Systems Theory of First Order Linear State Space Models Modeling and Simulation Hence with the convolution integral we find and x(t) = e at x(0) + e a(t-τ) bu(τ)dτ y(t) = c e at x(0) + c e a(t-τ) bu(τ)dτ as the explicit expressions for the state and the output responses as a function of the initial state x(0) and the input function u(t). The first term in the above expression is the zero input response while the second term is the zero state response. Thus in this case the general response is the sum of the zero input response and the zero state response.
.4. Systems Theory of First Order Linear State Space Models It is convenient to introduce some terminology. The expression G(s) = (c*b) / (s-a) is called transfer function of the system: the transform of the zero state space output response is the product of the transfer function and the transform of the input function. The expression g(t) =c*e at *b is called the weighting or impulse response function of the system; note that the transfer function is the transform of the weighting function. The denominator polynomial of the transfer function d(s) = s-a is called the characteristic polynomial..
.4. Systems Theory of First Order Linear State Space Models Modeling and Simulation The denominator polynomial of the transfer function d(s) = s-a, the characteristic polynomial is due to several boundaries. If a < 0, i.e the zero of the characteristic polynomial is negative, then the zero input response always tends to zero as t, for any initial state; the state equations are said to be stable. If a > 0 the state equations are said to be unstable. It its clear that the state equation indicated, with one state variable, is a minimal realization for the system, assuming that c*b 0. Any state model for the same system with more than one state variable would be certainly not be minimal
.4. Systems Theory of First Order Linear State Space Models Modeling and Simulation Example: A two-compartment model is used to study ingestion, distribution and metabolism of a drug in the individual. It provides the background information of the mechanism of action of drugs in general pharmacological terms and the significance of pharmacoki-netic parameters that determine the efficacy of drugs. In particular, the drug is ingested, e.g. orally as medication, the drug enters the gastrointestinal tract from where it is then distributed throughout the bloodstream of the individual to be metabolized and eliminated. The primary interest of studying compartment models is to govern how input ingestion rate and/or the initial concentration of the drug in the body affects the individual. State variable X(t) is of great interest, because it is ac-ces-sib-le to analysis by taking blood samples.
.4. Systems Theory of First Order Linear State Space Models Modeling and Simulation Example: Supposing that the drug is taken orally, it enters the gastrointestinal tract, is absorbed into the circulation and distributed throughout the body to be metabolized and finally eliminated. Compartment X describes the gastrointestinal tract and the gastrointestinal vascular bed (circulation) of the individual. Compartment X stands for the bloodstream (between the distribution and elimination processes) of the individual. K and K 0 represent the distribution and elimination constants, respectively. Start is at time zero X (t) denote the mass of drug in compartment X (t) be the mass of drug in compartment. If the ingestion rate of the drug U(t) > 0, we find the plausible assumption for the twocompartment model, that the rate of change of the mass of drug in the gastrointestinal tract is equal to the rate at which the drug is ingested minus the rate at which the drug is distributed from the gastrointestinal tract to the bloodstream: dx dt = U ( t) drug distribution rate compartment to
.4. Systems Theory of First Order Linear State Space Models Example: In the case of first order kinetics, drug distribution rate from compartment to is assumed to be proportional to the mass (or concentration) of the drug in the first compartment. If K > 0 is the corresponding proportionality constant, then the equation above becomes: dx () = U ( t) K X ( t) dt where K X is the inflow rate of drug distribution from the first compartment. Compartment is described by a flow rate equation that balances the inflow and outflow rates descri-bed by in the equation above. With respect to first order kinetics, the outflow rate of compartment is proportional to X. Thus the equation above becomes: K 0 being the elimination constant. Modeling and Simulation dx dt dx dt = inf low = K X rate outflow ( t) K 0 X rate ( t) ()
.4. Systems Theory of First Order Linear State Space Models Modeling and Simulation Example: The equations constitute the linear pharmacokinetic model. In a matrix vector format the constant coefficient linear differential equation model is (t > 0): dx dt dx dt = K K 0 X X K 0 ( t) U ( t) + ( t) 0 (3) with U( t) K3 t = A e (4) The pharmacokinetic model described in the equation above is a second-order linear model. The first differential equation is uncoupled from the second differential equation, the second differential equation is coupled with the first differential equation. It should be noted that this observation is important, since mathematical models should not be excessively difficult for analytical studies.
.4. Systems Theory of First Order Linear State Space Models Modeling and Simulation Example: The homogeneous differential equation of () is dx + K ( X t) dt = which can be solved using the approach Substituting (6) into (5) yields X ( t) t = e λ 0 (5) (6) λ e λ t λ t + K (7) e = 0 with the solution λ = K, which can be used in (6) X ( t) K = e t (8)
.4. Systems Theory of First Order Linear State Space Models Modeling and Simulation Example: The general solution is X ( t) K = c e Consider the initial value c = A we obtain t X( t) K = A e t (9) Substituting (9) into () results in dx dt K t = K A e K t 0 which can be rewritten after multiplying by as follows dx dt e K t + X ( t) K 0 e K t = X ( K ) A e ( K K 0 (0) ) t
.4. Systems Theory of First Order Linear State Space Models Modeling and Simulation Example: The left side of (0) can be rewritten using the product rule for the calculus of differential equations, which has the form d ( ) K t X t e dt ( 0 ) hence, (0) can be integrated as follows: X ( t) e K = K A t ( K 0 K ) t e dt + c or X ( t) e K = K K A K e t ( K K ) t 0 0 + c which results in X ( t) e + c e K t k0 = K K 0 A K t
.4. Systems Theory of First Order Linear State Space Models Modeling and Simulation Example: With the initial value X (0) = 0 we obtain 0 = K K 0 A K + c K c = K 0 A K and finally X K A K t k0 ( t) = ( e e K K 0 t )
.4. Systems Theory of First Order Linear State Space Models Modeling and Simulation Example: The model of the pharmaceutical kinetics can be implemented directly in common simulation systems. For the simulation software CSMP the model formalization is as follows: INITIAL PARAMETERS K=.0, K0=0.5 CONSTANT A=0., K3=0., X0=00., X0=00. DYNAMIC XDOT = - K X + U XDOT = K X K X X = INTGRL (X0, X!DOT) X = INTGRL (X0, XDOT) U = A EXP(-K3 TIME) TIMER DELT=0., OUTDEL=0., FINTIM=0.0 PRTPLT U, X, X END PARAMETER A=00.0 END STOP
.4. Example Radioactive Decay of Cesium Consider a tank containing a certain quantity of a radioactive material, e.g. isotope Cesium 3. Such radioactive material decays, the mass of the material changes as radiation is emitted. On the basis of statistical assumptions about the radioactive decay process it is usually assumed that the rate at which the mass decays is proportional to the mass itself. Assuming that radioactive material can be added to the tank at the rate r, the change in the mass m of the radioactive material in the tank is described by dm/dt = - K*M + r where K is a constant depending on the properties of the radioactive material. This simple ODE is in state form: to be consistent with the notation used previously define the state variable x = m and input variable u = r. The output variable y is the mass m of cesium in the tank. Thus the state model is dx/dt = - K*x + u y = x Modeling and Simulation
.4. Example Radioactive Decay of Cesium Our objective in the example is to analyze the response properties of the process so that the relation between the rate at which radioactive material is added to the tank and the mass of radioactive material in the tank can be clarified. In this simple example it is possible to determine an analytical expresion for the general response function. Assuming the initial state is M and u(t) for t > 0 is the input function, then the transform of the state equation leads to sx M = - K*X + U where U(s) and X(s) are Laplace transforms of u(t) and x(t) so that and X = M / (s+k) + u / (s+k) y(t) = x(t) = e -Kt M + e -K(t-τ) u(τ)dτ
.4. Example Radioactive Decay of Cesium Thus the impulse response or weighting function is g(t) = e -Kt and the transfer function is G(s) = / (s+k)
.4. Example Radioactive Decay of Cesium Since K > 0 the system is stable. Now consider the response to constant input u(t) = R for t > 0, where radioactive material is adder at a constant rate. The response is thus and y(t) = e -Kt M R/K* e -Kt + R/K * e -Kt y(t) R/K as t i.e. the mass of radioactive material in the tank tends to a constant value independent of the initial mass M.
.4. Example Temperature in a Building Temperature in a Building is considered where the outside air surrounding the building is at a generally different temperature than the building. The temperature inside the building is considered to be uniform throughout, say at temperature T. The air outside the building is considered to be at a uniform and constant Temperature T a. Heat can be generated within the building by operation of a furnace, which is assumed to supply heat uniformly to the building at the rate Q. Our interest is to relate the rate at which the furnance supplies heat to the building, considered as the input,to the temperature of the building, considered as the output.
.4. Example Temperature in a Building The basic physical relationship here is that the total rate of change of heat within the building, given by m*c P (dt/dt) where m: mass of the air in the building, and C: heat capacity, equals the rate at which heat is supplied by the furnace Q plus the rate at which heat is transferred to the building from the surrounding air. But the rate of heat transfer into the building from the surrounding air is proportional to the temperature difference between the building and the surrounding air, namely K a (T-T a ) for constant K a.
.4. Example Temperature in a Building Thus, the rate of change of temperature is given by or or m*c P * dt/dt = - K a (T-T a ) + Q dt/dt = - K a /(m*c P )*(T-T a ) + Q/(m*C P ) dt/dt = - /τ (T-T a ) + K*Q where τ = K a / m*c P, and K = / m*c P
.4. Example Temperature in a Building Assuming that the surrounding air is at constant temperature T a it is convenient to choose the temperature difference x = T-T a as the state variable the input variable u = Q, and the output variable y = T; thus the state model is given by dx/dt = - /τ *x * k*u y = x + T a Note that since y = x + T a these equations are not of the form considered previously; however with minor modifications this ideas are applicable here.
.4. Example Temperature in a Building It s a relatively simple matter to determine an exact analytical expression for the time response by taking the transform of the state equation. Assuming the initial state is (T 0 -T a ) and the input is u(t) for t > 0 then sx - (T 0 -T a ) = - /τ * X + K*U where U(s) and X(s) are L-transforms of u(t) and x(t) thus X = (T 0 -T a )/ (s + /τ) + K*U / (s + /τ) x(t) = e -t/τ (T 0 -T a )+ e -(t-λ)/τ K*u(λ)dλ consequently the output response is y(t) = T a + e -t/τ (T 0 -T a ) + e -(t-λ)/τ K*u(λ)dλ
.4. Example Temperature in a Building Due to the presence of the constant Ta ii is not possible to define an impulse response function or a transfer function with input u and output y. g(t) = e -t/τ K is the impulse response function, for the related system with input u and output y-t a Similarly the transfer function from u to y-t a is G(s) = K / (s + /τ)
.4. Example Temperature in a Building The zero input response is y(t) = T a + e -t/τ (T 0 -T a ) Thus, if no heat is supplied to the building by the furnace then the temperature of the building tends to the temperature of the surrounding air, no matter what is the initial temperature in the building. The rate at which this change occurs is characterized by the parameter τ, which is called time constant of the building
.4. Example Temperature in a Building Now consider the case where the furnace is in continual operation and transfers heat to the building at a constant rate so that u(t) = Q for t > 0. Then, from the above expression for the output response it is easily determined that y(t) = T a + e -t/τ (T 0 -T a ) + τ*k*q (- e -t/τ ) Thus if heat is transferred to the building by the furnace at a constant rate the temperature of the building satisfies y(t) T a + τ*k*q as t That is, the ultimate temperature of the building exceeds the outside temperature by the quantity τ*k*q
.4. Example Temperature in a Building Consider τ= 5, T a = 0, K = 0.05, than the temperature in the building during a hour time period (00 time steps), if the initial temperature in the building is assumed being 0 o F, can be simulated. What is the temperature in the building during a hour time period if the initial temperature in the building is 65 o F and the furnace, which supplies heat to the building, is switched on and of every 30 minutes?
.4. Example RC Network In this case study example a simple electrical network is examined Which contains a single ideal resistor and a single ideal capacitor. The input is voltage e i applied at the indicated terminals; it is assumed that the applied voltage source e i has zero impedance. The output of the network is the voltage e o across the resistor. Let Q denote the charge on the capacitor and let I denote the current through the capacitor; then the total voltage drop in the single loop is Q/C + R*I e i = 0 with C as capacitance and R as resistance of the circuit. The charge and current relationship is dq/dt = I
.4. Example RC Network We choose the charge Q on the capacitor as state variable x; the input variable u = e i, and the output variable y = eo = RI. Thus the state model for the RC-network is given by dx/dt = - /RC * x + /R * u y = - /C * x + u The objective is to relate the response properties of the output voltage to the initial state of the network and to the input voltage. The response properties of the system are now examined. It is possible to determine an expression for the general response as follows. Taking the transform of the state equation, assuming Q o is the initial state and u (t), t > 0 is the input, obtain sx Q 0 = - /RC * X + /R * U where U (s) and X (s) are Laplace transform of u (t) and x (t), so that X = Q 0 / (s+/rc) + U / [R(s+/RC)]
and consequently.4. Example RC Network and x(t) = e -t/rc Q 0 + e -(t-τ)/rc [u(τ)]/r *dτ y(t) = - Q 0 /C* e -t/rc /RC c e -(t-τ)/rc u(τ)dτ + u(t) The transfer function for the network is given by Since RC > 0 the system is stable. G(s) = s / [s + /RC ] Now consider the response to a sinusoidal input voltage u (t) = e i wt for t > 0. Then the output is given by cos y(t) = - Q 0 /C* e -t/rc e i /RC c e -(t-τ)/rc cos ωτ dτ + e i cos wt
.5 Second Order Linear State Space Models Linear second order state space models can be expressed in the form dx dt = ax u + ax + b, dx dt = ax, + ax + bu y = c x + c x with u as the input variable, y as the output variable and x and x as state space variables. The state space model is defined by the constants a, a, a, a, b, b, c, c. A block diagram interconnection with the Laplace transform /s is denoted to the integration operator. It should be noted that the implicit feedback structure of the state space equations is obvious from the block diagram.
.5 Second Order Linear State Space Models
.5 Second Order Linear State Space Models Supposing that U(s) denotes the Laplace transform of u(t), Y(s) denotes the Laplace transform of y(t) and X(s) denotes the Laplace transform of x(t). Then from the state space equations we obtain sx = a X + a X + b U sx = a X + a X + b U Y = c X + c X where the capital letters denote transforms of the corresponding variables.
Thus.5 Second Order Linear State Space Models X X = = [( s a) b + ab ] U ( s a)( s a) aa [ a b + ( s a ) b ( s a)( s a) aa ] U and Y = c[( s a ) b + a ( s a b ] + c )( s a [ a b ) a + ( s a a ) b ] U. Consequently G( s) = c [( s a ) b + a ( s a b ] + c )( s a [ a b ) a + ( s a a ) b ] is the so called transfer function; the transform of the zero state output response is the product of the transfer function and the transform of the input function.
.5 Second Order Linear State Space Models The denominator polynomial of the transfer function d( s) = ( s a a a )( s a) is called the characteristic polynomial of the system. The zeros of this polynomial are called the characteristic zeros of the system; they are also the poles of the transfer function. It can be shown that if the real part of the characteristic zeros, which may be complex, are negative then the zero input response always tends to zero; the system is said to be stable. If this property does not hold then the system is said to be unstable. Assuming that there are no common factors between the numerator polynomial and the denominator polynomial in the transfer function, it is clear that the state equations indicated, with two state variables, is a minimal state realization. One state variable would clearly not be sufficient to obtain a state realization while any state realization with more than two state variables would not be minimal. Modeling and Simulation
.5 Second Order Linear State Space Models To make the analysis somewhat simpler consider a special set of second order linear state equations, with constant parameters ξ, w 0,b defined by dx dt dx dt y = = = x x, ξw. 0 x w 0 x + bu,
.5 Second Order Linear State Space Models Such a special form occurs quite commonly and can be analysed rather easily. The parameter is usually referred ξ to as the damping ratio of the system and is referred to as the natural w 0 frequency of the system. The transfer function is given by G ( s) = s + b ξw 0 s + w 0 and the characteristic polynomial is d ( s) = s + ξw s + 0 w 0.
.5 Second Order Linear State Space Models Assuming ξ > 0 the system is always stable. If 0 < ξ > the characteristic zeros are complex; if ξ > the characteristic zeros are real. The zero input response is first examined; in this rather simple case it is not too difficult to determine the exact state responses using the method of Laplace transforms. Taking the transforms of the state equations obtain sx sx x x (0) (0) = = X, ξ w X 0 w X 0, so that X X = = ( s + sx s ξw s + 0 ) x ξw (0) w + ξw 0 0 s (0) + x 0 s + 0 w x(0). + w 0 (0),
.5 Second Order Linear State Space Models The state responses can be determined by taking the inverse transforms. The state response x (t), if 0 < ξ <, is given by and if the state response is given by ( ) ( ); sin (0) (0) cos (0) ) ( 0 0 0 0 0 0 t w e w w x x t w e x t x w t w t ξ ξ ξ ξ ξ ξ + + = > ξ ( ) ( ). 0 0 0 0 0 0 (0) (0) (0) (0) ) ( t w t w e w x w x e w x x w t x + + + + = ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ
.5 Second Order Linear State Space Models Also x ( t) = ( dx dt)( t). Even in this case the zero input response is a rather complicated function of the initial state x (0) and x (0). In order to make this dependence more explicit it is convenient to plot, for a fixed initial state one state function x ( t) versus the other state function x ( t). Curves in the phase plane corresponding to solutions of the state equations are called trajectories. Some typical trajectories in the phase plane are shown for two different values of the damping ratio in the following Figure.
.5 Second Order Linear State Space Models
.5 Second Order Linear State Space Models If the damping ratio is less than one the trajectories are of the form of spirals since the characteristic zeros are complex; if the damping ratio exceeds one then the trajectories are of nodal type since the characteristic zeros are negative. The arrows on the trajectories indicate the change in the values of the state variables as time increases. Since each trajectory is parameterized by time, the time parameter could be indicated along the trajectory; this is usually omitted however. Phase plane is a rather simple graphical way of illustrating the qua-litative dependence of the zero input response on the initial state. In this case the trajectories in the phase plane were determined from the time responses directly; this is usually very difficult. In many cases it is possible to determine, the form of the trajectories without use of the time responses. Use of the phase plane as a tool for understanding the zero input response is particularly valuable for systems which can be described by the two state variables x and x.
.5 Second Order Linear State Space Models Now the response for several standard input functions should be determined. First consider a constant input function u ( t) = u for t 0. Conditions for an equilibrium state are that dt dx = 0 and dx dt = 0 so that x = bu w0, x = 0 define an equilibrium state. Thus if (0) and then x x and for all t; and 0. for all If ( 0) = bu w ( t) = bu w 0 0 x ( t) = 0 y t 0; ( t) = bu w0 0 < ξ < then the zero state response can be determined as so that y ( t = bu w as t. If ξ = then the zero state response can be determined as so that as ) y ( t ) 0 = y( t) = b u w + bu w 0 w 0 bu e w 0 + w 0 ξw0t buξ e ξ w 0 0 cos ξw0t ( w0 ξ t) sin( w0 ξ t), ( ) ξ ξ ξ b u b u ( ) ξ + ξ ξ e w 0 e w ξ + 0 ξ ξ ξ t t x = 0 e ( ) ( ) ξ w t w ξ t and e sin w t ξ w 0 t 0 cos 0 0 ξ
.5 Second Order Linear State Space Models If ξ > so that the characteristic zero are real, then the responses are purely exponential in form. If 0 < ξ < the response is said to be under damped; if ξ> the response is said to be over damped; if the response is said to be critically damped. ξ = The form of some typical responses to a constant input, showing the dependence on the damping ratio, are shown in the following figures. The damping ratio ξ = 0.707 is generally considered to correspond to the fastest response; it represents a compromise between overshoot and rise time. * Transient response of the nonlinear dynamic system showing the influence of the damping factor * Impulse response functions, showing the influence of the damping ratio * Amplitude (top) and phase (bottom) as function of the damping ratio
.5 Second Order Linear State Space Models
.5 Second Order Linear State Space Models
.5 Second Order Linear State Space Models
.5. Second Order Linear State Space Models Example Modeling and Simulation Two tanks contain a salt solution; assume V and V are the Volumes of the two tanks and that the two tanks are interconnected. The solution of the first tank is pumped to the second tank at the constant flow rate Q ; the solution in the second tank is pumped back to the first tank at the constant flow rate Q. Assume the net flow is from the first to the second tank. Assuming that each tank remains full the external flow rate into the first tank and out of the second tank is necessarily Q Q. It is assumed that there is a salt solution in each tank, at a uniform concentration due to complete mixing within each tank. The external inflow added to the first tank contains a salt solution at a certain concentration C which is taken to be the input variable. The output is taken to be the concentration of the salt withdrawn from the second tank. The system may be characterize the output concentration as a function of the input concentration and the initial amount of salt in each tank.
.5. Second Order Linear State Space Models Example Modeling and Simulation Let S and S be the mass of salt in each tank; then the total rate of change of mass of salt in the first tank is the rate at which salt is added to the first tank minus the rate at which salt is withdrawn from the first tank, I.e. ds /dt = - Q *(S / V ) + Q *(S / V ) + (Q Q )*C Similarly, the total rate of change of mass of salt in the second tank is the rate at which salt is added to the second tank minus the rate at which salt is withdrawn from the second tank, i.e., ds /dt = Q *(S / V ) - Q *(S / V ) - (Q Q )*(S / V ) The above two equations describing the relation between physical variables in the process have been determined from the physical considerations, namely that mass is conserved. It is assumed that Q > 0, Q > 0 and that Q Q > 0. Physically, it is required that C > 0 and that S > 0 and S > 0 always. Clearly the masses of salt in each tank serve as state variables so define x = S and x = S ; the input variable u is the concentration C of salt added to the first tank; the output variable y is the concentration of salt in the second tank, S / V.
.5. Second Order Linear State Space Models Example Modeling and Simulation The state equations x and x for the physical process are dx /dt = - (Q / V )*x + (Q / V )*x + (Q Q )*u dx /dt = (Q / V )*x -(Q / V )*x y = ( / V )*x The above equations describing clearly a linear second order system. Consider the mixing of a salt solution as described may be Q = 0, V = 5, Q = 5 and V = 0 Assume C(t) = 0 for t> 0, and the initial amounts of salt in the two tanks are S = 5 and S = 0. Assume C(t) = 0. for 0 < t < 0, and C(t) = 0.0 for t > 0, the initial amounts of salt in the two tanks are S = 0 and S = 0. Consider V = 0, V = 5 and Q Q =
.5. Second Order Linear State Space Models Example Modeling and Simulation The operational considerations for a general DC motor will be investigated. A DC motor is essential a device for converting electrical energy into mechanical, or rotational, energy. The basic ingredients of the motor are an armature circuit and a field circuit; these circuits are coupled in such a way that a torque is produced on the motor shaft. i f i a v f T e ω v a The voltage applied to the field and armature circuits are v f and v a respectively; i f and i a are the current in the field and armature circuits respectively. The torque generated by the motor is T e and ωis the motor speed. In the field circuit -v f + R f * i f + L f * di f / dt = 0 where R f and L f are the field resistance and inductance.
.5. Second Order Linear State Space Models Example Modeling and Simulation In the armature circuits there is resistance R a and inductance L a ; there is a voltage drop in the armature circuit which is proportional to the product of the motor speed and the field current; thus -v a + R a *i a + L a *di a / dt + B*ω*i f = 0 where B is an electromechanical constant for the motor. The field current is clearly independent of the armature current by the armature current does depend on the field current and also the motor speed is indicated. Finally, the torque T e generated by the motor is assumed to be proportional to the product of the field current and the armature current to the equation T e = B*ω*i f In terms of deriving a realistic model for the motor it is necessary to consider the rotational motion of the motor shaft. It may be assumed that the inertia of the motor shaft, as well as any external load, is J; and that there is a viscous damping torque which is proportional to the motor speed where c is the damping coefficient. Thus J*dω/dt= -c*ω + T e.
.5. Second Order Linear State Space Models Example Modeling and Simulation The equations derived constitute the basis for the model of the DC motor. The external input variables are the applied field and/or armature voltages; the variables of interest are the field and/or armature currents and the motor speed. A careful analysis of the above equations is quite difficult since several of the equations involve products of the variables and hence nonlinear equations, which is really a task for simulation. Two cases are possible: -constant armature current -Constant field current Let s assume that armature current is maintained at a constant value. Such operation is referred to as a field controlled DC motor. From the previous equations it follows that and -v f + R f *i f + L a *di f / dt = 0 J*dω/dt= -c*ω + B*I*i f while i a = I are the descriptive equations for the motor. It is an easy matter to choose the state variables as the field current x = i f and the motor speed x = ω. The input variable u is the applied field voltage v f ; the output variable y is the motor speed ω.
.5. Second Order Linear State Space Models Example Modeling and Simulation The state equation can be written as: dx /dt = -R f /L f * x + u/l f dx /dt = - c/j * x + (B*I a / J) * x The state equations derived a second order linear state equations for the field controlled DC motor. Consider the following parameters as a priori information available R f = 3.3 R a = 0.75 L f = L a = 0. v f = 0cos360t
.6 Higher Order Linear State Space Models Consider linear state equations of the form dx dt = a u x + a n xn + bn, dx dt n = a x + a xn + b u n nm n, y = c x + c n x n with u as the input variable, y as the output variable and x,, x n as state variables, where n >. In case where more than two state variables are required, a deteiled analysis of the response properties of the system is usually difficult; simulation has become quite involved and the dependence of the output response on the initial state involved and the initial function may be quite complicated, but can be investigated using simulation. The transfer function is necessarily the ratio of two polynominal functions G(s) = n(s) / d(s)
.6 High Order Linear State Space Models Case Study Example: Automobile Suspension System The mass m represents one quarter of the mass of the automobile frame m, represents the effective mass of the wheel. The constant k is the stiffness of the suspension spring, k is the stiffness constant for the tire and µ is the damping constant of the shock absorber. Assuming there is some reference level for the ground the car and wheel assembly define an equilibrium level. Then as measured from this equilibrium level, q denotes the vertical displacement of the automobile frame, i.e., of the mass m, and q denotes the vertical displacement of the wheel, i.e.; of the mass m.
.6 High Order Linear State Space Models Case Study Example: Automobile Suspension System The input is assumed to be the variable q i which is the vertical displacement of the bottom of the tire due to the road irregularities, as measured from the ground reference level. The bottom of the tire is assumed to remain in contact with the ground. Using the simple spring-mass-damper model shown the mathematical equations which describe the motion of the suspension system are obtained by applying Newton s law to each mass in the system. This leads to the equations m m d q dt d q dt = = k k ( q q ) µ ( q q ) + k ( q q ) dq dt i dq dt µ dq dt. dq dt,
.6 High Order Linear State Space Models Case Study Example: Automobile Suspension System It is natural to choose the displacements and velocities of each mass as the state variables, i.e., x = q, x = dq /dt, x 3 = q, x 4 = dq /dt. The input variable u is the displacement of the road surface q i ; the output variable y is the displacement of the automobile frame q. The resulting state model is dx dt dx dt dx dt dx dt 3 4 y = x = = x = = x 4,, k m k m. ( x x ) ( x x ) ( x u ) ( x x ) + ( x x ) 3 3 µ m k m, 3 4, µ m 4,
.6 High Order Linear State Space Models Case Study Example: Automobile Suspension System The automobile suspension model is fourth order linear state equations. It is exceedingly difficult to determine exact analytical expressions for the response functions in terms of the given parameters. However, it is possible to determine certain qualitative features of the response properties. The transfer function of the system can be determined by talking the transform of the four state equations, assuming the initial state is zero; then by solving the resulting four algebraic equations an expression for the transform of the output can be obtained as a function of the transform of the input, given by G ( µ s + k ) ( ) k s = ( )( ) ( ). m s + µ s + k m s + µ s + k + k µ s + k Thus the characteristic polynomial is ( m s + µ s + k )( m s + µ s + k + k ) ( s ). d ( s ) = µ + k