Dynamic modelling J.P. CORRIOU. Reaction and Process Engineering Laboratory University of Lorraine-CNRS, Nancy (France) Zhejiang University 2016

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1 Dynamic modelling J.P. CORRIOU Reaction and Process Engineering Laboratory University of Lorraine-CNRS, Nancy (France) Zhejiang University 216 J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

2 Reference book "Process Control - Theory and Applications", Jean-Pierre CORRIOU Springer-Verlag, London (24) 75 pages ISBN Available at : http ://link.springer.com/book/1.17/ J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

3 Context In the development of a model, the first question that the engineer of the scientist must ask is : a model for what use? Models depend on the scientific domain, on the time of development, on their degree of sophistication, on the mathematical and physical degree of knowledge of the developers. Examples : a model to represent the weather over a large part of the world, a model to represent the brain activity, a model to represent a complete plant, a model to represent a part of the plant, a model to represent more or less complicated experiments. Our concern : main models used in chemical engineering and control. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

4 Classification of models Steady-state or dynamic, Empirical (black-box) or based on first principles (conservation laws), Continuous or discrete dynamic model, Deterministic or stochastic (probabilistic), Linear or non linear, Lumped-parameter models (ODEs) or distributed-parameter models (PDEs). J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

5 System representation External disturbances Measured Unmeasured Manipulated variables Process (states) Measured outputs Unmeasured outputs FIGURE : Input-output block diagram representation of a process J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

6 Contents Continuous systems : Part 1 : State-space modelling. Part 2 : Transfer functions. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

7 Part I : State-space model State-space multivariable system modelled by a set of algebraic and differential equations { ẋ = f(x, u, t) (1) y = h(x, t) x state vector of dimension n, u input vector (or control variables vector) of dimension n u y output vector of dimension n y (in general, n u n y ). f is in general non-linear. x(t) depend only on initial conditions at t and on inputs u(t) between t and t. h can be non-linear. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

8 State-space model In the presence of disturbances { ẋ = f(x, u, d, t) y = h(x, t) (2) d disturbance vector of dimension n d SISO system : 1 input, 1 output MIMO system : several inputs, several outputs. In general, n u n y Square system : same number of inputs and outputs, n u = n y. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

9 Typical inputs To simplify the study of systems, some typical inputs are often used : Step function : a unit step function is defined as f = 1 if t >, f = if t. Its response is called a step response. Impulse function : a unit impulse function is defined as : f = δ (theoretical Dirac). Its response is called an impulse response. Sinusoidal function : f = a cos(ω t +φ). Its response is referred to as a frequency response. Ramp : f = kt. This determines the behaviour of the process output to an input with constant rate of change (constant velocity). Parabolic function : f = kt 2. This is used whenever the response to a constant acceleration is desired. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

10 Typical inputs 1.5 Step Dirac Pulse Rectangular Pulse Sinusoid Ramp Parabolic function FIGURE : Different types of inputs starting at time t = 7 after steady state J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

11 Particular cases In general, in chemical engineering, manipulated inputs appear linearly. { ẋ = f(x, t)+g(x, t) u y = h(x, t) Models are called affine with respect to the input vector. (3) An important class is linear state-space models { ẋ(t) = Ax(t)+Bu(t) y(t) = Cx(t)+Du(t) A is the state matrix of dimension (n n), B is the control matrix of dimension (n n u ), C is the output matrix of dimension (n y n), D is the coupling matrix of dimension (n y n u ) which is very often equal to zero. When D is different from zero, it is said that the output is directly driven by the input. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85 (4)

12 Generalized linear state-space model Descriptor form { Eẋ(t) = Ax(t)+Bu(t) y(t) = Cx(t) + Du(t) E is called the descriptor matrix of dimension (n n). (5) If E is invertible, the system can be rewritten under the usual form { ẋ(t) = A x(t)+b u(t) y(t) = Cx(t)+Du(t) (6) If E is singular, this corresponds to an differential-algebraic system. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

13 Linearization Linearization is often performed around a steady-state to simplify the representation. It allows linear control. Consider the linearization of a set of differential equations of the most general form ẋ = f(x, u, t) (7) around a steady-state operating level x for a nominal input u. This is achieved by Taylor series expansion ẋ i = f i (x, u, t)+ n ( ) fi n u ( ) fi δx j + δu k x j x=x,u=u u k x=x,u=u j=1 k=1 (8) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

14 Linearization δx j = x j x j, with x j, is the j component of steady-state vector x δu k = u k u k, with u k, is the k component of the nominal input vector u δ indicates any deviation with respect to the nominal operating condition δx and δu are respectively deviations of the state and of the input with respect to the steady state of the process. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

15 Linearization We have : ẋ i = δẋ i + ẋ i, and ẋ i, = f i,. Resulting set of linear ordinary differential equations δẋ i = n ( ) fi n u ( ) fi δx j + δu k (9) x j x=x,u=u u k x=x,u=u j=1 k=1 Similarly δy i = n ( ) hi n u ( ) hi δx j + δu k (1) x j x=x,u=u u k x=x,u=u j=1 k=1 J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

16 Linearization Final matrix form { ẋ(t) = Ax(t)+Bu(t) y(t) = Cx(t)+Du(t) (11) Warning : x(t), u(t), y(t) are deviation variables (steady state is assumed at t = or negative times). [ ] [ ] [ ] [ ] fi fi hi hi A = ; B = ; C = ; D = x j u j x j u j (12) A and B are respectively the Jacobian matrices of f with respect to x and u C and D are respectively the Jacobian matrices of h with respect to x and u. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

17 Stability of a Linear State-Space System Given the linear state-space system ẋ(t) = Ax(t) (13) this system is stable is the eigenvalues of the matrix A are located in the left-hand complex half-plane. If A is reduced to a single element, the solution is the exponential x(t) = b exp(at) (14) thus A must be negative so that the system is stable. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

18 Typical models in chemical engineering Surge Tank F h F FIGURE : A surge tank with varying level In general, F, F and h are time-varying. F volume flow rate. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

19 Surge tank Steady state General mass balance F ρ = Fρ (15) (inlet mass / unit time) = (outlet mass / unit time) + (time rate of change of mass in the system) Dynamic model of the tank F ρ = F ρ+ dm dt with m = ρv. If ρ = ρ (constant liquid density), if cross-sectional area S not depending on liquid level, it gives dh dt fundamental model of level control. (16) = F /S F/S (17) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

20 Surge tank with valve on the inlet stream F h Single-input single-output (SISO) F One input : u = F one output y = h only one state : x = h State-space model ẋ = u/s F/S y = x Linear model (cross-sectional area S constant). Flow rate F is an external disturbance. (18) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

21 Surge tank with valve on the outlet stream F h F Single-input single-output (SISO) One input : u = F one output : y = h only one state : x = h State-space model ẋ = F /S u/s y = x Linear model (cross-sectional area S constant). Flow rate F is an external disturbance. (19) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

22 Isothermal Continuous Chemical Reactor (CSTR) F, ρ, T F 1, ρ 1, T C A V, C A, C B ρ, T F, ρ, T C A, C B Control of level and concentration J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

23 Isothermal Continuous Chemical Reactor (CSTR) Overall mass balance d(ρ h) dt = F ρ /S + F 1 ρ 1 /S F ρ/s (2) and densities in different streams assumed equal. Balance for component A in transient regime (rate of A entering) = (rate of A exiting) (rate of A produced) + (rate of accumulation of A), hence F C A = F C A R A V + d(v C A) dt (21) with R A production rate of A. It represents the number of moles of A produced per unit volume and unit time. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

24 Isothermal Continuous Chemical Reactor (CSTR) When R reactions designated by i occur simultaneously, the rate of production of a component A j is equal to n r R j = ν ij r i (22) i=1 where each reaction rate r i is in general positive. Stoichiometric coefficient ν ij > if A j is produced by reaction i, and ν ij < if A j is consumed by reaction i. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

25 Isothermal Continuous Chemical Reactor (CSTR) Consider the case of one first-order chemical reaction : A B without heat of reaction reaction rate r A is equal to r A = k C A production rate of A is equal to R A = r A stoichiometric coefficient ν ia = 1 Balance with respect to A F C A = F C A + k C A V + d(v C A) dt (23) This balance must be considered together with the global mass balance d(ρ h) = F ρ /S + F 1 ρ 1 /S F ρ/s (24) dt with ρ i =constant. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

26 Isothermal Continuous Chemical Reactor (CSTR) Assumptions : Manipulated inputs are inlet volumetric flow rate F 1 and inlet concentration C A control vector u = [F 1, C A ] T Control of reactor level and concentration output vector y = [h, C A ] T State vector x = [h, C A ] T J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

27 Isothermal Continuous Chemical Reactor (CSTR) State-space model ẋ 1 = F /S F/S + u 1 /S 1 ẋ 2 = [F (u 2 x 2 ) x 2 u 1 ] k x 2 S x 1 y 1 = x 1 y 2 = x 2 (25) MIMO model, nonlinear. Disturbances are flow rates F and F. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

28 Non-Isothermal Continuous Chemical Reactor (CSTR) F, ρ, T F 1, ρ 1, T 1 C A Heat-conducting fluid T j V, C A, C B ρ, T F, ρ, T C A, C B Control of level, concentration and temperature J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

29 Non-Isothermal Continuous Chemical Reactor (CSTR) Heat transfer between the jacket and the reactor contents Q = U Sex(T j T) (26) U overall heat transfer coefficient, Sex available heat exchange area, T j mean temperature in the jacket, T temperature of the reactor. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

30 Non-Isothermal Continuous Chemical Reactor (CSTR) General energy balance (variation of internal energy / unit time) = (inlet enthalpy by convection / unit time) (outlet enthalpy by convection / unit time ) + (rate of heat transfer and mechanical energy) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

31 Non-Isothermal Continuous Chemical Reactor (CSTR) Energy balance of the reactor du dt F in and F out total inlet and total outlet molar flow rates. h specific molar enthalpy of each stream = F in h in F out h out + Q (27) h in = j x j,in h j,in h out = j x j,out h j,out (28) x j,in and x j,out are the inlet and outlet mole fractions h j,in and h j,out specific enthalpies of component j in the inlet and outlet streams Specific molar enthalpy h of a pure component can be expressed with respect to its enthalpy of formation H f (T ref ) at a reference temperature T ref according to if no change of state between T ref and T. T h = H f (T ref )+ C p dτ (29) T ref J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

32 Non-Isothermal Continuous Chemical Reactor (CSTR) Overall thermal balance (m C p + m r C r ) dt dt = in F in ρ in C p,in (T in T)+ Q V R r i h i (3) m mass of the reactor contents, C p mean specific heat m r mass of the reactor wall and its accessories, C r corresponding heat capacity h i heat of reaction i Consider first-order chemical reaction A B with heat of reaction h A B reaction rate with Arrhenius term r = k exp( E/(RT)) C A, highly nonlinear Energy balance for the considered reactor (m C p +m r C r ) dt dt i=1 = F ρ C p (T T)+F 1 ρ 1 C p1 (T 1 T)+ Q V r h A B (31) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

33 Non-Isothermal Continuous Chemical Reactor (CSTR) Application to the considered reactor Control of level, concentration and temperature Output vector y = [h, C A, T] T Control vector u = [F 1, C A, Q] T State vector x = [h, C A, T] T. State-space model of the considered reactor (MIMO, nonlinear) ẋ 1 = F /S F/S + u 1 /S ẋ 2 = x 3 = y 1 = x 1 1 [F (u 2 x 2 ) x 2 u 1 ] k exp( E/(R x 3 )) x 2 S x 1 1 [ F ρ C p (T x 3 )+u 1 ρ 1 C p1 (T 1 x 3 ) ρ S C p x 1 + m r C r k exp( E/(R x 3 )) S x 1 x 2 h T + u 3 ] y 2 = x 2 y 3 = x 3 J.P. Corriou (LRGP) Dynamic modelling Zhejiang University 216 (32) 33 / 85

34 Distributed-Parameter Systems If process variables depend simultaneously on time and spatial variables, the process is described by partial differential equations. Case of : heat exchangers chemical tubular reactor chromatography column packed distillation column packed absorption column J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

35 Co-current heat exchanger T e, T e (z,t) T T(z,t) T N T e,n z = z = L J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

36 Counter-current heat exchanger T e,n T T e (z,t) T(z,t) T N z = z = L J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

37 Heat exchangers Consider an external tube in which a fluid circulates at velocity v e with density ρ e, heat capacity C pe, exchanging heat with a fluid circulating in an internal tube at velocity v, with density ρ, heat capacity C p. Assume that no fluid undergoes a phase change. L length of the heat exchanger. Internal and external cross-sections S e and S assumed constant. S surface submitted to the heat transfer Heat transfer coefficients h on the internal side and h e on the external side. Energy balances related to convection effects. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

38 Co-current heat exchanger Te, T Te(z, t) T(z, t) TN Te,N z = Energy balance for the internal tube z = L T(z, t) t T(z, t) = v + a[t e (z, t) T(z, t)] ; a = hs (33) z ρsc p Energy balance for the external tube T e (z, t) t = v e T e (z, t) z + a e [T(z, t) T e (z, t)] ; a e = h es ρ e S e C pe (34) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

39 Co-current heat exchanger Discretization of spatial derivatives to obtain a system of ordinary differential equations. Length L discretized in elements of length z = L/N dt i = v T i T i 1 + a(t e,i T i ) ; i = 1,...,N dt z dt e,i T e,i T e,i 1 = v e + a e (T i T e,i ) ; i = 1,...,N dt z (35) 5 T internal T external 45 Temperature (C) z (m) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

40 Model for control of a co-current heat exchanger Problem : number of differential equations 2N. Need to consider low N. Example : N = 3 Control of temperature T(L, t) by acting on the temperature T e (, t). T(, t) is a disturbance. Resulting model dt 1 dt dt 2 dt dt 3 dt dt e,1 dt dt e,2 dt dt e,3 dt = v T 1 T z = v T 2 T 1 z = v T 3 T 2 z T e,1 T e, = v e z T e,2 T e,1 = v e z T e,3 T e, = v e z + a(t e,1 T 1 ) + a(t e,2 T 2 ) + a(t e,3 T 3 ) + a e (T 1 T e,1 ) + a e (T 2 T e,2 ) + a e (T 3 T e,3 ) (36) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

41 Model for control of a co-current heat exchanger State vector x(t) = [ T 1 T 2 T 3 T e,1 T e,2 T e,3 ] T (37) Manipulated input : u(t) = T e,, disturbance d = T, controlled output y(t) = T 3 Matrix form v z a a v v z z a a v v ẋ(t) = z z a a a e ve z v e ae z a e ve z v e ae z a e ve + v e z v z u(t)+ d(t) z ae x(t) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85 (38)

42 Counter-current heat exchanger Energy balance for the internal tube T(z, t) t T(z, t) = v + a[t e (z, t) T(z, t)] ; a = hs (39) z ρsc p Energy balance for the external tube T e (z, t) t = v e T e (z, t) z + a e [T(z, t) T e (z, t)] ; a e = h es ρ e S e C pe (4) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

43 Counter-current heat exchanger Same type of discretization. dt i = v T i T i 1 + a(t e,i T i ) ; i = 1,...,N dt z dt e,i T e,i+1 T e,i = v e + a e (T i T e,i ) ; i =,...,N 1 dt z (41) 5 45 T internal T external Temperature (C) z (m) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

44 Model for control of a counter-current heat exchanger State vector (with N = 3) x(t) = [ T 1 T 2 T 3 T e, T e,1 T e,2 ] T (42) Manipulated input : u(t) = T e,3, disturbance d = T, controlled output y(t) = T 3 Matrix form v z a a v v z z a a v v ẋ(t) = z z a ve z v e ae z a e ve z v e ae z a e ve x(t) z ae v z a + u(t)+ d(t) a e v e z J.P. Corriou (LRGP) Dynamic modelling Zhejiang University 216 (43) 44 / 85

45 Isothermal tubular chemical reactor z z +dz v(z) v(z +dz) z = C(z) ρ(z) v(z) z = L Turbulent flow plug flow assumption. Assume first-order chemical reaction A B and reaction rate r A = k exp( E/(RT)) C A. Model of a control volume (CSTR) between z and z + dz : global mass balance, mass balance for a given component, energy balance. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

46 Isothermal tubular chemical reactor z z +dz v(z) v(z +dz) Global mass balance z = C(z) ρ(z) v(z) z = L v(z) Sρ(z) = v(z + dz) Sρ(z + dz)+ (S dz ρ(z)) t (44) or ρ t + (ρ v) z Mass balance for component A over the control volume = (45) v(z) S C A (z)+sn A (z) = v(z + dz) S C A (z + dz)+sn A (z + dz)+ k C A S dz + (S dz C A) t (46) or C A t + v C A z + k C A ( ) C A D A = (47) z z J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

47 Isothermal tubular chemical reactor Discretization as a series of n perfectly mixed continuous stirred tank reactors ρ i 1 v i 1 = ρ i v i + z dρ i dt v i 1 C A,i 1 = v i C A,i + k/s C A,i + z dc A,i dt (48) Complex model for control if N large. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

48 Part II : Transfer functions Laplace transform of a function f(t) defined by L[f(t)] = F(s) = f(t) exp( st) dt (49) assuming that the function or signal f(t) is zero for t <. This implies that deviation variables are used. The Laplace transformation is a linear mapping L[a 1 f 1 (t)+a 2 f 2 (t)] = a 1 L[f 1 (t)]+ a 2 L[f 2 (t)] (5) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

49 Linearization Example : F(t) C A (t) is nonlinear function. Taylor series expansion in the neighbourhood of the steady state truncated at first order. F(t) C A (t) = F s C s A + F s (C A (t) C s A )+Cs A (F(t) F s )+(ǫ 2 ) (51) For a general function f of n variables x 1,..., x n, Taylor series expansion in the neighbourhood of the steady state truncated at first order gives ( ) f s f(x 1,...,x n ) f(x1 s,..., x n s)+ i (x i x x i s ) ( ) i f s (52) δx i δf i x i δx i = (x i x s i ) is a deviation variable. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

50 Laplace Transformation The Laplace transformation is a linear operation L[a 1 f 1 (t)+a 2 f 2 (t)] = a 1 L[f 1 (t)] + a 2 L[f 2 (t)] (53) The Laplace transform of a first-order derivative of a function is L[ df(t) ] = s F(s) f() (54) dt If f(t) is a deviation variable with respect to the initial steady state, its initial value becomes zero : f() =, and the previous equation simply becomes L[ df(t) ] = s F(s) (55) dt The Laplace transform of the n th -order derivative of a function is L[ dn f(t) dt n ] = s n F(s) s n 1 f() s n 2 f (1) () f (n 1) () (56) If f(t) is a deviation variable, its initial value and successive derivatives up to the (n 1)-th-order become zero so that the previous formula becomes L[ dn f(t) dt n ] = s n F(s) (57) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

51 Laplace Transformation The Laplace transform of the integral of a function is t L[ f(x) dx] = 1 F(s) (58) s The final value theorem is Applicable only for stable systems. lim f(t) = lim s F(s) (59) t + s The Laplace transform of a delayed function is L[f(t t )] = exp( s t )L[f(t)] (6) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

52 Laplace Transformation TABLE : Laplace transform of some common functions Signal f(t) (t ) Transform L[f(t)] = F(s) Convolution product f(t) g(t) F(s) G(s) Derivative : df(t) s F(s) f() dt Integral : t f(x) dx 1 s F(s) Delayed function : f(t t ) exp( s t ) F(s) Dirac unit impulse : δ(t) 1 Step of amplitude A A s Exponential : exp( a t) 1 s + a 1 τ exp( t/τ) 1 τ s + 1 J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

53 Laplace Transformation f(t) g(t) h(t) = f(t) g(t) F(s) G(s) H(s) = F(s)G(s) A signal f(t) excites a linear time-invariant system with impulse response g(t), the response of the system h(t) is equal to the convolution product of f(t) by g(t) denoted by h(t) = f(t) g(t) = The Laplace transform H(s) of the output is equal to f(τ) g(t τ) dτ (61) L[f(t) g(t)] = F(s) G(s) (62) The Laplace transform of a convolution product is equal to the product of the Laplace transforms of the functions. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

54 Laplace Transformation Consequence of the convolution product : when a signal f(t) excites two linear systems in series g 1 (t) and g 2 (t), the response h(t) is equal to h(t) = f(t) (g 1 (t) g 2 (t)) (63) its Laplace transform H(s) is L[f(t) (g 1 (t) g 2 (t))] = F(s) G 1 (s) G 2 (s) (64) The Laplace transform of the impulse response of two linear systems in series is equal to the product of the Laplace transforms of the individual impulse responses. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

55 Transfer function Consider a linear single variable system whose dynamic behaviour is described by in terms of deviation variables by an ordinary differential equation of order n linking the input and the output b u(t)+b 1 du(t) dt + +b m d m u(t) dt m = a y(t)+a 1 dy(t) dt u and y are deviation variables. This system is of order n. Its relative order is n m. Assume that it is initially at steady state. Laplace transformation of (65) gives + +a n d n y(t) dt n, m n (65) (b + b 1 s+ + b m s m ) U(s) = (a + a 1 s+ +a n s n ) Y(s) (66) The transfer function of the system is Y(s) U(s) = G(s) = b + b 1 s+ +b m s m a + a 1 s + +a n s n (67) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

56 Obtaining the transfer function of a system To derive the transfer function of a process from a theoretical model Using the conservation principles, write the dynamic model describing the system, Linearize equations using Taylor series expansion, Express the equations in terms of deviation variables by subtracting the steady-state equations from the dynamic equations Operate Laplace transformation on the linear or linearized equations, Obtain the ratio of the Laplace transform of the output over the Laplace transform of the input. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

57 Application to the Surge Tank Valve on the inlet stream. The state-space model is No linearization necessary. It is linear. ẋ = u/s F/S y = x (68) Y(s) = X(s) = 1 S s U(s) 1 S s F(s) (69) Two transfer functions, both pure integrators. The process transfer function with respect to the input is G u (s) = 1 S s The load or disturbance transfer function with respect to the disturbance F is G d (s) = 1 S s (7) (71) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

58 Block diagram of the Surge Tank From Y(s) = X(s) = 1 S s U(s) 1 S s F(s) (72) obtain the block diagram representing the influence of the input and of the disturbance. F(s) U(s) 1 Ss 1 Ss + Y(s) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

59 Application to the Isothermal Chemical Reactor Nonlinear model. Linearization ẋ 1 = F /S F/S + u 1 /S 1 ẋ 2 = [F (u 2 x 2 ) x 2 u 1 ] k x 2 S x 1 y 1 = x 1 y 2 = x 2 (73) ẋ 1 = δf S δf S + δu 1 S ẋ 2 = 1 S(x1 s [F s )2 (us 2 x 2 s ) x 2 s us 1 ]δx 1+ 1 S x1 s [(u2 s x 2 s )δf + F s (δu 2 δx 2 ) x2 s δu 1 u1 s δx 2] k δx 2 (74) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

60 Application to the Isothermal Chemical Reactor Given the linear model ẋ 1 = δf S δf S + δu 1 S ẋ 2 = 1 [ F s S(x1 s (u s )2 2 ] xs 2 ) xs 2 us 1 δx1 + (75) 1 [ (u s S x1 s 2 x2 s )δf + F s (δu 2 δx 2 ) x2 s δu 1 u1 s δx ] 2 k δx2 Laplace transformation sx 1 (s) = 1 S [ F (s) F(s)+U 1 (s)] sx 2 (s) = 1 [ F s S(x1 s (u s )2 2 ] xs 2 ) xs 2 us 1 X1 (s) k X 2 (s) (76) + 1 [ (u s Sx1 s 2 x2 s ) F (s)+f s (U 2(s) X 2 (s)) x2 s U 1(s) u1 s X 2(s) ] J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

61 Application to the Isothermal Chemical Reactor F(s) F(s) [ ] [ ] Gd11 Gd21 Gd12 Gd22 [ ] U1(s) U2(s) [ ] G11 G G21 G22 [ ] Y1(s) Y2(s) System of transfer functions 1 S s [ ] [ ] Y1 (s) X1 (s) x s ( ) 2 k = = Y 2 (s) X 2 (s) Sx 1 s s + 1 F s [ ] U1 (s) Sx s U s + F s 1 Sx1 s + k s + F 2 (s) s Sx1 s + k 1 S s 1 (77) ( x + 2 s Sx1 s k s + F s F s + ) S s Skxs 1 x2 F s F s k (s)+ Sx s + F s 1 s F(s) s Sx1 s + k s + F s Sx1 s + k J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

62 Impulse Response and Transfer Function of a System Consider a linear system y(t) = u(t) g(t) (78) If this system is excited by a Dirac impulse : u(t) = δ(t), the output becomes y(t) = g(t) (79) In the Laplace domain, U(s) = 1 and Remarks : Y(s) = G(s) (8) A Dirac impulse is not realizable in practice, only an impulse of finite duration is possible. A simple impulse input contains poor characteristics with respect to frequency excitation and introduces difficulty in process identification. In chemical engineering, the impulse response technique (by means of a tracer) is often used to get the residence time distribution. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

63 Principle of superposition When the input can be decomposed into a sum of inputs (e.g. a rectangular pulse is the sum of a positive and a negative step, occurring at different times), the global output is the sum of the responses to the individual inputs U(s) = i U i (s) = Y(s) = G(s) U(s) = i G(s) U i (s) = i Y i (s) The principle of superposition results directly from the properties of linearity of the Laplace transform. The principle of superposition is not valid for a nonlinear system. (81) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

64 Experimental Determination (Identification) of a System Transfer Function Consider a general system described by dy dt Possible means of identification : The input is a Dirac impulse. = f(y(t),u(t),d ) (82) The input is a step or a succession of small amplitude positive and negative steps such as a pseudo-random binary sequence (PRBS). The input is sinusoidal. Identification is performed in the frequency domain. The Laplace variable s is replaced by s = j ω (ω angular frequency in rad/s). After a sufficient time, response y(t) G(j ω) exp(j ω t) = G(s) exp(st) for sufficiently large t (83) It suffices to vary the frequencyω of the input signal over a large range. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

65 Poles and Zeros of a Transfer Function Consider a system described by the differential equation dy(t) d n y(t) a y(t)+a 1 + +a n dt dt n du(t) d m u(t) = b u(t)+b 1 + +b m dt dt m (84) Transfer function expressed as the ratio of two polynomials G(s) = N(s) D(s) = b + b 1 s+ +b m s m a + a 1 s + +a n s n with : n m (85) Roots of the numerator polynomial N(s) are called the system zeros or zeros of the transfer function. Roots of denominator polynomial D(s) are called the system modes or poles of the transfer function. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

66 Stability by poles The behaviour of the system if largely determined by the poles of its transfer function. s 3 Im s 7 s 6 D(s) polynomial with real coefficients. Poles are real or complex conjugate. s 2 s 1 s 4 Re If real, negative (s 1 ), zero (s 4 ) or positive (s 5 ). s 5 If conjugate, negative (s 3 and s3 ), zero (s 7 and s7 ) or s 3 s 7 s 6 Consider a single pole s i. The response is positive real part (s 6 and s 6 ). If s i real, response is exponential. If s i complex, decompose s i = s R + j s I, response is y(t) = c i exp(s i t) (86) y(t) = exp(s R t) exp(j s I t) = exp(s R t)[cos(s I t)+jsin(s I t)] (87) if s i is negative real or complex with negative real part, stable pole. if s i is positive real or complex with positive real part, unstable pole. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

67 Different types of responses 2.5 x y Positive real pole y 5 1 Negative real pole y y Complex pole with positive real part Complex pole with negative real part J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

68 Linear Systems in State Space General SISO linear system in the state space { ẋ(t) = A x(t)+bu(t) y(t) = C x(t)+d u(t) (88) Laplace transformation s X(s) x() = A X(s)+B U(s) (s I A) X(s) = x()+b U(s) (89) hence X(s) = (s I A) 1 x()+(s I A) 1 B U(s) (9) Laplace transform of the output [ ] Y(s) = C X(s) + D U(s) = C(s I A) 1 x()+ C(s I A) 1 B + D U(s) (91) If x() =, transfer function of the system G(s) = Y(s) U(s) = C(s I A) 1 B + D (92) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

69 Solution with Matlab Assume that a system is given by its transfer function G(s). Assume that an input of any type is given u(t). It results U(s). Old method : calculate Y(s) = G(s)U(s), invert Y(s) to obtain the analytical solution δy(t). Numerical method (in Matlab) : from G(s), calculate the state space matrices A, B, C, D, consider the state space model { ẋ(t) = A x(t)+bu(t) y(t) = C x(t)+du(t) (93) where u(t) is known, integrate the differential equations to obtain the numerical solution δx(t) and then δy(t). J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

70 Natural and Forced Responses Consider G(s) system transfer function subject to an input U(s). Except for possible time delays, G(s) is a rational fraction. U(s) can also be expressed as a rational fraction. Decomposition G(s) = Ng(s) D g(s) ; U(s) = Nu(s) D u(s) (94) Y(s) = G(s) U(s) = Ng(s) Nu(s) D g(s) D u(s) = N 1(s) D g(s) + N 2(s) D u(s) = Yn + Y f Response = Natural response + Forced response (95) δy(t) = δy n(t)+δy f (t) y n(t) characteristic of the system. y f (t) characteristic of the type of input. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

71 Dynamic Behaviour of First-Order System System described by first-order differential equation τ dy dt + y(t) = K p u(t) (96) U(s) K p τs+1 Y(s) Corresponding transfer function G(s) = Kp τ s + 1 (97) τ time constant, K p steady-state gain. Assume u(t) is a step function with amplitude A U(s) = A s (98) Y(s) = G(s) U(s) = Kp τ s + 1 A s = A Kp A Kp τ s τ s + 1 = Y f(s)+y n(s) (99) y(t) = A K p (1 exp( t/τ)) = A K p A K p exp( t/τ)) }{{}}{{} y f (t) y n(t) (1) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

72 Dynamic Behaviour of First-Order System y Response of a first-order system (K p = 1,τ = 2) to a unit step function Time τ corresponds to the time necessary for the system response to reach 63.2% of its asymptotic value for a step input. After 3τ, the response reaches 95.2%. Examples of such systems : Systems storing mass, energy or momentum, Systems showing resistance to the flow of mass, energy or momentum. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

73 Dynamic Behaviour of Integrator y System described by differential equation U(s) K p τs+1 Y(s) dy dt = K p u(t) (11) Response of a pure capacitive system (K p = 1) to a unit step function Corresponding transfer function G(s) = Kp s (12) K p steady-state gain. Laplace transform of the output of such a system to a step function with magnitude A Time Time domain response y(t) Y(s) = A Kp s 2 (13) y(t) = A K p t (14) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

74 Dynamic Behaviour of Second-Order System Second-order system described by τ 2 d 2 y(t) dt 2 + 2ζτ dy(t) dt with corresponding transfer function + y(t) = K p u(t) (15) G(s) = K p τ 2 s 2 + 2ζτ s+1 (16) τ natural period of oscillation, ζ damping coefficient, K p steady-state gain. Examples of such systems : Two first-order systems in series, Intrinsic second-order systems, e.g. mechanical systems having an acceleration. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

75 Dynamic Behaviour of Second-Order System Poles of the transfer function are roots of τ 2 s 2 + 2ζτ s+1 = they are 1 s i = τ ( ζ ± ζ 2 1) if : ζ 1 1 τ ( ζ ± j 1 ζ 2 ) = ω n( ζ ± j 1 ζ 2 ) = σ ± j ω a if : ζ 1 (17) The position of the poles depends on the damping ratio ζ. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

76 Dynamic Behaviour of Second-Order System y Time FIGURE : Normalized response of a second-order system to a unit step function for different values of the damping coefficient ζ (=.25 ; 1 ; 1.3 resulting in oscillatory underdamped response to overdamped response) (K p = 1,τ = 1) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

77 Dynamic Behaviour of Second-Order System y 1.5 Rising time Stabilization time 1 A C 5% T a.5 B Time Overshoot = A /B Decay ratio = C/A = (overshoot) 2 Rise time : time necessary to reach the asymptotic value for the first time. First peak reach time : time necessary for the response to reach the first peak. Settling time : time necessary for the response to remain in an interval between ±ǫ (±5%, or ±2%) of the asymptotic value. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

78 Higher-Order Systems Examples : n First-order processes in series (multi-capacitive). For example, a plug flow reactor. Processes with time delay. Processes with inverse response. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

79 Processes with Time Delay Common problem : a transportation lag. (a) (b) FIGURE : Case (a) : In situ sensor. Case (b) : sensor placed in the exit pipe, inducing a transportation lag J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

80 Processes with Time Delay u(t) n-order System y(t) Time Delay y(t t d ) Consider first-order system with a time delay L[y(t t d )] L[u(t)] = K p exp( t d s) τ s+1 (18) y(t t d ) is the delayed output. Padé approximants can be used to transform the nonlinear term in a rational fraction. Processes with time delay pose important difficulties in control. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

81 Processes with Inverse Response Assume two first-order systems with opposite influences. Process 1 U(s) K 1 τ 1 s+1 K 2 τ 2 s+1 Process 2 + Y(s) FIGURE : Representation of a system with inverse response To produce an inverse response, it is necessary and sufficient that the plant transfer function exhibits a positive real zero or a complex zero with a positive real part. Examples exist in distillation columns, fixed-bed reactors. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

82 Processes with Inverse Response 5 4 y1 3 yg 2 y 1-1 y Time FIGURE : Inverse response of the system to a unit step input (K 1 = 5,τ 1 = 2, K 2 = 1.5,τ 2 = 2) J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

83 Identification of a Linear Model Different possibilities : The analytical form of the response is known. Minimize J = i (y mod (t i ) y exp (t i )) 2 (19) with respect to the parameters of the model. In general, nonlinear optimization. Method of Moments J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

84 Identification by Method of Moments The Laplace transform of the impulse response g(t) of a system is its transfer function G(s) = exp( s t) g(t) dt (11) The n-th-order moment of a function f(x) is defined by M n(f) = x n f(x) dx (111) The first two derivatives of G(s) with respect to s G (s) = t exp( s t) g(t) dt ; G (s) = t 2 exp( s t) g(t) dt (112) are related to the moments of the impulse response function by G() = g(t) dt ; G () = t g(t) dt ; G () = t 2 g(t) dt (113) G(), G (), G () are respectively the zero-, first- and second-order moments of the impulse response g(t). J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85

85 Identification by Method of Moments Application : Consider a first-order model with time delay G(s) = K p exp( t d s) τ s+1 (114) It yields G() = K p ; G () = K p (τ+t d ) ; G () = K p (2τ 2 +2τ t d +t 2 d ) (115) from which the three unknown parameters K p, τ and t d can be determined. J.P. Corriou (LRGP) Dynamic modelling Zhejiang University / 85