Why do we need a model?

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Morris Preston
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1 Why do we need a model? For simulation (study the system output for a given input) Ex. Thermal study of a space shuttle when it enters the atmosphere (étude thermique concernant la rentrée dans l atmosphère d une navette spaciale) For design (compute the system parameters to have a desired output for a given input) Ex. Design of electrical, mechanical or chemical installations (dimensionnement d une installation électrique,...) For prediction (forecast the future values of the output) Ex. weather forecasting ; flood forecasting (prévision météo, prévision de crue) For control (model-based controller design) Ex. Pole placement controller design for tracking and disturbance rejection Plants, Systems and Models (Chapter 1) System Identification Fall / 25

2 How can we find a model? First principle modeling Based on physical laws, physical models (modèles de connaissance) G(s) = K s(τs +1) (continuous-time models, pedagogical interest) System identification Based on input/output measured data (modèles de représentation) G(z) = bz z 2 + a 1 z + a 2 (discrete-time models, practical interest) Plants, Systems and Models (Chapter 1) System Identification Fall / 25

3 How can we find a model? Physical modeling R m L m u m i m E m M θ J c f E m = K m θ M m = K m i m (t) Assumptions : only viscous friction, backlash and self inductance L m neglected (frottement visqueux, néglige les jeux et l inductance de l induit) (J m + J c ) θ(t)+f θ(t) =K m i m (t) u m (t) =R m i m (t)+k m θ (J m + J c )R m θ(t)+(fr m + K 2 m ) θ(t) =K m u m (t) [ (Jm + J c )R m s 2 +(fr m + K 2 m )s] Θ(s) =K m U m (s) Θ(s) U m (s) = K m (J m + J c )R m s 2 +(fr m + K 2 m)s model : G (s) = K s(τs +1) Plants, Systems and Models (Chapter 1) System Identification Fall / 25

4 How can we find a model? System identification Cement Plant : Crusher & Separator Loop fraîches q(t) ciment c(t) Broyeur Crusher Séparateur refus Physical model (modèle de connaissance)? Identification : Apply a specific input q(t) and measure the output c(t) and represent it as a function of preceding values of input and output. c(t) =G[q(t), q(t 1), q(t 2),...,c(t 1), c(t 2),...] bz (modèle de représentation) G (z) = z 2 + a 1 z + a 2 Plants, Systems and Models (Chapter 1) System Identification Fall / 25

5 Physical or identified model? 1) Physical models (modèles de connaissance) G(s) = K m (J m + J c )R m s 2 +(fr m + K 2 m )s Direct relation with physical parameters High order, approximative, need complete process knowledge, physical parameters should be known 2) Identified models (modèles de représentation) bz G (z) = z 2 + a 1 z + a 2 Appropriate for controller design, simple and efficient Limited validity (operating point, type of input), sensors, measurement noise, unknown model structure Plants, Systems and Models (Chapter 1) System Identification Fall / 25

6 Type of models dynamic/static monvariable/multivariable deterministic/stochastic lumped parameter (à paramètres localisés)/distributed parameter (à paramètres répartis) linear/nonlinear time-invariant (stationnaire)/time-variant (nonstationnaire) causal/noncausal zero initial condition (à repos) /nonzero initial condition LCSR System : Linear Time-Invariant Causal System with zero initial condition Plants, Systems and Models (Chapter 1) System Identification Fall / 25

7 c +2c c 0 c =3q(t θ) =3 q(t θ)+3q 0 Plants, Systems and Models (Chapter 1) System Identification Fall / 25 Example : Cement mill (Moulin à ciment) q(t) System c(t) ċ(t)+2c 2 (t) =3q(t θ) c(0) = c 0 System : dynamic, monovariable, deterministic, lumped parameter, nonlinear, time-invariant, causal, nonzero initial condition. dynamic/statique? monovariable/multivariable? deterministic/stochastic? lumped parameter(à paramètres localisés)/distributed parameter? linear/nonlinear? time-variant/time-invariant (stationnaire)? causal/noncausal? zero/nonzero initial condition? Linearization? Linearization Taylor Series : f (x) f (x 0 )+f (x 0 )(x x 0 ) c 2 (t) =c c 0(c(t) c 0 ), c = c(t) c 0, q(t) =q(t) q 0

8 Type of representations input/output representation (représentation externe) state-space representation (représentation interne) time-domain representation (représentation temporelle) frequency-domain representation (représentation fréquentielle) continuous-time representation (représentation continue) discrete-time representation (représentation discrète) Example input/output time-domain continuous-time representation : y(t) =H[u(τ)] <τ t state-space time-domain discrete-time representation : x(k +1)=f [x(k), u(k), k] state equation x(k 0 )=x 0 y(k) =g[x(k), u(k), k] output equation Plants, Systems and Models (Chapter 1) System Identification Fall / 25

9 Input/output models Time-domain continuous-time representation y(t) =H[u(τ)] <τ t H : a causal linear operator Consider the impulse input signal at τ (impulsion de Dirac) : δ(t τ)dt =1 ; Convolution integral : u(t) = y(t) =Hu(t) =H u(τ)δ(t τ)dτ f (t)δ(t τ)dt = f (τ) u(τ)δ(t τ)dτ = δ(t-τ) t 0 τ Hδ(t τ)u(τ)dτ Define impulse response : g(t) Hδ(t) g(t τ) =Hδ(t τ) y(t) = g(t τ)u(τ)dτ = g(t) u(t) =u(t) g(t) Plants, Systems and Models (Chapter 1) System Identification Fall / 25

10 Input/output models Time-domain discrete-time representation y(k) =H[u(h)] < h k H : a causal linear operator For an LSCR system : k y(k) = g(k j)u(j) =g(k) u(k) =u(k) g(k) j=0 g(k) : discrete { impulse response (response to discrete impulse δ(k)) 1 k =0 δ(k) = 0 k 0 Frequency-domain representation Continuous-time : y(t) =g(t) u(t) Y (s) =G (s)u(s) Y (jω) =G(jω)U(jω) Discrete-time : y(k) =g(k) u(k) Y (z) =G (z)u(z) Y (e jω )=G (e jω )U(e jω ) Plants, Systems and Models (Chapter 1) System Identification Fall / 25

11 Relation between g(k) andg(t) δ(k) δ (t) δ(t) T 2T kt T t 0 T t δ(k) Surface = T Surface = 1 δ(k) δ*(t) g*(t) g(k) N/A G(s) A/N G(z) δ(t) g(t) g(t) G(s) A/N kt lim T 0 1 T g (t) =g(t) g(k) Tg(t) kt Plants, Systems and Models (Chapter 1) System Identification Fall / 25

12 Continuous- to discrete-time Continuous-time approach Continuous-time model G(s) continuous-time controller K(s) discrete-time controller K(z) for implementation (T small as possible) Discrete-time approach Continuous-time model G(s) discrete-time model G(z) (T according to Shannon s theorem) discrete-time controller K(z) Remark Discrete- to continuous-time conversion is used when G(z) is available (by identification) and we need G(s) for continuous-time design method or physical parameter identification. Plants, Systems and Models (Chapter 1) System Identification Fall / 25

13 Relation between G(s) andg(z) sortie u 1 (k) = u 2 (k) u 1 (t) u 2 (t) y 1 (k) y 1 (t) y 2 (t) y 2 (k) 0 T t 0 T t u(t) G(s) y(t) u 1 (k) =u 2 (k) y 1 (k) y 2 (k) G 1 (z) = Y 1(z) G 2 (z) = Y 2(z) U 1 (z) U 2 (z) A/N u(k) G(z) For a given G(s), G(z) depends on input u(t) A/N y(k) Plants, Systems and Models (Chapter 1) System Identification Fall / 25

14 Find G(z) fromg(s) G(z) = Y (z) U(z) = Z{y(k)} Z{u(k)} = Z{L 1 [G(s)U(s)] kt } Z{L 1 [U(s)] kt } u(t) is unit step (U(s) = 1/s) G 1 (z) = Z{L 1 [G(s)/s] kt } 1 1 z 1 =(1 z 1 )Z{L 1 [G(s)/s] kt } u(t) is unit impulse (U(s) = 1) U(z) =Z { L 1 [1] kt } = Z { δ(t) kt } = lim G 2 (z) = T Z{L 1 [G(s)] kt } T 0 1 G 1 (z) is not equal to G 2 (z) T Z{δ(k)} 1 T Plants, Systems and Models (Chapter 1) System Identification Fall / 25

15 Find G(z) fromg(s) Conversion with Zero-Order-Hold Input is a linear combination of steps. u(k) u(t) y(t) y(k) N/A G(s) A/N G(z) G(z) = Z{L 1 [G(s)/s] kt } 1 1 z 1 =(1 z 1 )Z{L 1 [G(s)/s] kt } G(s) G(s) s L 1 γ(t) sampling γ(k) Z G(z) 1 z 1 (1 z 1 ) G(z) Plants, Systems and Models (Chapter 1) System Identification Fall / 25

16 Find G(z) fromg(s) Example Find G(z) fromg(s) = 1 s +1 T =0.2 byzohmethod. [ } G(z) = (1 z 1 )Z {L 1 1 s(s +1)] { [ kt 1 = (1 z 1 )Z L 1 s 1 } (s +1)] [ kt ] = (1 z 1 1 ) 1 z e T z 1 = 1 1 z 1 1 e T z 1 = (1 e T )z 1 1 e T z 1 = 0.18z z 1 Plants, Systems and Models (Chapter 1) System Identification Fall / 25

17 Find G(z) fromg(s) Conversion using impulse response G(z) = T Z{L 1 [G(s)] kt } Example G(s) L 1 g(t) sampling T Z g(t) kt g(k) G (z) Find G(z) from G(s) = 1 T =0.2 by impulse response method. s +1 { [ } 1 G(z) =T Z L 1 1 = T s +1] kt 1 e T z 1 = z 1 Plants, Systems and Models (Chapter 1) System Identification Fall / 25

18 Find G(z) fromg(s) Zero-Pole matching (Transformation des pôles et des zéros) s Im ω s 2 z = e Ts Im z Three-step procedure : 0 R e R e ω s 2 1 Find discrete poles and zeros : p d = e Tpc z d = e Tzc. 2 Add zeros at z = 1 to have the same number of poles and zeros. 3 Adjust the static gain Plants, Systems and Models (Chapter 1) System Identification Fall / 25

19 Find G(z) fromg(s) Example Find G(z) from G(s) = 1 s +1 and T =0.2 by zero-pole matching. 1 Find discrete poles and zeros : p c = 1 p d = e Tpc = e 0.2 = Add one zero at z = 1. 3 Adjust the static gain. 1+z 1 G(z) =α z 1 K c = lim sg(s) 1 s 0 s = lim G (s) =1 s 0 K d = lim (1 + z 1 1 2α )G (z) = lim G (z) = z 1 1+z 1 z z 1 K c = K d α =0.09 G (z) = z 1 Plants, Systems and Models (Chapter 1) System Identification Fall / 25

20 Find G(z) fromg(s) by numerical integration 1. Forward integration Suppose that G(s) = Y (s) U(s) = 1 s then ẏ(t) =u(t). y(k +1) y(k) T = u(k) Example z 1 T Y (z) Y (z) =U(z) U(z) = Find G(z) from G(s) = 1 s +1 T z 1 s : z 1 T and T =0.2 by forward integration. s : z 1 T G(z) = 1 z 1 T +1 = T z 1+T = 0.2z z 1 Plants, Systems and Models (Chapter 1) System Identification Fall / 25

21 Find G(z) fromg(s) by numerical integration 2. Backward integration (intégration rétrograde) Example ẏ(t) =u(t) y(k) y(k 1) T 1 z 1 Y (z) =U(z) Y (z) T U(z) = Find G(z) from G(s) = 1 s +1 s : z 1 Tz G(z) = = u(k) T 1 z 1 s : z 1 Tz and T =0.2 by backward integration 1 z 1 Tz +1 = Tz z 1+Tz = z 1 Plants, Systems and Models (Chapter 1) System Identification Fall / 25

22 Find G(z) fromg(s) by numerical integration 3. Trapezoid integration (intégration trapézoidale) { y(k+1) y(k) T = u(k) y(k+1) y(k) T = u(k +1) y(k +1) y(k) T = 1 [u(k)+u(k +1)] 2 z 1 T Y (z) =1 Y (z) T (z +1) (1 + z)u(z) = 2 U(z) 2(z 1) s : 2 z 1 T z +1 Example Find G(z) from G(s) = 1 s +1 s : 2 T z 1 z +1 G(z) = 1 2(z 1) T (z+1) +1 = and T =0.2 by trapezoid integration. Tz + T 2z 2+Tz + T = z z 1 Plants, Systems and Models (Chapter 1) System Identification Fall / 25

23 Find G(z) fromg(s) by numerical integration Remark Forward integration : Backward integration : z = Trapezoid integration : z =1+Ts(z = e Ts 1+Ts) 1 1 Ts (z = ets = 1 e Ts 1 1 Ts ) z = 1+Ts/2 1 Ts/2 (z = ets = ets/2 1+Ts/2 e Ts/2 1 Ts/2 ) Plants, Systems and Models (Chapter 1) System Identification Fall / 25

24 Find G(s) fromg(z) The inverse of all methods can be used for conversion from discrete to continuous-time models. zoh impulse response backward integration Tustin transformation { [ ]} G(z) G(s) =sl Z 1 1 z 1 G(s) = 1 T L{Z 1 [G(z)]} G(s) =G(z) z= 1 1 Ts G(s) =G(z) z= 1+Ts/2 1 Ts/2 Plants, Systems and Models (Chapter 1) System Identification Fall / 25

25 Comparison of models Externe Continue Continue Interne c y(t) = h[u(τ), τ ] - < τ t c y(k) = h[u(h), h] - <h k. x(t) = f[x(t), u(t), t] x(0) = x 0 y(t) = g[x(t), u(t), t] c x(k+1) = f[x(k), u(k), k] x(0) = x 0 y(k) = g[x(k), u(k), k] Temporelle lsc lsc lsc lsc g ij (t) t y i (t) = g ij (t-τ) u j (τ) d τ - g ij (k) k y i (k) = g ij (k-h) u j (h) h=-. x(t) = Ax(t) + Bu(t) x(0) = x 0 y(t) = Cx(t) + Du(t) x(k+1) = A x(k) + Bu(k) x(0) = x 0 y(k) = Cx(k) + Du(k) lscr G ij (s) = L[ g ij (t)] Y(s) = [G ij (s)] U(s) lscr G ij (z) = Z[ g ij (k)] Y(z) = [G ij (z)] U(z) lscr -1 X(s) = (s I - A) B U(s) Y(s) = C X(s) + D U(s) lscr X(z) = (zi - A) -1 B U(z) Y(z) = C X(z) + D U(z) Plants, Systems and Models (Chapter 1) System Identification Fall / 25

LTI Systems (Continuous & Discrete) - Basics

LTI Systems (Continuous & Discrete) - Basics 1. A system with an input x(t) and output y(t) is described by the relation: y(t) = t. x(t). This system is (a) linear and time-invariant (b) linear and time-varying