z Transforms IIT Bombay Consider discrete time uniformly - PowerPoint PPT Presentation

Automation Lab z Transforms IIT Bombay Consider discrete time uniformly sampled discrete signal = f k k , ,.... { ( ) : 0 1 } f k z - Transform of { ( )} is defined as ∞ { } ∑ Ζ ≡ = − k f k f z f k z ( ) ( ) ( ) = k 0 where z is a complex va riable. Inverse transform is given as 1 ∫ = π − k f k f z z dz 1 ( ) ( ) 2 i f z where contour integral encloses all singularit ies of ( ). = ≡ ≡ Ts z e Note : and T sampling interval, s Laplace operator 4/10/2012 System Identification 43

Automation Lab z Transforms : Properties IIT Bombay Linearity [ ] [ ] [ ] Ζ α + β = α Ζ + β Ζ g k f(k) ( ) f(k) g(k) Time Shift { } − − Ζ = n n q f z f z (k) ( ) ⎡ ⎤ { } − n 1 ∑ − Ζ = − n n j q f z f z f j z ⎢ ⎥ (k) ( ) ( ) ⎣ ⎦ = j 0 Final Value Theorem Initial Value Theorem - z f z 1 If (1 - ) ( ) does not have any poles lim = f f z on or outside the unit circle, then (0) ( ) → ∞ z lim lim = - f k z f z 1 ( ) (1 - ) ( ) → ∞ → k z 1 4/10/2012 System Identification 44

Automation Lab z Transforms : Examples IIT Bombay Example 1 = α ≥ y(kT) with k 0 (step function) { } − − Ζ = α + α + α + z z 1 2 y(kT) ...... α = − − z 1 1 Example 2 = ≥ y(kT) kT with k 0 (ramp) { } − − Ζ = + + + Tz Tz 1 2 y(kT) 0 2 ...... − − = + + T z z 1 2 ( ....) Tz = − z 2 ( 1 ) 4/10/2012 System Identification 45

Automation Lab Pulse Transfer Function Matrix IIT Bombay + = + x k Φ x k Γ u k ( 1 ) ( ) ( ) Taking z - transform on bothe sides of difference equation ∞ ⎡ ∞ ⎤ ∑ ∑ − − + k = k − x k z z x k z x ( 1 ) ⎢ ( ) (0) ⎥ ⎣ ⎦ = = k k 0 0 ⎡ ⎤ ⎡ ⎤ ∞ ∞ ∑ ∑ − − = + k k Φ x k z Γ u k z ⎢ ( ) ⎥ ⎢ ( ) ⎥ ⎣ ⎦ ⎣ ⎦ = = k k 1 42 0 4 43 4 1 42 0 4 43 4 x z u (z) ( ) = x When (0) 0 we have = + z x Φ x Γ u z (z) (z) ( ) [ ] = I Φ x Γ u z Rearrangin g z - (z) ( ) [ ] − ⇒ = x I Φ 1 Γ u z (z) z - ( ) 4/10/2012 System Identification 46

Automation Lab Pulse Transfer Function Matrix IIT Bombay = y k Cx k ( ) ( ) Taking z transform on both the sides ⎡ ∞ ⎤ ⎡ ∞ ⎤ ∑ ∑ − − = k k y k z C x k z ⎢ ( ) ⎥ ⎢ ( ) ⎥ ⎣ ⎦ ⎣ ⎦ = = k k 1 42 0 4 43 4 1 42 0 4 43 4 y z x (z) ( ) ⇒ = y z Cx z ( ) ( ) [ ] − = = x z I Φ 1 Γ u z y z Cx z Combining ( ) z - ( ) with ( ) ( ) we have { } [ ] − = = 1 y z C I Φ Γ u z G z u z ( ) z - ( ) ( ) ( ) Pulse Transfer Function [ ] Γ − = − 1 G z C z I Φ ( ) 4/10/2012 System Identification 47

Automation Lab IIT Bombay Example: Quadruple Tank System Pulse transfer function model Sampling Time (T) = 5 sec ⎡ ⎤ v z ( ) ⎡ ⎤ 0.2 ⎡ ⎤ 1 h z ⎢ ⎥ ( ) ⎢ 0.01138 z + 0.01034 ⎥ 1 ⎢ ⎥ ⎢ ⎥ z - 0.9233 ⎢ ⎥ 2 ⎢ ⎥ z - 1.734 z + 0.749 = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ v z 0.006045 z + 0.005614 0.1528 ( ) ⎢ ⎥ ⎢ ⎥ h z 2 ⎣ ⎦ ⎢ ⎥ ( ) ⎣ ⎦ 1 2 3 2 ⎢ ⎥ 2 z - 1.793 z + 0.8009 z - 0.9462 ⎣ ⎦ 1 4 4 4 4 4 4 4 4 4 2 4 4 4 4 4 4 4 4 4 3 1 2 3 y z ( ) G z ( ) u (z) Developed using discretization of linearized mechanistic model 4/10/2012 System Identification 48

Automation Lab IIT Bombay Time domain difference Equation -d - 1 -d - 2 b z + b b z + bz = − = d y z z u z u z 1 2 1 ( ) ( ) ( ) + + 2 - 1 - 2 z a z + a 1 a z + b z 1 2 1 2 d = dead time [ ] [ ] + = - - -d- -d- a z 1 a z 2 y z b z 1 + z 2 u z 1 + ( ) ( ) 1 2 1 Taking Inverse z transform on both sides, we get + − − = − − − − y k a y k y k b u k d u k d ( ) ( 1 ) + a ( 2 ) ( 1 ) + b ( 2 ) 1 2 1 2 Linear Difference Equation Model = − − y k y k y k ( ) - a ( 1 ) - a ( 2 ) 1 2 + − − − − u k d b u k d b ( 1 ) + ( 2 ) 1 2 = k where 2 , 3 , 4 ,...... = = y y Initial Conditions : ( 0 ) ( 1 ) 0 4/10/2012 System Identification 49

Automation Lab IIT Bombay Practical Difficulty and Remedy Practical Difficulty A reliable mechanistic model may not be available for the system under consideration. Instead of starting from well developed nonlinear mechanistic model, can we estimate parameters of linear perturbation directly from operating data? Remedy 1.Inject known input perturbations in the system 2.Record measured outputs generated in response to the perturbations 3. Estimate dynamic model parameters using optimization 4/10/2012 System Identification 50

Automation Lab 4 Tank Experimental Setup IIT Bombay Quadruple Tanks Setup Tank 4 Tank 3 Tank 1 Tank 2 Control Control Valve 1 Valve 2 4/10/2012 System Identification 51

Automation Lab Identification Experiments IIT Bombay on 4 Tank Setup Input 2 Input 1 Output 2 Output 1 4/10/2012 System Identification 52

Automation Lab 4 Tank Setup: Input Excitations IIT Bombay Manipulated Input Sequence 1 Input 1 (mA) 0 -1 0 200 400 600 800 1000 1200 1 Input 2 (mA) 0 -1 0 200 400 600 800 1000 1200 Time (sec) 4/10/2012 System Identification 53

Automation Lab 4 Tank Setup: Measured Output IIT Bombay Meaured Output 5 Level 1 (mA) 0 -5 0 200 400 600 800 1000 1200 4 2 Level 2 (mA) 0 -2 -4 -6 0 200 400 600 800 1000 1200 Time (sec) 4/10/2012 System Identification 54

Automation Lab Experimental Data for Modeling IIT Bombay Experiment al data collected Operating Measured Outputs Steady State { } ≡ Y Y Y N Y ( 0 ), ( 1 ),....., ( ) N ≡ Y U ( , ) Known and/or manipulate d Inputs { } ≡ U U U N U ( 0 ), ( 1 ),....., ( ) N Generate Perturbati on Variables Measured Outputs = − y k Y k Y ( ) ( ) { } ≡ y y y N Y ( 0 ), ( 1 ),....., ( ) = − u k U k U N ( ) ( ) Known and/or manipulate d Inputs { } ≡ u u u N U ( 0 ), ( 1 ),....., ( ) N 4/10/2012 System Identification 55

Splitting Data for Automation Lab IIT Bombay Identification and Validation Input and output signals 5 0 y1 -5 0 500 1000 1 0.5 u1 0 -0.5 0 500 1000 Samples Identification Data Validation data 4/10/2012 System Identification 56

Automation Lab IIT Bombay Quadruple Tank: Grey Box Linear Model Let us treat elements of matrices in the discrete Linear state space model as unknowns α α β β ⎡ ⎤ ⎡ ⎤ 0 0 1 2 1 2 ⎢ ⎥ ⎢ ⎥ α α β β γ ⎡ ⎤ 0 0 0 0 0 ⎢ ⎥ ⎢ ⎥ Φ = Γ = = 3 4 3 4 C 1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ α β γ ⎣ ⎦ 0 0 0 0 0 0 0 5 5 2 ⎢ ⎥ ⎢ ⎥ α β ⎣ ⎦ ⎣ ⎦ 0 0 0 0 6 6 Note: We are assuming that structure of matrices is known i.e. we know where zero and non-zero elements are Define parameter vector [ ] T Θ = α α β β γ γ ... ... 1 6 1 6 1 2 4/10/2012 System Identification 57

Automation Lab Model Parameter Estimation IIT Bombay { } = − U u u u N Given set ( 0 ), ( 1 ),.... ( 1 ) , N Θ x ˆ guess for and initial state (0) we can generate state sequence by recursivel y using linear difference equation ( ) = Θ y x ˆ ˆ (0) C ) (0) = Φ Θ + Γ Θ x x u ˆ ˆ (1) ( ) (0) ( ) (0) ( ) (1) = Θ y x ˆ ˆ (1) C ) = Φ Θ + Γ Θ x x u ˆ ˆ (2) ( ) (1) ( ) (1) ( ) (2) = Θ y x ˆ ˆ (2) C ) = Φ Θ + Γ Θ x x u ˆ ˆ (3) ( ) (2) ( ) (2) ( ) (3) = Θ y x ˆ ˆ (3) C ) 4/10/2012 System Identification 58

Automation Lab Model Parameter Estimation IIT Bombay ≡ W Define prediction error Let Diagonal weighting + = − ε y k y k matrix wit h ve elements ˆ (k) ( ) ( ) Θ x ˆ Estimate and initial state (0) by solving following optimizati on problem ( ) Min N [ ] ∑ T Θ = ˆ x ε k W ε k ˆ , ( 0 ) ( ) ( ) Θ x ˆ , ( 0 ) = k 0 S ubject to + = Φ Θ + Γ Θ x x u ˆ ˆ (k 1) ( ) (k) ( ) (k) = Θ y x ˆ ˆ (k) C ( ) (k) = where k 0,1,2,.... ....N - 1 4/10/2012 System Identification 59