unsw ceic3000 w5 process modeling and analysis

unsw ceic3000 w5 process modeling and analysis - CEIC3000...

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CEIC3000 Process Modelling and Analysis Week 5: Nonlinear Analysis Session 1, 2012 Lecturer: A/Prof. J. Bao 1 Phase-plane Analysis (continued) If the eigenvalues of matrix A are complex, the phase plane plots will have the behaviors called “focus” or “center”. Example 1 (Unstable Focus (Spiral Source)) Consider the following system of equations: ˙ x 1 = x 1 + x 2 ˙ x 2 = 2 x 1 + x 2 (1) We have: A = [ 1 2 2 1 ] (2) with eigenvalues λ 1 = 1 + 2 i and λ 2 = 1 2 i. This system is unstable because the real portion of the complex eigenvalues is positive. Example 2 (Center) Consider the following system of equations: ˙ x 1 = x 1 x 2 ˙ x 2 = 4 x 1 + x 2 (3) We have: A = [ 1 1 4 1 ] (4) with eigenvalues λ 1 = 0 + 1 . 732 i and λ 2 = 0 1 . 732 i. Since the real part of the eigenvalues is 0 , this system has a periodic solution, resulting in a phase-plane plot where the equilibrium point is a center, as shown in Figure . The phase-plane plots of second order systems will have the following behaviors: Sinks (stable nodes): λ 1 < 0 and λ 2 < 0, where λ 1 , λ 1 R Saddles (unstable): λ 1 < 0 and λ 2 > 0, where λ 1 , λ 1 R Source (unstable nodes): λ 1 > 0 and λ 2 > 0, where λ 1 , λ 1 R Spirals: λ 1 and λ 2 are complex conjugates. If Re( λ 1 ) < 0 then stable otherwise unstable Center: λ 1 and λ 2 are complex conjugates with Re( λ 1 ) = 0 1
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Figure 1: Phase-plane plot for Example 1 Figure 2: Phase-plane plot for Example 2 2
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1.1 Generalisation of Phase-plane Behaviour Let us generalise the phase-plane analysis results for second-order linear systems: ˙ x = Ax (5) where A = [ a 11 a 12 a 21 a 22 ] (6) The eigenvalues can be found by solving det ( λI A ) = 0 : det ( λI A ) = ( λ a 11 ) ( λ a 22 ) a 12 α 21 = λ 2 ( a 11 + a 22 ) λ + a 11 a 22 a 12 α 21 = λ 2 tr( A ) λ + det ( A ) (7) Therefore the quadratic formula can be used to find the eigenvalues: λ 1 = tr ( A ) (tr ( A )) 2 4 det ( A ) 2 (8) λ 2 = tr ( A ) + (tr ( A )) 2 4 det ( A ) 2 (9) We have: 1. At least one eigenvalue will be negative if tr ( A ) < 0. 2. If tr ( A ) 2 > 4 det ( A ) , the eigenvalues will be real. (a) If det ( A ) > 0: i. If tr ( A ) > 0, λ 1 > 0 , λ 2 > 0 : Node source ii. If tr ( A ) < 0, λ 1 < 0 , λ 2 < 0 : Node sink (b) If det ( A ) < 0 , then one of the eigenvalues is positive and one is negative: saddle 3. If tr ( A ) 2 < 4 det ( A ) , the eigenvalues will be complex In this case, Re ( λ i ) = tr ( A ) . (a) If tr ( A ) > 0: Spiral source (b) If tr ( A ) < 0 : Spiral sink Example 3 A = [ 1 0 0 4 ] tr ( A ) = 5 det ( A ) = 4 Therefore, it is a stable node. 3
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Figure 3: Dynamic behaviour diagram for second-order linear systems. (From Bequette Process Dynamics: Modelling, analysis and simulation. Prentice Hall, 1998) 4
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Figure 4: Slope marks for the vector field (From Bequette Process Dynamics: Modelling, analysis and simulation. Prentice Hall, 1998) 1.2 Slope Marks for Vector Fields A qualitative assessment of the phase-plane behaviour can be obtained by plotting the slope marks for the vector field. Consider: ˙ x 1 = a 11 x 1 + a 12 x 2 (10) ˙ x 2 = a 21 x 1 + a 22 x 2 (11) We can divide (11) by (10) to find how x 2 changes with respect to x 1 : dx 2 dx 1 = a 21 x 1 + a 22 x 2 a 11 x 1 + a 12 x 2
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