unsw ceic3000 w5 process modeling and analysis

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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 , 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 Figure 1: Phase-plane plot for Example 1 Figure 2: Phase-plane plot for Example 2 2 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 > , λ 2 > 0 : Node source ii. If tr ( A ) < 0, λ 1 < , λ 2 < 0 : Node sink (b) If det ( A ) < , 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 − 4 ] tr ( A ) = − 5 det ( A ) = 4 Therefore, it is a stable node....
View Full Document

This note was uploaded on 03/26/2012 for the course CHEM ENG CEIC at University of New South Wales.

Page1 / 14

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online