{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

midsol04 - F 2 =-2 2 5-1 1-1 2 F 3 =-2-2-2-1-1 1-1 2 G 1...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem 1. (15 points) Consider the system ˙ x 1 = x 1 ( x 1 + x 2 ) ˙ x 2 = - x 3 + 4 + x 2 - x 2 2 - x 1 x 2 ˙ x 3 = x 2 + 1 - ± 1 2 + u ² x 3 Denote the state vector as X = [ x 1 , x 2 , x 3 ] T . For u = 0, fnd all the equilibria o± the system. Then fnd the linearization at all those equilibria. (This model is inspired by a model o± the rotating stall instability in axial ²ow compressors that are used in gas turbine/jet engines. The variables, (very) roughly, are: x 3 -pressure rise, x 2 -mean ²ow through compressor, x 1 -the amplitude o± the rotating stall instability, and u - control through downstream valve/throttle.) Solution: Equilibria o± the system: x 1 ( x 1 + x 2 ) = 0 - x 3 + 4 + x 2 - x 2 2 - x 1 x 2 = 0 x 2 + 1 - x 3 2 = 0 From the top equation we have: 1) x 1 = 0 x 2 2 + x 2 - 2 = 0 x 2 = 1 and x 2 = - 2 x 3 = 4 and x 3 = - 2 2) x 1 + x 2 = 0 x 2 - 2 = 0 x 2 = 2 x 3 = 6 , x 1 = - 2 So, we have 3 equilibria: X 1 = [0 1 4] T , X 2 = [0 - 2 - 2] T , X 3 = [ - 2 2 6] T . ˙ X = f ( X, u ) , ∂f ∂X = 2 x 1 + x 2 x 1 0 - x 2 1 - x 1 - 2 x 2 - 1 0 1 - 1 / 2 - u , ∂f ∂u = 0 0 - x 3 Now we can write ±or equilibrium X i : δ ˙ X i = ∂f ∂X ³ ³ X = X i δX i + ∂f ∂u ³ ³ X = X i δu = F i δX i + G i δu , i = 1 , 2 , 3. F 1 = 1 0 0 - 1 - 1 - 1 0 1 - 1 / 2
Background image of page 1

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

View Full Document Right Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: , F 2 = -2 2 5-1 1-1 / 2 , F 3 = -2-2-2-1-1 1-1 / 2 G 1 = -4 , G 2 = 2 , G 3 = -6 1 Problem 2. (10 points) Consider the circuit 1 2 3-+ +-Find the state representation for this system. Solution: KVL at node 2: y = v C 1 + u KCL at node 2: C 1 dv C 1 dt + v C 1 R 1 + y-v C 2 R 2 = 0 KCL at node 3: C 2 dv C 2 dt-y-v C 2 R 2 = 0 We get C 1 ˙ v C 1 + 1 R 1 v C 1 + 1 R 2 ( v C 1 + u-v C 2 ) = 0 C 2 ˙ v C 2 + 1 R 2 ( v C 2-v C 1-u ) = 0 y = v C 1 + u ˙ ± v C 1 v C 2 ² = "-1 C 1 ³ 1 R 1 + 1 R 2 ´ 1 C 1 R 2 1 C 2 R 2-1 C 2 R 2 # ± v C 1 v C 2 ² + ±-1 C 1 R 2 1 C 2 R 2 ² u y = µ 1 ¶ ± v C 1 v C 2 ² + u 2...
View Full Document

{[ snackBarMessage ]}

Page1 / 2

midsol04 - F 2 =-2 2 5-1 1-1 2 F 3 =-2-2-2-1-1 1-1 2 G 1...

This preview shows document pages 1 - 2. Sign up to view the full document.

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