Download Solutions to Computer-Aided Control System Design Exam: Problem 1 and 2 - Prof. Craig A. W and more Exams Aerospace Engineering in PDF only on Docsity! AOE 4004 Computer-Aided Control System Design Exam #1 Solutions Problem 1. (45 points) Figure 1 shows two equal masses which are coupled to a fixed frame through linear springs. The masses are coupled to each other through a nonlinear spring. The kinetic and potential energy are T = 1 2 mẋ21 + 1 2 mẋ22 V = 1 2 kx21 + 1 2 kx22 + 1 4 kn (x2 − x1) 4 . There are no damping or control forces in this system. (a) Use Lagrange’s equations to derive the equations of motion. (You should obtain two second order, nonlinear, ordinary differential equations.) (b) Rewrite the equations derived in part (a) as four first order differential equations and linearize them about the equilibrium at the origin to obtain a system of the form ẋ = Ax. m k x m kn k 1 x 2 Figure 1: Sketch for Problem 1. Solution. To apply Lagrange’s equations to the Lagrangian L = T − V , we compute ∂L ∂ẋ1 = mẋ1 and ∂L ∂ẋ2 = mẋ2 and also ∂L ∂x1 = − ( kx1 − kn (x2 − x1) 3 ) and ∂L ∂x2 = − ( kx2 + kn (x2 − x1) 3 ) . The equations d dt ∂L ∂ẋ1 − ∂L ∂x1 = 0 d dt ∂L ∂ẋ2 − ∂L ∂x2 = 0 give mẍ1 + kx1 − kn (x2 − x1) 3 = 0 mẍ2 + kx2 + kn (x2 − x1) 3 = 0. Define z = [x1, ẋ1, x2, ẋ2] T . We have d dt z1 z2 z3 z4 = z2 − k m z1 + kn m (z3 − z1) 3 z4 − k m z3 − kn m (z3 − z1) 3 . Linearizing about the equilibrium zeq = 0 gives ż ≈ 0 1 0 0 − k m − 3kn m (z3 − z1) 2 0 3kn m (z3 − z1) 2 0 0 0 0 1 3kn m (z3 − z1) 2 0 − k m − 3kn m (z3 − z1) 2 0 eq z = 0 1 0 0 − k m 0 0 0 0 0 0 1 0 0 − k m 0 z Note that the linearized dynamics represent two completely decoupled mass-spring systems. For small oscillations, the nonlinear spring connecting the two masses has little effect on the dynamics. Problem 2. (30 points) Using the semilog paper on the last page, sketch the Bode diagram for the following transfer function: G(s) = 1, 200, 000 ( (s + 0.5) (s2 + 2s + 100) (s + 600) ) −40 −20 0 20 40 60 M ag ni tu de ( dB ) 10 −2 10 −1 10 0 10 1 10 2 10 3 10 4 −180 −135 −90 −45 0 45 90 P ha se ( de g) Bode Diagram Frequency (rad/sec)