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Problem 6 (40 points) An underwater vehicle's motion in the yaw plane is described by the following simplified set of linearized equations: L m-Y; -Y. 0 0v Y U (Y-m)U 0 0 v Y U2 uudr -Ný I =- N, 0 0 r NU uv NU ur 00 r = N,s.U2 uuSr r (1.1) 0 0 10 y 1 0 0 U y 0 − 0 0 0 1 0 1 0 0 y 0 Yvdot = - 291.6078; Yrdot = - 16.4006; Mass = 265.3515kg; Nvdot = - 16.4006; Izz = 271.4616kg - m2; Nrdot = - 214.2764; Yuv = - 101.8686; Yur =91.6398; Nuv =- 233.5181; Nur =- 159.2130; Yu2dr = 61.8201; Nu2dr = - 44.6410; In compact form, Equation(1.1) can be written as Mx = Cax+BS, =Cax+ BOrudder where x =[v r y y]' . Equation(1.2) can be expressed in the traditional state space form as A11 A12 0 07 bị b (1.2) *= M-'Cax+M 'BS, = Ax+ Bô rudder 11 A21 A22 0 0 b2 0 1 0 0 A34 x+ 8 (1.3) rudder 0 1 0 0 0 A block diagram of a control system is shown in Figure 1. K Srudder Vemd 4 desired 2 e + 5 + Krp 4 H1(s) 4 H2(s) Figure 1: Schematic of yaw loop control for underwater vehicle (20 points) Compute the transfer functions H1 = rudder (s) i(s) and H2 = Smudder(s) ų(s) in terms of the components of the A and B matrices. Keep in variable form. (25 points) For a specified speed U of 20 knots and commanded yaw, l cmd , of 10 deg, verify that the rudder, Srudder, does not exceed a magnitude of 15 degrees for the following gains: K5 =1, Kwp =0.86,Kmp =- 0.3822. Use Simulink for this purpose and comment on the results and include a plot of the yaw command, the yaw response, and the rudder response on the same set of axes. 2

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