Question

1. Test the controllability and observability of the system

2. write the MATLAB code of the solution

MECH621 FINAL PROJECT- Project The dynamics of a controlled submarine are significantly different from those of an aircraft, missile, or surface ship. This difference results primarily from the moment in the vertical plane due to the buoyancy effect. Therefore, it is interesting to consider the control of the depth of a submarine. The equations describing the dynamics of a submarine can be obtained by using Newton's laws and the angles defined in Figure 1. To simplify the equations, we will assume that e(t) is a small angle and the velocity v is constant and equal to 25 ft/s. The state variables of the submarine, considering only vertical control, are x1(t) = 0 (t),xz(t) = e(t) and xz(t) = a(t), where act) is the angle of attack al foci Velocity v Center of gravity 8(1 Control surface Fig.1: Submarine depth control Thus the state vector differential equation for this system, when the submarine has an Albacore type hull, is *(t) = 0.01 0 1 0 [x1(1) -0.11 0.12 (0) + 0.07 -0.31 |x3(0) | u(t) 0.1 [xi(t) y(t) = [1 0 0 *r(t) |x3(0)|

Answer 1

Ans The state space representation can be used to obtain the transfer function. we know that Transfer function Gp(s) = Y(S) output u(s) Input General representation of state space medel is. X = AX BU } Y = cx +Du Apply Laplace tranform to SxCs) Ax(s) + BUCS) - 06) Y(S) - CX (S) + Do(s) + C) XCS) (51-A) = BUCs) xCs) = (52-A)" BUCS) d) sub (d) in (c) YGS) = C($I_A) BUCS) +D NGS) + U (S) (CCSI-A)'B +) YCS) YCS) Jos) c(SI-A)%B+D. (51 A)" is is the The transfer function matrin PCs). Ø (s) = (SIA)

6) Given BO - XLt) = f O -0.01 -0.11 0.12 nicy "C) & 904 1-ool uly 0.07 -0.3 -0.3 혹 B A X MCH YCH = (1 (10 00 C my CY M3 CH 7 Transfer Anatvin 0(5) - (57-4]? adi (SIA) Al SI ک 1 (SIAL=1s -oool stooll 0.12 0.07 S+03 (53A= ${$+0.1) (st03).–6-02) (0-12) -1 (-0.01(+0.3) کم to to 0.415+0.033 -0.0084) +0.018+ 0.003 . 510.4155+ 0.02465 +0.015+0.003. ISIA] = 53+0.415 +0.0346 5 +0.003 o S adi/ SI-AL 0:12 -0.11 Stoll -0-01 Stool X 0.02 0 0.07. S+0.3 Vio S D S -0.01 Stoll -0,01 Sto.ll 0.12

$10.415 +0.0246 -$+0-3) 0012 0.018 +0.003 s(s+03) -0.123 -7x103 -0.07s s +0.1157001 p(s)= adi (SIA) (SIA) s10.415+0.0246 -(5+0.3) 0:12 - $310.415+0.03485 0:01 5 +0.003 s(s+0:3) -0.125 to.003 -0.075 s+0./1570.01 -7x103 from state space representation. n = 2(t) ring --0-017,C4 -0.11n C4 +0.1273 CH -oluch öz = 0.07x,C4 -0.3YCH + 0.lucy y (H = MICH Note D > Integrated since ☆ & ý are associated with input we have to place ☆ & ų near the input. -ool ult) ule ng 93 012 B M₂ ycy. -0.11 ܨܬf - ccur -001 0.07 Mely.