EECE 460 Homework, Spring 2007

F. Mrad EECE 460 Spring 2007 Homework # 1 Solution (Exercises 3.7, 8, 9, 11 & 14) B-3-7-. When D(s) = 0, the block diagram of the system can be simplified as follows: The closed-loop transfer function Cp(s)/R(s) can be given by Ge Gr Ge Ge Crls) _ 1+G,G62 4 _ Ge Gi Ga Gs R6) pe 92 Gt G2 Gs Ha /t GjG2H1+ Ge, G2 Gate 1+ GiGa Hy When R(s) = 0, the block diagram of the system shown in Figure 3-76 can be modified as follows: ; ; | DG) Ge +|4,| +L) Ce) Hence Sb) _ Go G; Go Gs I DE) (+6293 Gy (Ge th + HE '+ GG. Gz Ge He + G, Gal F. Mrad EECE 460 Spring 2007 B3-8. There are infinitely many state-space representations for this system. We shall give two of the possible state-space representations. State-space tr 1: From Figure 3-77, we obtain YO. a Stz uN 2 6) +e £ Sp pstpsreZ which is equivalent to Yr pyr ytzysadrzu Comparing this equation with the standard equation Gt tay t4ay= dt hbathithn we obtain =p, 42=/, 43= 2, be =O, 4=0, b= 1, by =z F. Mrad EECE 460 Spring 2007 Define %) =9- fot X= Y- fp ~fUu=%,-fa % =, ~ far where fo = be =o Aah ~4, fo =0 fz = b.- 4, fy -42 fo =/ fs = b3 — 4% fa 4a By 43 fe = F-p Then, state-space equations can be given by ep fe ¢ Oily) 14 4] =] 9 Qe / |e] +] Ba | ee % -z -/ —p 3 Bs % yuls o Olle + faX X3 or 4 J / oO x o RIl=) oO. hhel + / 4%] |-? -/ -- Ile z~p & x yall 0 Olle % State-space representation 2: Since St+2 = StPt+z-p = /+ r-P S+pP S+p S+p we can redraw the block diagram as shown below. F. Mrad EECE 460 Spring 2007 U e-p | % VIL “se O41 From this block disgram we get the following equations: Va=0-X, + x3 X3 = 2-? U- xX; S+P —_ *~2 - 0-X, + Xs s * a Xo s from which we obtain 4a + pXs = (2-P)4-@-p) x, Xa MA +H % = 22 Rewriting, we have %, = 2% = Ut AztH Xy = — (2-p) X, —p%s +(2-p)a = %, 4, Qo ff OTTx, 0 4 =} -/ 0 / ait] / [4 %| [P-? 9 —plixg| | 2- P F. Mrad EECE 460 % = C7 Qo og] XR % Spring 2007 B39. eee oy . a+sy+2gaX% Define %) = 4 %=% %=Y Then %t3t3 + 2%=K LHX Xam or x o ft On a “I=]° 0 Sfx tol” 4] fo -2 3G} [7 % a=[/ © o]l% a3 B-3-11 A= we [Fc]. a[2], ee a The transfer function G(s) of the system is given by a= cerarany al i) ] -3 / stl -/ 2 = [’ 2] SS (S+5)(s4/) +3 3 sts | |S _ / [v 2] 25-3 ] _ _ /254+59 ~ St4+ 65th SS+3/ S*+6St+ A MATLAB solution to this problem is given below. A=[5 -13 -1]) B= [2;5]; C=[l 2]; D=0; (num, den] = ss2tf{A,B,C,D) F. Mrad EECE 460 Spring 2007 B-3-14. (a) MK EAL RL xX) __/ UG) ms*+k (b) Define the displacement of a point between springs ky and k2 as y. Then the equations of motion for this system become mx +h, (X-y) mee hy =A (*-g) From the second equation, we have Aythey =hkrx% or -—# x += kt ke i hike Mat or