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SolutionToHW1

# SolutionToHW1 - 1.1 A system has an input ntr and an output...

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Unformatted text preview: 1.1 A system has an input ntr) and an output y(i), which are related by the information provided below‘ Classify each system as linear or nonlinear and time-invariant or timenvarying, a] y(t) = 0 for all I. Solution: Linear, Time-Invariant b} y{1]=a,a¢0,forall.r. Solution: Nonlinear,Time-invariant -—-=_,_.. c) y(t)=—3u(r)+’_’. Solution: Nonlinear,Time-Invariant -—m-_.- d) J'U} Solution: Nonlinear, Time-Invariant W} NU) Solution: Ngrﬁlucar,Time-Invariant f) Eifl+3e"i’(fl+yif)=uitl. Solution: Linear, Time Varying g) y(:)+(1— y2(r))jr(t)+tn3y(t) ='u{z), w = constant e 0. Solution: Nonlinear. Time Varying h} y{f)=u(f-3)- Solution; Linear, Time-Invariant 5) ji(f)+ WWO) = 0_ Solution: Nonlinear‘TimeJnvariant ___.. ..___—- I j) yU) = j 6‘14“ - TJdT_ o 1 Solution: Linear, Time-Varying. {Often considered time-invariant for signals constrained to I)- U . Can be made time-invariant if lower limit of integration is -m .) k) yUc + 2) = -0.4y(.k +1)« you + 3n}: + 2) — ”(11). k I) yUc) = 25in(iT}e_'ru(k — 1'), T = constant a o. 1&0 Solution: Linear, Time-Varying: (Often considered time-invariant for signals canstralned to k z: 0 . Can be made time-Invariant if lower limit of summation is —oo .) 1.2 Figure P12 shows a model commonly used for automobile suspension analysis. In it, the uneven ground specifies the position of the wheel's contact point. The wheel itself is not shown, as its mass is considered negligible compared to the mass of the rest of the car. Write a differential equation and a state variable description for this system, considering the height of the car, x0}, to be lite output, and the road height, y(_tJ , to be the input. PLZ Solution: The force equation is . B K B. K mx“: :3 '—' K _. _ :>"+_.j+_x=_ 1+._ + . Ej'orces (y x)+ {y x2) mg x m m m} my g Let gnu—ﬂ, then §+£t§+—§'= £,i'+£y. K m m m in Let £1 = .5 and £2 = 4;, then we can form similar equation and simulation diagram ofFig. 1.7 in the text; = K B + Mr) 5,; _; ’E 52 1 El x(t‘) = [13 E] m m L9 Given the state variable description of the system in terms of LU] below, change the state variables and write new state equations for the variables §1(r}=3xl(r)+2x2(r}, and EH“) = 7xl(f)+5x2(:)‘ gm _ 11 m x.(r) + 02‘0“} 122(1) ' —21 ~13 3:30) 1 y(r)=[2 3][x'[r)]+zi(r} x30) Solution: , From £10] = 3x1(r)+ 21-20) and a5!) = 7x,(:}+5x2(r) EH: Elli] ball: ill; Ill: il'liiiilli :llw 3 2 " J.-[:)=[2 2][7 5:1 [:li:)]:|+h{t] Thus, Simpliﬁ above equations in) , -3 ' 0 £10) 2 [mHn -4][é2m]+[5]“m 1.10 For the state variable description of the system in terms of no“) , 5:10) 18 9 13 x§(r} *1 5:20) = so 23 35 mg) + o ufr) 5:30) —65 —31 —45 x3“) I change the state variables and write new state equations for variables §1(:')= —4x1{r)—2x1(1)—3x3{£} §2(:J=15xl(t)+ 7x2{!)+10x3(1) and :30} = -5x1(f]—- 2x2UJ ~3x3(£) Solution: «*IU) 18 9 13 x10) —l .920) = 50 23 35 14:26) + 0 111(1) EAX+bu :230} —65 —31 ~46 x30} 1 x1“) y(!)=[5 —5 51x20) 5 53(1) Now,let g, —4 —2 —3 x1 £,:.=,2=157'10x2 5M): 5,; “5 -—2 -—3 1:; So, x: M"§ and 5c: M'li. Thus, M"E = And—lg + bu g: MAM’le, + Mbu y=CM"§ 3 wow”; Deﬁne —2 1 0 32:14AM": 0 -2 0 Sam: 0 o ‘1 E =C‘M" =[—45 —25 ~40] 1.]! The robot shown in Figure Pl.11 has the differential equations of motion given. Symbols with»:2 , I, , 12:31 , and g are constant parameters, representing the characteristics of the rigid body links. Quantities B, and d: are the coordinate variables and are functions of time. The inputs are 1, and 12 . Linearize the two equations about the operating point Ell = 9. == 6] :0, d2 : 3, and d2 = 3: = n. PL}! (mlcf + +1: +m1d§ )é] -. irrigdzélﬁz +(mli’. +mztig )gcosEJ, = 11 ____ f .. .1 . nlgdg—in2d331‘+rttlgsme! = 13 Solution: Consider separate terms in the ﬁrst equation: 0 0 mzdgél z HEW + Zinzgzéflaéjwz —3)+ held-EL dd)! —0} =9m2i§l §1=0 ' 91=0 2 4'0 ’0 2m1d3é3133 2-. 2%" ili‘ilzidz-3 +2’”2W2=o(d2 “ 3) (not? + 1] +12%: is already linear. d1=0 31:0 0 @143 5" / +2n12d ' we. — mar} idaétlapoﬁi’: _ 0} = o J3=0 ‘ (31:0 9 ”nag-20561:: mlﬂlgcose‘laFo jmlfigsd/na’l/W;0(Bl "0) = ”1,913 mldzgcosﬁl 2-. mzdzgcoSBIIZII-s: + mggcosellelmw‘l -3) - mzdzgsinell:;:t;[81- 0) = 3m2g+ Hodge-3mg = ”12053 _.__.- SO,(m1€12”rt“+r’2+9m3}95+mlflg+mzdzg=‘rl . :5 But. this is not linear, because of the mzﬂlgterrn. We can eliminate constant terms with a change of variables. At equilibrium, the original DE. gives (”1191 + 3m1)g = 1,3 . "equilibrium" Deﬁne, 59: 59, 41;. =9, 50 59. Eélﬁél Eé. Gd; 5d; Larza': d2 —3 so 532 \$32,529: Earl 51' 31:: ~11: =11 ~mlflgn3ng and also, . d2 =¢Sd2 '43 “£1 =51! +mlelg+3mgg Then, (mlff +1] +11 +9m2)6ﬁ1 +:rrr,1?1g+nag-Em!2 +3m2g =511+mlﬂlg+3m3g Leaving I (m.e,1+ :5 +12 +931»:2 we}1 +ngﬁd2 :5“ Second equation ":25“; is already linear. magsinel a: mggsin e'IeFo + ngcosel 91:0(61_0)= mlgei 6b.; + mléf|_ (d2 -3)~2m2d2311dz-3(Ql — 0) = n Ell-o 91*0 31:0 in; (£2612 =5 —H12d20t2 So, 1112513 +m2g6. = r2 ...
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