{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 6
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}