{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Exam 2 2001 - .1 2gu.a w £{u(t = £{tu(t = 3‘...

Info icon This preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

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

View Full Document Right Arrow Icon
Image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: .1 +2gu..a+w,, £{u(t)} = £{tu(t)} = 3‘, £{e‘“‘u(t)} = fi; For linear motion, (Sum of Forces) 2 (Mass)*(Acceleration). For rotational motion (Sum of Torques) = (Moment of Inertia)*(Rotational Acceleration). For viscous friction at slow speeds, Pd,” = —fd,ng 2: (velocity). For a linear spring, Farms = —k 2: (displacr‘mzrnt). Proportional Control C(s) =-- K. Proportional-Derivative Control C(s) = 1\'(s + a). Proportional-Integral Control C(s) = K + f. Proportional-IntegraLDerivative Control C(s) = K(1 + + b5). Lead Control C(s) = Kfifi, a < b. Lag Control C(s) = Ki11 :1 > b. s+b‘ Lead-Lag Control C(s) = b < a < c < d. Asymptotes for Root Locus are at angles of $130” from a centroid at 2% Controllability Matrix (2 _—. [13 AB A23...]"—'" C . . . . ' _ CA Observabihty Matrm O —- CA.) Aokermann’s Controller Formula: K [O 0 O 0 1]C"d(.—1) 0 O Ackermann‘s Observer Formula: L = d(.-4)O’1 0 1 Control Canonical Form: A = C=[an_1 an_2 an] Cor-4 1 2 Parts 20 Points Consider three stable transfer functions H1(s), H2(s) and 173(5) 1.1 2 Points If W3) : (Hi(8) + H2(8) + H3(3))U(s), where u is the input, is the transfer function {78% stable? Why or Why Not? y?) 1L9 {Hamper pancfibfl f‘) ffqblgr 666406 If eqch O’C "‘9 transfer Huge.) qrc Maw, he“ the rem; y w, 1)? hwan 67 Me :wu 1.2 10 Points If H1, H2 and H3 have state-space realizations of {A1, Bl, Cl, D1}, {A2, B2, 02, D2} and {A3, 83, Ca, D3} respectively, write a state-space realization for Y(s). 2 3 Parts 20 Points Consider the transfer function Y(s) = films) + figULs) + s—i—gU(s) 2.1 10 Points Write this transfer function in state—space form, using any realization. Modal anMc‘qu Em r. 4 0 o l x '5 (1‘1 C Xl’ l U C U ‘3 I y: [1 Auk l'DUl 2.2 5Points For your realization‘ what are the eigenvalues of the “A” matrix? 2.3 5 Points Is your realization observable from Y? C l 13 5 e: (A : ‘2 -11“? JelCflJ‘zfl (“N 2 J7 Obtamb’e 3 2 Parts 20 Points 1 0 0 Consider a system with state-space realization :i:(t) = I: 0 -1 1 ] $03) + 0 0 —1 o [(1)]u(t),y(t) = [1 1 1]z(t). 3.1 10 Points Is this system controllable? Why or why not? I .: 8 ‘l .. o C [ MAB} e fill 491:0 ram anew/e I “I 1 “VA ufeh‘J Meal/y :W/f 3.2 10 Points Is this system observable? Why or why not?. g: C; a l5“! I t) M: — Jen/q ’9 11d 4 3 Parts 25 Points Consider a system with transfer function H (s) = 2H4 4.1 5 Points Find a state-space realization for the system. 1 v _L« A - at e I g 3"— __ . ‘ : ’ CV Hf) mq (3+1) 9+; 34 0D XZ”J)( 7" U V; l): J 4.2 10 Points Augment your state space to allow for zero-error tracking of a step reference input. Rfixzf 4&30 1‘ Cg~ Li}: [’33 33+ [it : [3 gm 4.3 10 Points y : [o {IQ}, Design a feedback controller which places your poles at s = ——1 :l: 3'. My ii :5 (3 M :4 5+ ~‘ :: l 5) C I J) S -01) & arm-:5 n: +1 _ it lit 24/5]:le all [Fifi [l7 K: [:0 HUG“) =C0‘1[%;}[::3 snags Cm: :: — ‘ l [farm [8 13+ géfl l 5; UR}? 1pc} 5 2 Parts 15 Points Consider the system with transfer function H (s) = % 5.1 5 Points Draw the root locus for the system in proportional feedback. 5.2 10 Points Consider the controller 0(3) = 5,174 (chosen to track a. sinusoid). Is the system H (5)0 (s) in proportional feedback ever stable? Why? ...
View Full Document

{[ snackBarMessage ]}