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:i:3 = . Z4 :i:4 ii = _1_ [_m2L2gsin9cos9 + (I + mL2)(F + mL82 - ky)]= L\(9)· . .

_1_ [_m2 L2g sin Zt COSZt + (I + mL2)(u + mLzi - kz4)]= L\(ZI) . (b) Set u =0 and :i:i =O.

o = f'2 o = (~+M)mgLsinit-mLcosit(mLi~-ki4) o = i4 . o = _m2L2gsinilCosit+(I+mL2)(mLi~-ki4)

The equilibrium points are given by (iI, 0, i3, 0) where it = 0, ::1:11', ••• , and :3 is arbitrary.

(e) Take it =O. Linearization at Z = : and u =0 results in the matrix

~ ~ !]·0 0 1, o 0 -C4

where the positive constats Cl to C4 are given by

(m+M)mgL mLk m2L2g k(l+mL2) Cl = A(O) • C2 =L\(O)' ca = L\(O) , C4 = L\(O)

The characteristic equation is given by

8 [83 + (482 !"" C18 + C2Ca - CtC4] = 0 From Routh-Hurwitz criterion, we can see that the matrix has eigenvalues in the right-half plane. Bence, the equilibrium point is unstable.

(d) The linearized system is given by i:, = h, + Bu,

where z, =Z -:, UI =U, A is given in part (c) and B =[0, -mLItJ.(O). 0, (1 + mL2)/L\(0)]T. It can be verified that (A. B) is controllable. Proceed to design K to stabilize the matrix (A + BK)using any pole placement algorihm .

.~) O1;SES2-~- eASc-o c.0NT+lL'U.E f.2.• 11.11,. . _._-' . . . . . ~ = [ :~ ] =[~ : ~ ~] ~'= C~ The pair (A, C) is observable. Design H such that A +XC is' observable and implem~t the observer-based controller.

~ -:+t-m::-GAAt- CO""~K~ , Start by designing a state feedback integral controller using 11 - ~J as the controlled output.

o . 0 [ .! ~ o C2 -C5o ] . :] =rank o 0 o 1 ! =4

Coll~~') .] [ 81.. -A 0 rank [ sIn+" - A B =rank -c· :] ~ ~Ov\~ 81"

Cb~ For s::/: 0,

k [ sIn - A O· :] =p+rank [81n -·A B ]ran -c .I" For s = 0,

rank [SIn - A 0 B ] = rank [AC· . -C sI" 0

Hence, (A, B) is controllable if and only if (A, B) is controllable .and the rank condition (11.19) is satisfi • 10.35

pCC;SCQ-.! B 11'46 1 (1 - jw)6

ruHcrl6 H G(;w) = (1 + ;w)6 = (1 + w2)6

~K()1...J:C-? 1 + 6(-j!ol) + 15(-;w)2 + 20(-;",,)3 + 15(-;w)4 + 6(_;w)5 + (_;w)6

(I + ",,2)6

= 1 - ISw2+ lSw4 - w6 + ;[-&1 + 2Ow3 - &15)

(I +w2)6

. 1 1m[G(jw)] = 0 ~ -6+2Ow2 - &14 = 0 ~w2 = 3 or w2 = 3

Re[G(j3)] = ~, Re [G (;1)] = - ~ From Example 10.16, we know that \)(11) = 4/7(11. For w2 = 3, the equation 1 + \)(II)Re[G] =0 has no solution. For ",,2 = i, the equation has a unique root II = 27/16'1t. Thus, we expect that the system will have a periodic solution with amplitude close to 27/16'1t and frequency close to 1/../3 rad/sec .

• 10.36

G(' ) j"" + 6 -;(6 + ;w)(2 - ;",,)(3 - jw) -w(24 +w2) - j(36 - w2)

JW = jw(j"" + 2)(;w +3) = ""(4+",,2)(9+,,,,2) = w(4+w2)(9+w2)

1m[G(;",,)] =0 ~ 36-",,2 =0 ~w =6 Re[G(j6)] =-1/30 and the equation 1 + \)(II)ReG =0 has the unique solution II = 2/1h. Thus, we expect that the system will have a periodic solution with amplitude close to 2/15'1t and frequency close to 6 rad/sec.

• 10.31

G(" ) _ jt.J. _ ;w _ -w2 +;w(I_",,2)

J"" - -",2 _ ;w + 1 - 1 _ w2 +;"" - (1 _ w2)2 + ",,2

1m[G(jw)l =0 ~ "" = I Re[G(j)] ~ -1 and the equa~ion 1 + \)(II).ReG = 0 has the unique solution II = (8/5)1/4. Thus, we expect that the system will have a periodic solution with ·amplitude close to (8/5)1/4 and freque~cy close to 1 rad/sec.

• 10.38

G(" ) _ . 5(1 + j4w) _ 5(1 + 4;"")(~- j",,)2 _ 5(4 + ISw2) + j2Ow(3 - w2) JW - 4(j",,)2(2+;w)2 - -4w2(4+!tf)2, - -4w2(4+w2)2

1m[G(;w)] = 0 => w(3- w2 ) = 0 ~ w = v'3 Re[G(iv'3)] =-5/12. The equation 1 + \)(II)ReG =0 yields \)(11) =12/5. (1)

4 12 10 \)(11) =2+- = - ~ II =

"'II 5 'It (2)

0, for II =:; 1

{\)(11) = iV1 _ (1)2 foro~1

• 11'4 .'

.(0) =:; 4/7(0. Hence, .(0) = 12/5 baS no solution. (3) The describing function is given in the solution of Exercise 10.34. Using a computer, it can be verified that \)(0) = 12/5 has no solution.