Practice Problems

# Practice Problems - E6.1 E6.2 The Routh array is 53 1 K...

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Unformatted text preview: E6.1 E6.2 The Routh array is 53 1 K —|— 2 42 3K 5 41 5 9 40 4 where _ 3K(K —|— ‘2) — 5 _ 3K ' For stability. we require K :“> 0 and b > 0. Therefore, using the condition that b > 0, we obtain 3) 3K2 + 611’ — 5 :3» 0 . and solving for K yields K :> 0.63 and K 4: —2.63. Vii-Te select K :> 0.53. since we also have the condition that K > 0. The Routh array is .3- 1 2 .92 9 24 .41 —2/3 0 so 24 The system is unstable since the ﬁrst column shows two sign changes. E6.13 The roots of q(s) are 31 : —4, 32 : —3, 33,4 : —1 :tj2 and 35,5 : :tj0.5. The system is marginally stable. The Routh array is 56 1 31.25 37.75 15 s5 9 61.25 14.75 34 24.44 66.11 15 53 31.999 9.2273 9 52 99 15 31 0 0 so The auxillary equation is 6052 + 15 : D . Solving the auxillary equation yields two roots at 5152 : ::j0.5. After accounting for the row of zeros, the completed Routh array veriﬁes that the system has no poles in the right half-plane. E7.1 (a) For the characteristic equation r H n' + 4 1 + Ix : 0 . the root. locus is shown in Figure E'?’.1. 4 _ _ 3 _ _ 2 _ _ 1 _ _ “E: 0 — o l/ a — E \ -1 _ _ -2 _ _ -3 _ _ .4 — _ -4 -3 -2 -1 0 1 2 3 4 Real Axis (b) The system characteristic equation can be written as (1 + my? + (2 + 4K).-,- + 2 : 0 . Solving for s- yields _ _-<1+ 2K) _ (1 + K) __ 2(1 + If) 'When (2 + 411’)? — 8(1 —— K) = 0? then we have two roots at 5152 = Solving for K yields K = 0.31. (c) When K : 0.31, the roots are —(1 + 0.62) _ 31,2 : _ —1.24 . (1.31) (d) When K : 0.31, the characterisitc equation is 32 + 2.4725 + 1.528 : (s + 1.24)2 : 0 . Thus, to“ : 1.24 and C : 1, the system is critically damped. The settling time is T3 2 4 sec. E7.5 (a) The root locus is in Figure EYE). The breakaway points are 03,1 : —13.0 T 052 : —5.89 . (b) The asymptote centroid is 033m : —18 , and - _ __ o Qa-sym — ——90 - (c) The gains are K1 : 1.57 and K2 : 2.14 at the breakaway points. 40 30 20 10 ImagAxis 0 -‘IO -2 0 -30 410 FIGURE E7.5 2 2 ——10 Root locus for 1 + KmﬁE—agm : 0. E7.10 (a) The characteristic equation is K(s + 2) _ 1+—S(8+1)—0 . Therefore, K _(32 + s) (s + 2) and £__32+45+2_0 ds— (s+2)2 — ‘ Solving 32+4s+2 : 0 yields 3 : —O.586 and —3.414. Thus, the system breakaway and entry points are at s : —0.586 and s : —3.414. (b) The desired characteristic polynomial is (s+2+aj)(s+2—aj) 252+4s+4+o2 :0, where a is not speciﬁed. The actual characteristic polynomial is 52+(1+K)s+2K=0. Equating coeﬂicients and solving for K yields K : 3 and a : Thus, when K : 3, the roots are 512 : —2 :: x555. (c) The root locus is shown in Figure E710. 15 _ K:3,s=-2+1.414j _ Imag Axis Real Axis ...
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Practice Problems - E6.1 E6.2 The Routh array is 53 1 K...

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