自动控制原理英文版课后全部 - 答案 - 图文

G1?K1?(s?zi)sq1?(s?pj)ji,G2?K2?(s?zk)sq2?(s?pl)lk,H?K3?(s?zm)sq3?(s?pn)nms?(s?p)(s?p)?ppE??KKK?(s?z)(s?z)(s?z)KKK?zzz1?1?ss?(s?p)(s?p)(s?p)?pppsq2?q3lnq2?q3lnl,nl,n123ikm123i,k,mi,k,mq1?q2?q3jlnq1?q2?q3jlj,l,nj,l,n1?K2K3?(s?zk)(s?zm)k,m1?K2K3?zkzmk,mikmnsq1?q2?q3?pjplpn?sq1K2K3?zkzmpj?j,l,nk,m,jsq1?pj(sq2?q3?plpn??zkzmK2K3)?jl,nk,msq1?q2?q3?pjplpn?K1K2K3?zizkzmj,l,ni,k,mK1K2K3?zizkzmi,k,msq1?pj??jK1?zii??1limG1(s)s?0

q1 is the type(not order) of G1, when q1>0, the steady state error is zero. Please note the difference between type and order Discussion: When the order of

G2His equal to the one of G1G2H,ess will depend on the specific

condition. And when the order of G2His higher than the order of G1G2H, limit.

Module9

Problem 9.1 Solution

esswill have no

Im Re

Problem 9.2 Solution：

In this subject, there’re two open-loop poles at s=0, -4 and no zeros. Inspecting the real axis, it may be seen that the argument equation is satisfied only for -4 <

st < 0. Figure SP9.21

shows the locus so far.

Im

Fig SP9.21Re going to Since there are two poles and no zeros, there must be two branches of the locus infinity. A trial point stis indicated in Fig SP9.22 and the argument of each component of ?40the open-loop transfer function is shown. In this subject, the argument equation becomes

K GH(s)=s(s+4)

K

?GH(s)=?=-1800 s(s+ 4) st Im 0=?K??????s??????(s+4)??=??-180 0 = -? s ?????? (s+ 4) ??? ? 2 ? = ?? -180

?(s+4)?s

Fig SP9.22 Re0?40 -?s??????(s+4)????2?=??-180 Figure SP9.23 shows in which directions the two branches of the locus go to infinity. ???900

Im

Re Fig SP9.23Promblem 9.3 Solution

In Fig.P9.3, we can see that the open-loop poles are 0, －3,－4 ?40So the open- loop transfer function is G(s)H(s) =

K错了

s(s?3)(s?4)To calculate the gain, the magnitude equation can be used K = ss?3s?4

When s = -2?2j, K ? 96.4 17.89 When s = - 2.5, K ? 89.4 1.875

Promblem 9.4 Solution:

The open-loop transfer function is

GH(s)?K

s(s?5)(s?6)Draw the root locus. We should have ?1??2??3??n?360??180?

Imagining st to be on different portions of the real axis, only when st??6 or ?5?st?0, the argument equation will be satisfying. In order to the branches go to radius centered at Only when the determine the direction that infinity, a circle of infinity the origin is considered. roots go to infinity in a

direction at 60? to the real axis, the phase angle equation will be satisfied, that is

Consider K as a positive number: (a) When the real part ???2 Substituting s??2?bj gives (?2?bj)(3?bj)(4?bj)??24?5b2?(?2?b2)bj Thus

?24?5b2?0;(?2?b2)b?0 b=0

Then the character equation will be

(?2?bj)(3?bj)(4?bj)??24?5b2?(?2?b2)bj??24 So the point is s=-2

The gain at this location is K＝24

(b) When the imaginary part ???1 When ??1

Substituting s?a?j gives

(a?j)(5?a?j)(6?a?j)?a3?11a2?27a?11?(3a2?22a?29)j Thus

a3?11a2?27a?11?0;3a2?22a?29?0

a=-1.72

When ???1, see the picture, the answer will be the same as that above. So the point is s??1.72?j

The gain at this location is

KK????1

a3?11a2?27a?1129.99K＝30

Problems 9.5

9.5 A unity-feedback position control system shown in Fig.P9.5 has an actuator and load