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G(S) = ~S+ ) s+3~s+5~
1) Represent the plant in phase-variable, state-space form.
2) Design a phase-variable,full-state feedback controller to yield 15tYo overshoot
with a seffling time of 1 s.
Solution. For 15r7o overshoot, from Eq. (5.75), we find that < - 0.5169. Then,
O)n = 4/< = 7.7384, tort - co,_ ~-=Cz= 6.6245, and cr : c * O)n = 4.0. Therefore,
the donunant poles are at -4 +_ j6.6245. Because the given system is a third-order
system, we must choose one more pole. Let this pole be located at -2.1 so that it
nearly cancels the zero at -2 justifying the second-order approximation.
by
LINEAR SYSTEMS, THEORY, AND DESIGN: A BRIEF REVIEW 527
Therefore, the characteristic equation that gives the desired response is given
(s +4 - j6.6245)(s +4 + j6.6245)(s + 2.1) = 0
S3 + lO.lS2 + 76.6s + 125.7543 - 0
The next step is to express the given plant in state-space, phase-variable form.
For this purpose, let us decompose the transfer function into two blocks (one for
the numerator and another for the denominator) in cascade as schematically shown
earlier in Fig. 5.41b.
For the first block,
Gi(s) = (: (,))
1
= ~s+ ) s+3~s+5~
1
= S3 + 9S2 + 23s+ 15
or
(S3 + 9S2 + 23s + 15)xi(s) -. r (s)
Taking the inverse Laplace transform,
d3A
~ + 9% + 23dl;'+15xi - r (t)
Let xi = X2 and X2 - X3 so that
1
C~;i:l= ~,s 0. _9l[x::l+[ilr,t,
i -23
Consider the second block. Following a sinular procedure,
y(t)=[50 25 1] [x_:
The phase-variable form with full-state feedback system is given by
0
F~ii:l= _(150l
5+kl)
1
O
-(23 + k2)
[x::] + [{l.,t,
The characteristic equation of the given system is given by
S3 + (9 + k3)S2 + (23 + k2)S +(15 +ki) = 0
-(9-.k3,]
-0.05]
-0.10
0.05 ']
[-1 -;,3 :]
528 PERFORMANCE, STABILI-fY, DYNAMICS, AND CONTROL
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