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X2 -. X3
X3 = -biX3 - b2X2 - b3X1 + kr(t)
(5.248)
(5.249)
(5.250)
In matrix form,
/li:l=[-g3 :,2 _:,l['::l+[jlkr(t) (5.251)
which is the required phase-variable representation.
Consider the second block. This block gives the output matrix. We have
y(s) = (S2 + OIS + a7.)xi(s)
Taking the inverse Laplace transform, we get
y(t) : XI + aixi + 2xi
- X3 + a1X2 + a2xl
-[a2 ai ,,[x-::]
(5.252)
(5.253)
(5.254)
(5.255)
The root-locus method discussed earlieris essentially a pole-placement method
in frequency-domain analyses. The term pole refers to the poles of the closed-loop
transfer function. When we consider higher order systems greater than two, the
classical PD or PI type ofcontrollers willnot be able to place all the poles as desired
because there are only two free variables at our disposalin PD or PI controllers.
Therefore, for higher order systems, the pole-placement method of the state-space
approach becomes very attractive because it can place all the closed-loop poles
arbitrarily, but subject to the conditions that all the states are available for feedback
and the given plant satisfies the controllability condition.
Consider the system represented in state-space form as given by
x N Ax + Bu (5.256)
y = Cx (5.257)
With full-state feedback, u : r (t) - Kx, where r (t) is the m x 1 input vector and
K is an m x n matrix of feedback gains. Then,
/ = (A - BK)x + Br(t) (5.258)
y - Cx
(5.259)
LINEAR SYSTEMS, THEORY, AND DESIGN: A BRIEF REVIEW 519
Fig. 5.42 Schematiic diagram of pole-placement method.
The block diagram implementation of the full-state feedback control system is
shown in Fig. 5.42.
For simplicity, consider a system with single input, which means K is of dimen-
sion 1 x n. Then, the design procedure is as follows.
1) Represent the given plant in phase- variable form.
z = ApZ + BpU
(5.260)
2) Feed back each state variable to the input of the plant with gains kt so that
Ap - BpK =
00
10
Ol
0
0
0
-(ao+ki) -(al+k2) . . -(an_i+kn)
3) Write down the characteristic equation for the plant as follows
(5.261)
}sl - (Ap - Bp K)l = Sn + (an _ I + kn)sn -1 + .. .+ (ai + k2)S
+ (ao + ki) : 0 (5.262)
4) Decide on all the closed-loop pole locations that give the desired system
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