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response, i.e., the desired characteristic equation is given by
Sn + dn _lSn -1 + ... + d] s + do - 0 (5.263)
5) Equate coefficients of the two charactenstic equations
so that
dn -i - an -i + kn, . - ', di -. ai + k2, do - ao + ki (5.264)
ki - do - ao, k2 = di - ai, . .., kn - dn -i - an -1 5.265)
t)
1~
520 PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
and
K -[ki k2 . . kn] (5,266)
The full-state feedback law in the transformed z-space is given by
or, in the original state-space,
u - -Kz +r(t)
(5.267)
u - -K Px + r(t) (5.268)
so that the given system with full-state feedback is given by
x - Ax + Bu
= (A - KPx)x+ Br(t)
(5.269)
(5.270)
6) Perform a simulation to verify the design.
The advantages of expressing the given plant in the phase-variable form is that
equations fort~ie gains k: are uncoupled and kr can be easily obtained as given
i:Eq. (5.265). However, if the plant is not controllable, then it is not possible to
represent it in the phase-variable form. For such a case, the above design procedure
remains same except for the fact the equations for k, will be coupled. Then the
gains ki have to be obtained by solving the n coupled algebraic equations.
5- t0-10 Dual Phase-Variable Form
The state-space representation, which is in the form,
where
x-
A-
x = Ax + BU
-an_i 1 o 0 . 0
-an_2 0 1 0 , o
. .
-a2 O O . 1 0
-ai . . . 1
-ao . . . - .
B-
(5.271)
(5.272)
is said to be in dual phase-variable form. Similar to the phase-variable form, the
elements of the first column of the matrix A in dual phase-variable form constitute
the coefficients of the characteristic equation as follows:
s" + a,r_isn-l + an~2Sn-2 + ... + a} s + ao - O (5.273)
Furthermore, this form of representation of the system matrix A is ver)r useful in
the design of state observers, which we will be discussing a little later.
LINEAR SYSTEMS, THEORY, AND DESIGN: A BRIEF REVIEW 521
5.10-11 Conversion of Transfer Function Form to Dual
Phase-Variable Form
We willillustrate this method with the help of an example. Consider once again
the system (see Fig. 5.41) given by
G(s)= k+~b++b+a+)b3 (5.274)
Rewrite this in the following form:
k+ 2k+ 2k
G(s)-.-\ - (5.275)
l -+b+~+b
y(s)
= r(s~ (5.276)
or
y(s)(l+b +b +b)=f(s)(k+ k+ k) (5.277)
Then,
y(s) = 1 [-b,y(s) + kf (s) + 1 ([-(s)kai - b2y(s)]
s
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