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JOURNAL OF COMPUTERS, VOL. 8, NO. 6, JUNE 2013 1563
MVC 3 problem. After deformation, the problem can be
solved by technology of LMI:
If and only if ∃W, X ≥ 0 , λ > 0 and Y i , U j , i = 1... p ,
j = 1... m and the following optimization as (18) can be
solved,
p
m
min qY i i+
λ rU j j
XYUWi , , , = 1 j=
1
∑ ∑ (18)
s.t. (13)-(17).
Then, according to the change of λ , optimized curve of
control outputs variance Var( y)
and manipulative variable
variance Var( u ) can be obtained. This optimized curve
can be used as the benchmark to evaluate controller
performance.
According to optimization objective function:
T
T
J = E( y Qy) + λ E( u Ru)
= Trace( Q ∑ y + λR
∑u)
.
T
T
= QTrace( C ∑ xC ) + λRTrace( K ∑x
K )
T
T
≤ QTrace( CXC ) + λRTrace( KXK )
T
T
Let CXC < Y , KXK < U , thus, minimizing
QTrace( Y ) + λRTrace( U ) can ensure J be minimized
too. Object function can be rewritten as
3
MVC
p
m
i i λ j j
i= 1 j=
1
J = ∑qY + ∑ rU . (13)-(17) can be obtained as
the same as LMI of advanced MVC 3 .
The MVC 3 benchmark curve is the lower limit of the
controller performance. That is, all linear controllers can
only be located in operating area above the curve.
Comparing actual run-time steady state outputs and
manipulated variables covariance with the MVC 3
benchmark cure, closer distance means better
performance. In practice, a benchmark
3
MVC
p m
i= 1
i i
j=
1
j j
J = ∑qY + ∑ rU can be calculated as λ = 1, then,
judge controller performance by the ratioη which is the
value of benchmark J 3 divided by actual operation
MVC
steady state covariance J arh .η closes to 1 means better
performance.
D. Extended MVC 3 and Infinite Horizon MPC
• Solution of infinite horizon MPC
For linear system, solving infinite horizon MPC is
equivalent to solving the LQR control problem as follows,
∞
min ∑ x( k) Qx( k) + u ( k) Ru( k)
(19)
u( k ) k = 0
T
x( k + 1) = Ax( k) + Bu( k)
s.t.
.
y( k) = Cx( k)
T
Let
T
p
Q= C DC,( D= ∑ qϕ ϕ ) , and let feedback
i=
1
T
i i i
controller be u( k) = Kx ( k)
, then,
T −1
T
K =− ( B PB + R)
B PA , where,
T T T −1
T
P = A PA − A PB( B PB + R)
B PA + Q .
• LQG control problem
LQG control problem can be described as:
1 T
T
T
min lim ∑ E[ y( k) Qy( k) + u( k) Ru ( k)]
(20)
u T
( k) T→∞ k = 0
x( k + 1) = Ax( k) + Bu( k) + Ew( k)
s.t.
,
y( k) = Cx( k)
where feedback controller can be solved by
T −1
T
K =− ( B PB + R)
B PA , here, PQis , the same as the
solution of infinite horizon MPC. Rewrite the object
function in (21) as,
1 T
T
T
J = lim ∑ E[ y( k) Qy( k) + u( k) Ru( k)
T →∞ T k = 0
T
T
= lim E[ y( k) Qy( k) + u( k) Ru( k)]
k→∞
.
= Trace ∑ + ∑
{ Q y R u}
p
m
T
T
∑qiϕi yϕi ∑ rjϕ j uϕ
j
i= 1 j=
1
= ∑ + ∑
• Extended MVC 3 control problem without
constraints
Extended MVC 3 control problem can be described as:
p m
qY i i+
rU j j
∑ x i= 1 j=
1
min ∑ ∑ (21)
s.t. ( A+ BK) ∑ ( A+ BK) + E∑ w E = ∑
x
T T
i ∑ x ϕi = i, = 1
ϕ C C Y i p
T
T
ϕ jK ∑ x K ϕ j = U j, j = 1… m .
The goal is to get a feedback gain matrix K by minimize
p m
the objective function ∑qY
+ ∑ rU .
T
i i j j
i= 1 j=
1
Assumption 1. R > 0 , Q ≥ 0 .
Assumption 2.The pairs ( A, B ) and ( A, Q)
is stabilizable
and detectable, respectively.
T
Assumption 3.The pair ( A, G∑ w G ) is controllable.
Let hypotheses 1-3 hold; then the solution to of MVC 3
problem as (21) is coincident with the solution of infinite
horizon MPC as (19). Hypotheses 1-2 indicate that the
solutions of problems (19) and (20) are unique and
stabilizing. Hypotheses 1-3 ensure that the solution of
problem (21) is unique and stabilizing. From the
construction of problem (20), it is equivalent to problem
(21), in the sense that the linear feedback generated by
T
x
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