TELKOM
NIKA Indonesia
n
Journal of
Electrical En
gineering
Vol. 12, No. 9, September
2014, pp. 67
1
1
~ 672
4
DOI: 10.115
9
1
/telkomni
ka.
v
12i9.459
3
6711
Re
cei
v
ed O
c
t
ober 1, 20
13;
Revi
se
d Apr
8, 2014; Acce
pted May 6, 2
014
Control Strategy Analysis on Preventive Maint
e
nance
Hong
sheng
Su, Yongqiang Kang
*, Juandi Li
Dept. of Electri
c
al Eng
i
ne
eri
n
g, Lanzh
ou Ji
a
o
tong U
n
iv
ersity, La
nzho
u, Ch
ina
88 W
e
st Annin
g
Roa
d
, Lanz
h
ou 73
00
70, Chi
n
a
*Corres
p
o
ndi
n
g
author, e-ma
i
l
: kang
yo
ng
137
@16
3
.com
A
b
st
r
a
ct
In view
of the
l
i
mitatio
n
s that
the o
p
timal
ex
am
ini
ng
ite
m
d
o
not
hav
e dy
n
a
mic re
al-ti
m
e
speci
a
lit
y
and ca
n
’
t refl
ect the actual
state of devi
c
es in t
he tra
d
itio
nal
preve
n
t
ive mainte
na
n
c
e(PM) mo
de
l
of
repa
irab
le
devi
c
es, in this
pa
per a r
eal-ti
m
e
control
proj
ect on the c
heck
i
ng rate
of PM i
s
prop
osed
ba
se
d
on the state of devic
es. The
differential equations us
ed to describe t
he dy
nam
i
c behavior of
the system
ar
e
establ
ishe
d, a
nd so
me per
forma
n
ce i
n
d
e
xes of
mai
n
tenanc
e syste
m
s i
n
clu
d
in
g
the steady-s
tate
availability, and the
mean time to failure (MTTF)
, and
as
well as the av
erage time of
st
aying in each s
t
ate
are c
a
lcu
l
ate
d
. The c
ontro
l st
rategy
on
the
checki
ng r
a
te
i
s
then
pr
opos
ed
an
d th
e a
d
aptab
ility
an
d t
h
e
stability of the
corresp
onding control system ar
e analy
z
ed.
The essenc
e
of the method is to achi
eve t
he
expected steady-state behav
ior by
control
l
i
ng the
dynam
i
c
behavio
r
of the system
, w
h
ich w
ill
ens
ur
e
relia
bl
e co
mp
le
tion of the t
a
sk
and r
educ
e th
e mai
n
tena
nce
cost me
anti
m
e. Rese
ar
ches
indic
a
te that t
h
e
prop
osed
meth
od is very effe
ctive to improv
e the utili
z
a
t
i
o
n
of devices a
n
d
provi
de the
o
r
e
tical su
pp
ort for
the practica
l ap
plicati
ons.
Ke
y
w
ords
:
re
pair
abl
e dev
ice
,
reliab
ility, che
cking rate, dyn
a
mic contro
l
Copy
right
©
2014 In
stitu
t
e o
f
Ad
van
ced
En
g
i
n
eerin
g and
Scien
ce. All
rig
h
t
s reser
ve
d
.
1. Introduc
tion
The
equi
pme
n
t mainte
nan
ce i
s
the i
m
p
o
rtant
safe
gu
ard
to
kee
p
it
in
a g
ood
st
ate. To
improve
its utilization and
prolong
its
servi
c
e life, the m
a
intena
nce transformation to ai
m at
pre
c
ise gua
ra
ntee obje
c
tively requires t
hat rele
vant system in operation must be
chan
ged fro
m
fixed perio
di
c preventive
maint
ena
nce (PM) to d
y
namic mai
n
tenan
ce st
ra
tegy base
d
on
equipm
ent actual state [1].
At prese
n
t, preventive maintenan
ce st
ra
tegy
has re
gu
lar mainten
a
n
c
e and m
a
intenan
ce
based on
state (CBM
) [2-6]. The
sh
ortco
m
ing of
regula
r
mai
n
tenan
ce i
s
that the optimal
examining
times or fre
q
u
ency
will
not
cha
nge
on
ce
determi
ned
in
advan
ce. It i
s
n’t dyn
a
mic
and
real
-time a
n
d
ca
n’t refle
c
t
the a
c
tual
stat
e of equi
p
m
ent. And a
m
ong th
e
CBM model
s,
the
informatio
n d
a
ta is difficult
to collect, a
nd the
ac
t
ual
state
of equi
pment i
s
diffi
cult to
estima
te
and the mo
d
e
l to be re
so
lved is more
compli
cate
d,
so that they are difficult to be practi
ca
lly
applie
d.
In view of the above issu
es, the pa
per star
ts from
analyzi
ng the
state of mai
n
tenan
ce
system,
and
then
esta
blish
e
s th
e dyn
a
m
i
c e
quatio
ns
use
d
to
de
scribe the
op
era
t
ion p
r
o
c
e
s
s
o
f
the sy
stem,
and a
nalyzes the be
havio
ur of th
e sy
stem ba
sed
o
n
state t
r
an
si
tion matrix. T
h
e
perfo
rman
ce
i
ndexe
s
of m
a
intenan
ce
sy
stem that
in
clu
de the
ste
ady
-state
availab
ility, the mean
time to failure
(MTTF
) an
d
the avera
ge ti
me of st
aying
in ea
ch
state
are give
n out
. The che
ckin
g
rates a
r
e
sel
e
cted a
s
con
t
rol variable
and cont
rol t
he dynami
c
behavio
ur of
the system to
achi
eve the
expecte
d ste
ady-state
ai
ms. In t
he e
nd, the ada
p
t
ability and the sta
b
ility of the
corre
s
p
ondin
g
control syst
em are a
naly
z
ed.
2. Model Des
c
ription
In orde
r to e
s
tabli
s
h the l
i
fecycle m
o
d
e
l of
r
e
pa
irab
le
d
e
v
ic
es
, w
e
do
th
e
fo
llo
w
i
ng
ass
u
mptions
[7-8].
H
y
pothesis 1:
Whethe
r d
e
vice is in
worki
ng stat
e or
sto
r
ag
e st
ate, it is not existed for
the failure tha
t
can’t be det
ected o
u
t.
H
y
pothesis 2:
The probability of device from state
S
i
at
time
t
to
state
S
j
at
time
tt
Evaluation Warning : The document was created with Spire.PDF for Python.
ISSN: 23
02-4
046
TELKOM
NI
KA
Vol. 12, No. 9, September 20
14: 67
11 – 672
4
6712
is only prop
ortional to the time interval
t
, the tran
sfer ra
te is a con
s
ta
nt which d
o
e
s
not depend
on the time
t
an
d
t
.
H
y
pothesis 3:
Mainte
nan
ce will n
o
t ch
ange failu
re rate of the device
s
.
H
y
pothesis 4:
PM will not
cause the fai
l
ure of sy
stem—during P
M
syst
em
can still work
norm
a
lly
.
Based o
n
ab
ove assumpti
ons, the lifecycle
model of
repairable d
e
vice ca
n be
denote
d
usin
g the stat
e transitio
n di
agra
m
as
sho
w
n in Figu
re
1. Its symboli
c
meani
ng is
belo
w
.
2
v
1
u
1
1
μ
ρ
22
(1
)
ρ
μ
1
1
(1
)
ρ
μ
2
λ
1
λ
2
u
1
v
22
ρ
μ
Figure 1. State Tran
sition
Di
ag
ram of Repairable
De
vices
In Figure 1,
S
1
denotes that device is
in
good state
in the ware
h
ouse;
S
2
denotes that
device i
s
in
worki
ng
state;
S
3
denote
s
th
at device i
s
in
pr
eventive m
a
intena
nce st
ate;
S
4
denot
es
that device i
s
in co
rrec
tiv
e
maintena
n
c
e state.
i
μ
is the maintena
nce
rate of d
e
vice.
Amon
g
them,
1
μ
is the maintena
n
c
e rate of d
e
vice
in pre
v
entive maintenan
ce stat
e;
2
μ
is the
maintena
nce
rate of device in
co
rre
ct
ive maintena
nce state;
1
1
μ
ρ
i
s
the trans
fer rate from
preventive m
a
intena
nce st
ate to storag
e state;
1
1
(1
)
ρ
μ
is the tran
sfe
r
rate from p
r
e
v
entive
maintena
nce state to worki
ng state;
2
2
μ
ρ
is the tran
sfer ra
te from corre
c
tive mainten
ance state
to storage
state ;
2
2
(1
)
ρ
μ
is
the trans
fer rate from
correc
tive maintena
nce
state to
workin
g
state.
i
λ
is the
state transition probab
ility
during the ti
me interval [
t
,
tt
]. Among
them,
1
λ
is
the probabilit
y from worki
ng
state to
st
orage state;
2
λ
is the p
r
ob
ab
ility from storage
state to
workin
g state
.
Let
3
11
vu
λ
,
4
22
vu
λ
,
1
v
and
1
u
denote re
sp
e
c
tively the
failure rate and the
che
c
king
rat
e
of the
stored devi
c
e,
2
v
and
2
u
den
ote
re
spe
c
tively the failure rate and
the
che
c
king rate
of the workin
g device.
3. Reliabilit
y
Anal
y
s
is
Acco
rdi
ng to Figure 1 and
reliability theo
ry [9-10], we
have:
1
12
3
4
23
1
1
22
1
2
12
3
4
24
11
2
2
1
3
12
3
12
1
4
12
4
12
2
d(
)
(
)
()
()
()
()
d
d(
)
()
)
(
)
(
1
)
()
(
1
)
(
)
(
d
d(
)
()
()
()
d
d(
)
(
)
()
()
d
xt
μ
xt
x
t
x
t
x
t
ρ
ρμ
λλ
λ
t
xt
x
tx
t
x
t
x
t
λρ
μ
ρ
μ
λλ
t
xt
ux
t
u
x
t
μ
xt
t
xt
vx
t
v
x
t
μ
xt
t
(1)
12
3
4
()
()
(
)
(
)
1
x
t
xt
x
t
xt
;
0
t
(2)
Evaluation Warning : The document was created with Spire.PDF for Python.
TELKOM
NIKA
ISSN:
2302-4
046
Control Strategy Anal
ysis
on Pre
v
enti
v
e
Maintenan
ce
(Hon
gshen
g Su)
6713
whe
r
e
1
()
x
t
denot
es the
pro
b
a
b
ility of being
in the state
S
1
at time
t
, and
2
()
x
t
denote
s
the
probability in
S
2
, and
3
()
x
t
is the probability in
S
3,
and
4
()
x
t
is the probability in
S
4
.
Con
d
u
c
ting the Lapl
ace transfo
rmatio
n on (1
), then we have:
1
11
2
3
4
23
1
1
22
1
2
21
2
3
4
24
11
2
2
1
31
2
3
12
1
412
4
12
2
()
(
0
)
(
)
()
()
()
()
)(
1
)
(
1
)
(
0
)
()
(
)
(
(
)
(
)
(
)
(
)
()
()
()
(
)
()
()
()
μ
sx
ρ
xs
xs
x
s
x
s
ρμ
xs
λλ
λ
sx
xs
x
s
λ
xs
ρμ
xs
ρμ
xs
λλ
su
u
μ
xs
x
s
x
s
x
s
sv
v
μ
xs
x
s
xs
xs
(3)
Theorem 1:
As the
checki
ng rate is constant, the
steady-state availab
ility of the
system
is:
1
12
01
2
23
4
21
1
2
li
m
(
)
(
)
(
)
t
μμ
d
AA
t
x
x
μμ
μ
μ
dd
d
(4)
Whe
r
e
11
2
3
2
211
12
1
2
uv
u
v
ρρ
ρ
ρ
d
λλ
λ
,
21
2
1
1
2
2
21
1
2
21
22
u
u
u
u
uv
u
v
uv
ρ
ρ
d
λλ
,
31
2
1
2
2
1
12
2
1
12
11
vv
v
u
v
v
u
v
u
v
ρ
ρ
d
λλ
,
2
4
12
1
2
1
11
2
2
uu
v
v
ρρ
ρ
ρ
λ
λ
d
.
Proof:
From
(3), we c
a
n have:
31
2
12
1
41
2
12
2
1
()
(
)
()
1
()
(
)
()
uu
x
sx
s
x
s
s
μ
vv
x
sx
s
x
s
s
μ
.
(5)
From (3) a
nd
(5), then:
(0
)
()
()
ss
s
Ax
xx
(6)
Whe
r
e,
12
34
=
aa
aa
A
,
1
2
()
=
()
()
x
s
s
x
s
x
,
1
2
(0)
(0)
=
(0
)
x
x
x
,
11
1
ρ
ρ
,
22
1
ρ
ρ
,
11
12
12
1
23
12
μ
μ
uv
ρρ
a
λλ
s
μ
s
μ
,
22
12
12
2
1
12
μ
μ
uv
ρρ
a
λ
s
μ
s
μ
,
1
2
11
12
3
2
12
μ
v
ρ
μ
u
ρ
a
λ
s
μ
s
μ
,
2
2
12
12
4
14
12
μ
v
ρ
μ
u
ρ
a
λλ
s
μ
s
μ
.
Then,
41
2
2
31
1
2
()
(
0
)
(
0
)
det
()
(0
)
(
)
(
0
)
de
t
sa
x
a
x
s
s
ax
s
a
x
s
IA
x
IA
(7)
2
14
1
4
2
3
det
(
)
s
sa
a
s
a
a
a
a
IA
(8)
Acco
rdi
ng to Lapla
c
e tra
n
sformation, an
d we have:
0
()
l
i
m
()
s
s
s
x
x
.
(9)
Evaluation Warning : The document was created with Spire.PDF for Python.
ISSN: 23
02-4
046
TELKOM
NI
KA
Vol. 12, No. 9, September 20
14: 67
11 – 672
4
6714
Combi
n
ing
(7
), (8) a
nd (9
), we can obtai
n:
1
12
2
2
12
1
23
4
21
1
2
()
()
μμ
uv
ρρ
λ
x
μμ
μ
μ
dd
d
,
23
12
1
1
12
2
23
4
21
1
2
()
()
μμ
uv
ρρ
λλ
x
μμ
μ
μ
dd
d
Acco
rdi
ng to (4), the sy
ste
m
steady-stat
e
availability can b
e
prove
n
, immediatel
y.
Theorem 2:
As the
ch
ecking rate i
s
co
nstant, the
m
ean time
to f
a
ilure
(M
TTF
) of the
sy
st
em i
s
:
12
11
4
2
2
2
3
1
11
2
1
2
1
23
1
4
1
2
11
2
2
1
11
1
1
(
)
(0
)
(
)
)
(0)
MT
T
F
=
()
(
)
(
)
(
)
μ
uu
x
μ
uu
x
h
λλ
λ
h
λλ
λ
μ
uu
u
u
ρρ
ρ
ρ
λλ
λ
λ
λ
λ
(10
)
Whe
r
e
11
2
4
12
11
uu
ρ
ρ
h
λλ
λ
,
21
2
3
21
11
uu
ρ
ρ
h
λλ
λ
.
Proof:
Let
S
4
be the absorbing state, an
d
12
3
()
()
()
()
Rt
x
t
x
t
x
t
,
2
0
μ
, and then,
=0
0
12
3
0
M
TTF
=
(
)
d
(
)
=
(
)
()
()
Rt
t
R
s
s
s
xs
x
s
x
s
Then the formula (1
0) can
be obtaine
d, immediately.
Theorem 3:
As the
che
c
ki
ng rate is con
s
tant, the to
ta
l avera
ge tim
e
that sy
stem
stay in
state
S
1
and state
S
2
before enterin
g the
abso
r
bin
g
st
ate is:
12
12
4
1
2
3
12
2
1
11
1
1
31
4
1
2
12
2
1
21
1
1
1
()
(
0
)
(
)
(
0
)
T=
)(
)
(
)(
)
(
uu
x
u
u
x
ρρ
ρ
ρ
λλ
λ
λ
λ
λ
uu
u
u
λρ
ρ
ρ
ρ
λλ
λ
λ
λ
(11)
Proof:
Le
t
X
be the total time of staying in state
S
1
and
S
2
b
e
fore ente
r
in
g the
absorbi
ng
sta
t
e. To get th
e di
stributio
n
of
X
, we mu
st
ded
uct th
e time of
stayin
g in
state
S
3
.
Acco
rdi
ng to
the p
h
ysi
c
al
meani
ng
of derivative, l
e
t the d
e
riva
tive on
t
in t
he differentia
l
equatio
n of state
S
3
be zero [11], then:
1
12
3
23
1
1
1
2
12
3
24
11
1
12
3
12
1
d(
)
(
)
()
()
()
d
d(
)
()
)
(
)
(
1
)
(
)
(
d
0(
)
(
)
(
)
xt
μ
xt
x
t
x
t
ρ
λλ
λ
t
xt
x
tx
t
x
t
λρ
μ
λλ
t
ux
t
u
x
t
μ
xt
(12)
After Lapla
c
e
transfo
rmatio
n we have:
1
11
2
3
23
1
1
1
2
21
2
3
24
11
1
12
3
12
1
()
(
0
)
()
()
()
()
)(
1
)
(
0
)
()
()
(
(
)
(
)
0
()
()
()
μ
sx
ρ
xs
x
s
x
s
x
s
λλ
λ
sx
xs
x
s
λ
xs
ρμ
xs
λλ
uu
μ
xs
x
s
x
s
(13)
Writing a
s
ma
trix formula, a
nd then:
(0
)
()
()
ss
s
Cx
xx
(14)
Whe
r
e
12
34
=
cc
cc
C
,
1
2
()
=
()
()
x
s
s
x
s
x
,
1
2
(0
)
(0
)
=
(0)
x
x
x
,
1
3
1
21
)
(
cu
λρ
λ
,
2
1
2
1
cu
ρ
λ
,
3
2
1
1
cu
ρ
λ
,
4
4
2
11
)
(
cu
λρ
λ
-
.
Evaluation Warning : The document was created with Spire.PDF for Python.
TELKOM
NIKA
ISSN:
2302-4
046
Control Strategy Anal
ysis
on Pre
v
enti
v
e
Maintenan
ce
(Hon
gshen
g Su)
6715
Then we ca
n get:
41
2
2
31
1
2
()
(
0
)
(
0
)
det
()
(0
)
(
)
(
0
)
det
sc
x
c
x
s
s
cx
s
c
x
s
IC
x
IC
(15
)
2
14
1
4
2
3
de
t
(
)
s
sc
c
s
c
c
c
c
IC
(16)
Let
12
()
(
)
()
()
Tt
P
X
t
x
t
x
t
, then, th
e total average time of staying in state
S
1
a
nd
state
S
2
before enterin
g the
abso
r
bin
g
st
ate is:
34
1
2
1
2
2
=0
0
12
0
14
2
3
()
(
0
)
(
)
(
0
)
T(
)
d
(
)
=
()
(
)
cc
x
c
c
c
x
Tt
t
T
s
s
s
xs
x
s
cc
c
c
After related
para
m
eters a
r
e su
bstituted
into t
he formula above, th
e formula (11
)
is proven.
4. Contr
o
l Strategy
Desig
n
To make the
dynamic b
e
havior of the
system
a
c
hi
eve the expe
cted ste
ady-state aim,
we may
cont
rol the che
cki
ng rate
s of m
a
intena
nce system, whi
c
h
are the i
n
verse of me
an ti
me
betwe
en che
c
ks. Thi
s
co
ntrol mod
e
is di
ffe
rent from the tradition
al method
s [12].
Acco
rdi
ng to (1) a
nd (2
), we have:
1
12
4
1
21
1
1
11
1
1
22
11
1
1
2
12
4
2
2
12
1
1
2
1
2
11
1
1
1
2
4
12
4
12
2
d(
)
()
(
)
(
)
(
)
)
(
)
(
)
(
)
(
d
d(
)
(
)
(
)
)(
)
(
)(
)
(
)(
)
(
d
d(
)
()
()
(
)
d
xt
μμ
μ
μ
vx
t
x
t
x
t
u
t
x
t
ρρ
ρ
ρ
ρμ
λλ
t
xt
μ
μ
xt
v
μ
xt
μ
xt
u
t
xt
μ
ρ
ρλ
ρ
ρ
ρ
λ
t
xt
vx
t
v
x
t
μ
xt
t
(17)
Writing
(17
)
a
s
the Matrix form, then:
10
1
0
d(
)
()
()
()
()
()
()
d
t
tt
t
t
t
t
t
x
Ax
U
x
b
A
x
X
u
b
(18)
Whe
r
e,
21
11
1
2
13
1
11
1
22
11
1
1
2
21
22
23
1
2
1
1
2
11
1
1
2
31
32
3
3
1
2
2
()
==
)
(
μμ
μ
aa
a
v
ρρ
ρ
ρμ
λλ
μ
aa
a
μ
v
μ
μ
ρ
ρλ
ρ
ρ
λ
aa
a
v
v
μ
A
,
1
2
4
()
()
=
(
)
()
x
t
tx
t
x
t
x
,
1
2
()
0
0
()
=
0
()
0
00
1
xt
tx
t
X
,
1
0
2
()
()
=
(
)
0
ut
tu
t
u
,
1
0
2
()
0
0
()
=
0
()
0
00
0
ut
tu
t
U
,
1
1
1
2
1
1
3
==
0
μ
ρ
b
b
μ
ρ
b
b
.
Proje
c
ting the
system (1
) o
n
to a plane of
1
x
and
2
x
, we then have:
12
()
()
1
xt
x
t
;
0
t
(19
)
As
1
x
or
2
x
tends
to zero, to en
sure the bo
u
ndne
ss of
()
t
u
, where
T
12
()
()
()
tu
t
u
t
u
,
Evaluation Warning : The document was created with Spire.PDF for Python.
ISSN: 23
02-4
046
TELKOM
NI
KA
Vol. 12, No. 9, September 20
14: 67
11 – 672
4
6716
we divide th
e
area
of
1
x
an
d
2
x
into four
sets which
are denote
d
wi
th
①
~
④
a
s
shown in
Figure 2, whe
r
e
01
.
Figure 2. Valid Area
s of Co
ntrol Varia
b
le
Theorem 4
:
In the area
①
,
1
1
x
,
2
1
x
, and
12
1
xx
. We do the ru
le
control for
()
t
u
below.
12
4
1
1
2
4
11
11
12
12
13
13
11
12
13
1
1
2
12
4
2
1
2
4
21
21
22
2
2
23
2
3
2
1
22
23
2
1
(
)
()
(
)
()
(
)
()
()
()
()
=
()
1
(
)
()
(
)
()
(
)
()
()
a
k
x
t
a
k
xt
a
k
xt
b
k
x
k
x
k
x
ut
xt
t
ut
a
k
x
t
a
k
xt
a
k
xt
b
k
x
k
x
k
x
xt
u
(20
)
whe
r
e
1
2
4
x
x
x
x
,
11
12
13
21
22
23
31
32
33
kk
k
kk
k
kk
k
K
,
x
is the expected st
eady-state probability,
K
matches
()
IK
is non
sing
ula
r
and
1
()
(
)
IK
I
K
is conv
erge
nt,
I
is
unit matrix.
Und
e
r the a
c
tion of the con
t
rol rule (4.4),
t
he steady-st
a
te availability of the system is:
01
2
1
2
()
()
A
xx
x
x
(21
)
Proof:
Kno
w
n from
the
T
heorem
1, th
e de
si
red
st
ate eq
uation
sh
ould
po
ssess the
following form.
m1
m1
d(
)
()
d
t
t
t
x
Ax
b
(22
)
And so, the st
eady-state value of the system can be
written as:
T
m1
12
4
()
xx
x
xx
x
Whe
r
e
m1
x
is the expected steady
-state probability,
41
2
12
2
1
()
x
vx
v
x
μ
.
Let
m1
A
K
,
m1
bK
x
and Co
mbining
(4.2) and (4.6
), we can
get (4.
4
), and
(4.5)
by
combi
n
ing (4.
2
), (4.4) and
(4.6). To ensure
that the
system is
asymptotically steady,
K
must
match that
()
IK
is non
singul
ar
and
1
()
(
)
IK
I
K
is conve
r
gent [13
-
1
4
], and
31
k
,
32
k
and
33
k
match the foll
owin
g co
ndition.
12
4
2
1
2
4
31
31
32
32
33
3
3
31
32
33
(
)
()
(
)
()
(
)
()
0
a
k
x
t
a
k
xt
a
k
xt
b
k
x
k
x
k
x
(23
)
End.
Evaluation Warning : The document was created with Spire.PDF for Python.
TELKOM
NIKA
ISSN:
2302-4
046
Control Strategy Anal
ysis
on Pre
v
enti
v
e
Maintenan
ce
(Hon
gshen
g Su)
6717
In the area
②
, in orde
r to ensure th
at
1
()
ut
is bound
ed,
we let
1
()
ut
be consta
nt, but
2
()
ut
remain u
n
ch
ange
d. So, the control rul
e
for
()
t
u
is
:
24
1
11
1
2
13
1
1
2
12
4
2
1
2
4
21
21
2
2
22
23
2
3
21
2
2
2
3
2
1
()
()
=
1
()
(
)
()
(
)
()
(
)
()
()
aa
x
a
x
b
ut
x
t
ut
a
k
x
t
a
k
xt
a
k
xt
b
k
x
k
x
k
x
xt
u
(24
)
So, the dynamic beh
avior
of the system
becom
es
m2
m
2
d(
)
()
d
t
t
t
x
Ax
b
(25
)
Whe
r
e,
24
1
12
13
12
13
1
m2
2
1
22
23
31
3
2
33
1
ax
a
x
b
a
a
x
kk
k
kk
k
A
,
1
1
m2
1
2
4
21
22
23
12
4
31
32
3
3
=
μ
ρ
kx
k
x
k
x
kx
k
x
k
x
b
31
k
,
32
k
and
33
k
match
(23). In ord
e
r
to ensu
r
e t
hat
the syste
m
is asympt
otically stea
d
y
,
m2
A
matche
s that
m2
()
IA
is non
sing
ula
r
and
1
m2
m2
()
(
)
IA
I
A
is conv
erge
nt.
Thus, the
ste
ady-state val
ue of the syst
em is:
T
2
12
4
()
m
xx
x
xx
x
In the same
way, In the area
③
, we set
2
()
ut
as co
nsta
nt, but
1
()
ut
remain u
n
ch
ang
ed.
So, we have:
12
4
1
1
2
4
11
1
1
12
12
1
3
13
11
12
1
3
1
1
2
14
2
22
2
1
23
2
1
(
)
()
(
)
()
(
)
(
)
()
()
()
=
()
1
ak
x
t
a
k
x
t
a
k
x
t
b
k
x
k
x
k
x
ut
xt
t
ut
aa
x
a
x
b
x
u
(26
)
So, the dynamic beh
avior
of the system
becom
es:
m3
m3
d(
)
()
d
t
t
t
x
Ax
b
(27
)
Whe
r
e,
11
12
13
m3
1
4
2
2
1
21
23
23
2
31
32
33
1
kk
k
aa
x
a
x
b
a
x
kk
k
A
,
12
4
11
12
13
m3
1
1
12
4
31
32
33
=
kx
k
x
k
x
μ
ρ
kx
k
x
k
x
b
31
k
,
32
k
and
33
k
match
(23). In ord
e
r
to ensu
r
e t
hat
the syste
m
is asympt
otically stea
d
y
,
m3
A
matche
s that
m3
()
IA
is non
sing
ula
r
and
1
m3
m3
()
(
)
IA
I
A
is conv
erge
nt.
Thus, the
ste
ady-state val
ue of the syst
em is:
Evaluation Warning : The document was created with Spire.PDF for Python.
ISSN: 23
02-4
046
TELKOM
NI
KA
Vol. 12, No. 9, September 20
14: 67
11 – 672
4
6718
T
m3
12
4
()
xx
x
xx
x
.
In the same
way, In the area
④
, we set the c
ont
rol rule for
()
t
u
as
:
24
1
11
12
13
1
1
2
14
2
2
2
21
23
2
1
()
()
=
()
1
aa
x
a
x
b
ut
x
t
ut
aa
x
a
x
b
x
u
(28
)
So, the dynamic beh
avior
of the system
becom
es:
m4
m
4
d(
)
()
d
t
t
t
x
Ax
b
(29
)
Whe
r
e,
24
1
1
2
13
12
1
3
1
m4
1
4
2
21
21
2
3
23
2
31
32
3
3
1
1
ax
a
x
b
a
a
x
aa
x
a
x
b
a
x
kk
k
A
,
1
1
m4
1
1
12
4
31
32
33
=
μ
ρ
μ
ρ
kx
k
x
k
x
b
31
k
,
32
k
and
33
k
match
(23). In ord
e
r
to ensu
r
e t
hat
the syste
m
is asympt
otically stea
d
y
,
m4
A
matche
s that
m4
()
IA
is non
sing
ula
r
and
1
m4
m
4
()
(
)
IA
I
A
is conv
erge
nt.
Thus, the
ste
ady-state val
ue of the syst
em is:
T
m4
12
4
()
xx
x
xx
x
Und
e
r the act
i
on of the con
t
rol rule
s (20
)
, (24), (26
)
an
d (28), the form of the motion equatio
n o
f
the s
y
s
t
em is
as
follows
.
mm
d(
)
()
d
ii
t
t
t
x
Ax
b
(
1
,2
,
3
,4
)
i
Then, solving
the above eq
uation, we g
e
t:
mm
m
0
()
e
x
p
(
)
(
0
)
e
x
p
(
)
d
t
ii
i
tt
t
xA
x
A
b
A
cco
rdi
ng t
o
m
i
A
, we can
kno
w
that the system
is asymp
t
otically stead
y. So,
11
mm
m
m
m
()
e
x
p
(
)
(
0
)
ii
i
i
i
tt
xA
x
A
b
A
b
Thus, the
ste
ady-state val
ue of the syst
em is:
1
mm
m
()
ii
i
xA
b
x
Then, we h
a
ve:
mm
m
(
)
exp(
)
(
0)
ii
i
tt
xA
x
x
x
Evaluation Warning : The document was created with Spire.PDF for Python.
TELKOM
NIKA
ISSN:
2302-4
046
Control Strategy Anal
ysis
on Pre
v
enti
v
e
Maintenan
ce
(Hon
gshen
g Su)
6719
Und
e
r th
e
co
ndition that th
e initial valu
e
(0)
x
an
d the
exp
e
cted
ste
ady-state valu
e
m
i
x
are
given o
u
t, it is quite
obviou
s
that
the mo
tion
e
quation
of th
e sy
stem i
s
only rel
a
ted
to
m
ex
p(
)
i
t
A
. So, we use the method of
resolvent mat
r
ix to solve
m
ex
p(
)
i
t
A
[13]. Thus, we
have:
11
mm
exp(
)
(
)
ii
tLs
AI
A
Firstly, we co
ndu
ct the followin
g
definitions.
12
4
1
12
1
3
1
1
wa
x
a
x
b
x
,
21
4
2
21
23
2
1
wa
x
a
x
b
x
32
1
11
22
33
11
33
11
22
22
3
3
12
21
23
32
31
13
11
22
33
12
2
3
31
13
32
21
13
31
22
12
21
33
32
23
11
()
(
)
(
)
p
s
s
s
k
k
k
s
kk
kk
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
kk
kk
k
k
k
k
32
21
1
22
33
22
33
22
33
12
2
1
23
32
13
31
1
22
33
32
23
12
23
31
21
33
13
32
21
31
22
()
(
)
()
(
)
(
)
ps
s
s
w
k
k
s
w
k
k
k
k
a
k
k
k
a
k
wk
k
k
k
a
k
k
k
k
a
k
k
k
k
32
32
2
11
33
11
33
11
33
12
21
23
32
31
13
2
11
33
13
31
23
12
31
32
11
21
13
32
12
33
()
(
)
()
(
)
(
)
ps
s
s
k
k
w
s
w
k
k
k
k
k
a
a
k
k
k
wk
k
k
k
a
k
k
k
k
a
k
k
k
k
32
41
2
1
2
2
33
33
33
12
21
32
2
3
13
3
1
21
1
1
3
31
33
32
23
12
21
33
12
2
3
31
13
32
21
()
(
)
(
)
ps
s
s
w
w
k
s
w
k
w
w
k
a
a
k
a
a
k
wa
k
w
k
k
a
w
a
a
k
a
a
k
a
k
a
2
1
22
33
22
33
23
32
()
(
)
qs
s
s
k
k
k
k
k
k
,
2
2
11
33
11
33
13
31
()
(
)
qs
s
s
k
k
k
k
k
k
2
3
11
22
11
22
12
21
()
(
)
qs
s
s
k
k
k
k
k
k
,
4
21
33
23
31
()
(
)
qs
k
s
k
k
k
5
31
22
21
32
()
(
)
qs
k
s
k
k
k
,
6
12
33
13
32
()
(
)
qs
k
s
k
k
k
7
32
11
12
31
()
(
)
qs
k
s
k
k
k
,
8
13
22
12
23
()
(
)
qs
k
s
k
k
k
9
23
11
13
21
()
(
)
qs
k
s
k
k
k
,
1
12
33
13
32
()
(
)
f
sa
s
k
a
k
2
21
1
33
33
13
31
()
f
ss
s
w
k
w
k
a
k
,
31
32
12
31
()
f
sk
s
w
a
k
4
13
22
12
2
3
()
(
)
f
sa
s
k
a
k
,
51
23
13
21
()
f
sk
s
w
a
k
2
61
1
22
22
12
21
()
f
ss
s
w
k
w
k
a
k
,
2
12
2
3
3
33
23
3
2
()
g
ss
s
w
k
w
k
a
k
2
21
33
23
3
1
()
(
)
g
sa
s
k
a
k
,
32
31
21
32
()
g
sk
s
w
a
k
42
13
23
12
()
g
sk
s
w
a
k
,
5
23
11
21
1
3
()
(
)
g
sa
s
k
a
k
2
62
2
11
11
21
12
()
g
ss
s
w
k
w
k
a
k
,
12
13
12
23
()
hs
a
s
w
a
a
21
23
13
21
()
+
hs
a
s
w
a
a
,
2
31
2
1
2
12
21
()
hs
s
s
w
w
w
w
a
a
.
And then, we
have:
14
7
11
1
25
8
1
m1
11
1
36
9
11
1
(
)
()
()
()
()
()
()
(
)
()
ex
p
(
)
()
()
()
()
()
()
()
()
()
q
s
qs
qs
p
s
ps
ps
qs
q
s
q
s
tL
p
s
ps
ps
qs
q
s
q
s
p
s
ps
ps
A
,
11
4
22
2
22
5
1
m2
22
2
33
6
22
2
()
()
()
()
()
()
()
(
)
()
ex
p
(
)
()
()
()
()
()
()
()
()
()
qs
f
s
f
s
ps
ps
ps
qs
f
s
f
s
tL
ps
ps
ps
qs
f
s
f
s
ps
ps
ps
A
,
14
4
33
3
25
5
1
m3
33
3
36
6
33
3
()
()
()
(
)
()
()
()
()
()
e
xp(
)
(
)
()
()
(
)
()
()
(
)
()
()
g
sq
s
g
s
ps
ps
ps
g
sq
s
g
s
tL
ps
ps
ps
g
sq
s
g
s
ps
ps
ps
A
,
11
1
444
22
2
1
m4
444
33
3
444
()
()
(
)
()
(
)
()
(
)
()
()
e
xp(
)
()
(
)
()
()
()
()
()
(
)
()
g
sf
s
h
s
ps
ps
ps
g
sf
s
h
s
tL
ps
ps
ps
g
sf
s
h
s
ps
ps
ps
A
.
Evaluation Warning : The document was created with Spire.PDF for Python.
ISSN: 23
02-4
046
TELKOM
NI
KA
Vol. 12, No. 9, September 20
14: 67
11 – 672
4
6720
Based o
n
the above discu
ssi
on
s, unde
r t
he condition
that the initia
l value
(0)
x
and the
expecte
d ste
ady-state val
ue
m
i
x
are give
n out, we ca
n get the motion equation
()
t
x
of the
system by co
mbining
1
mm
m
()
ii
i
xA
b
x
and
m
ex
p(
)
i
t
A
.
5. Stabilit
y
Anal
y
s
is
As the ch
ecki
ng rate is
con
s
tant, we hav
e the linear
system as follo
ws.
d(
)
()
d
t
t
t
x
Ax
(30
)
Whe
r
e,
21
11
11
1
22
11
1
2
12
1
2
1
2
11
1
1
2
12
2
()
=)
(
μμ
μ
vu
ρρ
ρ
ρμ
λλ
μ
μ
v
μ
u
μ
ρ
ρλ
ρ
ρ
λ
vv
μ
A
Acco
rdi
ng to
the linea
r co
ntrol theo
ry [13]
and [14],
if all eigenval
ues
of the m
a
trix
A
posse
ss
the negative real pa
rts, a
nd then
syst
em (5.1) i
s
asymptoticall
y
steady. In
other word
s,
the
matrix
A
must match the foll
owin
g co
nditions.
(1)
()
IA
is non
sin
gular.
(2)
1
()
(
)
IA
I
A
is conve
r
gent.
As the ch
ecki
ng rate is n
o
t con
s
tant, we
hav
e the linear time-varying s
y
s
t
em as
follows
.
d(
)
()
(
)
d
t
tt
t
x
Ax
(31
)
Whe
r
e,
21
11
11
1
22
11
1
2
12
1
2
1
2
11
1
1
2
12
2
((
)
)
()
=
(
)
)
(
μμ
μ
vu
t
ρρ
ρ
ρμ
λλ
μ
t
μ
v
μ
ut
μ
ρ
ρλ
ρ
ρ
λ
vv
μ
A
1
()
ut
and
2
()
ut
are bo
u
nded for
[0
,
)
t
, the real num
bers
1
δ
and
2
δ
are existent and
sati
sfy:
1
1
0(
)
ut
δ
;
2
2
0(
)
ut
δ
.
In orde
r to qu
ote lemma
s, we cond
uct d
e
finitions a
s
follows.
()
t
A
: It denotes th
e matrix that is made u
p
of
the element
s that are the d
e
rivative of
everyone el
e
m
ent on
t
in the matrix
()
t
A
.
: It denotes th
e norm of
vectors o
r
matri
c
es.
*
A
: It denotes th
e gone tra
n
sp
osition mat
r
ix of matrix
A
.
()
R
e
: It denotes th
e real pa
rt.
Acco
rdi
ng to [15], [16] and [17], we have
the followin
g
lemma
s.
Lemma 1:
If the eigenval
u
e
of the n order
squ
a
re m
a
trix
()
t
A
that are
1
()
t
λ
,
2
()
t
λ
,
()
n
t
λ
matc
h:
()
(
)
2
0
ij
tt
δ
λλ
Re
(,
1
,
2
,
,
)
ij
n
Evaluation Warning : The document was created with Spire.PDF for Python.