TMHP51 Servomechanisms (HT2012)
Lecture 05
Frequency domain
Linearization
Operational point
Magnus Sethson
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Model Analysis
f
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t
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2
Taylor series expansion
T (x1 , . . . , xd ) =
1
X
n1 =0
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...
1
X
(x1
nd =0
n1
a1 ) . . . (x1
n 1 ! . . . nd !
ad )
nd
✓
n1 +...+nd
@
f
@xn1 1 . . . @xnd d
◆
(a1 , . . . , ad )
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Linearization around an Operating Point
“A great and common engineering tool”
f (x) ⇡ f (p) + rf |p · (x
p)
Two variable example:
@f (x, y)
f (x, y) ⇡ f (a, b) +
@x
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(x
(a,b)
@f (x, y)
a) +
@y
(y
b)
(a,b)
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Non-linear Hydraulic Servo System Model
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p
A
B
ẋ
F
=
M
ẍ
>
L
p
p
p
L
t
p
>
>
<
q = Ap ẋp + 41 Vet ṗL
>
q
>
>
: q = Cq wxv 1 (Ps pL )
⇢
xv : Input signal
xp : Output signal
FL : External disturbance
xp , ẋp , pL : Internal states
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Linearized Model (Flow Equation)
@q
q=
@xv
(xv0 ,pL0 )
r
@q
xv +
@pL
pL
(xv0 ,pL0 )
@q
1
= Cq w
(Ps pL0 ) = Kq
@xv
⇢
@q
Cq wxv0
p
=
= Kc
@pL
2 ⇢(Ps pL0 )
q = Kq x v
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K c pL
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Linearized Model (Other Linear Parts)
q = Ap
Ap p L
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Bp ẋp
FL = M t
Vt
ẋp +
4 e
ẍp
ṗL
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Laplace Transform
F (s) = L [f (t)] (s) = L [f (t)] (i!) =
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Z
1
f (t)e
st
dt
0
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Simple Servo System in Frequency Domain
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>
>
<
>
>
:
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PL A p
Bp s Xp
Q = Ap s X p +
Q = Kq Xv
FL = Mt s
1 Vt
4 es
2
Xp
PL
K c PL
f
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page 9.
Internal Leakage: There are two sources of internal leakage;
Operation Point Coefficients
rvo or Proportional Valve frequency
nal amplitude, supply pressure, and
eters.The typical response varies with
ed by the change in frequency of the
figure 2. Note that Direct Drive Valve
system pressure.
Proportional Valve step response will
pressure and internal valve design
eries catalogs for specifications. Full
ill normally exhibit a straight line
ow saturation of the pilot stage.The
rtion will vary with the square root
sure.
GE
FIGURE 3
CHANGE IN CONTROL FLOW WITH
CURRENT AND LOAD PRESSURE
100
100% INPUT CURRENT
CONTROL FLOW–% RATED FLOW
80
75%
60
50%
40
25%
20
-100 -80
-60 -40
-20
-20
25%
-40
-60
-80 -100
-20
-40
50%
75%
-60
-80
100%
first, flow through the hydraulic amplifier (known as “tare flow”)
which is relatively constant, and second, flow around the spool
which varies with its position. Maximum internal
rleakage occurs
at null. See individual
Servo and Servo-Proportional1Valve catalogs
@q
L
for K
specifications.
(Ps pL0 )
= Cq w
q =
@xv pL =const
⇢
Spool Driving Forces: The maximum hydraulic force available
to drive the second-stage
spool will depend upon
the supply
Cq wx
@qL
v0
pressure,
multiplied
by
the
end
of
the
spool.
In
the
case
of
p
Kc =
=
Direct Drive Valves,
driving force is2created
@pL spool
⇢(Pbys the linear
pL0 )
xv =const
force motor and does not change with supply pressure.
@pL
Kq
2(Ps pL0 )
Kp =Gain: A measure of the change
= in control
= port pressures
Pressure
@xv isqvaried
xv0
as the input current
about the K
zero
c flow point. Pressure
L =const
gain is measured against a blocked load under no flow conditions.
Normally the pressure gain exceeds 30% of the supply pressure
for 1% change in rated current and can be as high as 100%.
Fthe
L,max
Null Bias: InputA
current
to
valve required to adjust the
⇡
p
output to zero flow. Most Moog
Inc. valves have mechanical
pL,max
adjustments which allow the null bias to be externally adjusted.
2
3 FL,max
pL,max ⇡ Ps ! Ap ⇡
3
2 Ps
pv = (Ps
pL ) (typically 70bar)
-100
LOAD PRESSURE DROP–% SUPPLY PRESSURE
5000
cs: Control flow to the load will
nations of load pressure drop and
figure 3.These characteristics closely
where:
QNL = no-load flow at
1,000 psi drop for
Servovalves and
150 drop for P.V.
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i
= actual/rated current (%)
PV = (PS – PR) – PL
qL = qN = KqN xv,max = Ap vmax
p
Ps
Kq,max = Kq0 = KqN p
pv
!
KqN
Ap vmax
=
xv,max
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Linear Flow Equation
RE 2
PONSE CHANGE
RESSURE
FIGURE 3
CHANGE IN CONTROL FLOW WITH
CURRENT AND LOAD PRESSURE
100
at other pressures
t 3,000 psi (210 bar)
3000
04000
PPLY PRESSURE (PSI)
100% INPUT CURRENT
CONTROL FLOW–% RATED FLOW
80
75%
60
25%
20
-100 -80
-60 -40
Null Bias: Inpu
output to zero
adjustments wh
50%
40
as the input cur
gain is measure
Normally the p
for 1% change
-20
-20
25%
-40
-60
-80 -100
-20
-40
50%
75%
-60
-80
100%
5000
-100
LOAD PRESSURE DROP–% SUPPLY PRESSURE
Q = Kq X v
K c PL
aracteristics: Control flow to the load will
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combinations of load pressure drop and
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s shown in figure 3.These characteristics closely
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Operation Null Point Coefficients
q L = pL = x v = 0
Valve coefficients
K
q0
Null
1
(4=-way,
Pcritical
)
Cq w
s = (Kq,maxcenter)
⇢
pL =0
@qL
Kc0 =
> 0 = (Kc,min )
L xv =0
coefficients (q@p=p
=x =0) – real
L
qL
xv
K q0
Kp0
p0
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@pL
=
@xv
ps q
Cq w
L
L =0
v
values
Kq0
=
= (Kp,max )
Kc0
K q max
pL 0
qL
pL
Kc0
K
@qL
=
@xv
r
pL
xv
0
K c min
Kq0
Kc 0
K p max
xv 0
qL 0
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Modeling Procedure, Summary
Linearization
Operation Point
Laplace Transform
𝛥-Drop
(by convention)
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pL Ap Bp ẋp FL = Mt ẍp
>
>
>
<
q = Ap ẋp + 41 Vet ṗL
>
q
>
>
: q = Cq wxv 1 (Ps pL )
⇢
8
>
>
<
>
>
:
8
>
>
<
>
>
:
p L Ap
Bp ẋp
FL = Mt ẍp
q = Ap ẋp +
1 Vt
4 e
q = Kq x v
K c pL
PL A p
Bp s Xp
q = Ap s Xp +
q = Kq X v
t
t
ṗL
FL = Mt s 2 X p
1 Vt
4 es
PL
f
K c PL
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2
P
A
B
sX
F
=
M
s
Xp
>
L
p
p
p
L
t
>
<
Q = Ap sXp + 14 Vet sPL
>
>
:
Q = K q X v K c PL
f
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Modeling Procedure, Coefficient Conditions
(x 6= 0)
(x = 0)
r
@qL
1
(4
way,
alveKqcoefficients
=
(Ps
= Cq w
@xv pL =const
⇢
critical
center)@qL
pL0 )
Kq0 =
@xv
Cq wxv0
@qL
@qL
Null coefficients
(q
=p
=x
p
Kc =
L = L
v=0) – real values
Kc0 =
@pL xv =const
2 ⇢(Ps pL0 )
@pL
qL
ps
K q0
Cq w
K q max
xv p 0
= Cq w
pL =0
r
1
Ps = (Kq,max )
⇢
> 0 = (Kc,min )
xv =0
L
Kc0
qL
pL
0
K c min
Kq0
pL
K p0
xv q 0 K c 0
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K p max
xv 0
L
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Next Lecture:
13:15, 2012-11-14, P34
Magnus Sethson
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