Dynamic model of an hydraulic clutch
Let us consider the following simplified dynamic model of an hydraulic clutch:
P Supply pressure
Q Flow rate through the valve Kv Valve proportional constant
Cm Hydraulic capacity of the cylinder P1 Pressure within the cylinder
A Section of the piston x Position of the piston
˙x Velocity of the piston mp Mass of the piston
b Friction coefficient of the piston Km Stiffness of the spring
Fm Spring force against the piston
Valve (Kv) mp
b, A Km
P R = 0 x
P1 Cm Qx
Q
The corresponding POG dynamic model is:
P
Q s
s -
Kv
?
?
s -
s -
1 Cms
6
6
P1
-
Qx
A
- A -Fx
-
1 b+mps
?
?
˙x
s -
s -
6
1 s
6
Km
6
Fm
-
x
0
s s
Let us introduce the following auxiliary variables:
G1 = Kv, G2 = 1
Cms, G3 = 1
b+ mps, G4 = Km s Using the Mason formula ones obtains the following transfer function:
G(s) = Fm(s)
P(s) = A G1G2G3G4
1 + G1G2 + A2G2G3 + G3G4 + G1G2G3G4 which, by substitution, simplifies as follows:
G(s) = AKmKv
Cmmps3 + (Cmb+ Kvmp)s2 + (A2 + CmKm+ Kvb)s + KmKv The corresponding differential equation is:
Cmmp...
Fm + (Cmb+ Kvmp) ¨Fm+ (A2 + CmKm+ Kv b) ˙Fm + KmKvFm = AKmKvP(t)
The system is characterized by the following static and differential equations:
Q = Kv(P − P1) CmP˙1 = Q − Qx
mpx¨ = Fx − b ˙x − Fm
Fm = Km x Fx = A P1 Qx = A ˙x
If the system equations are written without an order and without a method, it is difficult to get the dynamic order of the system or to understand if all the static and dynamic equations have been written. To cope with this problems it is useful to introduce a “state vector”:
x =
P1 ˙x x T
The system equations can now be rewritten as follows:
Cm P˙1 mpx¨
˙x
=
−Kv −A 0 A −b −Km
0 1 0
P1
˙x x
+
Kv
0 0
P y = Fm
where P is the input pressure and y = Fm is the output force. The system equations can also be rewritten as follows:
P˙1
¨ x
˙x
| {z }
˙x
=
−Kv
Cm − A
Cm 0
A
mp − b
mp −Km
mp
0 1 0
| {z }
A
P1
˙x x
| {z } x
+
Kv Cm
0 0
| {z }
B
|{z}P
u
y =
0 0 Km
| {z }
C
x
which, in compact form, is described as follows:
( ˙x = A x + B u
y = C x where
A is the system matrix B is the input matrix C is the output matrix
The final equilibrium point x0 can be computed setting ˙x = 0 and u0 = P : x0 = −A−1B u0, y0 = C x0
For the considered system it is:
x0 =
P1
˙x x
0
=
P
0
A P Km
, y0 = [Fm]0 = A P
The transfer function G(s) of a linear system described in the state space can be obtained using the following formula:
G(s) = C(sI − A)−1B
The obtained G(s) is equal to the G(s) obtained using the Mason formula.
If the system is nonlinear, the stability analysis is different. Let us suppose, for example, that, for the considered system, the valve and the spring are described by nonlinear static functions: Φv(∆P ) e Φm(x).
P
Q s
s -
Φv(∆P )
?
?
s -
∆P
s -
1 Cms
6
6
P1
-
Qx
A
- A -Fx
-
1 b+mps
?
?
˙x
s -
s -
6
1 s
6
Φm(x)
6
Fm
-
x
0
s s
The state space description of the nonlinear system is now the following:
P˙1
¨ x
˙x
=
−Φv(P1 −P)
Cm − A ˙x Cm A P1
mp − b ˙x
mp − Φm(x) mp
˙x
, Fm = Φm(x)
Setting x1 = P1, x2 = ˙x, x3 = x, x = [x1 x2 x3]T, u = u = P and y = Fm, the state space equations can be rewritten as follows:
˙x1
˙x2
˙x3
| {z }
˙x
=
Φv(u−x1)
Cm − A x2
Cm A x1
mp − b x2
mp − Φm(x3)
mp
x2
| {z }
f(x, u)
y=
Φm(x3)
| {z }
h(x)
↔
( ˙x = f (x, u) y = h(x)
where h(x) is a nonlinear function of the state vector x. In the general case the function h(x, u) also depends on the input vector u. Also in this case the final equilibrium point ¯x associated to the constant input u = ¯u can be determined imposing ˙x = 0:
f(¯x, u¯) = 0 →
Φv(¯u−¯x1) Cm = 0
Ax¯1
mp − Φm(¯x3)
mp = 0
¯
x2 = 0
→
¯
x1 = ¯u − Φ−v 1(0)
¯
x2 = 0
¯
x3 = Φ−m1(A ¯x1)
For nonlinear systems there may be several equilibrium points (also called “wor- king points”) associated with the same input vector u. A nonlinear system can be “linearized” in the neighborhood of an equilibrium point ¯x as follows:
˙x = ∂f(x, u)
∂ xT
(¯x,u)¯
(x − ¯x) + ∂f(x, u)
∂ uT
(¯x,¯u)
(u − ¯u) y = ¯y + ∂h(x)
∂ xT
(¯x)
(x − ¯x)
Setting ˜x = x − ¯x, ˜y = y − ¯y and ˜u = u − ¯u one obtains the following linearized dynamic model:
( ˙˜x = A ˜x + B ˜u
˜
y = C ˜x where
For the considered system it is:
A =
−Kv
Cm − A
Cm 0
A
mp − b
mp −Km
mp
0 1 0
, B =
Kv Cm
0 0
, C =
0 0 Km
where
Kv = −∂Φv(u − x1)
∂ x1
(¯x1,¯u)
= ∂Φv(u − x1)
∂ u
(¯x1,¯u)
= ∂Φv(w)
∂ w
w=Φ−1v (0)
Km = ∂Φm(x3)
∂ x3
x3=¯x3
dove x¯3 = Φ−m1(A (¯u − Φ−v 1(0)))
This obtained linearized model is equal to the same linear dynamic model initially considered.
