Page 539 - Mechatronics with Experiments

Printer: Yet to Come
October 9, 2014 8:41 254mm×178mm
JWST499-Cetinkunt
JWST499-c07
ELECTROHYDRAULIC MOTION CONTROL SYSTEMS 525
Example Consider a single axis electrohydraulic motion control system as shown in
Figure 7.90 and its block diagram representation in Figure 7.89. We assume that the cylinder
is rigidly connected to its base and load. Let us consider the small movements of the valve
around its null position, that is a case that is maintaining a commanded position. Determine
the cylinder positioning error for the following conditions:
1. the amplifier gain is K = 200 mA∕10 V = 20 mA∕V,
sa
3
2. the valve gain around the null position operation, K = 20 (in ∕s)∕200 mA =
q
3
0.1(in ∕s)∕mA,
3. the cross-sectional area of the cylinder is A = 2.0 in on both sides (rod is extended
c
to both sides),
4. the sensor gain is K = 10 V∕10 in = 1V∕in,
fx
5. the deadband of the valve is 10% of the maximum input current, i = 0.1 ⋅ i =
db max
0.1 ⋅ 200 mA = 20 mA.
As long as the error magnitude times the gain of the amplifier is less than the deadband,
there will not be any flow through the valve and no position correction will be made. Only
after the error times the amplifier gain is larger than the deadband of the valve will there be
flow through the valve and correction to the position error. Hence, the error in the following
range is an inherent limitation of this system, since it does not result in any flow or motion
in the system,

X e,max ⋅ K = i db (7.339)
sa
i db
X = (7.340)
e,max
K
sa
20 mA
= (7.341)
20 mA∕V
= 1 V (7.342)
which is 10% of the total displacement range.
Any positioning error less than that value, while maintaining a commanded position
(by the regulating motion of the valve around the null-position), is inherently not corrected,

X ≤ X e,max (7.343)
e
i db
≤ (7.344)
K
sa
≤ 1 V (7.345)
which is equal to 1.0 in in physical units since K = 1V∕in.
fx
Notice that if the amplifier gain is increased, say by 10, the positioning error due to
the same valve deadband is reduced by the same factor.
i db 20 mA
X ≤ = = 0.1 V (7.346)
e
K sa 200 mA∕V
which is equal to 0.1 in in physical units since K = 1V∕in.
fx
In order to ensure that positioning error due to the valve deadband is less than a
desired value, the amplifier gain should be larger than a minimum value as follows,
i
K ≥ db (7.347)
sa
X
e,max
20 mA
≥ (7.348)
X V
e,max