Force control of a servoactuator

VHUYRDPSOLILHUJDLQ V HUYRGULYHWUDQVIHUIXQFWLRQ

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8.3 Force control of a servoactuator

210 iii) Gain scheduling

The charging response is acceptable but the discharging response is more sluggish and could be improved by increasing the gain when discharging. Now consider gain scheduling and increasing the discharging gain by the ratio of the flow gains:

Flow gain ratio

3R 3

J GLVFKDUJLQ

FKDUJLQJ

J GLVFKDUJLQ

FKDUJLQJ

Therefore multiply the gain when discharging by a factor of 2.4, the result being as shown.

EDU3

The response is now very similar for both charging and discharging, and the duty cycle frequency may be increased to at least 10Hz, which actually may be too high in practice.

211

In practice a materials testing machine servoactuator will have a load cell attached to the actuator rod and it makes better dynamic sense to use this rather than pressure differential. In the author’s experience it has been more difficult to improve the closed-loop response with pressure differential control compared with force control to such an extent that even a variety of compensation techniques have proved fruitless.

For a force control system the linearised equations for a servovalve/cylinder are as previously used and as follows:

Y 3 3 3

GW 0G8 8

% )

TL DFW 5

G3 ȕ

9 5

$8 3 L

N ^" ^"

\ N ) IRUFH ORDG

VWUXFWXUH _P

The material stiffness is NPN/m and y is the structure deflection from its unloaded position. Servoactuator leakage has been included since this plays an important role in how the closed-loop system behaves.

(8.17) (8.18) (8.19)

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1349906_A6_4+0.indd 1 22-08-2014 12:56:57

212 Proportional control.

Consider the block diagram is as shown in figure 8.6.

NTL*D

V&5

⁸

9G ₃

V N_P

$ 0V

)

Figure 8.6 Force control for materials testing, P control

When a moving component with mass has a spring-type characteristic it is usual to add ⅓ of the mass of the component to the actuator/load cell mass.

The closed-loop linearised force and velocity transfer functions are as follows:

5 /&&

&5 5 5/

5 5 5 V&

. V

) )V

P Y P

P Y

5 /&&

&5 5 5/

5 5 5 V&

N V . GV

) 8V

P Y P

P Y

VY DFW

TL D

I 5

& 9

$ / 0

%Y 5Y 5$

* +

The mechanical capacitance is defined as

P P N

& $

The closed-loop transfer functions are 3^rd order and for the closed-loop system to be stable, given that 5Y5 may usually be neglected, then:

/ 5&5

. ^P ^Y ^P

(8.20)

(8.21)

(8.22)

(8.23)

(8.24)

213

Applying the Final Value Theorem to each transfer function then gives the following steady-state condition:

TL5$

DN I* +

TL5$

DN I* + .

. )

)

The following observations are made:

• this 3^rd order system can become unstable and a suitable gain . must be selected, and different to that given by (8.24) if servovalve dynamics are included

• the steady-state force will not achieve the demanded value unless K >>1 or flow losses are negligible such that 5 However, this is only an input calibration issue in practice since the input signal may be adjusted to give the required steady-state load force

• the steady-state velocity will be zero as required irrespective of the value of leakage resistance 5 PID control.

This is considered to ostensibly improve performance but results in a higher-order characteristic equation.

This means that explicit PID gain selection requires care and controller tuning information will be supplied by the experienced testing machine manufacturer. Consider therefore the system with PI control only as shown in figure 8.7.

V&5

9G ₃_Ɛ 8

V N_P

$ 0V

)

TL* V N .L

Figure 8.7 Force control for materials testing, PI control

The appropriate transfer functions now become:

*V . N V

V . GV )

*V . .V V )

)V _L _P ^L

5 /&&

&5 5 5/

5 5 5

V . V ..

^Y ^P

P P Y

(8.25)

(8.26)

(8.27)

214

Considering the Final Value Theorem now gives the following steady-state conditions:

) )G8

The following observations are made:

• the closed-loop characteristic equation is now 4^th order and the conditions to avoid closed- loop instability are not in a meaningful explicit design form. However, a suitable gain must be selected to avoid closed-loop instability

• the closed-loop response to a step input force demand will now result in no steady-state error in force and velocity and irrespective of any system leakage resistance 5

(8.28)

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215 Example 8.2

Consider a test system under force control with the test specimen to be deflected by 1mm prior to cyclic fatigue testing. Determine the possible significance of servovalve dynamics on the step response. Data are as follows:

/RDGFHOO )

Servoactuator bore 80mm diameter, double-rod 56mm diameter, full stroke 200mm.

The actuator piston is operating about its mid position Servoactuator leakage resistance 5DFW [1PPV Servovalve supply pressure 3V = 140bar

Servovalve flow equation is

3 [ 3

4 L ^V "

i mA, pressure N/m²

Servovalve dynamics damping ratio ȗ = 1.0, undamped natural frequency ȦQ

= 110Hz Structure stiffness NP

^[^1P Fluid bulk modulus ȕ [1P

Viscous damping %Y 1PV Moving mass M = 85kg Load cell gain +I 91 The calculations proceed as follows:

Mean load force )G NP\PHDQ

[ [1 Actuator net cross-sectional-area > @[ [ P

$ ʌ

Load pressure differential 3ƐR [[ [1P EDU To overcome losses and dynamic operation it is usual practice to select a supply pressure using:

3V [3ƐR EDU

So, a supply pressure of 140bar should be adequate to provide the additional forcec required for fatique testing, as discussed in the next example.

216

Now consider the terms needed to evaluate the linearised transfer function neglecting servovalve dynamics.

2/7)

5 /&&

&5 5 5/

5 5 5 V&

*V+V .

P Y P

P Y

[ P V P$

[

3 [ 3

N ^V ^R

1 / 3 2 12 act

1 / 3 2 10 6

2 4 2

v v

5 13 - 7

6 2 2

5 12 9

4 6 6

2 2

s m Nm 5x10 R

s m Nm 0.153x10 10

2.56 10 A

R B

/N m 28

. 2.x10 3

x10 2.56 k

C A

/N m 0.183x10 1.4x10

)(0.1) (2.56x10

C V

kg/m 3x10 x10 1

2.56 85 A

L M

x10

1 / 3 2 12 act

1 / 3 2 10 6

2 4 2

v v

5 13 - 7

6 2 2

5 12 9

4 6 6

2 2

s m Nm 5x10 R

s m Nm 0.153x10 10

2.56 10 A

R B

/N m 28

. 2.x10 3

x10 2.56 k

C A

/N m 0.183x10 1.4x10

)(0.1) (2.56x10

C V

kg/m 3x10 x10 1

2.56 85 A

L M

x10

1 / 3 2 12 act

1 / 3 2 10 6

2 4 2

v v

5 13 - 7

6 2 2

5 12 9

4 6 6

2 2

s m Nm 5x10 R

s m Nm 0.153x10 10

2.56 10 A

R B

/N m 28

. 2.x10 3

x10 2.56 k

C A

/N m 0.183x10 1.4x10

)(0.1) (2.56x10

C V

kg/m 3x10 x10 1

2.56 85 A

L M

x10

1 / 3 2 12 act

1 / 3 2 10 6

2 4 2

v v

5 13 - 7

6 2 2

5 12 9

4 6 6

2 2

s m Nm 5x10 R

s m Nm 0.153x10 10

2.56 10 A

R B

/N m 28

. 2.x10 3

x10 2.56 k

C A

/N m 0.183x10 1.4x10

)(0.1) (2.56x10

C V

kg/m 3x10 x10 1

2.56 85 A

L M

x10

1 / 3 2 12 act

1 / 3 2 10 6

2 4 2

v v

5 13 - 7

6 2 2

5 12 9

4 6 6

2 2

s m Nm 5x10 R

s m Nm 0.153x10 10

2.56 10 A

R B

/N m 28

. 2.x10 3

x10 2.56 k

C A

/N m 0.183x10 1.4x10

)(0.1) (2.56x10

C V

kg/m 3x10 x10 1

2.56 85 A

L M

x10

D TL

I* N 5$ * [ [ [ *

QHJOLJLEOH DQG

[

[ 5Y

[

The OLTF may now be evaluated:

*V+V

V [ V [ V

217

It is particularly useful to now use Matlab Control System Toolbox to investigate this OLTF. First consider the roots of the denominator and determined as follows:

a = [1.95e-6 2.35e-4 2.1 1];

roots(a);

-60 + M1036 -60 – M1036 -0.5

Clearly the denominator of the OLTF is characterised by a 1^st order component and a very lightly damped 2^nd order high frequency component. Now using the frequency response plotting facility assuming unity gain in the numerator:

a = tf([0 0 0 1],[1.95e-6 2.35e-4 2.1 1]);

bodeplot(a);

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218

The components of the OLTF may clearly be seen, and by zooming into a narrower frequency range the -180^o phase angle occurs when ȦG = 1038rad/s and with a gain of -48dB. Hence the onset of closed-loop instability occurs when the magnitude is lifted by +48dB.This gives:

+48 =20log(23.7*D) → *D = 10.6mA/V

Using the traditional approach the closed-loop characteristic equation is given by:

V [ V [ V

V E V E V E

E_R

The onset of closed-loop instability occurs when EE ERE

[

* [

P$9

*_D

The frequency of oscillation Ȧ EE [ UDGV+]

Alternatively in the frequency domain the characteristic equation is:

Ȧ [ MȦ

Ȧ [

219

The onset of instability occurs, by equating the Imaginary term to zero, which gives ȦG = 1038rad/s (165Hz) and from the Real term, when equated to zero, then *D = 10.6 mA/V.

The important conclusion at this stage is that the servovalve undamped natural frequency is 110Hz so it is clear that the linearised OLTF must also include servovalve dynamics.

The new OLTF is now 5^th order and is given by:

V [ V

[ V

V [ V

[

*V+V

V [ V

[ V

*V+V

ce = [4.1e-12 6.15e-9 7.04e-6 6.33e-3 2.1 1];

roots(ce);

-59.6 ± M1036, -0.5, -728, -653

The first three roots for the system only are clearly identified again and the other two real roots relate to the addition of the servovalve critically-damped dynamics.

gh = tf([0 0 0 0 0 1],[ 4.1e-12 6.15e-9 7.04e-6 6.33e-3 2.1 1]);

bodeplot(gh);

g1=tf([0 0 1],[2.1e-6 2.9e-3 1]);

g2=([0 0 0 1],[1.95e-6 2.35e-4 2.1 1]);

gh=g1*g2;

bodeplot(gh);

220

The condition for the onset of closed-loop instability, when the phase angle is -180^o, gives:

ȦG = 620rad/s (98.6Hz) *D = 64mA/V

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Dalam dokumen PDF The load pressure differential is defined as (Halaman 172-192)