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Lesson 18

Problem 18.1

The system below is a rotary
motion control system. The purpose of the system is to cause a large
load to rotate through a sequence of specified positions. In this case,
the sequence is a step input of 1 rad, followed by a ramp input of 1
rad/s, all with zero initial conditions. The design problem is to select
system parameters to obtain the desired motion subject to specified
performance criteria. (This scheme was actually used to position antiaircraft
guns aboard navy ships. Its basic principles can of course be considered
for any
rotary motion control system.)

System operation

Command voltage e</span><sub>V
is proportional to the desired position R (called the input reference).
Feedback voltage e</span><sub>B
is proportional to the measured position of the load.
The reference
input is compared to the measured position, and the voltage difference
is the input to a force motor.
The output
force of the motor acts on a rigid bar that translates (no rotation),
simultaneously imparting the same displacement x</span><sub>S to the spring damper system and to the valve
of a hydraulic amplifier.
The valve
opens, applying supply pressure to the piston, causing it to move and
stroke the pump. The piston position x</span><sub>P is affected also by the motion of the spring-damper
system.
The pump,
driven by an electric motor (not shown), pumps oil to a hydraulic motor,
which rotates the load towards the commanded position.

Figure 18.1: System for rotary motion control

(adapted from E.O. Doebelin, System Dynamics, Dekker,
1998).

Modeling

We are given that voltage e</span><sub>v
is proportional to the desired (reference) load position R,

. (1)

A potentiometer measures the load position   and produces a proportional
voltage e<sub>b,

. (2)

The output of the summing junction is voltage e</span><sub>1,
given by

. (3)

The force f<sub>m produced by the force-motor is proportional to e</span><sub>1,

. (4)

Letting x<sub>S > x<sub>A > x<sub>P, then applying the conservation of linear momentum
(COLM) to m<sub>S
yields,

. (5)

In this system m<sub>S is negligible. Assigning m<sub>S = 0 and substituting (1)-(4) into (5) yields

, (6)

where a new constant K K_mK<sub>v/k.  Applying the COLM to mass m<sub>A yields

, (7)

where a new constant 1 B/k.  The servo valve is modeled by

, (8)

where k<sub>VLV is the valve constant and the valve opening x<sub>V
is given by

. (9)

The servo piston is modeled by

. (10)

Combining (8)-(10) yields

, (11)

where a new constant 2 A<sub>P /k<sub>VLV.  To simplify the problem, we neglect
the dynamics of the pump, motor, and load, and simply relate the pump
input stroke to the motion of the load using

. (12)

Collecting eqations (7), (11), and (12) into one set
of equations, substituting (6) and rearranging yields the final system
model given by the following linear differential equations:

(13)

In this system, R is the input and the unknowns (outputs) are x<sub>A, x<sub>P,
and .

Problem statement

Create a Simulink model using (13). Run the simulation from a Matlab
m-file in which you can modify values for 1 and K. The time span for the simulation should be 3 or 4 seconds.
Known gains: 2 = 1/40 s, and K</span><sub>L = 1/50 in/(rad/s).
All displacements x are in inches.
All angular  displacements are in radians.
The input
to the system is a step input of 1 rad at t = 0, followed by a ramp input of 1 rad/s at t = 1.5 s. All initial conditions are zero. You can create
this input using the Ramp function under the Sources library in Simulink. Its parameter window is shown
below.

This
is the time the ramp starts step.

This
is the ramp slope.

This
is the initial unit step.

Design: select values for 1 and K such that the following performance criteria are met:

The step response
overshoot is no greater than 50%. (Try 20%?)
Position error R(t)(t) is 0.01 (try 0.5%?) no later than 1 second after
the onset of the step input and no later than 1 second after the onset
of the ramp input.
To help you
get started, limit your search such that and .

Results:

Plot R(t) and (t) on the same figure.
Plot R(t)(t) in a separate figure to show that the error is 0.01
in the time required.

What to turn in.

Your answer to the design
problem includes:

Final values
of 1 and K.
Simulation
results showing that the performance criteria for overshoot and position
error are satisfied.
A printout
of your simulation diagram and the m-file that runs the simulation.
(Write  comments in your m-file to describe the purpose of each
section of code)