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Revised 2018 G. Pang

THE UNIVERSITY OF HONG KONG

DEPARTMENT OF ELECTRICAL & ELECTRONIC ENGINEERING

Experiment C21: Servo Position Control System

Objectives: To study the characteristics and measure the parameters of a servo position

control system, and to investigate the effect of proportional and velocity

feedback.

Apparatus: Feedback servo trainer with a mechanical unit & an analogue unit;

Digital storage oscilloscope (DSO).

Introduction

Fig. 1 shows a servo position control system in which a load is driven by a motor through a

reduction belt mechanism. The angular position θ o of the load shaft is measured using an

output potentiometer. The aim of servo position control is to make θ o track a command

signal θ i set by an input potentiometer. In this experiment, we will study the mechanism of

a servo position control system and the effects of position and velocity feedback.

To begin with, we consider the system without velocity feedback, so that the system is simply

a proportional feedback system. In this case, the output position θ o is the only feedback

signal. θ o is compared with the reference input θ i to give an error signal e = θ i −θ o. The

purpose of the position feedback is to make θ o track θ i (i.e. to make e small). In the

proportional feedback scheme, the control signal u is directly proportional to the error signal

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e (hence termed proportional feedback). The proportional control should drive the motor in

a way so as to reduce the error. Notice that the proportional gain K1 of the error amplifier in

the forward path is the only adjustable parameter in a proportional control scheme, and this

does not provide enough freedom for obtaining a satisfactory transient response.

To improve the transient performance, the angular velocity of the motor is measured by a

tachometer (which is essentially a dc motor working as a generator) and velocity feedback

is added through an adjustable gain K2. The resulting system (as shown in Fig. 1) is said to

have proportional plus velocity feedback.

In order to study the characteristics of the servo position system, we need to obtain a

mathematical description for the system. In the analysis to follow, we will first consider the

dynamics of the motor-load mechanism (i.e. the open-loop system). Once we have derived

a relationship between the control signal u and the output θ o for the open-loop system, the

closed-loop system can be studied by means of block diagram analysis.

Motor torque-speed characteristics

An armature controlled motor can be represented as in Fig. 2. The motor is represented by

an armature resistance R in series with an idealized motor M which produces a torque T

proportional to the armature current I :

T = Ki I (1)

where Ki is the torque constant (Nm/A) of the motor.

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Fig. 2

The motor M also produces a back emf Vb proportional to the angular speed ωm (t) , i.e.,

Vb = Kb ωm (2)

where K

b is a constant (volts/rad/sec). It follows from Fig. 2 that

V = I R+Vb = I R+ Kb ωm (3)

Therefore,

𝐼 =V − 𝐾𝑏𝜔𝑚

R

(4)

Substituting (4) into (1) gives the torque-speed relationship of the motor:

𝑇 =𝐾𝑖

𝑅𝑉 −

𝐾𝑖𝐾𝑏

𝑅𝜔𝑚

(5)

For a constant V, the graph of T against ωm has a form shown in Fig. 3.

Fig. 3

Note that the constants Ki and Kb are related, as follows. Equating the mechanical power

generated by the motor to the electrical power gives

𝑇𝜔𝑚 = 𝑉𝑏𝐼 = (𝐾𝑏𝜔𝑚)(𝑇

𝐾𝑖)

(6)

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where the last equality follows from (1) and (2). This shows that

Ki = Kb (7)

Further details of the use of motors in servo systems can be found in [1].

Speed response

Let J be the total moment of inertia (referred to the motor shaft) of the motor together with the

load. The torque generated by the motor is used to accelerate the total moment of inertia and

to overcome viscous friction, giving

𝑇 = 𝐽�̈�𝑚 + 𝑏�̇�𝑚

(8)

where θm= angular position of motor shaft and b=viscous friction coefficient (Nm/rad/sec).

Since 𝜔𝑚 = �̇�𝑚, (8) can be written as

𝑇 = 𝐽�̇�𝑚 + 𝑏𝜔𝑚

(9)

Equating (5) and (9) gives

𝐽�̇�𝑚 + ( 𝑏 +𝐾𝑖𝐾𝑏

𝑅 ) 𝜔𝑚 =

𝐾𝑖

𝑅𝑉

(10)

which can be written as

𝜏 �̇�𝑚 + 𝜔𝑚 = 𝐾𝑚𝑉

(11)

where

𝜏 =𝐽𝑅

𝑏𝑅 + 𝐾𝑖𝐾𝑏

and (12)

𝐾𝑚 =𝐾𝑖

𝑏𝑅 + 𝐾𝑖𝐾𝑏

The speed response of the system is therefore governed by the differential equation (11). The

differential equation can be turned into a transfer function by taking Laplace transform with

zero initial conditions, giving

𝜔𝑚(𝑠)

𝑉(𝑠)=

𝐾𝑚

𝑠𝜏 + 1

Let u be the control signal applied at the input of power amplifier and ωt be the velocity signal

measured at the output of the tachometer (see Fig. 1). For the analysis to follow, it would be

more convenient to express the speed response in terms of u and ωt . Since ωt ∝ωm and V ∝u,

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if we absorb the constants of proportionalities into Km (and with an abuse of notation, using

the same symbol Km for the new constant), the transfer function from u to ωt is given by

𝜔𝑡(𝑠)

𝑢(𝑠)=

𝐾𝑚

𝑠𝜏 + 1

(13)

(13) is the speed response relating the power amplifier input to the tachometer output.

Position response

Note that the output position signal θo is related to the velocity signal ωt by an integral

relationship:

𝜃0(𝑡) = 𝐾0 ∫ 𝜔𝑡(𝑡)𝑑𝑡𝑡

0

(14)

where Ko is a constant. The corresponding transform relationship is

𝜃𝑜(𝑠)

𝜔𝑡(𝑠)=

𝐾𝑜

𝑠

(15)

From (13) and (15), the position response transfer function from the power amplifier input

u(s) to the position output θ o(s) is given by

𝜃𝑜(𝑠)

𝑢(𝑠)=

𝐾𝑜𝐾𝑚

𝑠(𝑠𝜏 + 1)

(16)

Closed-loop Transfer Function

Making use of the transfer functions derived in (13) and (16), the closed-loop system of

Figure 1 can be represented by the block diagram shown in Fig. 4.

Fig. 4

From Fig. 4, the closed-loop transfer function is given by:

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𝜃𝑜(𝑠)

𝜃𝑖(𝑠)=

𝐾𝑜𝐾1𝐾𝑚

𝜏

𝑠2 +1 + 𝐾1𝐾2𝐾𝑚

𝜏 𝑠 +𝐾𝑜𝐾1𝐾𝑚

𝜏

(17)

This is a second order system with a transfer function of the standard form

𝐺(𝑠) =𝜔𝑛

2

𝑠2 + 2𝜉𝜔𝑛𝑠 + 𝜔𝑛2

(18)

For a system with a transfer function given by (18), the undamped natural frequency ωn

determines the "bandwidth" of the frequency response (and hence how fast the output

responds to input signals) and the damping ratio 𝜉 determines the characteristics of the

transient response. The system is said to be

underdamped if 𝜉 < 1

critically damped if 𝜉 = 1

overdamped if 𝜉 > 1

A graph showing the step response of G (s) for a range of values of 𝜉 against normalized

time ωn t is given in Fig. 5. Note that the step response is oscillatory (with overshoot) if the

system is underdamped, and the oscillations may take a long time to settle down if 𝜉 is small.

The critically damped and the overdamped responses do not show any overshoot, but the

response could be slow if 𝜉 is large. In the design of a position servo control system, one

often aims at a critically damped system.

A direct comparison of (17) and (18) shows that undamped natural frequency ωn and the damping ratio 𝜉 of the closed-loop system are related to the gain constants of the system by

𝐾𝑜𝐾1𝐾𝑚

𝜏= 𝜔𝑛

2 1 + 𝐾1𝐾2𝐾𝑚

𝜏= 2𝜉𝜔𝑛

Clearly, ωn and 𝜉 can be set to any desired values by suitable choices of K1 and K2.

Fig. 5

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Experimental Procedures

Important:

(a) Always switch off the power supply before making connections on the Analogue Unit.

Check that all connections are correct before switching on the power.

(b) Make a note in your Log Book of the Set No. of the Apparatus.

(c) You should take a picture of the digital storage oscilloscope (DSO) when you see some

results along the experiment. Include the photos to your report with description.

In the following, potentiometer settings are given as percentages of the full range.

1. Familiarization of the Setup

Fig. 6a The Mechanical Unit

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Fig. 6b Photo of the Mechanical Unit

Fig. 7a The Analogue Unit

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Fig. 7b Photo of the Analogue Unit

(1a) Identify the following components on the Analogue Unit – this will help you understand

what you are doing in the experiment.

- Signal source: how to obtain an input signal of a desired amplitude?

- Error amplifier: how to set the amplifier gain?

- Power amplifier: where on the Analogue Unit is the point u shown in Fig. 1?

- Tachometer: where is the tachometer physically located? Where can its output be tapped?

- Input potentiometer: where is it physically located? Where can the signal θ i be tapped?

- Output potentiometer: where is it physically located? Where can the signal θ o be tapped?

(1b) Zero setting: Ensure that no connections are made on the Analogue Unit, switch the

power supply ON. The motor may revolve. Turn the power amplifier zero control knob

(clockwise and anticlockwise) on the Analog Unit and see that this control can be adjusted

to drive the motor in both directions. Adjust the zero control knob to its central position so

that the motor is stopped.

(1c) To check the motor: Hold the motor-check switch (on the Mechanical Unit) to

one side (+V) and then to the other side (-V) and check that the motor runs clockwise and

then anti-clockwise. Note that the output shaft speed (in rpm) can be read on the digital

display on the Mechanical Unit. (There is a reduction ratio of 32:1 between the motor shaft

speed and the output shaft speed. Hence a motor speed of 1000 rpm will show up as 31 rpm

on the digital display.)

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Find out what kind of instrumentation is used for angular position and speed

measurements on the Analogue Unit, and include your findings in the report.

2. Motor torque-speed characteristics in open-loop (no feedback connection yet )

(2a) Switch off the power. Connect up the Analogue Unit as shown in Fig. 8 with P3 initially

set to zero.

Fig. 8 Connections for Experiments 2 and 3

(2b) Switch on the power. Note that P3 enables a voltage in the range 10V to be applied

to the power amplifier. A continuous input voltage is applied by gradually increasing the

setting of P3 to run the motor speed up to 2000 rpm (62 rpm at the output shaft) with the

magnetic brake fully upwards. Measure the voltage at the input of the power amplifier.

(2c) Connect the voltmeter to measure the armature current as shown in Fig. 8.

Set the brake successively to 1

6,

2

6… ,1 along the scale by the side of the Mechanical Unit.

For each brake setting read the angular speed of the output shaft on the digital display

and record the armature current which has been converted to a voltage measurable at the

terminal labelled 1V/A (i.e., 1 volt per ampere) on the Mechanical Unit (see Fig. 6).

Plot the armature current versus the motor speed. For the plotting, scale your speed

readings to refer to the motor shaft and express the result in rad/sec.

Compare your results with the general form given in Fig. 3.

Assuming that the power amplifier has a gain of 3.0, determine the parameters R and Kb

from your graph. You should do the plotting and the calculations during the laboratory

session.

Note: For the rest of the experiment, leave the magnetic brake at its uppermost position.

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3. Steady-state speed response in open-loop (no feedback connection yet )

(3a)

Switch off the power

With the same connections as in Experiment 2, set P3 to 20. The motor would rotate

continuously with the input voltage. Use a DSO to measure the voltage u at the input of

the power amplifier and the speed signal ωt (in the form of a voltage) at the output of the

tachometer.

(3b)

Switch on the power.

Repeat for a few settings of P3(40, 60, 80, 100). The ratio 𝜔𝑡/𝑢 should be approximately

constant, and equal to the dc gain Km of the transfer function given by (13).

4. Transient speed response in open-loop using a square wave input command (no feedback

connection yet )

(4a)

Switch off the power

Connect the Analogue Unit as in Fig. 9 with P1 set to zero, P3 set to 25 and the input

shaft angle θi (on the Mechanical Unit) set to 0.

Note that the input potentiometer is simply used to generate a dc offset voltage in this

experiment.

Set the test signal frequency (using knobs on the Mechanical Unit) to 0.1 Hz.

Fig. 9 Connections for Experiment 4

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(4b)

Switch on the power. Adjust the input potentiometer and P3 so that a square wave with

lower and upper levels equal to 0 V and +5 V is obtained at the output of the error

amplifier. Then, increase P1 to 100.

Record the speed response at the output of the tachometer. The speed response should

have a form shown below.

From the speed response obtained, determine the time constant τ and the gain Km. As a

check, compare the value of Km with that obtained in Experiment 3.

Switch off the power

5. Position control with proportional closed-loop feedback using a square wave input

command

(5a) The step response of the position control system with proportional feedback will now be

investigated to show the effect of variation of forward path gain. Note that the error amplifier

can be configured as shown below.

Rf

𝑣0 = 𝑅𝑓

100𝐾(𝑣1 + 𝑣2)

The error amplifier serves as a summing node where two or more signals are summed

and then amplified by a gain factor K1 determined by the feedback resistor Rf . Connect

up the Analog Unit as shown in Fig. 10. Do not make the gray connections yet.

Identify and set the error amplifier feedback resistor to 100 kΩ, giving an amplifier gain

= 1. Initially set both P1 and P3 to zero. Turn the input potentiometer to zero position

(θi = 0). Set the signal frequency of the square wave to 0.1 Hz.

100 K

v 1

v 2

0 v

100 K

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Fig. 10 Connections for Experiments 5 & 6

(5b)

Switch on the power. Set P1 to 50. (What is the corresponding value of K1?)

Adjust P3 to 50 (to provide a +5V square wave).

Record the step response. Observe whether the response is overdamped.

(5c)

Repeat with P1 increased to 100 (P3 is still 50) and note the effect of increasing the forward

path gain.

(5d)

With P1 = 100, reduce P3 to zero (i.e. no square wave) and see if you can use the input

potentiometer to control the output potentiometer to any desired position (say 90 degrees ).

Is the tracking satisfactory?

(5e)

Switch off the power.

Set the feedback resistor to 330 kΩ and keep P1 at 100. (What is the value of K1 now?)

To avoid overloading the error amplifier, set P3 to 15.

Switch on the power and record the step response. This should now show some

overshoot as increasing the proportional gain has the effect of reducing the damping ratio.

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(5f)

Switch off the power. Keep P1 = 100 and P3 = 15, but set the feedback resistor to 1 M.

(What is the value of K1 now?)

Switch on the power and record the step response. Observe that the response should

become even more oscillatory.

6. Proportional and Velocity Feedback using a square wave input command

(6a)

Switch off the power.

Use the same connection diagram (Fig. 10) as in the last experiment, but now set the

error amplifier feedback resistor to 330 KΩ to give a gain = 3.3.

Make the gray connections to incorporate velocity feedback.

Initially set both P1 and P2 to zero. Set P3 to 50 and the signal frequency to 0.1 Hz.

(6b)

Switch on the power. Increase the setting of P1 to 100. A slightly oscillatory response

should be obtained.

Increase the setting of P2 (set P2 = 20 initially) to add a velocity component to the

feedback signal. Adjust P2 to give a critically damped response and measure the setting

of P2 for later analysis in your report.

Observe that increasing P2 further will cause the response to become more overdamped,

in which case the output moves relatively slowly into alignment with the input

potentiometer.

The photo bellows shows the connections you are expected to make.

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(6c)

Reduce P2 to zero and reverse the polarity of the velocity feedback (which can be

achieved by tapping the velocity signal at the output of the potentiometer before the

invertor). See the upper-right corner (in red) of the photo below for this connection.

Gradually increase the setting of P2. Observe that the response will become more and

more oscillatory as P2 is increased, and finally the system will maintain self-oscillation

without any input (i.e. with P3 set to zero). Explain why this is so in your report.

Report

Each student is required to submit a report for this experiment.

State the Apparatus Set No. on the Front Page of your report. There is no need to reproduce

the theory section of this document in your report. The theory section of your report should

be very brief (no more than two pages( to include):

(i) a derivation of the closed-loop transfer function (17) for the system shown in Fig. 4;

(ii) the relationship between K1 and K2 for the closed-loop system to be critically damped.

Submit the time responses and simulation results you obtained in the experiment with brief

comments and discussions as appropriate, and include any calculations you are required to

perform as indicated in the experimental procedures.

References

[1] B.C. Kuo and F. Golnaraghi, "Automatic Control Systems", 8th edition, Wiley, 2003.

[2] G.F. Franklin, J.D. Powell & A. Emami-Naeini, "Feedback control of Dynamic Systems",

5th edition, Prentice Hall, 2006.