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Assignment 3ANALOGUE SERVO Motor, TachogeneratorFUNDAMENTALS TRAINER and Brake Characteristics

33-002-USB 7-3-1

3 Motor, Tachogenerator and Brake Characteristics

The following Practicals are included in this assignment:

3.1 Steady-State Characteristics

3.2 Steady-State Characteristics Brake Load

3.3 Transient Response of Motor

3.4 Motor Time Constant

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7-3-2 33-002-USB

CONTENT The steady and transient characteristics of the motor areexamined, and the dependence of brake torque on setting andspeed is investigated.

EQUIPMENTREQUIRED Qty Designation Description

1 33-110 Analogue Unit

1 33-100 Mechanical Unit

1 Power Supply 15 V dc, 1.5 A;+5 V dc, 0.5 A(eg, Feedback PS446 or 01-100)

1 Oscilloscope, storage orlong-persistence, preferably withX-Y facility.(eg, Feedback 1810-01229)

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33-002-USB 7-3-3

OBJECTIVES When you have completed this assignment you will know:

That the steady speed of the motor is ideally proportional tothe applied voltage, less an amount proportional to loadtorque.

That a dc tachogenerator provides a signal representingspeed, independent of motor loading.

That the response of the motor to a change of input is notimmediate, but may be expressed as a time constant.

KNOWLEDGE LEVEL Before you start this assignment you should:

Have completed Assignment 1, Familiarisation.

Understand simple applications of operational amplifiers,preferably through completing Assignment 2, OperationalAmplifier Characteristics.

PRELIMINARYPROCEDURE The Analogue Unit and Mechanical Unit should be connected

together by the 34-way ribbon cable.

The Power Supply should be connected by 4 mm-plug leads tothe +15 V, +5 V, 0 V and 15 V sockets at the back of theMechanical Unit.

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7-3-4 33-002-USB

INTRODUCTION The motor and tachogenerator were used in Practical 1.3 todisplay the speed response characteristics as part of thegeneral familiarisation in Assignment 1. In the present assign-ment a more detailed and wider investigation is carried out.

The motor is a permanent magnet type and can be representedin idealised form as in Figure 7-3-1(a), where R a is thearmature resistance and T 1, T 2 are the actual motor terminals.

(a) (b)

Figure 7-3-1: Representation of a Motor in terms of an Ideal Motor.

If the motor is stationary and a voltage V a is applied, a currentIa flows which causes the motor to rotate. As the motor rotatesa back emf V b is generated. As the motor speeds up remember the general characteristics obtained in Practical 1.3

the back emf increases and I a falls. In an ideal (loss free)motor, the armature current falls to substantially zero and V bapproximately equals V a . Thus if V a is varied slowly in eitherpolarity, the motor speed is proportional to V a , and a plot ofmotor speed against V a would have the form of Figure 7-3-1(b).

In the 33-100 the armature voltage V a is provided by a poweramplifier. A power amplifier is necessary, because although thevoltages in the error channel may be of the same order as V a ,the motor current may be up to 1 A, while the error channeloperates with currents of less than 1 mA and could not drivethe motor directly. The amplifier has two input sockets, enablingthe motor rotation direction to be reversed for a given input.

The tachogenerator is a small permanent magnet machine andhence when rotated produces an emf proportional to speedwhich can be used as a measure of the rotation speed.

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33-002-USB 7-3-5

INTRODUCTION The motor and tachogenerator were used in Practical 1.3 todisplay the speed response characteristics as part of thegeneral familiarisation in Assignment 1. In the present assign-ment a more detailed and wider investigation is carried out.

The motor is a permanent magnet type and can be representedin idealised form as in Figure 7-3-1(a), where R a is thearmature resistance and T 1, T 2 are the actual motor terminals.

(a) (b)

Figure 7-3-1: Representation of a Motor in terms of an Ideal Motor.

If the motor is stationary and a voltage V a is applied, a currentIa flows which causes the motor to rotate. As the motor rotatesa back emf V b is generated. As the motor speeds up remember the general characteristics obtained in Practical 1.3

the back emf increases and I a falls. In an ideal (loss free)motor, the armature current falls to substantially zero and V bapproximately equals V a . Thus if V a is varied slowly in eitherpolarity, the motor speed is proportional to V a , and a plot ofmotor speed against V

a would have the form of Figure 7-3-1(b).

In the 33-100 the armature voltage V a is provided by a poweramplifier. A power amplifier is necessary, because although thevoltages in the error channel may be of the same order as V a ,the motor current may be up to 1 A, while the error channeloperates with currents of less than 1 mA and could not drivethe motor directly. The amplifier has two input sockets, enablingthe motor rotation direction to be reversed for a given input.

The tachogenerator is a small permanent magnet machine andhence when rotated produces an emf proportional to speedwhich can be used as a measure of the rotation speed.

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7-3-6 33-002-USB

The magnetic brake consists of a permanent magnet which canbe swung over an aluminium disc. When the disc is rotatededdy currents circulate in the area of the disc within the magnetgap, and these react with the magnet field to produce a torquewhich opposes rotation. This gives an adjustable torque speedrelation of the form of Figure 7-3-2, and provides a veryconvenient load for the motor.

Figure 7-3-2: Characteristic of Magnetic Brake

The overall characteristics of a motor may be considered fromtwo aspects, both of which can be related to the idealisedrepresentation of Figure 7-3-1(a). These aspects are:

Steady-state, which are concerned with constant or veryslowly changing operating conditions, and

Transient, corresponding with sudden changes.

Both are important in control system applications.

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7-3-8 33-002-USB

3.1 Practical 3.1 - Steady-State Characteristics

In this practical, the motor is operated in a range ofsteady-state conditions.

Arrange the system as shown in Figure 7-3-3, where P 3enables a voltage in the range 10 V to be applied to the poweramplifier.

Use the DVM on the 33-100 for voltage measurements. Foreach measurement set up the required steady state then switchbetween DVM and RPM.

By setting SW1 and varying P 3, make a plot of motor speedagainst amplifier input, in the range 10 V, scaling the verticalaxis in units of 1000 r/min. The plot should have the generalshape of Figure 7-3-4(a).

Initially the motor speed increases substantially linearly with thevoltage to the amplifier because the motor back emf V b, seeFigure 7-3-1(a), approximately equals the amplifier output, butfinally the amplifier limits before the full 10 V input is reached.

Tachogenerator The tachogenerator provides a voltage proportional to speed,which is required for various aspects of control systemoperation.

Plot the tachogenerator characteristics by setting the motorspeed to various values by P 3 and measuring the generatedvoltage. The plot should be a straight line with the general formof Figure 7-3-4(b).

An important parameter in the use of tachogenerators is thetachogenerator factor in volts per 1000 r/min.

Determine the tachogenerator factor by measuring the changein generated volts for a speed change of 1000 r/min.

The factor should be approximately 2.5 V per 1000 r/min.

Note:Since the reduction to the output shaft is 32:1, the motorspeed is calculated by multiplying the r/min readng by 32;eg, a reading of 31.25 = a motor speed of 1000 r/min.

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33-002-USB 7-3-9

Figure 7-3-5: Motor Characteristics Related to Load

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7-3-10 33-002-USB

3.2 Practical 3.2 - Steady-State Characteristics- Brake Load

Considering the idealised motor shown in Figure 7-3-5(a), whenthe motor is unloaded the back emf V b substantially equals theapplied voltage V a , the armature current being very small.

When the motor is loaded the speed falls, the back emf falls,and the armature current increases and the voltage drop in thearmature resistance V r (= I aRa) added to V b matches V a , that is:

V a = V r + V b

= I a R a + V b

Hence, if the motor is loaded so that the speed falls, thearmature current increases, the general characteristic being asthe solid lines in Figure 7-3-5(b). If the armature resistance islow, which is the situation for a normal motor, the currentincreases greatly, as shown dotted, for a small change inspeed. The proper operating range of the motor would be up toa load corresponding with a few percent drop in speed, perhapsto the point when the dotted current line crosses the speed line.

By adjusting P 3 set the motor speed to 2000 r/min (62.5 r/minat output), with the brake fully upwards. Connect the DVM tothe Armature Current (1volt/amp) output on the MechanicalUnit.

Set the brake to each of its six positions in turn and for eachsetting record and plot the speed and armature current. Theplot should have the general form of Figure 7-3-5(c).

Initially the brake has little effect, but then the speed fallssharply and the armature current increases. With greaterloading the back emf would become small and the currentwould be limited by the armature resistance.

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33-002-USB 7-3-11

Figure7-3-6: Connections for Practical 3.3

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7-3-12 33-002-USB

3.3 Practical 3.3 - Transient Response of Motor

The motor cannot change speed instantly due to the inertia ofthe armature and any additional rotating load (the brake disc inthe 33-002). This effect was shown in Practical 1.3 in thefamiliarisation assignment, and has very important conse-quences for control system design.

(a) (b)

Figure 7-3-7: Transient Characteristics of Motor

If Va for an ideal motor has a step form as in Figure 7-3-7(a),initially a large current will flow, limited only by the armatureresistance. As the motor rotates and speeds up the back emfincreases and the current is reduced to nearly zero in an idealmotor. This is shown in the left portion of Figure 7-3-7(b). If V ais suddenly reduced to zero, the back emf still exists, since themotor continues to rotate, and drives a current in the reversedirection dissipating energy and slowing the motor. This is

illustrated in the right-hand portion of (b).The 33-001 motor shows a speed characteristic approximatingto Figure 7-3-7(b), but the power amplifier is arranged to limitthe maximum armature current which does not show the idealpulse characteristic.

Connect the system as shown in Figure 7-3-6, which enablesthe motor to be driven from the test square-wave, and allowsthe speed to be displayed on the Y axis of an oscilloscope. It isconvenient to use an X-Y display.

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33-002-USB 7-3-13

Set P 3 to zero and the test signal frequency to 0.2 Hz.

Set the power amplifier zero adjustment to run the motor atmaximum speed in one direction.

Turn up P 3 and the square-wave signal will speed up and slowdown the motor.

Adjust P 3 until the motor is stationary for one half cycle. Thiscorresponds with V a in Figure 7-3-7(b).

The oscilloscope will now display the speed corresponding withVa in (b).

3.4 Practical 3.4 - Motor Time Constant

The delay in response of a motor is of great importance incontrol system design and is expressed as the time-constant.This is the time that would be required for the motor speed tochange between any steady values if the initial rate of speedchange was maintained. This is the dotted line in Figure

7-3-8(a), while the actual speed response is shown as a solidline.

(a) (b)Figure 7-3-8

It can be shown that the speed changes by 0.63 of the finalchange during the time constant. The time constant can bemeasured from a display of the speed against time.

Using the speed adjustment of Practical 3.3 and square wavefrequency of 0.2 Hz, the time across the trace is 2.5s.

Estimate the time constant by considering the initial slope andmaximum speed. The value should be in the region of 0.5s.

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7-3-14 33-002-USB

SUMMARY The motor, with no load, runs at a speed almost proportional tothe applied voltage.

The armature current increases with increasing load torque,causing a volt drop in the armature resistance. This effectivelyreduces the applied voltage, causing a drop in speed.

The magnetic brake provides a torque proportional to speedand dependent also on the overlap between the magnet andthe disc.

If the applied voltage is suddenly changed, the motor does notrespond instantly. Its time constant is defined as the time itwould take to reach its final speed if the initial acceleration weremaintained.

PRACTICALASPECTS The slowing up process associated with reverse current, shown

in Figure 7-3-7(b), assumes that the reverse current can bereturned through the source of V a . If this is not so the motortakes a much longer time to slow down.

If the armature resistance is low, as for a normal motor, theinitial armature current may be very large (dangerously so).Thus some starting equipment (a starter) is used to limit thecurrent while the motor is being run up to speed. This appliesespecially with large motors.

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33-002-USB 7-3-15

Notes

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Assignment 4ANALOGUE SERVOFUNDAMENTALS TRAINER Error Channel and Feedback Polarity

33-002-USB 7-4-1

4 Error Channel and Feedback Polarity


4.1 Feedback Polarity

4.2 Input and Output Rotation Directions

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7-4-2 33-002-USB

CONTENT The importance is shown of ensuring that feedback is negative.Care over polarities is also shown to be needed to ensurecorrect responses to input signals.




1 Power Supply 15 V dc, 1.5 A;+5 V dc, 0.5 A(eg Feedback PS446 or 01-100)

1 Oscilloscope, storage orlong-persistence, preferably withX-Y facility.(eg Feedback 1810- 01229)

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33-002-USB 7-4-3

OBJECTIVES When you have completed this assignment you will know:

That the polarity of the feedback in a closed-loop systemmust be negative for the system to work correctly.

That in a typical system there are several places where thepolarity of the feedback can be changed.

That care must be taken over polarities in order to ensurethat when the setting (or demanded value) is changed, thenthe output (actual or measured value) changes in theappropriate direction.


Be familiar with the equipment and preferably havecompleted Assignment 1, Familiarisation.

Understand the behaviour of a simple operational amplifiercircuit.




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7-4-4 33-002-USB

Figure 7-4-1: Closed-Loop Control System

Figure 7-4-2: Connections for Practical 4.1

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33-002-USB 7-4-5

INTRODUCTION A control system with input and output shafts, such as the33-002, requires some method of measuring the input andoutput shaft angles and determining the difference or errorbetween them. The error must then produce a voltage, or maybe measured directly as a voltage, suitable to drive the poweramplifier.

A very convenient method to measure the shaft angleselectrically is to attach a potentiometer to each shaft, as inFigure 7-4-1. The signals V i and V o can then be combined in an

operational amplifier to produce an error signal to operate thepower amplifier.

4.1 Practical 4.1 - Feedback Polarity

Make the connections shown in Figure 7-4-2, ignoring for themoment the connection shown as a shadow line, which givesthe circuit of Figure 7-4-1, setting P 1 to zero before connectingto the power amplifier.

Switch on the power supply.

Set the input potentiometer to 0. Use the power amplifier zeroadjustment to rotate the output shaft to set the 0 line on thescale to be horizontal and to the right. Check that this conditiongives a positive V o value. Is the value what you might expect?.Measure the maximum voltages from the potentiometer.

Slowly increase P 1 to 100 and the output shaft should rotateanti-clockwise and finally align vertically with the input.

Note that the output potentiometer, error amplifier, poweramplifier, motor and drive to the output shaft form a loop , andthe system is arranged to have negative feedback round theloop, which reduces the error. This is the usual operatingcondition.

Set P 1 to zero, keeping i at zero. Plug P 1 output to the lowerpower amplifier input, as shown shaded in Figure 7-4-1. Usethe power amplifier zero control to rotate the output shaft to setthe 0line on the scale to be horizontal and to the right.

Slowly turn P1 until the motor just rotates and the output shaft

will rotate clockwise and stop vertically downwards.

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7-4-6 33-002-USB

The system now has positive feedback round the loop andincreases the error.

The motor stops because the slider arm of the potentiometerhas moved into the gap between the ends of the track and V obecomes zero.

Turn P1 to 100 and use the power amplifier zero adjustment tomove the output shaft and the system will probably go into asustained oscillation.

The oscillation occurs because if the slider moves slightly to theleft the signal V o becomes 10 V, which drives the slider to theright, and then V o changes to +10, driving the slider to the left,and the whole process repeats.

Error Signal Polarity For correct system operation it is essential that the error signalrotate the motor in the appropriate direction to reduce the error

this is negative feedback. If the error signal rotates themotor to increase the error, this is positive feedback and the

system is useless.An outline of the 33-100 is represented in Figure 7-4-1, where itis assumed that the input shaft ( i) is set to 0so that the inputsignal V i is zero. The output shaft ( o) zero degree datum line isassumed to be horizontal and to the right, giving a positive V o.The two potentiometer signals are added in an operationalamplifier. With V i at zero, the amplifier output will be V o. Forthe system to operate correctly this signal must rotate theoutput shaft anti-clockwise and the amplifier must be

connected to the upper power amplifier socket. This will providenegative feedback.

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33-002-USB 7-4-7

Notes

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7-4-8 33-002-USB

(a) (b)

Figure 7-4-3: Circuits for Practical 4.2

Figure 7-4-4 - Connections for Practical 4.2

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33-002-USB 7-4-9

4.2 Practical 4.2 - Input and Output Rotation Directions

The previous Practical has shown that it is essential to havenegative feedback round the loop for the system to operateproperly. In addition, there are other considerations concerningrelative rotation directions of shafts.

Arrange the system as shown in Figure 7-4-2, using the solidconnections. Set the input ( i) to 0. Use the power amplifierzero adjustment with P 1 set to 100 to align o to 0.

Rotate the input ( i) slowly between 90 and note that theoutput is controlled, but rotates in the opposite direction.

The results show that the system is following, but with reversedshaft rotation. The reasons can be seen from Figure 7-4-3(a),representing the potentiometers and error amplifier. If i isrotated anti-clockwise V i is negative , hence the error signal V ewill increase positive , eventually causing V o to increase and V eto fall to zero.

To make the shafts rotate in the same direction, various

changes could be made, such as:(i) reverse the supply polarity to the input potentiometeror(ii) introduce a gain of 1 in the V i line

Either would cause V i to reverse in direction, hence V e wouldreverse and the output shaft move in the opposite direction.

Check that (i) or (ii) above would make the output shaft follow inthe same direction as the input motion by considering voltagesas shown in Figure 7-4-3(a).

It is possible to consider introducing either (i) or (ii) into theoutput potentiometer circuit. However, any one change willreverse the feedback polarity to give positive feedback,corresponding with the oscillatory situation of Practical 4.1, andthe system will not operate properly.

However, introducing two changes will cause the feedback toremain negative.

This can be arranged as in Figure 7-4-3(b), where a gain of 1is introduced in the V

o line and the input socket on the power

amplifier is reversed.

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7-4-10 33-002-USB

Assume that V i is negative, as in Figure 7-4-3(b), and check byconsidering voltages and motor rotation that the output shaftwould now align with the input.

Arrange the system of Figure 7-4-4, which corresponds withFigure 7-4-3(b), and check that the output shaft does follow theinput shaft rotation.

SUMMARY This assignment has shown that detailed consideration ofsignal polarity and motor rotation direction are essential toobtain the correct operation of a control system.

PRACTICALASPECTS It is very important to consider whether any alteration proposed

to a system may introduce an unexpected polarity change. Thismay happen with some redesign of the electronic circuitry, suchas introducing an additional operational amplifier for somereason. Also a change in the mechanical system may have thesame effect. If a single belt reduction is replaced with a singlespur gear pair, a rotational reversal between mechanical inputand output shafts will occur, giving a polarity reversal. This canbe compensated for by introducing an electrical phase reversal

at some point in the loop.Potentiometers, as used in the Mechanical Unit, give a verysimple and convenient method to provide an electrical signalfrom a shaft rotation. They have the disadvantage that a simplepotentiometer cannot provide a signal for a shaft rotating formore than one revolution. This is possible however using morecomplicated potentiometers.

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33-002-USB 7-4-11

Error accuracy An ideal potentiometer would provide a voltage from the sliderwhich varies linearly with rotation giving the heavy line inFigure 7-4-5, and if set at mid-track would give exactly 0V.

Figure 7-4-5: Potentiometer Characteristics

Practical potentiometers have a volts/rotation characteristic asshown dotted departing from the ideal. The extent of thedeparture non-linearity depends on the individualpotentiometer but is reduced by more expensive constructionmethods.

The effect of the non-linearity is that if two potentiometers areset to the same angle the voltages will be slightly different, thedifference varying with the angle. The control system willalways rotate the output shaft and potentiometer to give zeroerror. The effect is that the output shaft does not exactly followthe input shaft but has a misalignment error depending on theshaft angle. The error may vary from less than a degree toseveral degrees depending on the potentiometers used. Themisalignment can always be set to zero at some point(commonly 0 )

For rotations exceeding one revolution it is usual to employsynchros , which are energised by ac. These enable an acerror signal to be obtained, with the facility for continuousrotation. Then either the system must operate with an ac motoror the error signal must be converted to dc for a dc motorsystem. Synchros are more expensive than potentiometers, butare very accurate.

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7-4-12 33-002-USB

Notes

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Assignment 5ANALOGUE SERVOFUNDAMENTALS TRAINER The Influence of Gain

33-002-USB 7-5-1

5 The Influence of Gain

The following Practical is included in this assignment:

5.1 Step Response

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7-5-2 33-002-USB

CONTENT The influence is examined of the gain on the transient responseof a position servo.






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33-002-USB 7-5-3

OBJECTIVES When you have completed this assignment you will know thatvarying the gain in a closed-loop system:

Affects the speed and accuracy of response.

Affects the stability of the system.


Be familiar with the equipment and associated testequipment.

Understand the meaning and purpose of closed-loopcontrol.

Preferably have completed Assignment 4, Error Channeland Feedback Polarity.



The Power Supply should be connected by 4mm-plug leads tothe +15 V, +5 V, 0 V and 15 V sockets at the back of theMechanical Unit.

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7-5-4 33-002-USB

(a) (b)

Figure 7-5-1: A general system with (a) Test Signals and (b) Typical Responses

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33-002-USB 7-5-5

INTRODUCTION Previous assignments have investigated the characteristics ofportions of the system and the present assignment isconcerned with the performance of the complete system.

A general system has the form shown in Figure 7-5-1, where aninput x and an output y are compared to give an error e with therelation

e = x y

The process of comparison is represented by a conventionalsymbol as in the diagram, where the input and output may notnecessarily be voltage signals. For a purely electrical system,the comparator may be an operational amplifier. The erroroperates the forward path, which includes everything betweenthe error and the final output y. Thus the forward path maycontain a facility to convert the error to a voltage, followed by again and a power amplifier driving a motor, and then somereduction gear to operate the output shaft.

The design and performance characteristics of systems are

often considered in terms of the response to a step or rampinput (a), with possible responses shown at (b) where the inputis shown in shadow. The response to a sinusoidal input, thefrequency response, (amplitude and phase), also has a veryimportant application in control system design, but requiressignificant mathematical background theory and appropriatetest equipment.

An introduction to sinusoidal testing is provided in Assignments14 & 15.

In application, the system will probably operate from ageneralised input as at (a) bottom, which is not practical as adesign input, so that the simple inputs above are used fordesign purposes.

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7-5-6 33-002-USB

Figure 7-5-2: Effect of Varying Gain on Step Response

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33-002-USB 7-5-7

Step Response The step response of a system gives useful information aboutthe general system characteristics and is often required to meetsome requirements in the system performance specification. Togenerate the step response a very simple input is required, andfor these reasons much consideration is given to the stepresponse.

For a given system the form of the step response is greatlyaffected by the system gain. The gain essentially determineshow much power is applied to move the output for a given

error. For an electrical system, such as the 33-002, the gaindetermines the voltage applied to the motor for a given error.

A purely electrical system may be represented as in Figure7-5-2(a), where the input (V I), output (V o) and error (V e) are allvoltages and the forward path gain (G) is shown separately, thevoltage applied to the power amplifier being GV e .

If a step V i is applied, the initial value of the error is equal to V ias in (b), since V o is zero. If the gain is 1, the power amplifierinput is initially V i and as the motor rotates the output graduallyaligns with the input, with the motor slowing up as the errordecreases. If the gain is 2, the initial input to the poweramplifier is 2V i, causing the motor to move faster and althoughthe error decreases, the motor may overshoot the required finalposition due to the delay in the motor. This delay may beinvestigated in Assignment 3. When the motor finally stops theerror is reversed in sense, so that the motor reverses and thesystem aligns or may undershoot, but will finally settle. This isshown in (c). The gain values refer to relative gains associatedwith the comparator. The overall effective gain depends onmany factors, including the reduction ratio in the output system.

If the gain is increased further, the system may take severaloscillations to settle, as at (d). If there are two delays in thesystem the result may be a steady oscillation at the output, oreven an increasing oscillation. Systems with the characteristicsof (d) are useless for control purposes.

An additional effect that must be considered is the magnitudeof the input signal. In the hypothetical example describedabove, doubling the gain speeds up the response. However, ifVi increases the power amplifier drive (2V i) also increases andthe amplifier may limit and the motor may not be able to movefast enough to give the response at (c) for an increased V i.

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7-5-8 33-002-USB


(a) (b)

Figure 7-5-4: The Error Channel

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5.1 Practical 5.1 - Step Response

In this practical, the 33-002 step response is investigated toshow the effect of variation of forward path gain. It isconvenient to arrange increasing gain by increasing thefeedback resistor in the error operational amplifier as inFigure 7-5-4(a). This corresponds with the comparator andforward path gain for Figure 7-5-2, being a single unit as shownin Figure 7-5-4(b), and it is only possible to observe GV e , theoutput of the operational amplifier. If, for example, R f is 330k ,then G = 3.3.

Arrange the system with the solid connections of Figure 7-5-3,with the error amplifier feedback resistor 100 k , giving G = 1.

Set the test signal frequency to about 0.1 Hz and adjust P 3 toprovide a square wave input of 5 V (approx 50%).

Set P 1 to zero and arrange an X-Y display.

Turn up P 1 until the motor just rotates and the system will givea response similar to Figure 7-5-2(b).

Then increase P 1 until the system just overshoots and estimatethe time to alignment T a . It is useful to examine the output ofthe error amplifier which passes through zero at T a .

Finally set P 1 to 100 and note the reduced T a since the motormoves faster but gives increased overshoot.

When the gain is increased further, it is useful to examine theerror amplifier output, which is GV e , to check that the forwardpath is not being overloaded. The initial value of the errordepends on the total change of input, so that a 5 V input givesan initial error (V e) of 10 V.

Set the amplifier feedback resistor to 330 k giving G = 3.3.Connect the Y input to the error amplifier output and with P 1 at100 adjust P 3 until the peak amplifier error (GV e) is about 10 V.

The response should have the general form of the top ofFigure 7-5-2(d). The time to alignment is much reduced but theresponse is too oscillatory for a practical system.

Set the feedback resistor to 1 M giving G = 10, and repeat thetest, adjusting P 3 to limit the peak error to 10 V.

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7-5-10 33-002-USB

The response is now more oscillatory, but the time to initialalignment will be reduced from the value with G = 3.3 and canagain be estimated.

In the last two tests, the input has been reduced to prevent thesignals exceeding 10 V. However, if the input is increased tosay 5 V, then the theoretical initial output of the error amplifierwith G = 10, would be 50 V. If the amplifier could give 50 V andthe power amplifier accept an input of 50 V and drive the motorcorrespondingly fast, the system response would retain thesame form.

In practice, the error amplifier cannot provide outputs muchexceeding 10 V and hence limits; thus the motor runs atmaximum speed until the error is sufficiently reduced that theamplifier ceases to limit. The result of the limiting is that theresponse takes longer as the input is increased.

With G = 10, adjust P 3 to give 5 V. Observe that the errorlimits and that the response takes longer, and the oscillationtakes longer to die away.

These general results illustrate a fundamental problem forsimple control systems, that increasing the gain to give a fasterresponse leads to more marked oscillation, and no advantagemay be gained.

SUMMARY Testing of closed-loop systems is most conveniently done withspecial simple test signals. The most usual is a step signal.

The response to a step input is sluggish if the gain is low.

Increasing the gain causes a faster response, but if too much

gain is applied the response may overshoot and take severaloscillations before settling. Further increase in gain may causesustained oscillations to build up.

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Assignment 6ANALOGUE SERVOFUNDAMENTALS TRAINER Velocity Feedback

33-002-USB 7-6-1

6 Velocity Feedback


6.1 Simple Velocity Feedback

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7-6-2 33-002-USB

CONTENT A servo is studied with reference to the effect of velocityfeedback on system stability.






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33-002-USB 7-6-3

OBJECTIVES When you have completed this assignment you will know that:

Introducing velocity feedback into a closed-loop positioncontrol can make it more stable.

The improved stability may enable more gain to be used.

Excessive velocity feedback makes the system slow-acting.


Be familiar with the equipment.

Understand the effect of gain on the transient response of aservo and preferably have completed Assignment 5, TheInfluence of Gain




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7-6-4 33-002-USB

Figure 7-6-1: Velocity Feedback

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33-002-USB 7-6-5

INTRODUCTION Assignment 5 shows that increasing the gain increases theoscillatory tendency in the response, which is undesirable.There are good practical reasons for using a high gain. Animportant one is that, due to the brushes and other factors, allpractical motors have a constant friction (called coulombfriction ). Also usually an increased amount of friction force(called stiction ) has to be overcome to start the motor fromrest. Therefore a minimum voltage has to be applied to themotor before rotation starts. This means that there is a

minimum input below which the system will not respond; this istermed dead-band . If the gain is high the dead-band isreduced, which is advantageous, but the system may displayunwanted oscillation in the response.

The form of the system response with high gain can be muchimproved by applying a feedback signal to the inputproportional to the output shaft velocity . This arrangement istermed velocity feedback , and is illustrated in Figure 7-6-1.

In the diagram, it is assumed that the input and output signals

are available as voltages and a voltage V s proportional tooutput shaft speed is available from a tachogenerator. Afraction of this voltage K tVs is subtracted from V e to give V c,which is the control voltage applied to the power amplifiergiving:

Vc = Ve K tVs

If a step input is applied to the system the error will initiallyequal the input step and decrease as the motor speeds up. Asthe motor speed increases the velocity signal V s increases andsubtracts from V e to give a drive voltage V c, which is less thanthe error. This is illustrated in the lowest diagram, and themotor drive goes to zero and reverses before the error goes tozero. This means that the motor begins to slow up before theinitial alignment, greatly reducing or even preventing anyovershoot.

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7-6-6 33-002-USB


Figure 7-.6-3: Dead-beat Response

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33-002-USB 7-6-7

6.1 Practical 6.1 - Simple Velocity Feedback

Arrange the system as in Figure 7-6-2, initially omitting theconnection from P 2 to the tachogenerator and set P 1 to zero.Connect the oscilloscope for an X-Y display as indicated.

Set the error amplifier feedback resistor to 100k, giving G = 1,and set P 3 to give a V i of 5 V at 0.1 Hz. Turn P 1 to 100 and aslightly oscillatory response should be obtained.

Connect P 2 to one tachogenerator output and increase itssetting; if the overshoot decreases the velocity feedbackpolarity is correct. If the overshoot increases connect to theother socket.

When adjusting the tachogenerator feedback a dead beatresponse will be obtained when the system aligns in the leastpossible time, but with no overshoot as in Figure 7-6-3.Additional velocity feedback will cause an over-dampedresponse, in which the system slowly moves into alignment.

Increase P 2 until a deadbeat response is obtained, this occurswhen the motor just does not reverse. Additional feedback canbe obtained by disconnecting i and connecting thetachogenerator to that input as well.

Increase the feedback resistors to 330k, G = 3.3; 1 M , G =10, and in both cases deadbeat response should be obtainable.

These results show that velocity feedback is a very powerfultechnique to improve transient response when gain isincreased.

Note that reversing the polarity of velocity feedback can make asystem unstable.

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7-6-8 33-002-USB

SUMMARY Velocity feedback is feedback proportional to the velocity of theoutput shaft. A tachogenerator is often provided for thepurpose.

In a position servo, it serves the purpose of modifying the(position) error signal so that the modified signal goes to zerobefore the desired position is reached. This prevents the motorfrom accelerating right up to the desired position, so that inertiacauses it to overshoot.

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Assignment 7ANALOGUE SERVOFUNDAMENTALS TRAINER System Following Error

33-002-USB 7-7-1

7 System Following Error


7.1 Following Error

7.2 Velocity Feedback and Following Error

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7-7-2 33-002-USB

CONTENT A simple system follows a ramp with an error. This followingerror is investigated and the effect of velocity feedback on theerror studied.

EQUIPMENTREQUIRED QTY Designation Description



1 Power Supply 15 V dc, 1.5 A+5 V dc, 0.5 A(eg, Feedback PS446 or 01-100)


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33-002-USB 7-7-3

OBJECTIVES When you have completed this assignment you will know that :

A simple system follows a ramp with an error.

Increasing the gain reduces the error but leads to adeterioration of the transient response.

Velocity feedback can improve the transient response butincreases the following error.

KNOWLEDGE LEVEL Before you start this assignment you should :

Be familiar with the effects of gain and velocity feedback onsystem performance.

Preferably have completed Assignments 5, The Influence ofGain and 6, Velocity Feedback.




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7-7-4 33-002-USB

Figure 7-7-1: Steady Following Error

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33-002-USB 7-7-5

INTRODUCTION

SystemFollowing Error The step response of a system (considered in Assignment 5) is

an important general indication of system performance.Another important characteristic is the system response to asteadily changing input requiring the output to move at aconstant speed. This is sometimes termed the ramp responseand is represented in Figure 7-7-1, where all signals areassumed available as voltages.

This general situation could correspond with the requirementfor the cutting tool of a lathe to move at a constant speed alonga workpiece or a radar dish to sweep at a constant velocity.

If the output is to move at a steady speed then, when thesystem has settled, there must be an appropriate constantvoltage applied to the motor. This voltage can only be obtainedfrom the error. Since the error V e depends on the differencebetween input and output

Ve = V i Vo

the system will show a constant following error, that is theoutput always lags on the input. In terms of the two examplesmentioned above, this means that the cutting tool is neverexactly where it is commanded to be or the dish does not pointexactly where it is intended to point.

If the forward path gain G is increased, then the error for agiven speed falls inversely, that is doubling the gain halves the

error, however, the system response becomes more oscillatoryand the settling time may increase.

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7-7-6 33-002-USB

Figure 7-7-2: System for following error investigation

Figure 7-7-3: System connections for Practical 7.1

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33-002-USB 7-7-7

7.1 Practical 7.1 - Following Error

In order to be able to examine the following error and see howit changes with gain, it is necessary to arrange the system sothat the error V e is directly available as:

Ve = V i Vo

from the error operational amplifier as in Figure 7-7-2(a). Anyadditional gain G must be introduced as a separate amplifier,see Figure 7-7-2(a), so that for any gain value the direct error is

always available. If the system gain G is incorporated in theerror operational amplifier as in Figure 7-7-2(b) by using anoutput resistance GR, the amplifier output is GV e .

For the above reason, it is convenient to use the generalsystem of Figure 4-7-2(c). It is important to note that if anadditional operational amplifier is introduced to provide gain G,this also introduces an additional sign reversal in the loop,giving positive feedback unless some additional sign reversal isintroduced (see Practical 4.1 Feedback Polarity).

In the Controller section of the Analogue Unit, there is anoperational amplifier at the right-hand end which enablesconvenient values of gain to be introduced, and changing thepower amplifier input socket provides an additional signreversal.

The additional amplifier has input resistors of 100 k andfeedback resistors of 100 k , 500 k and 2 M hence withvarious different arrangements gains of 1, 2, 5, 20 and 25 canbe obtained.

Connect up the system of Figure 7-7-3, making an X-Y display

of the error.Apply an input of 5 V at 0.1 Hz and estimate the steadyfollowing error with controller operational amplifier gain of 1; ie,input and output resistance of 100 k and P 1 at 50.

Increase the gain to:2 : (2 x 100 k input in parallel and 100 k output)5 : (100 k input and 500 k output)

Note that the error decreases inversely with increasing gain,

but that the transient response deteriorates as expected.

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7-7-8 33-002-USB

Figure 7-7-4: Increased following error with velocity feedback

Figure 7-7-5: Additional connections (solid) to operationalamplifier of Figure 7-7-3 for velocity feedback.

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33-002-USB 7-7-9

Velocity(tachogenerator)Feedback andFollowing Error Since velocity feedback improves the step response transient,

it could be supposed that the following error transient responsewould also be improved. This is correct, but there is thedisadvantage that the use of velocity feedback increases thesteady following error.

The application of velocity feedback is represented inFigure 7-7-4, (see also Figure 7-7-1), where V

c is the control

voltage applied to the power amplifier. If the system is followingat the same steady speed the motor drive V c must remainconstant and hence the signal immediately before the amplifierG must have the same value as the original error (V eo ) withoutvelocity feedback. However, the actual error between V i and V o(Vev ) must increase so that when K tVs is subtracted, the originalerror is available as input to G, thus:

error with velocity feedback (V ev ) K tVs = original error (V eo )

or Vev = Veo + KtVs

7.2 Practical 7.2 - Velocity Feedback

To investigate the effect of velocity feedback on following error:

Connect the system of Figue 7-7-3, including the X-Y displayand the triangle input signal (shown as shadow connections inFigure 7-7-5).

Add the solid connections of Figure 7-7-5 which apply velocity

feedback, setting P 2 to zero.Set the controller amplifier gain to 5, and apply a triangle inputof 5 V at 0.1 Hz.

As P 2 is turned up to 100% the error transient will improve, butthe steady following error V ev will increase.

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7-7-10 33-002-USB

SUMMARY This assignment has investigated steady following error, whichcan be an important performance requirement for a controlsystem.

It has also been shown that increasing the system gain reducesthe steady following error but causes the transient response todeteriorate.

It has been shown that velocity feedback will improve thesteady following error transient; but will increase the followingerror.

In a later assignment, a method of improving the transientresponse without increasing the following error is investigated.

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Assignment 8ANALOGUE SERVOFUNDAMENTALS TRAINER Unstable System

33-002-USB 7-8-1

8 Unstable System


8.1 Additional Time Constant

8.2 Unstable System

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7-8-2 33-002-USB

CONTENT The effect of an additional time constant is investigated to showthat a system which requires high gain can become unstable.

EQUIPMENTREQUIRED QTY Designation Description





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33-002-USB 7-8-3

OBJECTIVES When you have completed this assignment you will know that :

A time constant can be represented by an operationalamplifier circuit.

An additional time constant causes a system transientresponse to deteriorate.

With high gain the system may become unstable.

KNOWLEDGE LEVEL Before you start this assignment you should :

Be familiar with the equipment and the general properties ofcontrol systems, preferably by completing Assignment 5,The Influence of Gain.




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7-8-4 33-002-USB

Figure 7-8-1: System characteristics due to delay.

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33-002-USB 7-8-5

INTRODUCTION

Unstable System Motor characteristics are investigated in Assignment 3,Practicals 3.3 and 3.4, and it is shown that there is a delay inthe speed response of a motor to a sudden change of supplyvoltage. If a step voltage is applied the speed response wouldbe generally as in Figure 7-8-1(a) (see also Figure 7-3-8).

It is also shown in the closed-loop step response investigation,Assignment 5 and Figure 7-5-2, that the delay in the motorresponse can cause the system to overshoot and then settlewith reducing oscillation as in Figure 7-8-1(b). An additionalseparate delay in the system can cause more markedovershoot because the motor has been able to move furtherbefore the drive is reversed. The additional delay may lead tosustained or increasing amplitude oscillation, as in Figure7-8-1(c).

Thus it is very important to avoid significant additional delay ina system even though various procedures used in a system,such as filtering to eliminate noise on signals or signalprocessing may introduce delays. If such delays becomecomparable with those inherent in the system then, at least, thetransient response will deteriorate.

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7-8-6 33-002-USB

Figure 7-8-2: Time constant and operational amplifier circuit

AdditionalDelay Most additional delays have the general characteristics of a

time-constant, represented by the RC circuit of Figure 7-8-2(a),showing the delay in the step response as the capacitorcharges through the resistor.

This circuit characteristic may be obtained by the operationalamplifier circuit of Figure 7-8-2(b) where there is a capacitor inparallel with the output resistor. If a step is applied, there is aconstant input current since the amplifier input is a virtual earthpoint.

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33-002-USB 7-8-7

When the step is applied:

the capacitor current (i c) = input current (i)

the resistor current (i r) = 0

Since the input current (i) is constant, thus as the capacitorcharges:

ic reduces

ir increases

and finally:

ic = 0

ir = i

This gives a voltage output which has a time constant form,with time constant RC, identical with that of the RC circuit of (a),except that the output has reversed polarity.

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7-8-8 33-002-USB

Figure 7-8-3: Circuit for Practical 8.1

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33-002-USB 7-8-9

8.1 Practical 8.1 - Additional Time Constant

In order to investigate the effect of an additional time constant,the error amplifier has a capacitor available to enable the circuitof Figure 7-8-2(b) to be made.

Connect the error amplifier as in Figure 7-8-3(a), and arrangean X-Y display.

Apply a square-wave of 5 V at 2 Hz and a time constantdisplay as in Figure 7-8-3(b) should be obtained.

Since:

C = 0.1 F (nominal)R = 100 k the time constant should be about 0.01 sec.

If the 100 k is increased the time constant increases, but thesteady gain also increases, so that the input must be reducedto prevent overloading, (output 10 V max). Check the effect ofR = 330 k and R = 1 M .

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7-8-10 33-002-USB

Figure 7-8-4: Circuit for Practical 8.2 - Unstable system

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33-002-USB 7-8-11

8.2 Practical 8.2 - Unstable System

Connect the system of Figure 7-8-4, giving a simple positioncontrol with an X-Y display.

Set the error feedback resistor to 100 k , without the 0.1 Fcapacitor. Apply a square wave 5 V at 0.1 Hz and set P 1 to100%.

Note that introducing the capacitor has little effect on thetransient, since the time constant (0.01 secs) is smallcompared with the approximate motor time constant(0.4 - 0.5 sec - see Assignment 3, Practical 3.4).

Set the feedback resistor to 330 k and P 1 to 100%. Thecapacitor will now have a marked effect on the transient.Reduce the input (P 3) to prevent overloading at error amplifieroutput.

Set the feedback resistor to 1 M and connect the capacitor.

Set the gain (P 1) and the square wave input to zero. Slowly turnup the gain, at the same time moving the input potentiometerabout 10. The error amplifier time constant, 0.1 sec, is nowappreciable compared to the motor time constant.

The system should eventually maintain self oscillation, with afrequency in the region of 1 Hz. Additional gain, if needed, canbe obtained from the Controller operational amplifier.

Check that the system can be stabilised by using velocity(tachogenerator) feedback.

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7-8-12 33-002-USB

SUMMARY Any additional delay in a system will cause the transientpressure to deteriorate because the motor can overshoot morebefore the drive signal is reversed.

An additional delay frequently has the characteristics of a timeconstant and can be represented by an operational amplifiercircuit.

The combination of additional delay and high gain can cause asystem to become unstable.

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Assignment 9ANALOGUE SERVOFUNDAMENTALS TRAINER Speed Control System

33-002-USB 7-9-1

9 Speed Control System


9.1 Closed-loop Speed Control

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7-9-2 33-002-USB

CONTENT A velocity servo (speed control system) is studied withreference to the effect of loading the output shaft with andwithout the loop closed .





1 Oscilloscope, storage or longpersistence, preferably with X-Yfacility.(eg, Feedback 1810-01229

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33-002-USB 7-9-3


Velocity feedback can be used (without position feedback)to enable a speed to be closely regulated.

The polarity of the feedback is important (as for positionfeedback).

The effectiveness of the control depends mainly on the gainemployed.


Be familiar with the equipment.

Preferably have completed Assignments. 5, The Influenceof Gain and 6, Velocity Feedback.



The power supply should be connected by 4 mm plug leads tothe +15 V, +5 V, 0 V and 15 V sockets at the back of theMechanical Unit.

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7-9-4 33-002-USB

Figure 7-9-1: Essential features of a Closed Loop Speed Control

Figure 7-9-2 : Speed Control System

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33-002-USB 7-9-5

INTRODUCTION The previous assignments have been concerned with positioncontrol, but an important aspect of closed-loop control is speedcontrol, which has many industrial applications, varying fromheavy industrial, such as paper mills or steel rolling mills, totape or video transport mechanisms.

The essential principle of closed-loop speed control is similar toposition control, as illustrated in Figure 7-9-1, except that thefeedback signal is now an output velocity signal V s , normallyfrom a tachogenerator, which is compared with a referencevoltage V r to give an error

Ve = Vr V s

In operation the reference is set to a required value, whichdrives the motor to generate V s , which reduces the error untilthe system reaches a steady speed.

If the motor is loaded, eg with the magnetic brake on the33-100, the speed falls; this tends to increase the error,

increasing the motor drive and thus reducing the speed fall fora given load. Note that this implies negative feedback aroundthe loop.

The speed fall with load, sometimes termed droop is a veryimportant characteristic in speed control systems.

The rotation direction can be reversed by reversing thereference voltage, though many industrial speed controlsystems are required to operate in one direction only.

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7-9-6 33-002-USB


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33-002-USB 7-9-7

9.1 Practical 9.1 - Closed-loop Speed Control

A speed control system which can be made with the 33-002corresponding with Figure 7-9-1, is given in Figure 7-9-2, withthe connection diagram as in Figure 7-9-3.

Arrange the system as in Figure 7-9-3. Set P 2 (tacho) to zeroand set the amplifier feedback resistor to 100K , this gives G =1. Set P 1 to 100. Set SW1 up to +10 and adjust P 3 to run themotor at 1000 r/min (31.25 r/min at output).

Turn up P 2 slightly, if the speed decreases the loop feedback isnegative as required. If the speed increases use the othertachogenerator polarity.

Note that if the system has negative feedback and both thetachogenerator polarity and the power amplifier input arereversed, the system still has negative feedback, but the motorruns in the opposite direction.

Set P2 to zero and plot the speed against brake setting to full

brake load. The general characteristic should be as in Figure7-9-4.

Figure 7-9-4: Speed Regulation with and without Closed-Loop Control

Note that the armature current, which increases with loadingcan be measured by correcting the DVM to the ArmatureCurrent socket on the Mechanical Unit.

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7-9-8 33-002-USB

Set P 2 to 100 and readjust P 3 to give 1000 r/min with the brakeoff. Replot the speed characteristic and error (PA input) up tofull brake load. Change the feedback resistor to 330 k (G = 3.3), adjust P 3 to give 1000 r/min with no load, and replotthe load characteristic. The droop should be reduced.

Repeat with G = 10, adjusting P3 as required, and the droopshould be less.

The tachogenerator output contains a ripple component whichwill be amplified in the forward path, and with high gain couldsaturate the power amplifier.

The ripple can be reduced by connecting a capacitor across theerror amplifier output resistor, which introduces a time-constant, (see Assignment 8 and Figure 7-8-2) and reduces theresponse to fast signals.

With G = 10, set the speed to 1000 r/min on no load andexamine the input and output ripple of the error amplifier using

the oscilloscope. Note the ripple amplification.

Then connect the 0.1 F capacitor across the output resistorand the ripple will be much reduced. A smaller capacitor couldbe used.

The introduction of a time-constant as above may be desirableto reduce ripple, but represents an additional delay in the loop,and could cause serious deterioration of the speed stepresponse.

With G = 10, apply a square wave at 0.1 Hz and adjust P3 togive a steady speed of 1000 r/min with no load.

Make an X-Y display with the tachogenerator output to showthe step response.

Re-connect the 0.1 F capacitor and note the change inresponse.

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33-002-USB 7-9-9

SUMMARY This practical has shown the general principle of speed control.Increasing the gain would give the system less speed fall at fullload. Theoretically, an infinite gain would give zero speed fall,but this is impractical. However, good results can be achievedwith a different control technique in the error channel.

PRACTICALASPECTS A problem that may arise in speed control systems, where the

full feedback signal is obtained from a tachogenerator, is thatthe generator output may have an appreciable ripplecomponent due to commutation. This component may beamplified in the system and cause saturation. The ripple can bereduced by filtering in the system, or better by specialiseddesign of the generator.

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7-9-10 33-002-USB

Notes

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Assignment 10ANALOGUE SERVOFUNDAMENTALS TRAINER Introduction to 3-Term Control

33-002-USB 7-10-1

10 Introduction to 3-Term Control


10.1 Derivative Measurement

10.2 Operational Amplifier Integrator

10.3 3-Term Controller Test

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7-10-2 33-002-USB

CONTENT To study the effects of combining various levels of proportional,integral and derivative control to produce the 3-term or PIDcontroller.





1 Oscilloscope, storage or longpersistence, preferably with X-Yfacility(eg, Feedback 1810-01229)

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33-002-USB 7-10-3


A very versatile control signal can be obtained bycombining components depending on the error, thederivative of the error, and on the integral of the error.

With capacitor input, an operational amplifier acts as adifferentiator.

The input must be modified, giving partial differentiation, forpractical reasons.

With a capacitor in the feedback path, an operationalamplifier can act as an integrator.


Understand the basic working of an operational amplifier,including the concept of virtual earth.

Preferably have completed Assignment 2, OperationalAmplifier Characteristics.




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7-10-4 33-002-USB

Figure 7-10-1: Derivative of Error

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33-002-USB 7-10-5

INTRODUCTION Previous assignments have considered system performanceand how this is affected by gain. In principle a higher gain leadsto improved performance in respect of reduction of dead-bandand following error and also to a reduction of droop withincreasing load for a speed control system. The disadvantageof high gain is that the transient response deteriorates, givingovershoots or oscillations. This can be corrected by the use ofvelocity (tachogenerator) feedback, but that increases thesteady following error.

A more general method to improve system performance is toarrange that the drive signal to the motor or other outputelement is a combination of the direct error, with components ofthe derivative (rate of change), and integral of the error. Sincethe final drive signal contains three components or terms, theprocess is called Three-Term Control.

Derivativeof Error When a step input signal is applied, the error signal will typically

respond as shown in the upper portion of Figure7-10-1(c). The derivative or rate of change of error correspondsgraphically with the slope of the error graph. If the derivative is

measured the general form will be as the lower diagram.Initially the slope is zero, reaches a maximum negative value(corresponding with rapidly decreasing error) shortly before thesystem initially reaches alignment and then falls to zero whenthe system reaches maximum overshoot.

If the derivative signal is added to the error signal, thecombination of error and derivative goes to zero before thesystem aligns (dotted V e graph). Thus if the power amplifierdrive signal V c is formed from V e by:

Vc = error + derivative of error

= V e +dV e

dt ,

then V c will reverse the motor drive before alignment of theoutput shaft is reached, much improving the transientresponse. The arrangement is illustrated in Figure 7-10-1(b),where the amount of derivative signal is adjustable by K d togive:

Vc

= Ve

+ Kd

dV e

dt .

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7-10-6 33-002-USB

The general effect of error derivative is similar to velocityfeedback (see Assignment 6 and Figure 7-6.1), but is obtainedby operating on the error only.

Figure 7-10-2: Integral Control

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33-002-USB 7-10-7

Integral of Error Suppose that a simple position control system is following aramp input, the steady error (V es ) will be exactly that required todrive the motor to make the output speed match the inputspeed. This is investigated in Assignment 7, and is illustrated inthe upper portion of Figure 7-10-2(a).

Suppose that the error is integrated by some method. At anytime the value of the integral is the area under the error. This isindicated for a particular time t 1 in Figure 7-10-2(a). If the errorhas reached the steady value V

es, the integrator output would

rise steadily.

If the integrator output is added to the error to form V c, as inFigure 7-10-2(b), the motor control voltage is given by:

Vc = Ve + K i'Ve dt

Initially when the ramp is applied, the error will occur as in theupper diagram, but the integrator output will gradually build up.This will cause the motor to speed up slightly, reducing theerror, but as long as there is any error the integrator output willcontinue to increase. The only situation that will give a constantintegrator output is that the error has fallen to zero , and this iswhat happens.

The integrator final output becomes V es , as in Figure 7-10.2(b), being the signal required to drive the motor at thespeed to match the input.

The use of an integrator in the forward path, which is calledintegral control , gives the characteristic that any steady error

is reduced to zero with the integrator output established atexactly the value required to provide the motor drive tomaintain the error at zero. Integral control is commonly appliedto speed control systems and if the measured speed does notexactly match the required speed due to loading, the integratorwill develop an output to reduce the error to zero.

A problem that may arise with integral control is that theintegrator takes some time to develop the required output and ifthe system operating condition suddenly changes, theintegrator requires time to readjust. This may lead to a slow

and undesirable oscillatory transient response.

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7-10-8 33-002-USB

Figure 7-10-3: Three-Term Control

Three-TermController As mentioned previously, a general controller combines integral

and derivative actions, with the direct error as inFigure 7-10-3 giving:

Vc = G (error + K d [derivative of error] + K i [integral of error])

V c = G V e + K ddV edt

+ Ki V e dt'

! #

$ &

where there is an adjustable overall gain G, and the integraland derivative components are individually adjustable. Thecontroller is often referred to as a P.I.D (signifying Proportional+ Integral + Derivative) controller.

The processes required in a Three-Term Controller, the

generation of derivative and integral signals and theircombination in adjustable proportions are very easily realisablein analogue form with operational amplifiers if the signals areavailable as voltages. However, it is possible to realise Three-Term Controllers in a non-electrical medium such aspneumatics.

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33-002-USB 7-10-9

Notes

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7-10-10 33-002-USB

Figure 7-10-4: Differentiation by Operational Amplifier

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33-002-USB 7-10-11

DIFFERENTIATIONBY OPERATIONALAMPLIFIER Suppose that a ramp voltage is applied to a capacitor as in

Figure 7-10-4(a). Since the voltage across the capacitor risessteadily a constant current must flow into the capacitor. Thecurrent is proportional to the capacitance C, being given by therelation

i = C dvdt

.

If the capacitor is used as the input element for an operationalamplifier as in (b), the amplifier input will be a virtual earth point(see Assignment 2 and Figure 7-2-3) and the amplifier outputwill be given by:

and since

finally

V o = iR o

i = C dvdt

,

V o = CR odv

dt ,

giving the constant output voltage of (b). This indicates thederivative of the input with a scaling factor CR o.

Although this circuit in principle measures the derivative, thereis a limitation in practical application. If the input signal containsnoise or disturbance components which are small but rapidlychanging, these may cause currents in the capacitorcomparable to those of slower changing signals for which thederivative is required. These unwanted components in the input

are emphasised and may even saturate the amplifier or somelater stage in the amplifying system.

The effect of unwanted rapidly changing high frequencycomponents can be limited by a resistor in series with thecapacitor as in Figure 7-10-4(c). If the input is changing slowlythe input current is largely determined by the capacitor, but ifthe input is changing fast the current is limited by the resistorgiving an overall gain of R o /R.

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7-10-12 33-002-USB

In frequency response terms the gain of the ideal differentiatorin (c) rises continuously with increasing frequency, noisecorresponding with high frequency components. Theintroduction of an input resistor, called a limited derivative,gives a gain initially rising with frequency, representing correctderivative action, but finally becoming constant preventingemphasis of high frequency components.

Figure 7-10-5(a): Connections for Practical 10.1

Figure 7-10-5(b): X-Y Display

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33-002-USB 7-10-13

10.1 Practical 10.1 - Derivative Measurement

The upper amplifier in the Controller is intended for use as adifferentiator. The two 1 F capacitors can be used in series togive 0.5 F, singly to give 1 F or in parallel to give 2 F.

Arrange the circuit of Figure 7-10-5(a), where the upperamplifier is connected as a limited differentiator with 1 Fcapacitor and adjustable input from the triangle test waveform.

Make an X-Y display between the triangle waveform and theamplifier output.

Set the test frequency to about 1 Hz, and turn up P 4 to 100. Adisplay should be obtained as in Figure 7-10-5(b), the steadyvalue being CR( dV / dt). The time-constant start of the waveformis due to the effect of the 10 k resistor.

For a frequency of 1 Hz, the test waveform changes through 20V in 0.5 second, hence

dvdt

= 40 V/ s.

Since the capacitor is 1 F, a voltage rate of change of 1 V/swould give i = 1 A, so 40 V/s gives i = 40 A, where i is thecurrent in Figure 7-10-4(a), (b).

If Ro = 200 k , the steady output voltage would be

40 A 200 k (i.e CR dv

dt ) = 8V

If the frequency is reduced the steady value will fallproportionately.

If the capacitance is reduced to 0.5 F (both capacitors inseries), the amplifier output will fall to 50%, but the time toestablish the steady value will also fall to 25%.

Set P 4 to zero, and using 1 F, connect directly to the amplifierinput socket. As P 4 is turned up a fast initial response will be

obtained but a large pulse or oscillation may appear on theoutput.

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7-10-14 33-002-USB

Figure 7-10-6: Operational Amplifier as Integrator

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33-002-USB 7-10-15

INTEGRATION BYOPERATIONALAMPLIFIER Conversely to Figure 7-10-4(a) and page 7-10-11, the voltage

across a capacitor is given by the integral of the current throughthe capacitor:

v = 1C i dt.'

This is illustrated in Figure 7-10-6(a). A constant current gives asteady increase of voltage, the voltage representing the areaunder the current /time plot.The illustrated current pulse waveform, comprising a positivepulse, a negative pulse and then zero current, gives a finalvoltage value, which is the overall time integral of the current.Integration can be obtained by an operational amplifier if thefeedback resistor is replaced by a capacitor as in Figure7-10-6(b). If a voltage V is applied an input current i will flowthrough R and the amplifier output will change to hold theamplifier input point (virtual earth point) substantially at zero. Thismeans that the amplifier output will steadily increase to maintainthe current i through the capacitor. The input current is

i = VR

.

The voltage across the capacitor will be

V c = 1C Vdt

'

giving the amplifier output as

V o = 1CR Vdt

'

which is the negative scaled integral of the input.

If the applied voltage V becomes zero, the current i is also zeroand the integrator holds indefinitely whatever output has beenobtained, since the amplifier does not ideally draw any current atthe virtual earth point. If the input is a general waveform theoutput is correspondingly scaled integral.It is often required to set the output of an integrator to zero beforethe start of integration, and this can be arranged by a switch,mechanical or electronic, connected across the capacitor, asshown dotted, which discharges the capacitor. Integration doesnot start until the switch is open, irrespective of a possible input

signal.

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7-10-16 33-002-USB

An ideal integrator will hold an accumulated signal indefinitely,however an operational amplifier may draw a very small currentat the virtual earth input, which will cause the output to drift veryslowly. The drift may or may not be important and depends on theamplifier used.


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33-002-USB 7-10-17

10.

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