me2142e speed or position control of a dc motor

8/10/2019 ME2142E Speed or Position Control of a DC Motor

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Bachelor of Technology Programme

ME2142E Feedback Control Systems

Report:

Speed/Position Control

of a DC Motor

Name: DOLCE GUSTO

Matric No.: A0123456VLab Group: 1

Date: 31st Sep 2014



Speed/Position Control of a DC Motor

2

Objectives

To become familiar with the operation of an armature-controlled DCmotor.

To study the transient and steady state response of a closed-loopspeed control system.

To study the transient response of a closed-loop position controlsystem.

Results: Tables

Table 1: Brake Setting = 0 (With Feedback) Gain = 0.1Motor Characteristics SPEED VS INPUT

Speed (rpm) 400 800 1200 1600 2000

V0 (mV) 39 82 123 168 212

*Speed range is 400rpm to 2000rpm in increments of approximately 400rpm.Measure V0 with a multi-meter. V0 is the voltage output from the operationalamplifier (OU150A in the figure).

Table 2: Brake Setting = 4 (With Feedback) Gain = 0.1

Motor Characteristics SPEED VS INPUT Speed (rpm) 400 800 1200 1580

V0 (mV) 49 104 160 243

* Speed range is 400rpm to 2000rpm in increments of approximately 400rpm.

Table 3: OPEN-LOOP LOAD – Speed Characteristics (No Feedback)

Brake Setting 0 2 4 6 8 10

Speed

(Gain = 0.03)1000 910 750 660 540 470

Speed

(Gain = 0.06)1000 920 750 670 550 480

Speed

(Gain = 0.1)1000 890 740 650 530 470

* For each gain setting, start with speed of about 1000 rpm at zero brakescale reading. Then apply the various brake settings and read the speed fromthe tachometer.




3

Table 4: CLOSED-LOOP LOAD – Speed Characteristics (With Feedback)

Brake Setting 0 2 4 6 8 10

Speed

(Gain = 0.03)1000 950 860 790 690 620

Speed

(Gain = 0.06)1000 960 880 830 750 700

Speed

(Gain = 0.1)1000 980 920 880 820 770

* For each gain setting, start with speed of about 1000 rpm at zero brakescale reading. Then apply the various brake settings and read the speed fromthe tachometer.

*Note

Gain = 0.03, Set the variable potentiometer to 3 in Attenuator Unit (AU 150B)Gain = 0.06, Set at 6 in Attenuator Unit (AU 150B)Gain = 0.1, Set at 10 in Attenuator Unit (AU 150B)

Table 5:Closed-Loop Speed Control Transient Response (With Speed Feedback)

Brake Setting

Brake 5 Brake 10

RemarksTime Constant time τ

(ms)

Time Constant time τ

(ms)

Gain = 0.03 36 24 -

Gain = 0.06 28 20 -

Gain = 0.1 28 24 -

* Time constant: Time to reach 63.2% of the steady state value from the startof the input step.

Table 6:Closed-Loop Position Control System (With Position Feedback)

Brake

Setting = 5

With Speed Feedback No Speed Feedback

RemarksOvershoot

%

Rise Time

(ms)

Overshoot

%

Rise Time

(ms)

Gain = 0.03 0 636 0 432 -

Gain = 0.06 0 528 0 252 -

Gain = 0.1 0 460 7.4 188 -

* Rise Time: Time to first reach 90% of the steady-state value from the start of

the input step.




4

Results: Graphs

0

400

800

1200

1600

2000

2400

0 50 100 150 200 250 300

M o t o S p e e d ( r p m )

Input Voltage (mV)

Graph 1: Motor Speed vs Input Voltage

Brake Setting = 0

Brake Setting = 4

0

200

400

600

800

1000

1200

0 2 4 6 8 10 12

S p e e d ( r p m )

Brake Setting

Graph 2: Speed vs Brake (3 Gains, Open-loop)

Gain = 0.03

Gain = 0.06

Gain = 0.1

Linear (Gain = 0.03)


Linear (Gain = 0.1)




5

0

200

400

600

800

1000

1200

0 2 4 6 8 10 12

S p e e d ( r p m )

Brake Setting

Graph 3: Speed vs Brake (3 Gains, Closed-Loop)

Gain = 0.03

Gain = 0.06

Gain = 0.1



Linear (Gain = 0.1)




6

Sample Calculation

Table 5 Gain = 0.06, Brake setting 10, with speed feedback

Identify Steady State Voltage Vss from Dual trace oscilloscope,Vss = 1.141mV

63.2% of Vss= 0.632 X 1.141= 0.721mV

Move the voltage cursor V2 to 0.721mV, move time cursor t2 tointersect with V2, we can get the Time Constant value:Time Constant τ = t2 – t1 = 144 – 124 = 20ms

Table 6:Brake setting = 5, with position feedback, no speed feedback, gain = 0.1 Identify Steady State Voltage Vss from Dual trace oscilloscope,

Vss = 3.812mV

90% of Vss= 0.9 X 3.812= 3.431mV

Move the voltage cursor V2 to 3.431mV, move time cursor t2 to

intersect with V2, we can get the Rise Time value:Rise Time = t2 – t1 = -248 – (-436) = 188ms

And, continue to identify the Overshoot Percentage:Vss = 3.812mVVovershoot = 4.094mV

% =ℎ −

× 100% =4.094 − 3.812

3.812 = . %




7

Discussion

1. Discuss the dif ferences in open-loop and closed loop control in

achieving a speed con trol sys tem and the effects of loop g ain and

load on the ou tput sp eed.

Observation and analysis for the Graph 2 (plotted results from table 3,open-loop speed control system):

The motor speed output drops significantly if there is adisturbance, which is the magnetic brake in this experiment. Thereduction of the motor speed is linearly proportional to themagnitude of disturbance, which in this experiment, when weadjust the magnetic brake to increase the disturbance, the motorspeed reduced significantly.

We recorded three gain values: 0.03, 0.06 and 0.1. From the

graph, we can see there is no much difference in between them,which the different gain setting of Op. Amp. has no effectagainst disturbance.

A characteristic of the open-loop control is that it does not usefeedback to determine if its output has achieved the desired goal of theinput. This means that the system does not observe the output of theprocesses that it is controlling.

From the block diagram, the error does not feedback to the controller,no matter how we adjust Kt, it has no effect on the output. K t has noeffect on the open-loop control. On the other hand, the brake has effecton the output speed, so the more the load caused by the brake is, theslower the output is.




8

Observation and analysis for the Graph 3 (plotted results from table 4,closed-loop speed control system):

The reduction of the motor speed is still linearly proportional tothe magnitude of disturbance, but the reduction is not as muchas open loop.

Different gain setting affects the reduction in motor outputspeed. The higher the gain setting value is, the lesser thesystem speed reduction is affected by the disturbance.

In a closed-loop control system, a sensor monitors the system outputand feedback the errors to a controller, adjusts the controller tomaintain the desired system output.

So the errors will feedback to the controller, and it’s compensated inthe output. So when enlarge the gain Kt, the compensation is alsoincreased; the speed will be dropped lesser. In the closed-loop control

system, the output is less sensitive to the outside disturbance. And themore load we apply, the slower the output is.




9

2. Discuss th e effect of loop gain and brake scale sett ings on the

transient respon se of the closed-loop s peed con trol system .

From table 5, we observe that:

The time constant decreases while the brake increases from 5to 10.

As the gain setting value increases, the time constant τ alsodecreases.

From the graph, the highest power of s is 1, so it is the first ordercontrol system:

()

() =

+ 1

The coefficient of s is time constant τ:

=

1 +

The smaller the time constant τ is, the faster the systems canresponse. If we increase the gain, the value of τ will be decreased.

As for disturbance TL, the transfer function between it and Ω is: ()

() =

+ 1

When KL is a constant, if we increase the brake, add in more load TL, τwill also decrease, system will response faster.




10

3. Discuss the effect of loop gain and veloci ty feedback on the

response of the closed loop po sit ion con trol system.

From table 6 (closed-loop, with position feedback):

With speed feedback, the rise time is significantly longer thanthose without speed feedback

For those without speed feedback, when the gain setting valueincreases to 0.1, there is an overshoot of 7.4% (please refer tothe appendix attached behind)

From the graph, the highest power of s is 2, so it is the second order

control system. There are two important parameter, ωn (naturalfrequency) and ζ (damping ratio).

When we increase the Kp, the damping ratio ζ will be decreased.When the damping ratio is equal or larger than 1, there will be noovershoot. When increase the Kp to large enough to force dampingratio to fall below 1, there will be an overshoot.

So when come to this experiment, the gain of 0.1 is large enough tocause this 7.4% overshoot. For the natural frequency, Kp is thenominator, so the natural frequency will increase when the value of Kp increases. Not only means it will oscillate faster, but also it willresponds faster.

me2142e speed or position control of a dc motor

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