control engineering lab 4

8/13/2019 control engineering lab 4

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University of Duhok

Faculty of engineering and applied science

Electrical and computer engineering

Control Lab

Exp #4

Names:

Ammar Sabri Andrawous

Hameed Ismaeel Ahmed

Steven wasfi dinkha

Barhav sarbast lazgeen

Discussion and Results:

Position Control (Exp 1&2)


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1. Write down your observations on what happened in exp 1. Why this

is an open loop system?

In this experiment the circuit is a normal input/output design the

output is controlled manually and based on observations, there is no

feedback so the error should be dealt with manually and the results arenot accurate, not having a feedback makes it an open loop system.

2. What can be the source of error (misalignment)?

The source can be:Physical, the devices used need maintenance to work optimally.

The controller might not be getting an accurate feedback or there is adelay.

The device has to have a misalignment at the beginning until thecontroller gets the feedback to know the problem and fix it.

There might be a problem with the controller

3. Compare the open loop and closed loop systems. Discuss theadvantage and disadvantage of these systems giving reasons.

Closed Loop System

In a closed loop control system, the input variable is adjusted by thecontroller in order to minimize the error between the measured output

variable and its set point. This control design is synonymous tofeedback control, in which the deviations between the measured

variable and a set point are fed back to the controller to generateappropriate control actions. The controller C takes the difference e

between the reference r and the output to c hange the inputs u to thesystem. This is shown in figure below. The output of the system y is fed

back to the sensor, and the measured outputs go to the reference value


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Open Loop System

On the other hand, any control system that does not use feedback informationto adjust the process is classified as open loop control. In open loop control,the controller takes in one or several measured variables to generate control

actions based on existing equations or models.

Advantages

The advantages of feedback control lie in the fact that the feedback control

obtains data at the process output. Because of this, the control takes intoaccount unforeseen disturbances such as frictional and pressure losses.Feedback control architecture ensures the desired performance by altering theinputs immediately once deviations are observed regardless of what causedthe disturbance. An additional advantage of feedback control is that byanalysing the output of a system, unstable processes may be stabilized.Feedback controls do not require detailed knowledge of the system and, inparticular, do not require a mathematical model of the process. Feedbackcontrols can be easily duplicated from one system to another. A feedbackcontrol system consists of five basic components: (1) input, (2) process beingcontrolled, (3) output, (4) sensing elements, and (5) controller and actuating

devices. A final advantage of feedback control stems from the ability to trackthe process output and, thus, track the systems overall performance.

Disadvantages

Time lag in a system causes the main disadvantage of feedback control. Withfeedback control, a process deviation occurring near the beginning of theprocess will not be recognized until the process output. The feedback control

will then have to adjust the process inputs in order to correct this deviation.This results in the possibility of substantial deviation throughout the entireprocess. The system could possibly miss process output disturbance and the

error could continue without adjustment. Generally, feedback controllers onlytake input from one sensor. This may be inefficient if there is a more directway to control a system using multiple sensors. Operator intervention isgenerally required when a feedback controller proves unable to maintainstable closed-loop control. Because the control responds to the perturbationafter its occurrence, perfect control of the system is theoretically impossible.Finally, feedback control does not take predictive control action towards theeffects of known disturbances.

For the open loop system:

Advantages


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Effective regardless of the origin of the disturbance.

Relatively simple to implement. A mathematical model of the exiting watertemperature is not needed.

Disadvantages

Corrective action taken only after there is a deviation in the output fromthe set point.

Can cause instability if incorrectly tuned

4. Give two examples of position control in real life

Solar cells (position follows the sun) and AC units (control temperature)

Speed Control (EXP 3&4)

5. Tacho v against rpm

Y=(1/Kg) *R12.16 = (1/Kg) * 1

Kg = 1/12.16

Kg = 0.082

6. Rpm against input V

1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 610

20

30

40

50

60

70

80


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7. Rpm against torque

Slope = (BP2-BP1)/(S2-S1)

= (2-0)/(940-1000)= 0.0333

8. Error against brake

0 1 2 3 4 5 6 7 8 9 10400

500

600

700

800

900

1000


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BP against speed

0 1 2 3 4 5 6 7 8 9 10

0.8

0.9

1

1.1

1.2

1.3

1.4

1.5

0 1 2 3 4 5 6 7 8 9 10750

800

850

900

950

1000


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At gain 10

Error against brake:

BP against speed

0 1 2 3 4 5 6 7 8 9 101

2

3

4

5

6

7

8

9

10

11

0 1 2 3 4 5 6 7 8 9 10700

800

900

1000

1100

1200

1300

control engineering lab 4

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