Comparing Open Loop and Closed Loop Control Systems

Task 1

Part 1:

1. Compare open loop and closed loop control systems

Open loop control system is where the output has not effect on the control action. Such as system is always free from feedback and the working relies on time. The best example of the open loop control system is the automatic washing machine. On the other hand, a closed loop system is where the output relies on the input. The system has at least one feedback loop that connects the input and the output. The best example of the closed loop system is the air conditioner.

2. Basic functions of:

Sensors: they provide means of gathering information on processes and operations being performed. They are ultimate used in transforming the physical stimulus into electrical signals which can easily be analysed and used in decision making.

Actuators: Upon receiving a control signal, the actuator responds through conversion of the signal’s energy into significant mechanical motion. Therefore, an actuator is essentially a component of the machines charged with the responsibility of controlling or moving the system.

3. Calculate which of the following motors gives the greater load acceleration.

For Motor A

Acceleration Torque

Ta = J x A

Where Ta : Acceleration Torque

J = Moment of inertia

A = Acceleration rate

This means that

A = Ta/J Given that Ta = 0.15 Nm and inertia is 2.5kgm^2 then A = 0.15Nm/2.5kgm^2 A = 0.06 m/s2

For Motor B

Load acceleration = Ta/J

= 0.25/5

Comparing A and B, it could be noted that motor A has more load acceleration compared to motor B

Part 2:

4. Determine the gear ratio necessary to obtain greatest load acceleration and then calculate the maximum load acceleration if the load inertia is 450kgm2 using the motor with the greater load acceleration.

Solution

Gear ratio for the greatest load acceleration is given by Gr = sqrt(JL/JM) Where JL is the load inertia and JM is the motor inertia Taking motor A with the highest load acceleration Gr = sqrt (450/2.5) Gr = 13.42 Maximum load acceleration αL = (TM x Gr)/ (JM x Gr2 + Jl) = (0.15 x 13.42)/(2.5 x 13.422 + 450) = 2.013/900.241 = 0.0022m/s2

5. Use diagrams to help describe how it is possible to control the speed of an AC servo-motor using a closed loop speed control system. Explain how the error signal is formed and discuss what fault conditions could cause the speed to drop abnormally.

The speed control of the AC servo motors is regarded as a highly demanded task achieved through the torque control or the velocity control mode. A control loop is always significant in the velocity model with a closed loop control system put to task as shown below.

From the diagram, it can be seen that the signal goes through the amplifier section. The low power level signal is then amplified to appropriate levels, which can trigger movements of the servo motor. The power is supplied from an AC source, which is also converted to DC while low level voltage is supplied to the integrated circuits. Devices such as tachometer or resolver are used to provide a signal to the controller. The feedback signal informs the controller as to whether the motor is performing a good work or not. In case errors are detected, then the controller will make appropriate corrections until the feedback signals becomes same to the command signal, which means that there is no error. The error signal is automatically triggered when the specific parameters such as the maximum pulse, the repetition rate or the minimum pulse are not adhered to by the system. Some of the faulty conditions that would trigger the error signal include high overload capacity among others.

Task 2:

Part 1:

The feedback transducer has a transfer function of 50mV per r/min and the transfer function for the motor/load is 100r/min per volt. The reference signal applied to the system is 100 V. Given that the allowance speed error is 40r/min, determine the amplifier gain (transfer function).

Solution

Av = (Output voltage)/Input voltage The input voltage = 100v The output voltage = 50mV per r/min x 100r/min per volt = 5000mv Converting this into Volts (5000 x 10)/1000 = 50v AV = 100v/50 = 2.

Part 2:

For the system in the above question the reference voltage is reduced to 80V. Determine the corresponding reference speed. For the system in the above question, the load speed is adjusted to 1900 r/min. Determine the corresponding reference voltage. Critically investigate the behaviour of the control system to compare the above 3 approaches.

Solution

If input voltage is 80V Corresponding speed is (100 x 80)/100 = 80r/min If load speed is adjusted to 1900 r/min Then the voltage across Is given by (1900 r/min)/(100r/min per volt) = 19V The corresponding reference voltage is therefore =Reference voltage – the voltage drop across the load = 80 – 19 = 61V

Task 3:

Part 1: Laboratory work

Part 2: Explain the procedure for this and talk about accuracy, resolution, tolerances, stability, sensitivity and response time for the measuring system. Critically evaluate the performance of an ideal measuring system compared to a real circuit.

Lab Experiment: To determine Instrument Characteristics of the Digital Voltmeter

Objective

The objective of this experiment is to determine the effect of internal resistance of digital voltmeter on instrument characteristics

Background

Voltmeters are designed with high input resistance for the purposes of reducing effect on the voltage measurements. This is true especially for the new digital meters. Commonly, modern digital voltmeters exhibit very high resistances, higher than the resistance boxes or the variable resistors. This observation has attracted significant studies on what such resistance can influence in terms of the response time, accuracy, resolution, stability and sensitivity of the voltmeter. Assuming that

RS = Variable resistor V0 = Output voltage I = current VR = voltage drop across variable resistor VM = Voltage drop across the digital voltmeter

Theoretically, RS should be very close to RM for the voltmeter to be accurate and sensitive at the same time. However, in practice, reasonably large RS would pave way for accurate measurements.

Procedure

a. Set the voltmeter to 20V and set the AC/DC switch for it to provide DC.

b. With power supply disconnected from any active circuit, switch on the power supply, put ON the output and adjust the power supply to give 10V

c. Turn OFF the output

d. Connect as shown below

e. Set the resistance RS to 0.

f. Turn ON the power supply and adjust the output voltage to 10V.

g. Record the figure shown on the voltmeter and denote it as V0

h. Increase RS to a large value

i. Record the value shown on the voltmeter as VM

Now, the set up can be adjusted to perform the following experiment. It should be noted that the voltmeter scale should not be altered

a. Set the voltmeter to 20V and set the AC/DC switch for it to provide DC.

b. With power supply disconnected from any active circuit, switch on the power supply, put ON the output and adjust the power supply to give 10V

c. Turn OFF the output

d. Connect as shown below

f. Put ON the power supply OUTPUT. This should be 10V

g. Record the value shown on the voltmeter

h. Put OFF the power supply output

i. Set R to 500 kilo ohms

j. Put ON the power supply output

k. Record the value shown on the voltmeter

l. Put OFF the power supply output

m. Set R to 1 Mega Ohms

n. Put ON the power supply output

o. Record the voltage

Note: Based on the recorded values, one is expected to compute theoretical values and draw comparison with measured values.

Task 4:

Part 1:

Write down the advantages and disadvantages of increased use of wireless and remote control in the GRACO Industry.

GRACO is regarded as a highly reputable engineering industry, with keen focus on dealing with design, development as well as manufacture of different industrial applications. As an Industrial Technician, I treat technology as one platform that ensures pursuit of efficient monitoring as well as control of different variables. This concern has paved way for increased attention towards wireless and remote control, which have a number of benefits and limitations at the same time. First, a wireless and remote control system enhances operational security. This is accompanied by a better overview, which attracts quicker reactions in case of the emergency situations. Secondly, wireless and remote control system provide a more effective way of working due to the ease of operation. This equally saves on time and additional costs. In addition, wireless and remote control systems attract lesser maintenance costs, which can be advantageous to professionals working in remote areas. Apart from the benefits, wireless and remote control system can be prone to insecurity issues such as cyber-attacks in the GRACO industry. This can be as a result of network infiltration among other forms of attacks. In addition, capital costs can be extreme especially where industrial robots are involved, which may lead to poor returns on investment.

Part 2:

Analyse appropriate analytical techniques used in your industry and justify your recommendations to improve the performance.

One of the appropriate analytical techniques used in the GRACO industry is the system analysis. System analysis is highly regarded as a process engaged in studying a procedure, for the purposes of identifying goals and purposes while developing systems and procedures that will attain them in a more efficient way. Besides, system analysis is closely seen as one of the problem solving technique known for breaking down the entire system into constituent pieces with the central mission of studying and understanding how well each component works, as well as interact with the rest before accomplishing the goal. The rationale behind system analysis cannot be delinked from operations research as well as requirement analysis which all necessary in the decision are making process.

What makes system analysis more appropriate in engineering and engineering-related activities is its closeness to systems design, processing and system boundaries. The observable properties under system analysis include organization, which denotes the structure and order, interaction, interdependence and integration. The use of system analysis presents a number of benefits to the users. The first benefit is linked to flexibility, efficiency and costs. Flexibility is aligned to the customized approach, which takes care of different data processing needs associated to the scale of operations and the nature of the system. While using system analysis, it helps the designers to identify significant opportunities as well as problems through evaluation of weaknesses and strengths. Apart from being a customized approach, the approach is profitable in the sense it aims at enhancing the performance of each component of the system. In this sense, it helps in minimizing errors while trying to fix problems where possible. System analysis also leads to better controls as well as better management.

Apparently, system analysis attracts quality enhancement while promoting frequent system checks. Apart from the advantages, system analysis suffers from major drawbacks such as time consumptions when it comes to large projects. It can also attract stressful errors and some of the adjustments are likely to violate normal procedures. For system analysis to work more effectively, it is recommendable to conduct continuous improvement analysis, which constantly checks the system and draws comparison against a more effective system with similar functions. It is also recommendable to assess the kind of technology introduced in system analysis.