SECTION A: Analogue Electronics II

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Question 1(a): Distinguish between linear wave shaping and non-linear wave shaping circuits.

Wave shaping circuits modify the shape of an input waveform based on their circuit components and properties. They are categorized into:

1. Linear Wave Shaping Circuits:

   • These circuits process signals without introducing distortion, meaning the output waveform maintains its shape but may be altered in amplitude or time delay.
   • Examples include RC integrators and differentiators that shape waveforms smoothly over time.
   • Used in signal conditioning, communication systems, and waveform generators.

2. Non-Linear Wave Shaping Circuits:

   • These circuits distort the shape of the input waveform using components like diodes and transistors.
   • Examples include clipper circuits (which limit voltage levels) and clamper circuits (which shift voltage levels).
   • Used in rectifiers, voltage limiters, and pulse generators.

📌 Key Difference:

• Linear circuits modify the waveform smoothly without introducing sharp changes.
• Non-linear circuits create abrupt changes in the waveform (clipping, shifting, rectification).

 

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Question 1(c): Class A Transistor Amplifier Power Calculations

Given Data:

• Vcc = 20V (Supply Voltage)
• VCEQ = 10V (Quiescent Collector-Emitter Voltage)
• ICQ = 600mA (Quiescent Collector Current)
• RL = 16Ω (Collector Load Resistor)
• AC Output Current = ±300mA

(i) Power Supplied by the DC Source (Input Power)

PDC=VCC×ICQP_{DC} = V_{CC} times I_{CQ}PDC​=VCC​×ICQ​ PDC=20V×0.6A=12WP_{DC} = 20V times 0.6A = 12WPDC​=20V×0.6A=12W

(ii) DC Power Consumed by Load Resistor

PRL(DC)=ICQ2×RLP_{RL} (DC) = I_{CQ}^2 times R_LPRL​(DC)=ICQ2​×RL​ PRL(DC)=(0.6)2×16=5.76WP_{RL} (DC) = (0.6)^2 times 16 = 5.76WPRL​(DC)=(0.6)2×16=5.76W

(iii) AC Power Developed Across Load Resistor

PRL(AC)=(Iac)2×RL2P_{RL} (AC) = frac{(I_{ac})^2 times R_L}{2}PRL​(AC)=2(Iac​)2×RL​​ PRL(AC)=(0.3)2×162=0.72WP_{RL} (AC) = frac{(0.3)^2 times 16}{2} = 0.72WPRL​(AC)=2(0.3)2×16​=0.72W

(iv) DC Power Delivered to Transistor

Ptransistor=PDC−PRL(DC)P_{transistor} = P_{DC} – P_{RL} (DC)Ptransistor​=PDC​−PRL​(DC) Ptransistor=12W−5.76W=6.24WP_{transistor} = 12W – 5.76W = 6.24WPtransistor​=12W−5.76W=6.24W

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Question 1(d)(i): Two roles of capacitors in multistage transistor amplifiers

1. Coupling Capacitors: Block DC voltage while allowing AC signals to pass between amplifier stages.
2. Bypass Capacitors: Improve gain by stabilizing the operating point and bypassing AC signals to ground.

 

Question 2(a): Three Application Areas of Operational Amplifiers (OP-AMPs)

Operational amplifiers (OP-AMPs) are widely used in electronic circuits due to their high gain and versatility. Three key application areas include:

1. Signal Conditioning: OP-AMPs are used for amplification, filtering, and level shifting in sensor circuits.
2. Mathematical Operations: Used in analog computers for summation, subtraction, differentiation, and integration.
3. Oscillators and Waveform Generators: Used in generating sine, square, and triangular waveforms in communication and testing circuits.

 

Question 3(a): State three applications of photodiodes.

Answer: Photodiodes are semiconductor devices that convert light into electrical current. They are commonly used in:

1. Optical Communication Systems – Used in fiber-optic receivers to convert optical signals into electrical signals.
2. Light Sensing Applications – Found in cameras, light meters, and automatic street lighting systems.
3. Medical Equipment – Used in pulse oximeters and blood analysis devices to measure oxygen levels and detect biological properties.

 

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Question 3(d): A phase shift oscillator uses 5 pF capacitors. Determine the value of R to produce a frequency of 800 kHz.

Answer: The formula for the frequency of a phase shift oscillator is:

f=12πRC6f = frac{1}{2pi R C sqrt{6}}f=2πRC6​1​

Given:

• f = 800 kHz = 800,000 Hz
• C = 5 pF = 5 × 10⁻¹² F

Solving for $R$:

R=12πfC6R = frac{1}{2pi f C sqrt{6}}R=2πfC6​1​ R=12π(800,000)(5×10−12)6R = frac{1}{2pi (800,000) (5 × 10^{-12}) sqrt{6}}R=2π(800,000)(5×10−12)6​1​ $$R\approx 45.84k\Omega$$

📌 Final Answer: The required resistor value is approximately 45.84 kΩ to achieve a frequency of 800

Section B: Digital Electronics  

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Question 4(a): State three advantages of digital-over-analogue systems.

Answer: Digital systems have several advantages over analogue systems, including:

1. Noise Immunity: Digital signals are less affected by noise and distortion compared to analogue signals, ensuring more reliable data transmission.
2. Easy Storage & Processing: Digital data can be easily stored, processed, and transmitted without significant loss in quality.
3. Higher Accuracy & Precision: Digital systems allow for higher accuracy in measurements and computations due to discrete signal representation.

 

 5(a): State each of the following Boolean laws for three literals x, y, and z.

(i) Distributive Law

The Distributive Law states:

1. AND over OR: $$=xy+xz$$
2. OR over AND:x+(yz)=(x+y)(x+z)

(ii) Associative Law  $$(x+y)(x+z)$$

The Associative Law states that the grouping of variables does not affect the result:

1. For AND operation: (xy)z- x (y z) 
2. For OR operation: (x+y)+z= x + (y + z) 

 6(a): State three areas of application of shift registers.

Shift registers are widely used in digital electronics for data storage and transfer. Three key applications include:

1. Data Storage:

   • Used in memory devices to store and retrieve data sequentially.

2. Serial-to-Parallel and Parallel-to-Serial Conversion:

   • Used in communication systems to convert serial data to parallel format (or vice versa), such as in UART (Universal Asynchronous Receiver-Transmitter).

3. Delay Line Applications:

   • Used in digital signal processing for introducing time delays in signals, such as in digital filters.

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 7(a): State four applications of analogue-to-digital converters (ADC).

Analogue-to-Digital Converters (ADCs) are used in many applications where analogue signals need to be converted into digital form for processing.

1. Digital Signal Processing (DSP):

   • Used in audio and video processing, such as converting microphone input into a digital format for storage and manipulation.

2. Microcontrollers and Embedded Systems:

   • ADCs enable microcontrollers to read sensor values (e.g., temperature, pressure, or light sensors).

3. Communication Systems:

   • Used in radio receivers and mobile phones to convert analogue signals (radio waves) into digital data.

4. Medical Devices:

   • ECG (Electrocardiogram) and MRI (Magnetic Resonance Imaging) machines use ADCs to convert biological signals into digital formats for analysis.

Question 7(d): Distinguish between fan-in and fan-out in logic families.

Term	Definition
Fan-in	The number of inputs a logic gate can handle. Higher fan-in means more inputs can be processed simultaneously.
Fan-out	The number of logic gates that a single gate’s output can drive without degradation in signal quality.

 

Question 8(a): Define each of the following with respect to memories:

(i) Capacity

Memory capacity refers to the total amount of data a memory device can store, usually measured in bits, bytes, kilobytes (KB), megabytes (MB), gigabytes (GB), or terabytes (TB).

📌 Example: A 256K × 8 RAM means it can store 256K words, each 8 bits wide, giving a total capacity of 256K × 8 = 2 megabits (Mb).

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(ii) Access Time

Memory access time is the time interval between the request for data and its availability. It is measured in nanoseconds (ns) or microseconds (µs) and indicates how fast a memory device retrieves data.

📌 Example:

• SRAM (Static RAM) has an access time of 4–10 ns (very fast).
• DRAM (Dynamic RAM) has an access time of 50–100 ns (slower but cheaper)

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Question 8(b): For a 256K × 8 RAM memory, determine the:

(i) Word Size

📌 Word size = Number of bits per word = 8 bits

(ii) Number of Address Lines

The number of address lines (A) required is given by:

2A=Total memory locations2^A = text{Total memory locations}

Since the memory has 256K words, we calculate:

2A=256K=256×1024=2621442^A = 256K = 256 times 1024 = 262144 A=log⁡2(262144)=18 address linesA = log_2(262144) = 18 text{ address lines}

📌 Number of address lines = 18

(iii) Capacity in Kilobytes

Since the memory is 256K × 8, the total capacity in bytes is:

256K×8/{8} = 256K { bytes} = 256 KB

📌 Capacity = 256 KB

 

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Section A: Electrical Power Generation and Transmission

1. (a) State three factors that determine the location of a thermal power plant and two merits of a nuclear power station.

Answer:
Factors determining the location of a thermal power plant:

1. Availability of fuel – The plant should be near coal, oil, or gas resources to reduce transportation costs.
2. Water supply – A sufficient supply of water is needed for cooling purposes.
3. Proximity to load centers – Being closer to demand reduces transmission losses and costs.

Merits of a nuclear power station:

1. High efficiency – Produces a large amount of power using a small quantity of fuel.
2. Low fuel cost – Nuclear fuel is more efficient than fossil fuels and requires less frequent refueling.

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 (b) Describe the following parts of a diesel power station: fuel supply system, exhaust system, and cooling system.

Answer:

• Fuel supply system – Stores and supplies fuel to the engine. It consists of a fuel tank, fuel pump, and filters.
• Exhaust system – Removes combustion gases and includes silencers to reduce noise pollution.
• Cooling system – Maintains the engine at an optimal temperature using water or air to prevent overheating.

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2. (a) State two limitations of Kelvin’s law.

Answer:

1. Does not consider voltage regulation – It focuses only on economic aspects, ignoring performance limitations.
2. Ignores future system expansion – The law does not consider increased loads over time.

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 (c) Explain three demerits of low power factor.

Answer:

1. Increased power losses – More current flows through the conductors, causing higher resistive losses.
2. Overloading of electrical equipment – Transformers, generators, and cables need to handle more current, reducing efficiency.
3. Higher electricity costs – Utilities charge extra fees for poor power factor due to increased demand on the system.

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3. (a) State four essential properties of switchgear.

Answer:

1. High reliability – Must operate correctly under normal and fault conditions.
2. Quick operation – Should respond quickly to faults to prevent damage.
3. High breaking capacity – Must handle high fault currents safely.
4. Thermal and mechanical strength – Should withstand high temperatures and mechanical forces during faults.

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 (b) Explain three factors affecting arcing resistance of a circuit breaker.

Answer:

1. Contact material – High-resistance materials like tungsten reduce arcing.
2. Contact separation speed – Faster separation reduces arc duration.
3. Medium used for arc extinction – Oil, air, vacuum, or SF6 gas helps to extinguish the arc.

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4. (a) State three types of overhead line supports.

Answer:

1. Wooden poles – Used for low-voltage and short-distance applications.
2. Steel towers – Used for high-voltage transmission lines due to strength and durability.
3. Concrete poles – Resistant to corrosion and used in urban areas.

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4. (b) Describe each of the following tests on overhead line insulators: puncture test, porosity test, mechanical test, and temperature test.

Answer:

• Puncture test – Applies high voltage to check insulation strength.
• Porosity test – Insulator is boiled in water, and any air bubbles indicate porosity.
• Mechanical test – Checks the insulator’s ability to withstand mechanical stress.
• Temperature test – Exposes the insulator to extreme temperatures to check for expansion or cracks.

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5. (a) State three reasons why copper is the commonly preferred material in overhead transmission lines and two advantages of suspension-type insulators over pin-type.

Answer:
Reasons why copper is preferred in overhead transmission lines:

1. High conductivity – Reduces energy losses.
2. Corrosion resistance – Lasts longer in different weather conditions.
3. Mechanical strength – Withstands tension and stress.

Advantages of suspension-type insulators over pin-type:

1. Higher voltage handling – Can be used for very high voltages by adding more units.
2. Less mechanical stress on poles – The string arrangement distributes the weight more evenly.

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Section B: Building Services

6. (a) Explain the stroboscopic effect and state three methods of minimizing the stroboscopic effect in fluorescent lamps.

Answer:
Stroboscopic effect: This occurs when a rotating or moving object appears stationary or moving slowly under fluctuating light, typically from fluorescent lamps.

Methods to minimize the stroboscopic effect:

1. Use of high-frequency ballasts – Reduces flickering.
2. Using multiple lamps – Operated on different phases to reduce synchronization.
3. Using electronic ballasts – Ensures stable lighting.

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 (c) Explain two factors considered while designing a lighting scheme and each of the following types of lighting schemes: direct lighting scheme, semi-direct lighting, and general lighting.

Answer:
Factors considered while designing a lighting scheme:

1. Illumination level – Should be appropriate for the purpose (e.g., brighter lighting for workspaces).
2. Energy efficiency – Choosing LED or energy-saving lamps to reduce costs.

Types of lighting schemes:

• Direct lighting scheme – 90% of the light is directed towards the working area.
• Semi-direct lighting scheme – 60–90% of light is directed downward while some is diffused.
• General lighting scheme – Provides uniform light distribution throughout the area.

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7. (a) State two types of sanitary appliances, citing an example in each case.

Answer:

1. Water closets (WCs) – Example: Toilet bowl.
2. Washbasins – Example: Sinks used for handwashing.

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 (b) Describe two characteristics of sanitary appliances.

Answer:

1. Hygienic design – Smooth surfaces to prevent bacterial growth.
2. Durability – Made of corrosion-resistant materials like porcelain or stainless steel.

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 (c) List two types of utility services.

Answer:

1. Water supply system – Provides clean water for domestic and industrial use.
2. Drainage system – Removes wastewater and prevents flooding.

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8. (a) State two groups of final sub-circuits.

Answer:

1. Lighting circuits – Supply power to lights.
2. Power circuits – Supply power to sockets and appliances.

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 (c) Explain each of the following factors to be considered when selecting a wiring system: cost, flexibility, and disposition.

Answer:

1. Cost – The system should be affordable and within budget.
2. Flexibility – It should allow for future expansion or modifications.
3. Disposition – The arrangement should ensure safety and convenience.

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 (d) Outline six steps considered when selecting cable size using a given load data.

Answer:

1. Determine the load current – Based on the power rating of connected devices.
2. Choose the appropriate conductor material – Copper or aluminum.
3. Check the voltage drop – Ensure it does not exceed acceptable limits.
4. Consider environmental conditions – Temperature, moisture, and chemical exposure.
5. Account for future expansion – Select a cable size that allows for increased load.
6. Ensure compliance with regulations – Follow electrical safety codes.

 

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Section A: Control Systems

1. (a) Define each of the following with respect to control systems: reference variable, controlled variable, and disturbance.

Answer:

• Reference variable: The desired value that a control system aims to maintain.
• Controlled variable: The actual output of the system that is adjusted to match the reference variable.
• Disturbance: Any external factor that affects the system’s output.

 (b) Classify the control systems in Table 1 as natural, man-made, or hybrid systems.

Answer:

• Automated traffic light: Man-made system
• Plant health monitoring system: Hybrid system
• Smart greenhouse: Hybrid system
• Air conditioning system: Man-made system
• Plant water absorption system: Natural system

  (c) (i) Explain the principle of superposition with reference to multi-input control systems.

Answer:

• The principle of superposition states that in a linear system with multiple inputs, the total response is the sum of the individual responses caused by each input acting alone.

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2. (a) Differentiate between source and sink modes, and branch and path gains with respect to Signal Flow Graph (SFG).

Answer:

• Source mode: A starting node providing input signals while Sink mode: An ending node where the final output is obtained.
• Branch gain: The gain along a single path.while Path gain: The total gain of a complete route from source to sink.

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 (d) State the elements of a mechanical translational system model.

Answer:

• Mass
• Damping
• Stiffness (spring constant)

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4. (a) State three demerits of using differentiators in analog computing amplifiers.

Answer:

• High sensitivity to noise.
• Instability due to high-frequency components.
• Requires careful component selection to avoid errors.

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5. (a) (i) State two conditions to be satisfied for a linear time-invariant system to be stable.

Answer:

• All poles of the characteristic equation must have negative real parts.
• The system response should not grow unbounded over time.

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Section B: Programmable Logic Controllers (PLCs)

6. (a) State three types of programmable logic controller (PLC) manufacturers and PLC programming languages.

Answer:

• PLC Manufacturers: Siemens, Allen-Bradley, Mitsubishi.
• PLC Programming Languages: Ladder Logic (LD), Function Block Diagram (FBD), Structured Text (ST).

(b) Explain hardware setup, parameter setting, and integration with reference to PLC configuration.

Answer:

• Hardware setup: Installing PLC components, wiring, and connections.
• Parameter setting: Configuring memory, input/output addresses, and timers.
• Integration: Connecting PLC with external devices like sensors and actuators.

(c) Outline four checks performed before installing industrial software.

Answer:

• Compatibility with the PLC hardware.
• Availability of required drivers and libraries.
• Verification of software license.
• Proper installation of required updates.

(d) Describe simulation and monitoring PLC programming tools.

Answer:

• Simulation tools: Allow testing of PLC logic before implementation.
• Monitoring tools: Enable real-time analysis of PLC execution and troubleshooting.

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7(a) State three factors affecting PLC scan time.

Answer:

• Number of instructions in the program.
• Processing speed of the PLC CPU.
• Number of I/O operations and communication tasks.

(b) Differentiate between timers and counters in PLC systems, citing one type each.

Answer:

• Timers: Used to delay actions (e.g., ON-delay timer).
• Counters: Used to count events (e.g., Up-counter).

 (c) Describe ‘rail’ with respect to PLC ladder programming.

Answer:

• The vertical lines in ladder logic diagrams that provide power flow direction, from left (power source) to right (output).

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8. (a) Define ‘data acquisition’ with respect to Supervisory Control and Data Acquisition (SCADA) system and list three devices used in SCADA data acquisition.

Answer:

• Data acquisition: The process of collecting and transmitting real-time data from industrial equipment for monitoring and control.
• Devices used: Sensors, Remote Terminal Units (RTUs), and Programmable Logic Controllers (PLCs).

(b) Explain three measures for enhancing SCADA data security and integrity.

Answer:

• Implementing strong authentication and encryption.
• Regularly updating software and firmware.
• Using network segmentation to limit access.

(d) State the function of network switch, network hub, repeater, and modem in industrial networks.

Answer:

• Network switch: Directs data to specific devices in a network.
• Network hub: Broadcasts data to all connected devices.
• Repeater: Amplifies signals for long-distance communication.
• Modem: Converts digital data to analog signals and vice versa for internet communication.

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