← MES Jr. Grade
Past Questions — 2024
In TDM systems, guard times are used to:
A)
Reduce bit error rate
B)
Prevent overlap of adjacent time slots
C)
Minimize modulation distortion
D)
Reduce phase jitter
In Time Division Multiplexing (TDM), multiple signals are transmitted over a single channel by assigning each signal a specific time slot. Due to synchronization errors, propagation delays, or clock drift between the transmitter and receiver, the timing of these slots can shift slightly. If two adjacent time slots overlap, it results in Inter-Symbol Interference (ISI) or data corruption. To prevent this, 'guard times' (small intervals of silence or idle time) are inserted between the time slots. These guard times act as a buffer, ensuring that even with minor timing misalignments, the data from one slot does not bleed into the next.
Why other options are incorrect:
- (a) Bit error rate (BER) is primarily affected by noise, interference, and signal attenuation, not specifically by guard times.
- (c) Modulation distortion is related to the hardware characteristics of the modulator/demodulator and the bandwidth of the channel.
- (d) Phase jitter is a high-frequency variation in the phase of a clock signal; while guard times mitigate the *effect* of jitter on slot boundaries, they do not 'reduce' the jitter itself.
The decimal equivalent of (3102.12)4 is:
A)
325.467
B)
210.375
C)
215.670
D)
190.325
To convert a number from base 4 to base 10 (decimal), we multiply each digit by 4 raised to the power of its position relative to the decimal point.
Given number: (3102.12)₄
Step 1: Calculate the integer part (3102₄)
- Digit 3 is at position 3: 3 × 4³ = 3 × 64 = 192
- Digit 1 is at position 2: 1 × 4² = 1 × 16 = 16
- Digit 0 is at position 1: 0 × 4¹ = 0 × 4 = 0
- Digit 2 is at position 0: 2 × 4⁰ = 2 × 1 = 2
Sum of integer part = 192 + 16 + 0 + 2 = 210
Step 2: Calculate the fractional part (.12₄)
- Digit 1 is at position -1: 1 × 4⁻¹ = 1 / 4 = 0.25
- Digit 2 is at position -2: 2 × 4⁻² = 2 / 16 = 1 / 8 = 0.125
Sum of fractional part = 0.25 + 0.125 = 0.375
Step 3: Combine the results
Total = 210 + 0.375 = 210.375
Comparing this result with the options:
- Option (a) 325.467 is incorrect.
- Option (b) 210.375 matches our calculation exactly.
- Option (c) 215.670 is incorrect.
- Option (d) 190.325 is incorrect.
The operating mode of 8086 after applying active low MN/MX’ is:
A)
Minimum mode
B)
Maximum mode
C)
Supervision mode
D)
User mode
The 8086 microprocessor has a pin named MN/MX' (Minimum/Maximum). This pin is used to determine the operating mode of the processor based on the logic level applied to it:
1. **Minimum Mode**: When the MN/MX' pin is tied to Vcc (Logic 1), the processor operates in Minimum Mode. In this mode, the 8086 generates all the necessary control signals (like ALE, RD, WR) itself, and it is typically used in single-processor systems.
2. **Maximum Mode**: When the MN/MX' pin is tied to Ground (Logic 0), the processor operates in Maximum Mode. The term 'active low' (indicated by the prime symbol ' or the bar over the signal) means the function is activated when the signal is at a low voltage level (0). In Maximum Mode, the 8086 requires an external bus controller (like the 8288) to generate the control signals, allowing for multi-processor configurations or co-processor integration.
Therefore, applying an active low signal to the MN/MX' pin puts the 8086 into Maximum Mode.
In 8086 ALU is operated on _____________ data.
A)
8-bit
B)
16-bit
C)
20-bit
D)
32-bit
The Intel 8086 is a 16-bit microprocessor. This means its internal architecture, including the Arithmetic Logic Unit (ALU), general-purpose registers (like AX, BX, CX, DX), and the internal data bus, are all designed to process data in 16-bit chunks. While the 8086 can perform 8-bit operations (by accessing the lower half of the registers, e.g., AL or AH), the native width and the maximum capacity of the ALU to operate on data in a single cycle is 16 bits.
Analysis of other options:
- a. 8-bit: This is the characteristic of the 8085 microprocessor, not the 8086.
- c. 20-bit: While the 8086 has a 20-bit address bus to access 1MB of memory, the ALU does not perform arithmetic operations on 20-bit data.
- d. 32-bit: This is characteristic of later processors like the 80386.
In an analog communication system, if the carrier frequency is 1 MHz, what is the upper sideband frequency for an AM signal with a modulation frequency of 5 kHz?
A)
995 kHz
B)
1.005 MHz
C)
1.0055 MHz
D)
5 kHz
In Amplitude Modulation (AM), the modulated signal consists of the carrier frequency and two sidebands.
1. **Given Data:**
- Carrier Frequency ($f_c$) = 1 MHz = 1000 kHz
- Modulating Frequency ($f_m$) = 5 kHz = 0.005 MHz
2. **Formula for Sidebands:**
- The Upper Sideband (USB) frequency is calculated as: $f_{USB} = f_c + f_m$
- The Lower Sideband (LSB) frequency is calculated as: $f_{LSB} = f_c - f_m$
3. **Calculation:**
- $f_{USB} = 1 ext{ MHz} + 0.005 ext{ MHz} = 1.005 ext{ MHz}$
- Alternatively, in kHz: $1000 ext{ kHz} + 5 ext{ kHz} = 1005 ext{ kHz} = 1.005 ext{ MHz}$
4. **Analysis of Options:**
- Option (a) 995 kHz is the Lower Sideband frequency ($1000 - 5$).
- Option (b) 1.005 MHz is the correct Upper Sideband frequency.
- Option (c) 1.0055 MHz is mathematically incorrect.
- Option (d) 5 kHz is simply the modulating frequency ($f_m$).
How many flags are there in 8085 microprocessor?
A)
3
B)
5
C)
7
D)
9
The 8085 microprocessor has a Flag Register (also known as the Program Status Word) which is an 8-bit register. However, only 5 of these bits are used as status flags to indicate the results of arithmetic and logical operations. These five flags are:
1. Carry Flag (CY): Set if there is a carry out from the most significant bit (D7).
2. Parity Flag (P): Set if the result has an even number of 1s.
3. Auxiliary Carry Flag (AC): Set if there is a carry from bit D3 to D4 (used for BCD arithmetic).
4. Zero Flag (Z): Set if the result of an operation is zero.
5. Sign Flag (S): Set if the result is negative (most significant bit D7 is 1).
The other 3 bits in the 8-bit register are undefined/unused, which is why the correct answer is 5, not 8.
What is the modulation index for FM if the maximum frequency deviation is 50 kHz and the modulating frequency is 5 kHz?
A)
5
B)
10
C)
50
D)
500
To find the modulation index for Frequency Modulation (FM), we use the standard formula:
Modulation Index (β) = Δf / fm
Where:
- Δf = Maximum frequency deviation
- fm = Modulating frequency
Given values from the question:
- Maximum frequency deviation (Δf) = 50 kHz
- Modulating frequency (fm) = 5 kHz
Calculation:
β = 50 kHz / 5 kHz
β = 10
Therefore, the modulation index is 10.
Analysis of options:
- Option (a) 5: Incorrect calculation.
- Option (b) 10: Correct calculation based on the FM modulation index formula.
- Option (c) 50: Incorrect; this is just the value of the frequency deviation.
- Option (d) 500: Incorrect; this would result from multiplication rather than division.
In PCM, the main factor deciding the number of quantization levels is:
A)
Noise power
B)
Desired signal-to-quantization-noise ratio
C)
Sampling rate
D)
Channel type
In Pulse Code Modulation (PCM), quantization is the process of mapping continuous analog values to a finite set of discrete levels. This process introduces a 'quantization error' or 'quantization noise'.
1. The quantization noise power is inversely proportional to the square of the number of quantization levels (L). Specifically, for a uniform quantizer, the quantization noise power is given by σ² = Δ²/12, where Δ (step size) = (Vmax - Vmin) / L.
2. The Signal-to-Quantization-Noise Ratio (SQNR) is the ratio of the signal power to this noise power. For a sinusoidal signal, the formula is approximately SQNR ≈ 6.02n + 1.76 dB, where n is the number of bits per sample (L = 2^n).
3. Therefore, to achieve a higher (better) SQNR, one must increase the number of quantization levels (L). The design requirement for the quality of the reconstructed signal (the desired SQNR) is the primary factor that dictates how many levels L are needed.
Why other options are incorrect:
- a. Noise power: While quantization noise is a result, the 'number of levels' is chosen to control this noise, not decided by an external noise power.
- c. Sampling rate: The sampling rate is determined by the Nyquist criterion (2 * maximum frequency) to avoid aliasing; it affects the bandwidth but not the number of quantization levels.
- d. Channel type: The channel type affects the transmission rate and error correction, but the quantization process happens at the encoder based on signal quality requirements.
The magnetron generates microwaves mainly by:
A)
Velocity modulation
B)
Cyclotron resonance
C)
Space-charge effect
D)
Electron bunching in crossed fields
A magnetron is a high-power vacuum tube that functions as a microwave oscillator. It operates using 'crossed fields,' meaning an electric field (between the cathode and anode) and a magnetic field (perpendicular to the electric field) are applied simultaneously.
1. **Mechanism**: Electrons emitted from the central cathode are accelerated toward the anode. However, the perpendicular magnetic field forces them into curved, cycloidal paths.
2. **Bunching**: Due to the interaction between the electrons and the RF field of the resonant cavities in the anode, electrons form 'spokes' or bunches. This process is known as electron bunching in crossed fields.
3. **Energy Transfer**: As these bunches rotate and pass the openings of the resonant cavities, they induce high-frequency electromagnetic fields, transferring their kinetic energy to the microwave field.
Why other options are incorrect:
- **a. Velocity modulation**: This is the primary mechanism for Klystrons, where an electron beam is modulated in velocity to create bunches.
- **b. Cyclotron resonance**: While the magnetron involves circular motion, it is not a cyclotron resonance device (like a cyclotron accelerator); it relies on the interaction of the bunch with the cavity walls.
- **c. Space-charge effect**: While space charge exists in all vacuum tubes, it is not the primary mechanism for microwave generation in a magnetron.
For its S matrix to be unitary, the circuit should be:
A)
Reciprocal
B)
Lossless
C)
Both
D)
None
In microwave network analysis, the Scattering matrix (S-matrix) relates the incident voltage waves to the reflected voltage waves.
1. **Unitary Property**: A matrix [S] is said to be unitary if [S][S]ᴴ = [I], where [S]ᴴ is the conjugate transpose (Hermitian) of the matrix and [I] is the identity matrix.
2. **Physical Significance**: The condition [S][S]ᴴ = [I] mathematically represents the conservation of energy. Specifically, it implies that the total power entering the network is equal to the total power leaving the network (no power is dissipated as heat within the device).
3. **Lossless Network**: A network that does not dissipate any power is called a lossless network. Therefore, for any lossless network, the S-matrix must be unitary.
Regarding the other options:
- **Reciprocal**: A network is reciprocal if its S-matrix is symmetric ([S] = [S]ᵀ). Reciprocity is independent of whether the network is lossless or lossy. A network can be reciprocal but lossy (not unitary), or non-reciprocal but lossless (unitary).
- **Both**: Since reciprocity is not a requirement for a matrix to be unitary, 'Both' is incorrect.
Which type of wave propagation is used for long-distance communication using the ionosphere?
A)
Ground wave propagation
B)
Sky wave propagation
C)
Space wave propagation
D)
Surface wave propagation
Step-by-step reasoning:
1. **Understanding the Requirement**: The question asks for the type of wave propagation used for long-distance communication that specifically utilizes the ionosphere.
2. **Analyzing Sky Wave Propagation**: Sky wave propagation (also known as ionospheric propagation) occurs when radio waves are transmitted toward the sky and are refracted (bent) back toward Earth by the ionized layers of the upper atmosphere (the ionosphere). This allows the signal to 'skip' over the horizon, enabling communication over thousands of kilometers, far beyond the visual line-of-sight.
3. **Evaluating Other Options**:
- **Ground wave propagation (a)**: These waves travel along the surface of the Earth. They are primarily used for low-frequency (LF) and medium-frequency (MF) communications (like AM radio) and are limited to shorter distances due to attenuation.
- **Space wave propagation (c)**: These waves travel in a straight line from the transmitter to the receiver (Line-of-Sight). They are used for very high frequencies (VHF, UHF, Microwave) and satellite communication. They penetrate the ionosphere rather than being reflected by it.
- **Surface wave propagation (d)**: This is essentially another term for ground wave propagation, where the wave follows the curvature of the Earth's surface.
4. **Conclusion**: Since the ionosphere is the key mechanism for reflecting signals back to Earth for long-range communication, Sky wave propagation is the correct answer.
In optical fiber communication, the most common laser source is:
A)
CO2 laser
B)
Nd:YAG laser
C)
Semiconductor laser
D)
Argon laser
In optical fiber communication, the light source must meet several critical criteria: it must be compact, efficient, capable of high-speed modulation (switching on and off rapidly), and emit light at wavelengths where optical fibers have low attenuation (typically in the near-infrared region, such as 1310 nm or 1550 nm).
1. Semiconductor Lasers (e.g., Laser Diodes): These are the most common sources because they are extremely small, can be directly modulated at GHz speeds, and their emission wavelengths can be precisely engineered to match the low-loss windows of silica fibers.
2. CO2 Lasers: These emit in the far-infrared region (~10.6 μm), where silica fibers are opaque (extremely high absorption), making them unsuitable for fiber communication.
3. Nd:YAG Lasers: While they emit in the infrared, they are bulky, expensive, and cannot be modulated at the high speeds required for data transmission compared to semiconductor lasers.
4. Argon Lasers: These emit primarily in the visible spectrum (blue/green), which suffers from high attenuation in standard optical fibers, and they are too large for practical communication hardware.
To reduce the memory access time we generally make use of:
A)
SDRAM
B)
Heaps
C)
Cache
D)
Higher capacity RAM
The goal of reducing memory access time is to bridge the 'processor-memory gap'—the difference between the high speed of the CPU and the relatively slow speed of main memory (RAM).
1. **Cache Memory (Correct):** Cache is a small, high-speed SRAM (Static RAM) located very close to or inside the CPU. It stores frequently accessed data and instructions. Because it operates at speeds close to the CPU clock, it significantly reduces the average memory access time by avoiding the need to access the slower main memory for every request (a concept known as 'locality of reference').
2. **SDRAM (Incorrect):** Synchronous Dynamic RAM is a type of main memory. While it is faster than older asynchronous DRAM, it is still significantly slower than cache memory and does not solve the fundamental latency issue between the CPU and RAM.
3. **Heaps (Incorrect):** A heap is a region of a computer's memory used for dynamic memory allocation in software programming. It is a logical memory management structure, not a hardware mechanism to reduce physical access time.
4. **Higher capacity RAM (Incorrect):** Increasing the capacity of RAM allows the system to hold more data and reduces the need to swap data to the hard disk (virtual memory), but it does not reduce the actual access time (latency) for a single piece of data retrieved from the RAM.
Compare TDM and FDM in terms of resource usage, efficiency, and applications.
Polarization of an antenna refers to the:
A)
Shape of the antenna
B)
Direction of the electric field of the radiated wave
C)
Material used in the antenna design
D)
Frequency of operation
In antenna theory and electromagnetics, polarization is defined as the orientation of the electric field (E-field) vector of an electromagnetic wave relative to the earth's surface or a specific reference axis as the wave propagates through space.
1. Why option (b) is correct: The electric field is the primary component that interacts with the receiving antenna. Depending on how the E-field oscillates (e.g., vertically, horizontally, or in a circle), the wave is classified as vertically polarized, horizontally polarized, or circularly polarized.
2. Why other options are incorrect:
- Option (a): The shape of the antenna (geometry) can influence the polarization, but it is not the definition of polarization itself.
- Option (c): The material (e.g., copper, aluminum) affects conductivity and efficiency, not the polarization of the radiated wave.
- Option (d): The frequency of operation determines the wavelength and bandwidth, but polarization is independent of the frequency value.
What is the purpose of modulation in communication systems?
A)
To eliminate noise from the signal
B)
To increase the frequency of the signal
C)
To encode information onto a carrier wave for transmission
D)
To amplify the transmitted signal
Modulation is the process of varying one or more properties (amplitude, frequency, or phase) of a high-frequency periodic waveform (called the carrier wave) with a modulating signal (the information or baseband signal).
Step-by-step reasoning:
1. Baseband signals (like voice or music) have low frequencies and cannot be transmitted efficiently over long distances because they would require impractically large antennas (antenna size is typically proportional to wavelength, λ = c/f).
2. By modulating the information onto a high-frequency carrier wave, the signal can be transmitted over the air with reasonable antenna sizes and can be multiplexed (multiple signals sent over the same medium at different frequencies).
Analysis of options:
- Option (a) is incorrect: Modulation does not eliminate noise; in fact, some modulation schemes are more susceptible to noise than others, although some (like FM) provide better noise immunity than others (like AM).
- Option (b) is partially true in a physical sense (the resulting signal has a higher frequency), but 'increasing the frequency' is the mechanism, not the 'purpose'. The purpose is the transmission of information.
- Option (c) is correct: This is the fundamental definition and purpose of modulation.
- Option (d) is incorrect: Amplification is handled by amplifiers (power amps), not by the process of modulation itself.
Write short notes on pipelining and its advantages.
Non uniform quantization includes:
A)
Compression
B)
Expansion
C)
Compression & Expansion
D)
None of the mentioned
Non-uniform quantization is a technique used in digital communication to improve the Signal-to-Quantization Noise Ratio (SQNR) for signals where the amplitude distribution is non-uniform (e.g., speech signals, where low-amplitude signals occur more frequently than high-amplitude ones).
1. **Compression**: To achieve non-uniform quantization, the input signal is first passed through a 'compressor'. This process increases the gain for low-amplitude signals and decreases the gain for high-amplitude signals, effectively 'compressing' the dynamic range. This ensures that small signals are quantized with finer steps, reducing the relative quantization error.
2. **Expansion**: After the signal has been quantized and transmitted/processed, it must be restored to its original dynamic range. This is done using an 'expander', which is the inverse process of compression.
Common examples of this process include $\mu$-law and A-law companding (Compression + Expanding). Therefore, non-uniform quantization inherently involves both compression at the transmitter and expansion at the receiver.
Which of the following is NOT a function of the control unit?
A)
Fetching instructions
B)
Decoding instructions
C)
Arithmetic operations
D)
Generating control signals
The Central Processing Unit (CPU) is primarily divided into two main components: the Control Unit (CU) and the Arithmetic Logic Unit (ALU).
1. The Control Unit (CU) acts as the 'manager' of the CPU. Its primary functions include:
- Fetching instructions from memory (Option a).
- Decoding those instructions to determine what operation needs to be performed (Option b).
- Generating control signals to coordinate the movement of data between the CPU, memory, and I/O devices (Option d).
2. The Arithmetic Logic Unit (ALU) is the component specifically designed to perform actual data processing. This includes:
- Arithmetic operations (Addition, Subtraction, Multiplication, Division).
- Logical operations (AND, OR, NOT, XOR).
Since arithmetic operations are performed by the ALU and not the Control Unit, option 'c' is the correct answer to the question 'Which of the following is NOT a function of the control unit?'
Which of the following is the correct sequence of operations in a microprocessor?
A)
Opcode fetch, memory read, memory write, I/O read, I/O write
B)
Opcode fetch, memory write, memory read, I/O read, I/O write
C)
I/O read, opcode fetch, memory read, memory write, I/O write
D)
I/O read, opcode fetch, memory write, memory read, I/O write
In the context of microprocessor architecture (specifically referring to the 8085 microprocessor, which is the standard for these types of engineering exams), the 'sequence of operations' refers to the priority or the standard order of machine cycles used to execute an instruction.
1. **Opcode Fetch**: This is always the first operation of every instruction cycle. The microprocessor must first fetch the operation code (opcode) from memory to determine what action to perform.
2. **Memory Read/Write**: After the opcode is decoded, the processor may need to read data from memory (Memory Read) or store a result back into memory (Memory Write) to complete the instruction.
3. **I/O Read/Write**: Finally, the processor may interact with external peripherals via input/output ports (I/O Read to get data from a sensor/switch, or I/O Write to send data to a display/actuator).
Option (a) correctly lists these in the logical order of execution: first fetching the instruction, then handling internal memory operations, and finally handling external I/O operations. Options (b), (c), and (d) are incorrect because they either place I/O operations before the opcode fetch or swap the logical flow of memory access.
Explain DAA and DAD instructions with example.
Outline the main blocks of a satellite microwave communication system and their functions.
Explain the various losses in a microstrip line.
Write a short note on link budget analysis and its significance in satellite communication.
A modulating signal of 2cos(5000)t is amplitude modulated over a carrier signal of 5cos(20000)t. Derive expressions for the modulation index, LSB and VSB frequencies, Bandwidth and the ratio of Side Band Power in the Total Power of AM wave.