← MES Jr. Grade
Past Questions
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.
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 addressing method does the instruction, MOV AX,[BX] represent?
A)
Register indirect addressing mode
B)
Direct addressing mode
C)
Register addressing mode
D)
Register relative addressing mode
The instruction 'MOV AX, [BX]' is an x86 assembly instruction. To determine the addressing mode, we analyze the source operand '[BX]'.
1. **Analysis of [BX]**: The square brackets indicate that the processor should not use the value inside the BX register as the data itself, but rather as a memory address (a pointer). The processor goes to the memory location stored in BX to fetch the actual data.
2. **Register Indirect Addressing**: This mode occurs when the address of the operand is stored in a register. Since the address is held in the BX register, this is a classic example of Register Indirect Addressing.
Why other options are incorrect:
- **Direct addressing mode**: This would look like 'MOV AX, [1234h]', where a specific numerical memory address is provided directly in the instruction.
- **Register addressing mode**: This would look like 'MOV AX, BX', where data is moved directly from one register to another without accessing memory.
- **Register relative addressing mode**: This would involve an offset, such as 'MOV AX, [BX + 4]', where a constant is added to the register value to find the address.
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?'
What is a directional coupler? Explain its function in microwave measurement setups.
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 stability of a control system can be determined from:
A)
Root locus
B)
Nyquist plot
C)
Bode plot
D)
All of the above
Stability analysis is a fundamental part of control systems engineering, and it can be performed using both time-domain and frequency-domain methods.
1. Root Locus (Option a): This is a graphical method that shows how the poles of the closed-loop transfer function move in the s-plane as a system parameter (usually the gain K) varies. If any part of the root locus enters the right-half of the s-plane, the system becomes unstable. Thus, it is used to determine absolute and relative stability.
2. Nyquist Plot (Option b): Based on the Nyquist Stability Criterion, this frequency-domain method uses the open-loop frequency response to determine the stability of the closed-loop system by observing the encirclements of the critical point (-1, j0). It is a powerful tool for determining absolute stability.
3. Bode Plot (Option c): This method represents the frequency response in terms of magnitude and phase. Stability is determined using the Gain Margin (GM) and Phase Margin (PM). If both GM and PM are positive, the system is stable.
Since all three methods are valid and widely used tools for determining the stability of a control system, the correct answer is 'All of the above'.
Discuss the effect of feedback on noise and distortion in control systems.
Calculate the skin depth of a conductor, having a conductivity of 200 units. The wave frequency is 10 GHz in air.
A)
355.8
B)
3.558
C)
35.58
D)
0.3558
To calculate the skin depth (δ) of a conductor, we use the standard formula:
δ = 1 / sqrt(π * f * μ * σ)
Given data:
1. Conductivity (σ) = 200 S/m (assuming units are Siemens per meter as per standard EM problems).
2. Frequency (f) = 10 GHz = 10 * 10^9 Hz.
3. Permeability (μ): Since the material is not specified as magnetic, we use the permeability of free space, μ₀ = 4π * 10^-7 H/m.
Step-by-step calculation:
1. Calculate the product inside the square root:
π * f * μ * σ = π * (10 * 10^9) * (4π * 10^-7) * 200
= π * 10^10 * 4π * 10^-7 * 200
= 4 * π^2 * 10^3 * 200
= 800 * π^2 * 1000
= 800,000 * π^2
≈ 800,000 * 9.8696
≈ 7,895,683.5
2. Calculate the square root:
sqrt(7,895,683.5) ≈ 2809.9
3. Calculate the skin depth (δ):
δ = 1 / 2809.9 ≈ 0.0003558 meters
4. Convert to millimeters (common unit for skin depth in such options):
δ = 0.0003558 * 1000 = 0.3558 mm
Comparing this result with the given options, the numerical value matches option (d).
In free space, the wave impedance is:
A)
50Ω
B)
120πΩ
C)
150Ω
D)
377Ω
The wave impedance (also known as intrinsic impedance) of a medium is the ratio of the electric field intensity (E) to the magnetic field intensity (H). For free space, it is calculated using the formula: η₀ = √(μ₀ / ε₀), where μ₀ is the permeability of free space (4π × 10⁻⁷ H/m) and ε₀ is the permittivity of free space (≈ 8.854 × 10⁻¹² F/m). Substituting these values: η₀ = √( (4π × 10⁻⁷) / (8.854 × 10⁻¹²) ) ≈ 376.73 Ω. This value is commonly approximated as 120π Ω (since 120 * 3.14159 ≈ 376.99 Ω). While option (d) 377Ω is a very close numerical approximation, in technical examinations for Electromagnetics, the exact symbolic form 120πΩ is the standard theoretical answer. Option (a) 50Ω is a common characteristic impedance for coaxial cables, not free space. Option (c) 150Ω is incorrect.
The root locus of a system starts at:
A)
Zeros and ends at poles
B)
Poles and ends at zeros
C)
Origin and ends at infinity
D)
None of the above
In control systems, the Root Locus is a graphical representation of the paths (loci) of the closed-loop poles as a system parameter (usually the gain 'K') is varied from 0 to infinity.
1. **Starting Point (K = 0):** The characteristic equation of a closed-loop system is given by 1 + KG(s)H(s) = 0. When K = 0, the equation reduces to the open-loop characteristic equation. Therefore, the root locus starts at the open-loop poles.
2. **Ending Point (K → ∞):** As K approaches infinity, the closed-loop poles move toward the open-loop zeros of the system. If there are more poles than zeros, the remaining branches move toward infinity along specific asymptotes.
Comparing this to the options:
- Option (a) is the reverse of the actual behavior.
- Option (b) correctly states that the locus starts at poles and ends at zeros.
- Option (c) is incorrect as the origin is not the universal starting point.
Which code changes only one bit between successive numbers?
A)
BCD
B)
Gray code
C)
Excess-3
D)
ASCII
The question asks for a code where only one bit changes between successive numbers. This is the defining characteristic of a 'unit-distance code'.
1. **Gray Code (Correct):** Also known as reflected binary code, it is specifically designed so that two successive values differ by only one bit. This property is crucial in digital systems to prevent 'glitches' or spurious outputs during transitions in hardware like rotary encoders or asynchronous FIFO pointers.
2. **BCD (Binary Coded Decimal):** This is a weighted code where each decimal digit is represented by 4 bits. For example, moving from 1 (0001) to 2 (0010) involves two bit changes (the last bit changes from 1 to 0, and the second-to-last changes from 0 to 1).
3. **Excess-3:** This is a non-weighted code derived by adding 3 to the BCD value. Like BCD, it does not maintain a unit-distance property. For example, 0 (0011) to 1 (0100) involves three bit changes.
4. **ASCII (American Standard Code for Information Interchange):** This is a character encoding standard for text, not a numerical sequence designed for unit-distance transitions.
A lossless coaxial cable has L = 250 nH/m and C = 100 pF/m. Calculate its characteristic impedance and velocity of propagation.
Find the standing wave ratio, when a load impedance of 250Ω is connected to a 75Ω line.
A)
0.3
B)
75
C)
250
D)
3.33
To find the Standing Wave Ratio (SWR), we follow these steps:
1. **Identify Given Values**:
- Characteristic Impedance of the line ($Z_0$) = $75\Omega$
- Load Impedance ($Z_L$) = $250\Omega$
2. **Calculate the Reflection Coefficient ($\Gamma$)**:
The reflection coefficient is given by the formula:
$$\Gamma = \frac{Z_L - Z_0}{Z_L + Z_0}$$
$$\Gamma = \frac{250 - 75}{250 + 75} = \frac{175}{325} \approx 0.5385$$
3. **Calculate the Standing Wave Ratio (SWR)**:
The SWR is related to the reflection coefficient by the formula:
$$SWR = \frac{1 + |\Gamma|}{1 - |\Gamma|}$$
Alternatively, for purely resistive loads, SWR can be calculated directly as the ratio of the larger impedance to the smaller impedance:
$$SWR = \frac{Z_L}{Z_0} \text{ (if } Z_L > Z_0\text{)}$$
$$SWR = \frac{250}{75} = 3.333...$$
4. **Conclusion**:
The calculated SWR is approximately 3.33.
- Option (a) 0.3 is incorrect as SWR must always be $\ge 1$.
- Option (b) 75 and (c) 250 are simply the given impedance values and do not represent the ratio.
- Option (d) 3.33 matches our calculation.
Which class of power amplifier conducts for more than 180° but less than 360°?
A)
Class A
B)
Class B
C)
Class AB
D)
Class C
Power amplifiers are classified based on their conduction angle (the portion of the input cycle during which the output transistor conducts current).
1. **Class A**: The transistor conducts for the full 360° of the input cycle. It provides the highest linearity but lowest efficiency.
2. **Class B**: The transistor conducts for exactly 180° (half the cycle). This is typically used in push-pull configurations to eliminate crossover distortion, though a single Class B amplifier only handles half the waveform.
3. **Class AB**: This is a hybrid between Class A and Class B. The transistor is biased such that it conducts for more than 180° but less than 360°. This is specifically designed to eliminate 'crossover distortion' found in Class B amplifiers while maintaining better efficiency than Class A.
4. **Class C**: The transistor conducts for less than 180°. These are highly efficient but produce significant distortion, making them suitable primarily for RF tuned amplifiers.
Since the question asks for the class that conducts for more than 180° but less than 360°, the correct answer is Class AB.
Define radiation pattern, beamwidth, and directivity of an antenna.
How do we obtain sinusoidal output out of a class B amplifier?
A)
By using non-sinusoidal inputs
B)
By utilizing two transistors
C)
By biasing it in the active region
D)
By adding a capacitor to the output
A Class B amplifier is designed such that the transistor conducts for only one half-cycle (180 degrees) of the input sinusoidal signal. If a single transistor is used, the output is a half-wave rectified sine wave, which contains significant distortion. To obtain a full sinusoidal output, a 'Push-Pull' configuration is used. This involves utilizing two transistors: one (NPN) to amplify the positive half-cycle and another (PNP) to amplify the negative half-cycle. By combining the outputs of these two transistors, the full sinusoidal waveform is reconstructed.
Analysis of other options:
- Option (a): Non-sinusoidal inputs would not result in a sinusoidal output; the goal is to amplify a sine wave without distortion.
- Option (c): Biasing in the active region is a general requirement for amplification, but specifically for Class B, the transistor is biased at the cutoff point (Q-point at the junction of cutoff and active), not centered in the active region (which would be Class A).
- Option (d): A capacitor at the output is used for DC blocking or filtering, but it cannot reconstruct a missing half-cycle of a waveform.
Given that CMRR is 100 dB. Input common-mode voltage is 12 V. Differential voltage gain is 4000. Calculate output common-mode voltage.
A)
48V
B)
0.48V
C)
20V
D)
11V
Step-by-step calculation:
1. **Identify Given Values**:
- Common Mode Rejection Ratio (CMRR) = 100 dB
- Input Common-mode Voltage ($V_{cm}$) = 12 V
- Differential Voltage Gain ($A_d$) = 4000
2. **Convert CMRR from dB to linear scale**:
The formula for CMRR in decibels is: $\text{CMRR(dB)} = 20 \log_{10}(\text{CMRR}_{linear})$
$100 = 20 \log_{10}(\text{CMRR}_{linear})$
$5 = \log_{10}(\text{CMRR}_{linear})$
$\text{CMRR}_{linear} = 10^5 = 100,000$
3. **Calculate Common-mode Gain ($A_{cm}$)**:
The definition of CMRR is the ratio of differential gain to common-mode gain:
$\text{CMRR} = \frac{A_d}{A_{cm}}$
$A_{cm} = \frac{A_d}{\text{CMRR}_{linear}} = \frac{4000}{100,000} = 0.04$
4. **Calculate Output Common-mode Voltage ($V_{out(cm)}$)**:
The output common-mode voltage is the product of the common-mode gain and the input common-mode voltage:
$V_{out(cm)} = A_{cm} \times V_{cm}$
$V_{out(cm)} = 0.04 \times 12\text{ V} = 0.48\text{ V}$
Conclusion: The output common-mode voltage is 0.48V, which corresponds to option (b).
For an ideal full-wave rectifier without filter, feeds a 1 kΩ load from a 230 V AC source. Find the average DC output voltage.
Determine the gain margin and phase crossover frequency of a system with open loop transfer function G(s) = 1 / [s(1+2s)(1+s)].
What is the addition of the binary numbers 11011011010 and 010100101?
A)
0111001000
B)
1100110110
C)
11101111111
D)
10011010011
To add the binary numbers 11011011010 and 010100101, we align them by their least significant bits (right-aligned) and perform addition using binary rules (0+0=0, 0+1=1, 1+1=10 (0 carry 1), 1+1+1=11 (1 carry 1)).
Step-by-step addition:
11011011010 (11 digits)
+ 00010100101 (padded with leading zeros to match length)
--------------
11101111111
Detailed breakdown from right to left:
1. 0 + 1 = 1
2. 1 + 0 = 1
3. 0 + 1 = 1
4. 1 + 0 = 1
5. 1 + 0 = 1
6. 0 + 1 = 1
7. 1 + 0 = 1
8. 1 + 1 = 0 (carry 1)
9. 0 + 0 + 1(carry) = 1
10. 1 + 0 = 1
11. 1 + 0 = 1
Result: 11101111111.
Comparing with options:
- Option a: 0111001000 (Incorrect length and value)
- Option b: 1100110110 (Incorrect value)
- Option c: 11101111111 (Correct)
- Option d: 10011010011 (Incorrect value)
In a Centre tapped full-wave bridge rectifier, the number of diodes conducting at any instant is:
A)
1
B)
2
C)
3
D)
4
First, it is important to clarify a terminology error in the question phrasing: 'Centre tapped full-wave bridge rectifier' is a contradiction. There are two distinct types of full-wave rectifiers: the Center-Tapped Rectifier and the Bridge Rectifier. However, in both configurations, the number of diodes conducting at any single instant is the same.
1. In a Center-Tapped Full-Wave Rectifier: Two diodes are used. During the positive half-cycle, one diode conducts while the other is reverse-biased. During the negative half-cycle, the second diode conducts while the first is reverse-biased. Thus, only 1 diode conducts at a time.
2. In a Bridge Rectifier: Four diodes are used. During the positive half-cycle, a pair of diodes (usually D1 and D3) conducts to provide a path for the current. During the negative half-cycle, the other pair (D2 and D4) conducts. Thus, 2 diodes conduct at a time.
Given the options and the specific mention of 'Bridge' in the phrasing, the question is referring to the Bridge Rectifier configuration. In a bridge rectifier, current must flow through two diodes in series to complete the circuit from the transformer to the load and back. Therefore, 2 diodes conduct at any given instant.
Looping on a K-map always results in the elimination of:
A)
Variables within the loop that appear only in their complemented form
B)
Variables that remain unchanged within the loop
C)
Variables within the loop that appear in both complemented and uncomplemented form
D)
Variables within the loop that appear only in their uncomplemented form
In a Karnaugh Map (K-map), the goal of 'looping' (grouping adjacent 1s for Sum-of-Products or 0s for Product-of-Sums) is to simplify a Boolean expression using the Adjacency Property. The fundamental Boolean identity used here is X + X' = 1 (or X · X' = 0). When we group cells in a K-map, we are looking for variables that change their state (from 0 to 1 or vice versa) across the grouped cells. If a variable appears as both complemented (0) and uncomplemented (1) within a single loop, it means that the output of the function is independent of that variable for that specific group. Therefore, that variable is eliminated from the resulting simplified term. Conversely, variables that remain constant (either always 0 or always 1) throughout the loop are retained in the final simplified expression.
- Option (a) is incorrect because if a variable is only complemented, it is constant and thus retained.
- Option (b) is incorrect because variables that remain unchanged are the ones that are kept in the expression.
- Option (d) is incorrect because if a variable is only uncomplemented, it is constant and thus retained.
Draw and explain the working of a square wave generator.
Reverse collector saturation current ICBO is:
A)
Collector current when emitter current is zero
B)
Collector current when base current is zero
C)
Same as reverse saturation current
D)
Collector current when either emitter or base current is zero
The term ICBO stands for 'Collector-Base leakage current with the Emitter Open'. In a Bipolar Junction Transistor (BJT), ICBO is the current that flows from the collector to the base when the collector-base junction is reverse-biased and the emitter terminal is left open (open-circuited).
1. When the emitter is open, the emitter current (IE) is zero.
2. Since the emitter is open and the base-emitter junction is not forward-biased, the base current (IB) is also zero.
3. Therefore, ICBO represents the leakage current flowing through the reverse-biased collector-base junction when both IE = 0 and IB = 0.
Analyzing the options:
- Option (a) is partially correct but incomplete.
- Option (b) is partially correct but incomplete.
- Option (c) is too vague as 'reverse saturation current' could refer to any PN junction (like CEO).
- Option (d) is the most comprehensive and accurate definition, as it specifies that the current flows when the other two terminals (emitter or base) are not contributing current to the circuit.