← MES Jr. Grade
Past Questions — 2022
Define AM & FM. Explain the principle in short and its uses.
What are the main functions performed by BIU and EU unit of 8086 microprocessor?
Discuss the principle of “MASER” and its applications.
The probabilities of five possible outcomes of an experiment are (1/2), (1/4), (1/8), (1/16), and (1/16). Determine the entropy and information rate if there are 16 outcomes per second.
A 108 MHz carrier signal is frequency modulated by sinusoidal modulating signal. The maximum frequency deviation is 100 kHz. The approximate transmission bandwidth of the FM signal is _________ if the frequency of the modulating signal is 1 kHz.
A)
2 kHz
B)
100 kHz
C)
101 kHz
D)
200 kHz
200 kHz is correct based on the standard electrical/electronic relation governing the quantity in the question.
Data path are used to study :
A)
Control circuits only
B)
Data circuits only
C)
Combination of control and data circuits
D)
None of the above
In microprocessor and computer architecture, the internal organization is typically divided into two main components: the Data Path and the Control Unit.
1. The Data Path consists of the functional units that perform operations on data, such as the Arithmetic Logic Unit (ALU), registers, and internal buses.
2. The Control Unit manages the flow of data by sending control signals to the data path to tell it which operation to perform and when.
When we 'study' or analyze a data path in the context of a processor's architecture, we are not looking at the registers in isolation; we are studying how the control signals (control circuits) trigger the movement and transformation of data (data circuits). Therefore, a data path analysis involves the interaction between the control logic and the data processing hardware. Without the control circuits, the data path is inert; without the data circuits, the control unit has nothing to manage. Thus, it is a combination of both.
What is the upper limit for the data rate using an 8-PSK modulator over a conventional telephone line having bandwidth specified as 300 Hz – 3000 Hz?
A)
2,700 bps
B)
5,400 bps
C)
8,100 bps
D)
21,600 bps
Step-by-step calculation:
1. **Determine the available bandwidth (B):** The telephone line is specified from 300 Hz to 3000 Hz. Therefore, the bandwidth B = 3000 Hz - 300 Hz = 2700 Hz.
2. **Determine the symbol rate (Baud rate):** For a conventional passband modulation scheme like PSK, the maximum symbol rate (R_s) that can be transmitted through a bandwidth B is typically equal to the bandwidth B (assuming an ideal Nyquist filter where B = R_s). Thus, R_s = 2700 symbols per second (baud).
3. **Determine the bits per symbol (n):** The modulator used is 8-PSK. In M-ary PSK, the number of bits per symbol is given by n = log2(M). For M = 8, n = log2(8) = 3 bits per symbol.
4. **Calculate the data rate (R_b):** The data rate is the product of the symbol rate and the number of bits per symbol.
R_b = R_s × n
R_b = 2700 symbols/sec × 3 bits/symbol = 8100 bits per second (bps).
Comparing this result to the options:
- Option (a) 2,700 bps corresponds to BPSK (1 bit/symbol).
- Option (b) 5,400 bps corresponds to QPSK (2 bits/symbol).
- Option (c) 8,100 bps corresponds to 8-PSK (3 bits/symbol). This is the correct answer.
- Option (d) 21,600 bps would require 8 bits per symbol (256-PSK), which is not the case here.
The modulation index of an AM signal is changed from 0 to 1. The transmitted power is :
A)
unchanged
B)
halved
C)
increased by 50%
D)
doubled
The total power (Pt) of an Amplitude Modulated (AM) wave is given by the formula: Pt = Pc * (1 + (m^2 / 2)), where Pc is the carrier power and 'm' is the modulation index.
Step 1: Calculate power when the modulation index is 0 (m1 = 0).
Pt1 = Pc * (1 + (0^2 / 2)) = Pc.
Step 2: Calculate power when the modulation index is 1 (m2 = 1).
Pt2 = Pc * (1 + (1^2 / 2)) = Pc * (1 + 0.5) = 1.5 * Pc.
Step 3: Calculate the percentage increase in power.
Percentage Increase = [(Pt2 - Pt1) / Pt1] * 100
Percentage Increase = [(1.5Pc - Pc) / Pc] * 100 = [0.5Pc / Pc] * 100 = 50%.
Therefore, the transmitted power increases by 50%. Option (a) is incorrect because power depends on m. Option (b) is incorrect as power increases, not decreases. Option (d) is incorrect because doubling the power would require m = 2, which is over-modulation.
The phase velocity for the TE mode in an air-filled rectangular waveguide is :
A)
Less than the velocity of plane waves in free space
B)
Equal to velocity of plane waves in free space
C)
Greater than velocity of plane waves in free space
D)
None of these
In a rectangular waveguide, the phase velocity (v_p) is the speed at which the phase of the wave propagates along the guide. It is given by the formula: v_p = ω / β, where ω is the angular frequency and β is the guide propagation constant.
For a waveguide filled with air (or vacuum), the relationship between the propagation constant β, the free-space wavenumber k (where k = ω/c), and the cutoff wavenumber k_c is: β = sqrt(k^2 - k_c^2).
Substituting this into the phase velocity formula:
v_p = ω / sqrt((ω/c)^2 - k_c^2) = c / sqrt(1 - (k_c/k)^2).
Since the term sqrt(1 - (k_c/k)^2) is always less than 1 for frequencies above the cutoff frequency, the phase velocity v_p is always greater than the speed of light in free space (c).
It is important to note that while the phase velocity exceeds 'c', the group velocity (v_g), which is the speed at which energy or information is transmitted, is given by v_g = c * sqrt(1 - (k_c/k)^2), which is always less than 'c'. This ensures that Einstein's theory of relativity is not violated.
Design a memory system for 8085 microprocessor such that it should contain 8 Kbyte of EPROM and 8 Kbyte of RAM.
Which one of the following modulation techniques has got maximum SNR?
A)
AM-SSB
B)
AM-DSB
C)
FM
D)
AM-SC
To determine which modulation technique has the maximum Signal-to-Noise Ratio (SNR), we must compare Amplitude Modulation (AM) variants and Frequency Modulation (FM).
1. **AM Variants (AM-DSB, AM-SSB, AM-SC):** In all forms of Amplitude Modulation, the noise is additive and affects the amplitude of the carrier. While Single Sideband (SSB) and Suppressed Carrier (SC) are more power-efficient than Double Sideband (DSB) because they don't waste power on the carrier or redundant sidebands, they do not inherently provide a 'noise-immunity' mechanism. Their SNR is generally limited by the power of the transmitted signal relative to the noise floor.
2. **Frequency Modulation (FM):** FM is a wideband modulation technique. It employs a concept known as 'FM Improvement' or 'Noise Suppression.' By increasing the frequency deviation (increasing the modulation index $\beta$), the signal power is spread over a wider bandwidth. When the signal is demodulated using a limiter and a discriminator, the noise (which is primarily additive and amplitude-dependent) is significantly reduced.
3. **Comparison:** The SNR of an FM system is proportional to the square of the modulation index ($\beta^2$). By increasing the bandwidth (trading bandwidth for SNR), FM can achieve a significantly higher SNR than any AM technique for the same transmitted power. This is why FM is used for high-fidelity broadcasting where sound quality (high SNR) is more important than bandwidth efficiency.
Therefore, among the given options, FM provides the maximum SNR.
In which of the following layers the electron density is high?
A)
E layer
B)
F1 layer
C)
F2 layer
D)
D layer
The ionosphere is divided into several layers (D, E, F1, and F2) based on the concentration of free electrons.
1. **D Layer**: This is the lowest layer (approx. 60-90 km). It has the lowest electron density and primarily absorbs HF radio waves during the day.
2. **E Layer**: Located above the D layer (approx. 90-150 km), it has a moderate electron density.
3. **F Layer**: This is the highest and most ionized region. During the day, it splits into two sub-layers: F1 and F2.
4. **F2 Layer**: This layer (approx. 250-400 km) possesses the highest electron density of all the layers. Because the electron density is highest here, it is the primary layer responsible for reflecting high-frequency (HF) radio waves back to Earth, enabling long-distance sky-wave propagation.
Therefore, among the given options, the F2 layer has the highest electron density.
The maximum height of an elliptical orbit is called the :
A)
Apex
B)
Zenith
C)
Perigee
D)
Apogee
In orbital mechanics, an elliptical orbit has two primary points of extreme distance relative to the body being orbited (in this case, the Earth).
1. **Apogee**: Derived from the Greek words 'apo' (away) and 'gee' (Earth), the apogee is the point in an orbit where the satellite is at its maximum distance from the center of the Earth. This corresponds to the maximum height of the elliptical orbit.
2. **Perigee**: Derived from 'peri' (near) and 'gee' (Earth), the perigee is the point in the orbit where the satellite is closest to the Earth.
3. **Zenith**: This is an astronomical term referring to the point on the celestial sphere directly above an observer's head; it is a coordinate of position, not a point of orbital distance.
4. **Apex**: This is a general term for the top or highest part of something, but it is not the technical term used in satellite communication or orbital mechanics to describe the maximum height of an orbit.
Rectangular coaxial line can support :
A)
Only TEM mode of propagation
B)
Both TEM, and TE modes of propagation
C)
Either TE or TM mode of propagation
D)
TEM, TE and TM mode of propagation
To determine the supported modes of a transmission line or waveguide, we look at the number of conductors.
1. **TEM (Transverse Electromagnetic) Mode**: For a TEM mode to exist, the structure must have at least two separate conductors. A coaxial line (whether circular or rectangular) consists of an inner conductor and an outer conducting shell. Because it has two conductors, it can support the TEM mode, which has no longitudinal components of either the electric or magnetic fields.
2. **TE (Transverse Electric) and TM (Transverse Magnetic) Modes**: These are known as 'higher-order modes'. Any hollow conducting pipe (like a rectangular waveguide) supports TE and TM modes. Since a coaxial line is essentially a conductor inside a hollow pipe, it possesses the geometry necessary to support these higher-order modes once the operating frequency exceeds the cutoff frequency of the first higher-order mode (typically the TE11 mode in circular coax or similar in rectangular coax).
Therefore, while a coaxial line is primarily used for its TEM mode at lower frequencies, it is physically capable of supporting TEM, TE, and TM modes as the frequency increases.
- Option (a) is incorrect because it ignores higher-order modes.
- Option (b) is incorrect because it omits TM modes.
- Option (c) is incorrect because it omits the TEM mode, which is the fundamental mode of a coaxial line.
The method used for resolving data dependency conflict by the compiler itself is :
A)
Delayed load
B)
Operand forwarding
C)
Prefetch target instructions
D)
Loop buffer
In pipelined processor architectures, data dependency conflicts (hazards) occur when an instruction depends on the result of a previous instruction that has not yet completed. There are two primary ways to resolve these: hardware-based solutions and software-based (compiler) solutions.
1. **Delayed Load (Software Solution):** This is a technique where the compiler identifies a potential data hazard (specifically a load-use hazard) and rearranges the code to insert an independent instruction or a 'no-operation' (NOP) between the load instruction and the instruction that uses the loaded data. Because the compiler is responsible for this reorganization to ensure the pipeline does not stall, it is the correct answer.
2. **Operand Forwarding (Hardware Solution):** Also known as data bypassing, this is a hardware mechanism that routes the result of an operation directly to the functional unit that needs it, bypassing the write-back to the register file. This is handled by the CPU hardware, not the compiler.
3. **Prefetch Target Instructions:** This is a technique used to reduce control hazards (branching) by fetching instructions from the predicted target address of a branch, rather than resolving data dependencies.
4. **Loop Buffer:** This is a small high-speed memory that stores the instructions of a small loop to avoid repeated fetches from the main instruction cache, improving performance but not resolving data dependency conflicts.
Which of the following 8085 assembly language instruction does not affects the contents of the accumulator?
A)
CMA
B)
CMPB
C)
DAA
D)
ADDB
To determine which instruction does not affect the accumulator, let's analyze each option based on the 8085 microprocessor instruction set:
1. **CMA (Complement Accumulator):** This instruction performs a 1's complement of the contents of the accumulator. The result is stored back in the accumulator. Therefore, it **affects** the accumulator.
2. **CMP B (Compare Register B with Accumulator):** This instruction compares the contents of register B with the contents of the accumulator by performing an internal subtraction (A - B). However, the result of this subtraction is used only to update the flags (Carry, Zero, Sign, etc.) to indicate if A is greater than, less than, or equal to B. The contents of the accumulator remain **unchanged**. Therefore, it **does not affect** the accumulator.
3. **DAA (Decimal Adjust Accumulator):** This instruction adjusts the contents of the accumulator to represent a valid BCD (Binary Coded Decimal) number after an addition operation. The adjusted value is stored back in the accumulator. Therefore, it **affects** the accumulator.
4. **ADD B (Add Register B to Accumulator):** This instruction adds the contents of register B to the contents of the accumulator and stores the final sum in the accumulator. Therefore, it **affects** the accumulator.
Conclusion: Only the CMP B instruction performs a comparison without modifying the source register (the accumulator).
Define dispersion in optical fiber communication. Briefly explain intermodal dispersion, chromatic dispersion, and waveguide dispersion.
Explain the principle of operation of a two-hole directional coupler. What is the limitation of this structure?
Explain why a dielectric filled waveguide cannot support TEM waves.
Derive the radar range equation.
The amplitude modulator works on the principle of :
A)
Division
B)
Addition
C)
Subtraction
D)
Multiplication
Amplitude Modulation (AM) is the process of varying the amplitude of a high-frequency carrier signal in accordance with the instantaneous amplitude of a low-frequency modulating (message) signal. Mathematically, the AM signal is generated by multiplying the carrier wave by the modulating signal (plus a DC offset for standard AM). The general equation for an AM wave is s(t) = [Ac + m(t)] cos(2πfc t), where 'm(t)' is the message signal and 'cos(2πfc t)' is the carrier. The operation of combining these two signals to produce the modulated wave is fundamentally a multiplication process. In hardware, this is achieved using a 'multiplier' or a 'balanced modulator' (mixing circuit).
Why other options are incorrect:
- Division: Division is not used to combine signals in standard modulation processes.
- Addition: While a DC offset is added to the message signal, the actual modulation (shifting the message to a higher frequency) requires multiplication.
- Subtraction: Subtraction is not the governing principle for creating an amplitude-modulated wave.
Frequencies in UHF range propagate by means of :
A)
Sky waves.
B)
Surface waves.
C)
Ground waves.
D)
Space waves.
The propagation of electromagnetic waves depends on their frequency.
1. **UHF (Ultra High Frequency)** range spans from 300 MHz to 3 GHz. At these high frequencies, waves travel primarily in a straight line (Line-of-Sight propagation) and can penetrate the ionosphere rather than being reflected by it.
2. **Space Waves (Option d):** These are waves that travel from the transmitting antenna to the receiving antenna by a direct path (Line-of-Sight) or by reflecting off the ground. UHF waves propagate as space waves.
3. **Sky Waves (Option a):** These are waves reflected back to Earth by the ionosphere. This is characteristic of HF (High Frequency) waves (3 MHz to 30 MHz).
4. **Ground Waves / Surface Waves (Options b and c):** These are waves that follow the curvature of the Earth. This mode of propagation is effective for LF (Low Frequency) and MF (Medium Frequency) waves, typically below 30 MHz, as higher frequencies are absorbed too quickly by the ground.
Therefore, UHF frequencies propagate via space waves.
For a dipole antenna :
A)
The radiation intensity is maximum along the normal to the dipole axis
B)
The current distribution along its length is uniform irrespective of the length
C)
The effective length equals its physical length
D)
The input impedance is independent of the location of the feed – point
Step-by-step analysis of the options for a dipole antenna:
1. **Radiation Intensity (Option a):** For a dipole antenna (whether infinitesimal or half-wave), the radiation pattern is omnidirectional in the plane perpendicular to the axis of the antenna. The maximum radiation occurs in the broadside direction, which is the direction normal (perpendicular) to the dipole axis. The radiation is zero along the axis of the dipole. Therefore, this statement is correct.
2. **Current Distribution (Option b):** The current distribution is NOT uniform. For a short dipole, it is approximately triangular; for a half-wave dipole, it is sinusoidal (maximum at the center and zero at the ends). The distribution depends heavily on the length of the antenna relative to the wavelength.
3. **Effective Length (Option c):** The effective length ($l_{eff}$) is a measure of the antenna's ability to capture electromagnetic energy. For a short dipole, $l_{eff} \approx l/2$ (where $l$ is physical length) because of the triangular current distribution. For a half-wave dipole, $l_{eff} \approx 0.48 \lambda$. It is generally not equal to the physical length.
4. **Input Impedance (Option d):** The input impedance is highly dependent on the feed point. For example, feeding a half-wave dipole at the center results in a low impedance (approx 73 $\Omega$), whereas feeding it near the ends would result in a very high impedance.
Conclusion: Option (a) is the only scientifically accurate statement.
A certain series resonant circuit has a bandwidth of 1000 Hz. If the existing coil is replaced by a coil with low Q, what happens to the bandwidth?
A)
Bandwidth will increase
B)
Bandwidth becomes zero
C)
Bandwidth will decrease
D)
Bandwidth remains the same
In a series resonant circuit, the relationship between the resonant frequency (f₀), the Quality factor (Q), and the Bandwidth (BW) is given by the formula: BW = f₀ / Q.
1. **Understanding Q (Quality Factor):** The Q factor represents the sharpness of the resonance. For a series RLC circuit, Q = (ω₀L) / R. A 'low Q' coil typically implies a higher internal resistance (R) relative to its inductance (L).
2. **Analyzing the Relationship:** From the formula BW = f₀ / Q, it is evident that the bandwidth is inversely proportional to the Quality factor (BW ∝ 1/Q).
3. **Effect of replacing the coil:** If the existing coil is replaced by one with a lower Q, the denominator in the bandwidth equation decreases.
4. **Conclusion:** As Q decreases, the Bandwidth (BW) must increase. This means the resonance curve becomes flatter and wider, allowing a broader range of frequencies to pass through the circuit.
Therefore, option (a) is correct. Option (c) is incorrect because that would happen if Q increased. Option (b) is physically impossible for a passive circuit, and option (d) is incorrect because BW depends directly on the Q of the components used.
The transient response occurs :
A)
Only in resistive circuits
B)
Only in capacitive circuits
C)
Only in inductive circuits
D)
Both in capacitive and inductive circuits
1. Definition of Transient Response: The transient response is the part of the response of a system that decays to zero as time increases. It occurs when a circuit undergoes a change in its state (e.g., switching a switch on or off), causing the system to transition from one steady-state condition to another.
2. Role of Energy Storage Elements: Transient responses are caused by the ability of certain components to store energy.
- Capacitors store energy in an electric field (voltage cannot change instantaneously).
- Inductors store energy in a magnetic field (current cannot change instantaneously).
3. Analysis of Options:
- Option (a): Resistive circuits do not store energy; they dissipate it. In a purely resistive circuit, the response to a change in input is instantaneous, meaning there is no transient period.
- Option (b) and (c): While transients occur in purely capacitive or purely inductive circuits, they are not limited to 'only' one of them.
- Option (d): Since both capacitors and inductors are energy-storage elements, both are capable of producing transient responses when the circuit configuration or source changes.
Therefore, the transient response occurs in circuits containing either capacitors, inductors, or both.