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Past Questions
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).
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.
Derive the radar range equation.
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.
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.
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.
Explain the principle of operation of a two-hole directional coupler. What is the limitation of this structure?
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 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.
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.
Explain why a dielectric filled waveguide cannot support TEM waves.
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.
Discuss the principle of “MASER” and its applications.
Design a memory system for 8085 microprocessor such that it should contain 8 Kbyte of EPROM and 8 Kbyte of RAM.
Define dispersion in optical fiber communication. Briefly explain intermodal dispersion, chromatic dispersion, and waveguide dispersion.
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 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.
Explain the various steps involved while executing CALL instruction with an example.
If the height of the waveguide is halved, its cut-off wavelength will :
A)
Halved
B)
Doubled
C)
Remain unchanged
D)
One forth of the previous value
Remain unchanged is the best answer because it directly satisfies the fact, principle, or condition asked in the question.
What are the main functions performed by BIU and EU unit of 8086 microprocessor?
Define AM & FM. Explain the principle in short and its uses.
The range of signed decimal numbers that can be represented by 7-bit 1’s complement representation is :
A)
-64 to +63
B)
-63 to 63
C)
-127 to +128
D)
-128 to +127
To find the range of signed decimal numbers for a given number of bits (n) using 1's complement representation, we use the following formula: Range = [-(2^(n-1) - 1) to +(2^(n-1) - 1)].
Step-by-step calculation for n = 7 bits:
1. Identify the number of bits: n = 7.
2. Calculate the magnitude limit: 2^(n-1) - 1 = 2^(7-1) - 1 = 2^6 - 1.
3. 2^6 = 64, so 64 - 1 = 63.
4. Therefore, the range is from -63 to +63.
Why other options are incorrect:
- Option (a) -64 to +63 is the range for 7-bit 2's complement representation.
- Option (c) and (d) refer to ranges for 8-bit representations (2^7 = 128).
In 1's complement, there are two representations for zero (+0 and -0), which is why the range is symmetric and slightly smaller than 2's complement.
In parallel-plate waveguide, what is the principal wave?
A)
TEM wave
B)
TE wave
C)
TM wave
D)
Combination of TE and TM waves having axial components of both electric and magnetic fields
A parallel-plate waveguide consists of two infinite conducting plates separated by a distance 'd'. Unlike hollow metallic waveguides (like rectangular or circular waveguides), a parallel-plate waveguide has two separate conductors. This configuration allows for the existence of a Transverse Electromagnetic (TEM) mode, where both the electric field (E) and the magnetic field (H) are entirely transverse to the direction of propagation (z-axis). In this mode, Ez = 0 and Hz = 0. The TEM wave is the principal or fundamental wave because it has no cutoff frequency (cutoff frequency = 0 Hz), meaning it can propagate at any frequency. While TE (Transverse Electric) and TM (Transverse Magnetic) modes can also exist in parallel-plate waveguides (as higher-order modes), the TEM mode is the primary and most characteristic mode of this specific structure.
Decimal 54 in hexadecimal and BCD number system is respectively :
A)
63, 10000111
B)
36, 01010100
C)
66, 01010100
D)
36, 00110110
Step 1: Convert Decimal 54 to Hexadecimal.
To convert decimal to hexadecimal, we divide the number by 16 and record the remainders:
54 ÷ 16 = 3 with a remainder of 6.
3 ÷ 16 = 0 with a remainder of 3.
Reading the remainders from bottom to top, the hexadecimal equivalent is 36₁₆.
Step 2: Convert Decimal 54 to BCD (Binary Coded Decimal).
In BCD, each decimal digit is represented by its own 4-bit binary equivalent:
Digit 5 in binary is 0101.
Digit 4 in binary is 0100.
Concatenating these two 4-bit groups, we get 0101 0100.
Comparing these results to the options:
- Hexadecimal: 36
- BCD: 01010100
This matches Option (b).
The memory devices which are similar to EEPROM but differ in the cost effectiveness is :
A)
CMOS
B)
Memory sticks
C)
Blue-ray devices
D)
Flash memory
Step-by-step reasoning: 1. EEPROM (Electrically Erasable Programmable Read-Only Memory) is a type of non-volatile memory that can be erased and reprogrammed electrically. However, traditional EEPROM requires erasing data byte-by-byte, which is slow and expensive for large capacities. 2. Flash memory is a specific type of EEPROM that allows data to be erased in large blocks rather than individual bytes. This block-erasing capability makes it significantly faster to write to and much more cost-effective to manufacture at high densities. 3. Analyzing other options: (a) CMOS refers to a fabrication process (Complementary Metal-Oxide-Semiconductor) used to make chips, not a specific memory device similar to EEPROM. (b) Memory sticks are physical form factors (like USB drives) that typically contain Flash memory, but 'Flash memory' is the actual device technology. (c) Blu-ray devices are optical storage media, which use lasers for reading/writing and are fundamentally different from the electrical architecture of EEPROM.