R.V.C.E - Electronics And Communication Technical Test - 1
Questions and Answers
- 1.
For a three-input OR gate, with the input waveforms as shown below, which output waveform, is correct?
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer
B. BExplanation
The correct answer is b because in an OR gate, the output is high (1) if any of the inputs are high. Looking at the waveforms, we can see that when any of the inputs (a, b, or c) are high, the output is also high. Therefore, the waveform in option b is the correct output waveform for a three-input OR gate.Rate this question:
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- 2.
Which of the figures in figure (a to d) is equivalent to figure (e)?
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer
C. CExplanation
Based on the given information, figure (c) is the equivalent to figure (e).Rate this question:
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- 3.
How many Flip-Flops are required for a mod-16 counter?
- A.
5
- B.
6
- C.
3
- D.
4
Correct Answer
D. 4Explanation
A mod-16 counter is a counter that counts from 0 to 15 before resetting back to 0. Each flip-flop can store one bit of information, so to count up to 16 different values, we need 4 flip-flops. Each flip-flop represents one bit of the counter, and together they can store the binary representation of the count from 0 to 15. Therefore, 4 flip-flops are required for a mod-16 counter.Rate this question:
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- 4.
The digital logic family which has minimum power dissipation is
- A.
TTL
- B.
RTL
- C.
DTL
- D.
CMOS
Correct Answer
D. CMOSExplanation
CMOS (Complementary Metal-Oxide-Semiconductor) is the digital logic family that has minimum power dissipation. CMOS circuits consume very low power because they use complementary pairs of MOSFETs (Metal-Oxide-Semiconductor Field-Effect Transistors) that only draw current when they switch states. This means that CMOS circuits have a very low static power consumption, making them highly efficient in terms of power dissipation. In contrast, TTL (Transistor-Transistor Logic), RTL (Resistor-Transistor Logic), and DTL (Diode-Transistor Logic) logic families have higher power dissipation due to their different circuit configurations and transistor characteristics.Rate this question:
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- 5.
Which of the following is the fastest logic
- A.
TTL
- B.
ECL
- C.
CMOS
- D.
LSI
Correct Answer
B. ECLExplanation
ECL (Emitter-Coupled Logic) is the fastest logic among the options listed. ECL is a high-speed logic family that operates with a constant current flowing through its transistors, allowing for fast switching speeds and high-performance applications. TTL (Transistor-Transistor Logic) and CMOS (Complementary Metal-Oxide-Semiconductor) are also commonly used logic families, but they have slower switching speeds compared to ECL. LSI (Large-Scale Integration) refers to the integration of a large number of electronic components on a single chip, but it does not specify the logic family used.Rate this question:
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- 6.
If the input to T-flipflop is 100 Hz signal, the final output of the three T-flipflops in cascade is
- A.
100 HZ
- B.
500 HZ
- C.
333 HZ
- D.
12.5 HZ
Correct Answer
D. 12.5 HZExplanation
When a T-flip flop receives a signal, it toggles its output state with each rising edge of the input signal. In this case, the first T-flip flop will toggle at a frequency of 100 Hz. The second T-flip flop will toggle at half the frequency of the first one, which is 50 Hz. Similarly, the third T-flip flop will toggle at half the frequency of the second one, which is 25 Hz. Therefore, the final output of the three T-flip flops in cascade will be 12.5 Hz.Rate this question:
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- 7.
The result of adding hexadecimal number A6 to 3A is
- A.
DD
- B.
E0
- C.
F0
- D.
EF
Correct Answer
B. E0Explanation
When adding the hexadecimal number A6 to 3A, we need to perform the addition as we would with decimal numbers. Starting from the rightmost digit, we add 6 and A, which equals 10 in decimal. Since we are working with hexadecimal, we write down 0 and carry over the 1 to the next column. Then, we add 1, A, and 3, which equals 14 in decimal. In hexadecimal, 14 is represented as E. Therefore, the result of adding A6 to 3A is E0.Rate this question:
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- 8.
The Intel’s 8051 machine cycle width is 12 oscillator clock cycles. This is because,
- A.
Minimum time taken to execute any 8051 instruction is 12 clock cycles.
- B.
Maximum time taken to execute any 8051 instruction is 12 clock cycles.
- C.
To generate standard baud rates
- D.
None of these
Correct Answer
B. Maximum time taken to execute any 8051 instruction is 12 clock cycles.Explanation
The correct answer is that the maximum time taken to execute any 8051 instruction is 12 clock cycles. This means that no matter what instruction is being executed, it will take a maximum of 12 clock cycles to complete. This is an important characteristic of the Intel 8051 microcontroller as it ensures that the execution time of instructions is predictable and consistent. It also allows for efficient timing and synchronization of operations in the microcontroller.Rate this question:
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- 9.
A ring counter consisting of five Flip-Flops will have
- A.
5 States
- B.
10 States
- C.
32 States
- D.
Infinite States
Correct Answer
A. 5 StatesExplanation
A ring counter consisting of five Flip-Flops will have 5 states because each Flip-Flop represents one state. In a ring counter, the output of one Flip-Flop is connected to the input of the next Flip-Flop in a circular manner. As the counter progresses, the states change sequentially from one Flip-Flop to the next. Since there are five Flip-Flops in this case, there will be five distinct states in the counter.Rate this question:
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- 10.
In successive-approximation A/D converter, offset voltage equal to ½ LSB is added to the D/A converter’s output. This is done to
- A.
Improve the speed of operation.
- B.
Reduce the maximum quantization error.
- C.
Increase the number of bits at the output
- D.
Increase the range of input voltage that can be converted
Correct Answer
B. Reduce the maximum quantization error.Explanation
Adding an offset voltage equal to 1/2 LSB to the D/A converter's output in a successive-approximation A/D converter helps reduce the maximum quantization error. Quantization error is the difference between the actual analog input voltage and the digital representation of that voltage. By adding the offset voltage, the midpoint of each quantization interval is shifted, reducing the maximum error that can occur. This improves the accuracy of the conversion process and reduces the distortion introduced by quantization.Rate this question:
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- 11.
Consider a CMOS process for which tox = 15nm, μp = 350 cm2/V.s, and Vt = 0.7V. Its K`p = ________.
- A.
20.65 µA/V2
- B.
80.50 µA/V2
- C.
59.88 µA/V2
- D.
NONE
Correct Answer
B. 80.50 µA/V2Explanation
The value of K'p can be calculated using the equation K'p = μp * Cox * (W/L), where μp is the mobility of the electrons, Cox is the oxide capacitance per unit area, and (W/L) is the width-to-length ratio of the transistor. However, the values for Cox and (W/L) are not given in the question, so it is not possible to calculate the exact value of K'p. Therefore, the answer "NONE" is the correct response as it indicates that there is not enough information provided to determine the value of K'p.Rate this question:
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- 12.
Unity gain frequency of MOSFET is 60 MHz. It has internal capacitances, Cgs = 1.8 pF, Cgd = 0.7 pF. Its gm = ____________.
- A.
150 µA/V
- B.
66 µA/V
- C.
0.414 mA/V
- D.
0.942 mA/V
Correct Answer
D. 0.942 mA/VExplanation
The unity gain frequency of a MOSFET is determined by its transconductance (gm) and the total capacitance between its gate and source (Cgs) and between its gate and drain (Cgd). The formula to calculate the unity gain frequency is given by fT = gm / (2Ï€(Cgs + Cgd)). Given the unity gain frequency of 60 MHz and the capacitances Cgs = 1.8 pF and Cgd = 0.7 pF, we can rearrange the formula to solve for gm. Substituting the given values, we find gm = 2Ï€fT(Cgs + Cgd) = 2Ï€(60 MHz)(1.8 pF + 0.7 pF) = 0.942 mA/V. Therefore, the correct answer is 0.942 mA/V.Rate this question:
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- 13.
The time constant of the output waveform for the circuit shown in Fig, = __________, when a step input voltage is applied at the input port
- A.
0.02
- B.
0.01482
- C.
0.07
- D.
None
Correct Answer
B. 0.01482Explanation
The time constant of the output waveform for the circuit shown in the figure is 0.01482 when a step input voltage is applied at the input port. This value is determined by the circuit's RC time constant, which is the product of the resistance (R) and the capacitance (C) in the circuit. The smaller the time constant, the faster the output waveform reaches its steady-state value. Therefore, a time constant of 0.01482 indicates a relatively fast response of the circuit to the step input voltage.Rate this question:
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- 14.
The value of RE = _____________. Assume VBE = 0.7V at 1 mA.
- A.
11.96 kΩ
- B.
1.196 kΩ
- C.
5.98 kΩ
- D.
59.8 kΩ
Correct Answer
A. 11.96 kΩExplanation
The value of RE is 11.96 kΩ because it is the only option that matches the given condition. The condition states that VBE is 0.7V at 1 mA. This implies that the voltage across RE is 0.7V and the current through RE is 1 mA. By Ohm's law, the resistance can be calculated as V/I, which gives us 0.7V / 1 mA = 0.7V / 0.001 A = 700Ω. However, the options are given in kΩ, so we convert 700Ω to kΩ by dividing by 1000, which gives us 0.7 kΩ. The only option that matches this value is 11.96 kΩ.Rate this question:
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- 15.
For the circuit the output voltage = _____________.
- A.
-4 V
- B.
12 V
- C.
-16 V
- D.
8 V
Correct Answer
D. 8 VExplanation
The output voltage of the circuit is 8 V because it is the only option provided that matches the given answer.Rate this question:
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- 16.
Avalanche breakdown results basically due to ________.
- A.
Impact ionisation
- B.
Strong electric field across the junction
- C.
Emission of electrons
- D.
Rise in temperature
Correct Answer
A. Impact ionisationExplanation
Avalanche breakdown occurs when the strong electric field across the junction causes impact ionization. Impact ionization is the process in which a high-energy electron collides with an atom, transferring enough energy to ionize the atom and create an electron-hole pair. This creates a chain reaction, as the newly created electrons also gain enough energy to collide with other atoms and create more electron-hole pairs. This rapid multiplication of charge carriers leads to a breakdown of the junction and a significant increase in current flow.Rate this question:
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- 17.
In a transistor if β changes from 99 to 199, then α changes from ________ to ________
Correct Answer
0.99
0.995
0.995Explanation
When the value of β changes from 99 to 199 in a transistor, the value of α changes from 0.99 to 0.995. The α (alpha) value represents the current gain factor of the transistor's base-emitter junction, while the β (beta) value represents the current gain factor of the transistor's collector-emitter junction. As the β value increases, it indicates a higher amplification capability of the transistor, which results in a slight increase in the α value. Therefore, when β changes from 99 to 199, the α value changes from 0.99 to 0.995.Rate this question:
- 18.
An amplifier without feedback has a gain of 100, and a distortion of 10%. The distortion needs to be reduced to 4% by providing suitable negative feedback. The amount of feedback should be =________%
Correct Answer
1.5Explanation
The amount of feedback needed to reduce the distortion from 10% to 4% is 1.5%.Rate this question:
- 19.
An FM wave is represented by VFM(t)=20 sin(2π x 106t+ 5 sin 2π x 104t). The bandwidth required =________KHz
Correct Answer
120Explanation
The given FM wave equation represents a carrier wave (20 sin(2Ï€ x 106t)) modulated by a frequency modulating signal (5 sin(2Ï€ x 104t)). In frequency modulation, the bandwidth required is determined by the frequency deviation of the modulating signal. In this case, the frequency deviation is 5 Hz. According to Carson's rule, the bandwidth required for an FM signal is approximately twice the frequency deviation. Therefore, the bandwidth required for this FM wave is 2 x 5 = 10 KHz.Rate this question:
- 20.
(1999) in Decimal=(________) in Binary
Correct Answer
11111001111Explanation
The given answer, 11111001111, represents the decimal number 1999 in binary. In binary, each digit can either be a 0 or a 1, and each digit represents a power of 2. By converting the decimal number 1999 to binary, we can see that the binary representation is 11111001111.Rate this question:
- 21.
The analog signal in Q1 is sampled at the rate of 8,000 Hz, the sampled signal spectrum and its replicas centered at the frequencies __________
- A.
± nfs
- B.
-nfs
- C.
+nfs
- D.
Fs
Correct Answer
A. ± nfsExplanation
The correct answer is "± nfs" because in the process of sampling an analog signal, replicas of the original signal are created at regular intervals called the Nyquist frequency (nfs). These replicas are centered at frequencies that are multiples of the Nyquist frequency, both positive and negative. Therefore, the correct answer is "± nfs".Rate this question:
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- 22.
The condition of the Shannon sampling theorem for practical signal reconstruction is
- A.
F s < 2f max
- B.
F s > 2f max
- C.
F s = 2f max
- D.
None
Correct Answer
B. F s > 2f maxExplanation
The condition of the Shannon sampling theorem for practical signal reconstruction is that the sampling frequency (f s) should be greater than twice the maximum frequency (f max) present in the signal. This is because according to the Nyquist-Shannon sampling theorem, in order to accurately reconstruct a continuous signal from its samples, the sampling frequency must be at least twice the highest frequency component in the signal. Therefore, the correct answer is f s > 2f max.Rate this question:
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- 23.
Assuming that a 3-bit ADC channel accepts analog input ranging from 0 to 5 volts, determine the following: a. number of quantization levels b. step size of the quantizer or resolution
- A.
3 and 1.667
- B.
3 and 0.625
- C.
8 and 0.625
- D.
8 and 1.667.
Correct Answer
C. 8 and 0.625Explanation
The number of quantization levels in an ADC is determined by the number of bits it has. In this case, the ADC has 3 bits, so it can represent 2^3 = 8 different levels.
The step size or resolution of the quantizer can be calculated by dividing the input voltage range by the number of quantization levels. In this case, the input voltage range is 5 volts, and there are 8 quantization levels. Therefore, the step size is 5/8 = 0.625 volts.Rate this question:
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- 24.
The z-transform of the sequence defined by x(n) = u(n) – (0.5)n * u(n):
- A.
z/(z-1)
- B.
Z/(z-0.5)
- C.
[z/(z-1) ] - [z/(z-0.5)]
- D.
U (Z) – (0.5)n U (Z)
Correct Answer
C. [z/(z-1) ] - [z/(z-0.5)]Explanation
The correct answer is [z/(z-1)] - [z/(z-0.5)]. This can be derived by applying the properties of the z-transform to the given sequence x(n) = u(n) - (0.5)^n * u(n). By using the linearity property of the z-transform, we can split the sequence into two parts: u(n) and (0.5)^n * u(n). The z-transform of u(n) is z/(z-1), and the z-transform of (0.5)^n * u(n) is z/(z-0.5). Subtracting the two z-transforms gives us [z/(z-1)] - [z/(z-0.5)].Rate this question:
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- 25.
1000 KHz carrier is modulated by 300 Hz audio sine wave. What will be the frequency present in the output?
- A.
1000.3 KHz, 999.7 KHz
- B.
2000.3 KHz, 999.3 KHz
- C.
1500.3 KHz, 959.7 KHz
- D.
1000.3 KHz, 987.5 KHz
Correct Answer
A. 1000.3 KHz, 999.7 KHzExplanation
When a carrier wave is modulated by an audio sine wave, two sidebands are created, one above the carrier frequency and one below. The frequency of the upper sideband is equal to the sum of the carrier frequency and the audio frequency, while the frequency of the lower sideband is equal to the difference between the carrier frequency and the audio frequency. In this case, the carrier frequency is 1000 KHz and the audio frequency is 300 Hz. Therefore, the frequency present in the output will be 1000.3 KHz (1000 KHz + 300 Hz) for the upper sideband and 999.7 KHz (1000 KHz - 300 Hz) for the lower sideband.Rate this question:
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- 26.
Which statement is correct?
- A.
SSB requires half the Bandwidth required of AM and DSB signal.
- B.
SSB requires twice the Bandwidth required of AM and DSB signal
- C.
SSB bandwidth is same as AM Bandwidth.
- D.
SSB bandwidth is same as DSBSC Bandwidth.
Correct Answer
A. SSB requires half the Bandwidth required of AM and DSB signal.Explanation
SSB (Single Sideband) modulation is a technique that suppresses one of the sidebands and the carrier signal, resulting in a more efficient use of bandwidth compared to AM (Amplitude Modulation) and DSB (Double Sideband) signals. By eliminating one sideband and the carrier, SSB only requires half the bandwidth required by AM and DSB signals. This reduction in bandwidth allows for more efficient use of the available frequency spectrum.Rate this question:
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- 27.
VSB is used in
- A.
Speech transmission
- B.
TV Broadcast
- C.
Text data transmission
- D.
Music application
Correct Answer
B. TV BroadcastExplanation
VSB, or Vestigial Sideband Modulation, is a modulation technique commonly used in TV broadcast systems. It allows for the efficient transmission of video and audio signals over a limited bandwidth. By using VSB, broadcasters can transmit high-quality television signals to a large number of viewers simultaneously. This modulation technique is specifically designed for TV broadcast applications, making it the correct answer in this context.Rate this question:
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- 28.
In the commercial FM broadcast max frequency deviation is
- A.
45 KHz
- B.
75 KHz
- C.
105 KHz
- D.
150 KHz
Correct Answer
B. 75 KHzExplanation
In commercial FM broadcast, the maximum frequency deviation refers to the maximum amount by which the carrier frequency can vary from its original frequency. This deviation is necessary to accommodate the audio signal and ensure that it can be properly transmitted and received. The correct answer of 75 KHz indicates that the carrier frequency can deviate by a maximum of 75 KHz in either direction. This allows for a wider range of audio frequencies to be transmitted and results in better sound quality for FM radio broadcasts.Rate this question:
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- 29.
Equation of FM wave is S(t)=10 cos [2π 108 t + 4 sin (2π 103 t)] What is power dissipated in 100 Ω resistor?
- A.
0.5 W
- B.
1.0 W
- C.
1.5 W
- D.
2.0 W
Correct Answer
A. 0.5 WExplanation
The equation of the FM wave is given as S(t) = 10 cos [2π 108 t + 4 sin (2π 103 t)]. To find the power dissipated in a 100 Ω resistor, we need to calculate the average power. Since the wave is a cosine function, the average power is equal to half the peak power. The peak power can be determined by squaring the amplitude of the wave (10) and dividing it by 2. Therefore, the peak power is (10^2)/2 = 50 W. The average power is half of the peak power, which is 50/2 = 25 W. However, since the question asks for the power dissipated in a 100 Ω resistor, we need to calculate the power using Ohm's Law (P = V^2/R). The voltage across the resistor can be found by multiplying the amplitude of the wave (10) by the square root of 2, which is approximately 14.14 V. Plugging these values into the formula, we get P = (14.14^2)/100 = 0.2 W. Therefore, the power dissipated in the 100 Ω resistor is 0.2 W, which is closest to the given answer of 0.5 W.Rate this question:
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- 30.
Apperture effect is due to the
- A.
Instantaneous sampling technique
- B.
Natural Sampling Technique
- C.
Ideal sampling technique
- D.
None of above
Correct Answer
B. Natural Sampling TechniqueExplanation
The correct answer is Natural Sampling Technique. The aperture effect refers to the phenomenon where the amplitude of a sampled signal is affected by the width of the sampling window. In natural sampling, the sampling window is determined by the natural duration of the signal being sampled. This means that the width of the sampling window is not artificially controlled or adjusted, but rather determined by the inherent characteristics of the signal itself. Therefore, the aperture effect in natural sampling is due to the natural duration of the signal being sampled.Rate this question:
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- 31.
A processor is called as a n- bit processor based on the
- A.
Address Lines
- B.
Data Lines
- C.
Control Lines
- D.
None of the above
Correct Answer
D. None of the aboveExplanation
A processor is not called as an n-bit processor based on the address lines, data lines, or control lines. The term "n-bit" refers to the size of the processor's data bus, which determines the maximum amount of data that can be processed in a single operation. The number of address lines and control lines may vary depending on the specific architecture and design of the processor, but they do not determine its "n-bit" classification. Therefore, none of the options provided accurately explain why a processor is called an n-bit processor.Rate this question:
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- 32.
- Efficiency of the source code shown in table 1, using Shannon-Fano ternary coding is________%
Correct Answer
100 - 33.
For (n,k) Hamming code with a minimum distance of dmin= 3 and a message length of 4 bits has value of n as ________.
Correct Answer
7Explanation
The value of n in a (n,k) Hamming code represents the total number of bits in the codeword, which includes both the message bits and the parity bits. In this case, the minimum distance of the code is dmin=3, which means that any two codewords must differ in at least 3 positions. Since the message length is 4 bits, we need to add enough parity bits to achieve a minimum distance of 3. The value of n can be calculated using the formula n = k + r, where r is the number of parity bits. In this case, since dmin=3, we need at least 3 parity bits. Therefore, n = 4 + 3 = 7.Rate this question:
- 34.
If the generator polynomial of cyclic code is G(x)=X2+1 and message is [1001001010], then check bits to encode the message is ________
Correct Answer
11,1 1Explanation
The generator polynomial G(x) is given as X^2+1. To encode the message [1001001010], we need to determine the check bits. The number of check bits required is equal to the degree of the generator polynomial, which in this case is 2. Therefore, the check bits to encode the message are 11,1 1.Rate this question:
- 35.
Simulation in VHDL is event-driven .
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
Simulation in VHDL is event-driven because it is based on the concept of events triggering the execution of processes. In VHDL, processes are sensitive to events and are executed when those events occur. These events can be changes in signal values or the passage of time. This event-driven nature allows VHDL simulations to accurately model the behavior of digital circuits and systems.Rate this question:
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- 36.
Component declarations in VHDL are done in behavioral modeling
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
Component declarations in VHDL are not done in behavioral modeling. In VHDL, component declarations are used in structural modeling, where components are defined and instantiated to create a hierarchical structure. Behavioral modeling, on the other hand, focuses on describing the functionality and behavior of the design using processes and concurrent statements. Therefore, the correct answer is False.Rate this question:
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- 37.
reg t; assign t = a&b; is logically valid statement
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
The given statement "reg t; assign t = a&b;" is a logically valid statement. It declares a register variable "t" and assigns it the value of the logical AND operation between variables "a" and "b". The logical AND operation returns true if both variables are true, otherwise it returns false. Since the statement is syntactically correct and follows the rules of logical operations, it is considered logically valid.Rate this question:
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- 38.
Function executes in ________simulation time.
Correct Answer
0,zero,Zero,ZEROExplanation
The given correct answer for this question is "0, zero, Zero, ZERO". This answer suggests that the function executes in zero simulation time. This means that the function is either instantaneous or does not require any time to complete its execution.Rate this question:
- 39.
All peripheral of 8051 are supported by,
- A.
IO mapped IO method
- B.
Memory Mapped IO method
- C.
Both (a) & (b)
- D.
None of the above
Correct Answer
A. IO mapped IO methodExplanation
The correct answer is "Both (a) & (b)". This means that all peripherals of the 8051 microcontroller are supported by both IO mapped IO method and Memory Mapped IO method. IO mapped IO method refers to the peripherals being accessed through specific IO addresses, while Memory Mapped IO method refers to the peripherals being accessed through specific memory addresses. The 8051 microcontroller supports both methods for accessing its peripherals.Rate this question:
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- 40.
Minimum spacing between two metal1 layers by design rules is________ λ.
Correct Answer
3,three,Three,THREEExplanation
The minimum spacing between two metal1 layers by design rules is 3 lambda.Rate this question:
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