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## Communications MCQ Set 1

1. Which is the process of encoding information from a message source in suitable manner for transmission?
a) Modulation
b) Demodulation
c) Encryption
d) Decryption

Answer: a [Reason:] Modulation is the process of encoding information from a message source in suitable manner for transmission. It translates the baseband message signal to a bandpass signal at frequencies that are very higher compared to the baseband frequency.

2. The bandpass signal is called the modulating signal.
a) True
b) False

Answer: b [Reason:] The bandpass signal is called the modulated signal and the baseband message signal is called the modulating signal. Modulation is done by varying the amplitude, phase or frequency of a high frequency carrier in accordance with the amplitude of the message signal.

3. AM signal have all information in ______ of the carrier.
a) Amplitude
b) Magnitude
c) Frequency
d) Power

Answer: a [Reason:] AM signals have all information in the amplitude of the carrier. Because, AM signal superimpose the exact relative amplitudes of the modulating signal onto the carrier.

4. In frequency modulation, the amplitude of modulated carrier signal is ______and its frequency is ______ by the modulating message signal.
a) Constant, constant
b) Varied, constant
c) Constant, varied
d) Varied, varied

Answer: c [Reason:] In frequency modulation (FM), the amplitude of modulated carrier signal is kept constant while its frequency is varied by the modulating message signal. FM is the most popular analog modulation technique used in mobile communication.

5. FM signals have all their information in ______ of the carrier.
a) Amplitude
b) Magnitude
c) Frequency
d) Power

Answer: c [Reason:] FM signal have all their information in the phase or frequency of the carrier. This provides a nonlinear and very rapid improvement in reception quality once a certain minimum received signal level, called FM threshold is achieved.

6. Amplitude modulation has ______ relationship between the quality of the received signal and the power of the received signal.
a) No
b) Non linear
c) Constant
d) Linear

Answer: d [Reason:] In amplitude modulation schemes, there is a linear relationship between the quality of the received signal and the power of the received signal. AM signals superimpose the exact relative amplitudes of the modulating signal onto the carrier.

7. FM has _______ noise immunity when compared to amplitude modulation.
a) Same
b) Less
c) No
d) Better

Answer: d [Reason:] FM offers many advantages over amplitude modulation which makes it a better choice for many mobile radio applications. Frequency modulation has better noise immunity when compared to amplitude modulation.

8. FM signals are _____ susceptible to atmospheric and impulse noise as compared to AM.
a) Largely
b) Less
c) Not
d) Better

Answer: b [Reason:] FM signals are represented as frequency variations rather than amplitude variations. Therefore, FM signals are less susceptible to atmospheric and impulse noise.

9. Burst noise affects FM systems. State whether True or False.
a) True
b) False

Answer: b [Reason:] If the FM received signal is above the FM threshold, burst noise does not affect FM system performance as much as AM systems. It is due to the fact that message amplitude variations do not carry information in FM.

10. AM has a modulation index. State whether True or False.
a) True
b) False

Answer: b [Reason:] Unlike AM, FM has a modulation index. Hence, bandwidth occupancy can be varied to obtain greater signal to noise performance. Thus, it is possible to tradeoff bandwidth occupancy for improved noise performance.

11. Which of the following is an advantage of AM systems in comparison to FM sysytems?
a) Occupy less bandwidth
c) Better noise immunity
d) Not susceptible to impulse noise

Answer: a [Reason:] AM signals are able to occupy less bandwidth as compared to FM signals, since the transmission system is linear. Small scale fading causes rapid fluctuations in the received signal. Thus, FM offers superior qualitative performance in fading when compared to AM.

12. An FM signal is a ______ envelope signal.
a) Varied
b) Small
c) Large
d) Constant

Answer: d [Reason:] An FM signal is a constant envelope signal. It is due to the fact that the envelope of the carrier does not change with changes in the modulating signal. Hence, the transmitted power of an FM signal is constant regardless of the amplitude of the message signal.

13. FM uses class______ amplifiers and AM uses class ______ amplifiers.
a) C, C
b) A, C
c) C, A
d) AB, C

Answer: c [Reason:] The constant envelope of the transmitted signal allows efficient class C amplifiers. However in AM, it is critical to maintain linearity between the applied message and amplitude of transmitted signal. Therefore, class A or AB are used for AM systems.

14. AM exhibits capture effect characteristics. State whether True or False.
a) True
b) False

Answer: b [Reason:] AM systems do not exhibit capture effect characteristics because all the interferers are received at once and must be discriminated after the modulation process. FM exhibits this characteristic. Capture effect is a direct result of rapid non-linear improvement in received quality for an increase in receives power.

15. Which of the following is a drawback of FM systems?
a) Burst noise
b) Susceptible to atmospheric noise
c) Wider frequency band

Answer: c [Reason:] FM systems have many advantages over AM systems. They also have certain disadvantages. FM systems require a wider frequency band in transmitting media in order to obtain the advantages of reduced noise and capture effect. It is generally several times as large as that needed for AM.

## Communications MCQ Set 2

1. In block codes, parity bits are _____ to block of messages.
b) Subtracted
c) Multiplied
d) Divided

Answer: a [Reason:] In block codes, parity bits are added to the block of message bits to make codewords or code blocks. They enable a limited number of errors to be detected and corrected without retransmission.

2. How many redundant bits are added in block codes for k information bits and n code bits?
a) n+k
b) n-k
c) k2
d) n2

Answer: b [Reason:] In block encoder, k information bits are encoded into n code bits. A total of n-k redundant bits are added to the k information bits for the purpose of detecting and correcting errors.

3. For (n,k) block codes, rate of the code is defined as _______
a) n2/k
b) k2/n
c) n/k
d) k/n

Answer: c [Reason:] The block code is referred to as an (n,k) code, and the rate of the code is defined as R=n/k. It is equal to the rate of information divided by the raw channel rate.

4. The ability of the block code to correct errors is a function of ______
a) Number of parity bits
b) Number of information bits
c) Number of code bits
d) Code distance

Answer: d [Reason:] The ability of a block code to correct errors is a function of the code distance. Block codes can be used to improve the performance of communication systems when other means of improvement are impractical.

5. The weight of a code is given by number of _______
a) Non-zero elements in the codeword
b) Zero elements in the codeword
c) Total elements in the codeword
d) Elements in parity bits

Answer: a [Reason:] The weight of a code is given by number of non-zero elements in the codeword. For a binary code, the weight is basically the number of ones in the codeword.

6. Which of the following is not a property of block code?
a) Linearity
b) Systematic
c) Cyclic
d) Non linearity

Answer: d [Reason:] Block codes are linear, systematic and cyclic in nature. Encoding and decoding techniques make use of mathematical constructs known as finite fields.

7. In systematic codes, parity bits are appended at the ____
a) Beginning
b) End
c) End
d) Odd places

Answer: b [Reason:] A systematic code is one in which the parity bits are appended to the end of the information bits. For an (n,k) code, the first k bits are identical to the information bits, and the remaining (n-k) bits of each code word are linear combinations of k information bits.

8. Which of the following is not an example of block code?
a) Hamming code
b) Cyclic code
c) Convolution code
d) BCH codes

Answer: c [Reason:] Hamming codes, cyclic codes and BCH codes are the example of block codes. Convolution code does not come in the category of block code. Some other examples of block codes are Reed Solomon codes, Golay codes and Hadamard codes.

9. Which of the following code is a class of non-binary BCH?
a) Hamming code
c) Golay code
d) Reed Solomon codes

Answer: d [Reason:] The most important and most common class of non binary is the family of codes known as Reed Solomon codes. BCH codes are among the most popular block codes that exist for a wide range of rates, achieve significant coding gains.

10. Which of the following linear codes achieve largest possible minimum distance?
a) Hamming code
c) Golay code
d) Reed Solomon codes

Answer: d [Reason:] RS codes achieve the largest possible minimum distance, dmin of any linear code. They are non-binary codes which are capable of correcting errors that appears in bursts.

11. CDPD stands for _____
a) Cellular Digital Packet Data
b) Cellular Decoded Packet Data
c) Cellular Demodulated Packet Data
d) Cellular Decoded Plane Data

Answer: a [Reason:] CDPD (Cellular Digital Packet Data) is a wide area mobile data service which uses uses unused bandwidth. It was mostly used in AMPS phones. Reed Solomon code in US CDPD uses m=6 bits per code symbol.

## Communications MCQ Set 3

1. Which of the following factor could not determine the performance of algorithm?
a) Structural properties
b) Rate of convergence
c) Computational complexity
d) Numerical properties

Answer: a [Reason:] The performance of an algorithm is determined by various factors. These factors are rate of convergence, computational complexity and numerical properties. The performance of algorithm does not depend on structural properties.

2. Rate of convergence is defined by _____ of algorithm.
a) Time span
b) Number of iterations
c) Accuracy
d) Complexity

Answer: b [Reason:] Rate of convergence is required as number of iterations required for the algorithm to converge close enough to the optimum solution. It enables the algorithm to track statistical variations when operating in non stationary environment.

3. Computational complexity is a measure of ________
a) Time
b) Number of iterations
c) Number of operations
d) Accuracy

Answer: c [Reason:] Computational complexity is the number of operations required to make one complete iteration of the algorithm. It helps in comparing the performance with other algorithms.

4. Choice of equalizer structure and its algorithm is not dependent on ________
a) Cost of computing platform
b) Power budget
d) Statistical distribution of transmitted power

Answer: d [Reason:] The cost of the computing platform, the power budget and the radio propagation characteristics dominate the choice of an equalizer structure and its algorithm. Battery drain at the subscriber unit is also a paramount consideration.

5. Coherence time is dependent on the choice of the algorithm and corresponding rate of convergence.
a) True
b) False

Answer: a [Reason:] The choice of algorithm and its corresponding rate of convergence depends on the channel data rate and coherence time. The speed of the mobile unit determines the channel fading rate and the Doppler spread, which is directly related to coherence time of the channel.

6. Which of the following is an algorithm for equalizer?
a) Zero forcing algorithm
b) Least mean square algorithm
c) Recursive least square algorithm
d) Mean square error algorithm

Answer: d [Reason:] Three classic equalizer algorithm are zero forcing (ZF) algorithm, least mean squares (LMS) algorithm and recursive least squares (RLS) algorithm. They offer fundamental insight into algorithm design and operation.

7. Which of the following is a drawback of zero forcing algorithm?
a) Long training sequence
b) Amplification of noise
c) Not suitable for static channels
d) Non zero ISI

Answer: b [Reason:] The zero forcing algorithm has the disadvantage that the inverse filter may excessively amplify noise at frequencies where the folded channel spectrum has high attenuation.

8. Zero forcing algorithm performs well for wireless links. State whether True or False.
a) True
b) False

Answer: b [Reason:] ZF is not often used in wireless links as it neglects the effect of noise altogether. However, it performs well for static channels with high SNR, such as local wired telephone links.

9. LMS equalizer minimizes __________
a) Computational complexity
b) Cost
c) Mean square error
d) Power density of output signal

Answer: c [Reason:] LMS equalizer is a robust equalizer. It is used to minimize mean square error (MSE) between the desired equalizer output and the actual equalizer output.

10. For N symbol inputs, LMS algorithm requires ______ operations per iterations.
a) 2N
b) N+1
c) 2N+1
d) N2

Answer: c [Reason:] The LMS algorithm is the simplest algorithm. For N symbol inputs, it requires only 2N+1 operations per iteration.

11. Stochastic gradient algorithm is also called ________-
a) Zero forcing algorithm
b) Least mean square algorithm
c) Recursive least square algorithm
d) Mean square error algorithm

Answer: b [Reason:] The minimization of the MSE is carried out recursively, and it can be performed by the use of stochastic gradient algorithm. This more commonly called the least mean square (LMS) algorithm.

12. Convergence rate of LMS is fast. State whether True or False.
a) True
b) False

Answer: b [Reason:] The convergence rate of the LMS algorithm is slow. It is slow due to the fact that it uses only one parameter i.e. step size that control the adaptation rate.

13. Which of the following does not hold true for RLS algorithms?
a) Complex
c) Slow convergence rate
d) Powerful

Answer: c [Reason:] Recursive least square (RLS) algorithm uses fast convergence rate as opposed to LMS algorithms. They are powerful, albeit complex, adaptive signal processing techniques which significantly improves the convergence of adaptive equalizer.

14. Which of the following algorithm uses simple programming?
b) FTF algorithm
c) Fast Kalman DFE

Answer: a [Reason:] Advantages of LMS gradient DFE algorithm are low computational complexity and simple programming. While fast tranversal filter (FTF) algorithm, Fast Kalman DFE and gradient lattice DFE uses complex programming.

## Communications MCQ Set 4

1. ______ is the maximum number of channels that can be provided in a fixed frequency band.
a) Channel capacity
c) Spectral capacity
d) Carrier capacity

Answer: a [Reason:] Channel capacity for a radio system can be defined as the maximum number of channels or users that can be provided in a fixed frequency band. It is useful in computer science, in electrical engineering, and in other disciplines evaluating the capacity of a channel or conduit.

2. Which of the following measures spectrum efficiency of a wireless system?
a) Channel capacity
c) Spectral capacity
d) Carrier capacity

Answer: b [Reason:] Radio capacity is a parameter which measures spectrum efficiency of a wireless system. This parameter is determined by the required carrier to interference ratio and the channel bandwidth.

3. If D is the distance between co-channel cells and R be the cell radius, co-channel reuse ratio is given by ____
a) D*R
b) D2/R
c) D/R
d) D/R2

Answer: c [Reason:] Let, D is the distance between co-channel cells and R be the cell radius. Then the minimum ratio of D/R that is required to provide a tolerable level of co-channel interference is called the co-channel reuse ratio.

4. __________ is the interference at a base station receiver that comes from the subscriber units in the surrounding cells.
a) Forward channel interference
b) Carrier interference
d) Reverse channel interference

Answer: d [Reason:] In a cellular system, the interference at a base station receiver will come from the subscriber units in the surrounding cells. This is called reverse channel interference.

5. In practice, TDMA systems improve capacity by _______- compared to analog cellular systems.
a) Three to six times
b) Equal capacity
c) Two time
d) Ten to twenty times

Answer: a [Reason:] In practice, TDMA systems improve capacity by a factor of three to six times as compared to analog cellular radio systems. Powerful error control and speech coding enable better link performance in high interference environment.

6. _____ allows subscribers to monitor neighbouring base stations.
a) TDMA
b) MAHO
c) FDMA
d) ACA

Answer: b [Reason:] Mobile assisted handoff (MAHO) allows subscribers to monitor the neighbouring base stations, and the best base station choice may be made by each subscriber. MAHO allows deployment of densely packed microcells, thus giving substantial capacity gains in a system.

7. Time division multiple access introduces ACA, which stands for ________
a) Acute carrier allocation
d) Acute channel allocation

Answer: c [Reason:] TDMA makes it possible to introduce adaptive channel allocation (ACA). ACA eliminates system planning since it is not required to plan frequencies for cells.

8. Which of the following standard had not adopted digital TDMA for high capacity?
a) USDC
b) GSM
c) PDC
d) IS-95

Answer: d [Reason:] Various proposed standards such as the GSM, US digital cellular (USDC), and Pacific Digital Cellular (PDC) have adopted digital TDMA for high capacity. IS-95 does not use TDMA.

9. Capacity of CDMA is bandwidth limited. State whether True or False.
a) True
b) False

Answer: b [Reason:] The capacity of CDMA systems is interference limited. It is bandwidth limited in FDMA and TDMA. Therefore, any reduction in the interference will cause a linear increase in the capacity of CDMA.

10. In a CDMA system, link performance for each user ______ as the number of users _______
a) Increase, decrease
b) Increase, increase
c) Decrease, increase
d) Decrease, decrease

Answer: a [Reason:] In a CDMA system, the link performance for each user increases as the number of users decreases. A straightforward way to reduce interference is to use multisectorized antennas, which results in spatial isolation of users.

11. Capacity of CDMA can be increased by operating in DTX, which stands for _______
a) Discrete transmission mode
b) Discrete transmission modulation
c) Discontinuous transmission mode
d) Digital transmission mode

Answer: c [Reason:] Capacity of CDMA can be increased by operating in DTX, which stands for Discontinuous transmission mode. It takes advantage of the intermittent nature of speech. In DTX, the transmiiter is turned off during the periods of silence in speech.

12. Frequency reuse factor for CDMA system is _____
a) One
b) Two
c) Zero
d) Ten

Answer: a [Reason:] The amount of out-of-cell interference determines the frequency reuse factor, f, of a CDMA cellular system. Ideally, each cell shares the same frequency and the maximum possible value of f (f=1) is achieved.

## Communications MCQ Set 5

1. US digital cellular system based on CDMA was standardized as ________
a) IS-54
b) IS-136
c) IS-95
d) IS-76

Answer: c [Reason:] A US digital cellular system based on CDMA was standardized as Interim Standard 95 (IS-95). It was standardized by US Telecommunication Industry Association (TIA) and promised increased capacity.

2. IS-95 was not compatible with existing AMPS frequency band.
a) True
b) False

Answer: b [Reason:] Like IS-136, IS-95 system was designed to be compatible with the existing US analog cellular system (AMPS) frequency band. Hence, mobile and base stations can be economically produced for dual mode operation.

3. Which of the following is used by IS-95?
a) DSSS
b) FHSS
c) THSS
d) Hybrid

Answer: a [Reason:] IS-95 uses a direct sequence spread spectrum CDMA system. It allows each user within a cell to use the same radio channel, and users in adjacent cell also use the same radio channel.

4. Each IS-95 channel occupies _______ of spectrum on each one way link.
a) 1.25 MHz
b) 1.25 kHz
c) 200 kHz
d) 125 kHz

Answer: a [Reason:] To facilitate graceful transition from AMPS to CDMA, each IS-95 channel occupies 1.25 MHz of spectrum on each one way link, or 10% of the available cellular spectrum for a US cellular provider.

5. IS-95 uses same modulation technique for forward and reverse channel.
a) True
b) False

Answer: b [Reason:] IS-95 uses different modulation and spreading technique for the forward and reverse links. On the forward link, the base station simultaneously transmits the user data for all mobiles in the cell by using different spreading sequence for each mobile.

6. IS-95 is specified for reverse link operation in _________ band.
a) 869-894 MHz
b) 849-894 MHz
c) 849-869 MHz
d) 824-849 MHz

Answer: d [Reason:] IS-95 is specified for reverse link operation in the 824-849 MHz band and 869-894 MHz for the forward link. The PCS version of IS-95 has also been designed for international use in the 1800-2000 MHz bands.

7. User data in IS-95 is spread to a channel chip rate of ________
a) 1.2288 Mchip/s
b) 9.6 Mchip/s
c) 12.288 Mchip/s
d) 0.96 Mchip/s

Answer: a [Reason:] User data is spread to a channel chip rate of 1.2288 Mchip/s (a total spreading factor of 128) using a combination of techniques. The spreading process is different for the forward and reverse links in the original CDMA specification.

8. _______ are used to resolve and combine multipath components.
a) Equalizer
b) Registers
d) Frequency divider

Answer: c [Reason:] At both the base station and the subscriber, RAKE receivers are used to resolve and combine multipath components, thereby reducing the degree of fading. A RAKE receiver exploits the multipath time delays in a channel and combines the delayed replicas of transmitted signal.

9. CT2 was the first generation of cordless telephones.
a) True
b) False