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Communications MCQ Set 1

1. IPv6 has a larger address space of _________
a) 216
b) 2128
c) 232
d) 28

View Answer

Answer: b [Reason:] The main advantage of IPv6 over IPv4 is its larger address space. The length of an IPv6 address is 128 bits, compared with 32 bits in IPv4. The address space therefore has 2128 or approximately 3.4×1038 addresses.

2. Three strategies used to handle transition from version 4 to version 6 are dual stack, tunneling and ________
a) Header Switching
b) Header Translation
c) Header Transfer
d) Header Transmission

View Answer

Answer: b [Reason:] Three strategies used to handle transition from version 4 to version 6 are dual stack, tunneling and header translation. Header translation techniques are more complicated than IPv4 NAT because the protocols have different header formats.

3. MTU stands for _________
a) Minimum Transfer Unit
b) Maximum Transfer Unit
c) Maximum Transport Unit
d) Maximum Transmission Unit

View Answer

Answer: b [Reason:] In computer networking, the maximum transmission unit (MTU) is the size of the largest network layer protocol data unit that can be communicated in a single network transaction.

4. In IPv6, real-time audio or video, particularly in digital form, requires resources such as ______
a) Fixed Bandwidth
b) Variable Bandwidth
c) High Bandwidth
d) Low Bandwidth

View Answer

Answer: c [Reason:] In IPv6, real-time audio or video, particularly in digital form, requires resources such as high bandwidth, large buffers, long processing times, and so on. A process can make a reservation for these resources beforehand to guarantee that real time data will not be delayed.

5. In practical IPv6 application, a technology encapsulates IPv6 packets inside IPv4 packets, this technology is called _______
a) Tunneling
b) Hashing
c) Routing
d) NAT

View Answer

Answer: a [Reason:] IPv6 tunneling enables IPv6 hosts and routers to connect with other IPv6 hosts and routers over the existing IPv4 Internet. The main purpose of IPv6 tunneling is to deploy IPv6 as well as maintain compatibility with large existing base of IPv4 hosts and routers.

6. Which one of the following descriptions about IPv6 is correct?
a) Addresses are not hierarchical and are assigned at random.
b) Broadcasts have been eliminated and replaced with multicasts.
c) There are 2.7 billion available addresses.
d) An interface can only be configured with one IPv6 address.

View Answer

Answer: b [Reason:] In IPv6, there’s no longer any broadcast, sending one packet to a large number of unspecified hosts. There’s only multicast, unicast and anycast. In IPv6 all nodes are required to support multicast.

7. The header length of an IPv6 datagram is ___________
a) 10bytes
b) 25bytes
c) 30bytes
d) 40bytes

View Answer

Answer: d [Reason:] IPv6 datagram has fixed header length of 40bytes. It results in faster processing of the datagram. Fixed length IPv6 header allows the routers to process the IPv6 datagram packets more efficiently.

8. In the IPv6 header, the traffic class field is similar to which field in the IPv4 header?
a) Fragmentation field
b) Fast switching
c) TOS field
d) Option field

View Answer

Answer: c [Reason:] This field enables to have different types of IP datagram. In an IPv6 packet, the Traffic Class byte is used in the same way as the ToS byte in an IPv4 packet. A ToS/Traffic Class byte includes a DSCP (Differentiated Services Codepoint) and precedence bits.

9. Which are the features present in IPv4 but not in IPv6?
a) Fragmentation
b) Header checksum
c) Options
d) All of the mentioned

View Answer

Answer: d [Reason:] All the features are only present in IPv4 and not IPv6. IPv6 no longer has a header checksum to protect the IP header, meaning that when a packet header is corrupted by transmission errors, the packet is very likely to be delivered incorrectly.

10. IPv6 is designed to allow extension of the________
a) Protocol
b) Dataset
c) Headers
d) Routes

View Answer

Answer: a [Reason:] IPv6 is designed to allow the extension of the protocol if required by new technologies or applications. IPv6 uses a new header format in which options are separated from the base header and inserted, when needed, between the base header and upper-layer data.

11. In IPv6, base header can be followed by, up to_______
a) Six Extension Layers
b) Six Extension Headers
c) Eight Extension headers
d) Eight Extension layers

View Answer

Answer: b [Reason:] The length of the base header is 40 bytes. However, to give more functionality to the IP datagram, the base header can be followed by up to six extension headers.

12. In an IPv6 datagram, M bit is 0, value of HLEN is 5, value of total length is 200 and offset value is_____
a) 400
b) 350
c) 300
d) 200

View Answer

Answer: d [Reason:] In an IPv6 datagram, M bit is 0, value of HLEN is 5, value of total length is 200 and offset value is 200. If the M bit is 0, it means there are no more fragments, the fragment is the last name.

Communications MCQ Set 2

1. Multiple access schemes are used to allow ____ mobile users to share simultaneously a finite amount of radio spectrum.
a) Many
b) One
c) Two
d) Ten-Fifteen

View Answer

Answer: a [Reason:] Multiple access schemes are used to allow many mobile users to share simultaneously a finite amount of radio spectrum. The sharing of spectrum is required to achieve high capacity by simultaneously allocating the available bandwidth to multiple users.

2. The technique that makes possible the task of listening and talking in communication system is called ________
a) Simplexing
b) Duplexing
c) Modulating
d) Multiple access technique

View Answer

Answer: b [Reason:] In conventional telephone systems, it is possible to talk and listen simultaneously. This effect is called duplexing and is generally required in wireless telephone systems.

3. Frequency division duplexing provides ____ distinct bands of frequencies for _____ user.
a) Two, two
b) One, two
c) Two, one
d) Two, many

View Answer

Answer: c [Reason:] Frequency division duplexing (FDD) provides two distinct bands of frequencies for every user. In FDD, any duplex channel actually consists of two simplex channels.

4. The forward band in FDD provides traffic from the mobile to base station.
a) True
b) False

View Answer

Answer: b [Reason:] The forward band in FDD provides traffic from the base station to the mobile. The reverse band provides traffic from the mobile to the base station.

5. The frequency separation between each forward and reverse channel changes throughout the system.
a) True
b) False

View Answer

Answer: b [Reason:]The frequency separation between each forward and reverse channel is constant throughout the system. It is regardless of the particular channel being used. A device called a duplexer is used inside each subscriber unit and base station to allow simultaneous bidirectional radio transmission.

6. Time division duplexing uses ____ to provide both a forward and reverse link.
a) Frequency
b) Time
c) Time and frequency
d) Cell spacing

View Answer

Answer: b [Reason:] Time division duplexing (TDD) uses time instead of frequency to provide both a forward and reverse link. In TDD, multiple users share a single radio channel by taking turns in the time domain.

7. TDD is effective for _____
a) Fixed wireless access and users are stationary
b) Dynamic wireless access and users are stationary
c) Fixed wireless access and users are moving
d) Dynamic wireless access and users are moving

View Answer

Answer: a [Reason:] TDD is effective for fixed wireless access when all users are stationary. This makes the propagation delay does not vary in time among the users. Because of rigid timing required for time slotting, TDD generally is limited to cordless phone or short range potable access.

8. In wideband systems, the transmission bandwidth of a single channel _____ coherence bandwidth of the channel.
a) Equal to
b) Not related to
c) Larger than
d) Smaller than

View Answer

Answer: c [Reason:] In wideband systems, the transmission bandwidth of a single channel is much larger than the coherence bandwidth of the channel. Thus, multipath fading does not greatly vary the received signal power within a wideband channel.

9. In narrowband system, the channels are usually operated using TDD.
a) True
b) False

View Answer

Answer: b [Reason:] In narrowband system, channels are usually operated using FDD. To minimize interference between forward and reverse links on each channel, the frequency separation is made as great as possible within the frequency spectrum.

10. Narrowband FDMA allows users to share the same radio channel allocating a unique time slot to each user.
a) True
b) False

View Answer

Answer: b [Reason:] In narrowband FDMA, a user is assigned a particular channel which is not shared by other users in the vicinity. However narrowband TDMA allows the users to share the same radio channel allocating a unique time slot to each user.

Communications MCQ Set 3

1. In linear modulation technique, ______ of transmitted signal varies linearly with modulating digital signal.
a) Amplitude
b) Frequency
c) Phase
d) Angle

View Answer

Answer: a [Reason:] In linear modulation technique, the amplitude of transmitted signal varies linearly with modulating digital signal. It is a form of digital modulation technique.

2. Linear modulation techniques are not bandwidth efficient. State whether True or False.
a) True
b) False

View Answer

Answer: b [Reason:] Linear modulation techniques are bandwidth efficient. They are used in wireless communication systems when there is an increasing demand to accommodate more and more users within a limited spectrum.

3. Which of the following is not a linear modulation technique?
a) OQPSK
b) π/4 QPSK
c) FSK
d) BPSK

View Answer

Answer: c [Reason:] OQPSK, π/4 QPSK and BPSK are the most popular linear modulation techniques. They have very good spectral efficiency. However, FSK is an non-linear modulation technique.

4. In BPSK, the _____ of constant amplitude carrier signal is switched between two values according to the two possible values.
a) Amplitude
b) Phase
c) Frequency
d) Angle

View Answer

Answer: b [Reason:] In binary phase shift keying (BPSK), the phase of a constant amplitude carrier signal is switched between two possible values m1 and m2. These two values corresponds to binary 1 and 0 respectively.

5. By applying cos(2πft), BPSK signal is equivalent to ________
a) Double sideband suppressed carrier amplitude modulated waveform
b) Single sideband suppressed carrier amplitude modulated waveform
c) Frequency modulated waveform
d) SSB amplitude waveform

View Answer

Answer: a [Reason:] The BPSK signal is equivalent to a double sideband suppressed carrier amplitude modulated waveform, where cos(2πft) is applied as the carrier. Hence, a BPSK signal can be generated using a balanced modulator.

6. BPSK uses non-coherent demodulator. State whether True or False.
a) True
b) False

View Answer

Answer: b [Reason:] BPSK uses coherent or synchronous demodulation. It requires the information about the phase and frequency of the carrier be available at the receiver.

7. DPSK uses coherent form of PSK. State whether True or False.
a) True
b) False

View Answer

Answer: b [Reason:] Differential phase shift keying uses noncoherent form of phase shift keying. Noncoherent form avoids the need for a coherent reference signal at the receiver. Noncoherent receivers are also easy and cheap to build.

8. In DPSK system, input signal is differentially encoded and then modulated using a _____ modulator
a) Amplitude
b) Frequency
c) BPSK
d) QPSK

View Answer

Answer: c [Reason:] In DPSK system, input binary sequence is first differentially encoded and then modulated using a BPSK modulator. The differentially encoded sequence is generated from input binary sequence by complimenting their modulo-2 sum.

9. The energy efficiency of DPSK is ______ to coherent PSK.
a) Superior
b) Same
c) Zero
d) Inferior

View Answer

Answer: d [Reason:] The energy efficiency of DPSK is inferior to that of coherent PSK by about 3 dB. But, it has an advantage of reduced receiver complexity.

10. QPSK has _____ the bandwidth efficiency of BPSK.
a) Twice
b) Same
c) Half
d) Four times

View Answer

Answer: a [Reason:] Quadrature phase shift keying (QPSK) has twice the bandwidth of BPSK. It is because two bits are transmitted in a single modulation symbol. The phase of the carrier takes on one of the four equally spaced values, where each value of phase corresponds to a unique pair of message bit.

11. QPSK provides twice the bandwidth efficiency and ____ energy efficiency as compared to BPSK.
a) Twice
b) Half
c) Same
d) Four times

View Answer

Answer: c [Reason:] The bit error probability of QPSK is identical to BPSK but twice as much data can be sent in the same bandwidth. Thus, when compared to BPSK, QPSK provides twice the spectral efficiency with exactly the same efficiency.

12. What is the full form of OQPSK?
a) Optical Quadrature Phase Shift Keying
b) Orthogonal Quadrature Pulse Shift Keying
c) Orthogonal Quadrature Phase Shift Keying
d) Offset Quadrature Phase Shift Keying

View Answer

Answer: d [Reason:] OQPSK stands for offset quadrature phase shift keying. It is a modified form of QPSK which is less susceptible to deleterious effects and supports more efficient amplification. OQPSK is sometimes also called staggered QPSK.

13. The bandwidth of OQPSK is _______ to QPSK.
a) Identical
b) Twice
c) Half
d) Four times

View Answer

Answer: a [Reason:] The spectrum of an OQPSK signal is identical to that of QPSK signal. Hence, both signals occupy the same bandwidth. The staggered alignment of the even and odd bit streams in OQPSK signal does not change the nature of spectrum.

14. QPSK signals perform better than OQPSK in the presence of phase jitter. State whether True or False.
a) True
b) False

View Answer

Answer: b [Reason:] OQPSK signal perform better than QPSK in the presence of phase jitter. It is due to the presence of noisy reference signal at the receiver.

15. Which of the following is not a detection technique used for detection of π/4 QPSK signals?
a) Baseband differential detection
b) IF differential detection
c) FM discriminator detection
d) Envelope detection

View Answer

Answer: d [Reason:] There are various types of detection techniques used for detection of π/4 QPSK signals. They include baseband differential detection, IF differential detection and FM discriminator detection.

Communications MCQ Set 4

1. Linear predictive coders belong to _______ domain class of vocoders.
a) Time
b) Frequency
c) Direct
d) Indirect

View Answer

Answer: a [Reason:] Linear predictive vocoders belong to the time domain class of vocoders. This class of vocoders attempts to extract the significant features of the speech from the time waveform.

2. Linear predictive coders are computationally simple.
a) True
b) False

View Answer

Answer: b [Reason:] Linear predictive coders are computationally intensive. But, they are the most popular among the class of low bit vocoders. With LPC, it is possible to transmit good quality voice at 4.8 kbps and poorer quality voice at even lower rates.

3. Linear predictive coding system models the vocal tract as _____ linear filter.
a) Pole and zero
b) All zero
c) All pole
d) No pole

View Answer

Answer: c [Reason:] The linear predictive coding system models the vocal tract as an all pole linear filter. The excitation to this filter is either a pulse at the pitch frequency or random white noise depending on whether the speech segment is voiced or unvoiced.

4. Linear predictive vocoders use _____ to estimate present sample.
a) Weighted sum of past samples
b) Multiplication of past samples
c) One past sample
d) Do not use past samples

View Answer

Answer: a [Reason:] The linear predictive coder uses a weighted sum of p past samples. Using this technique, the current sample can be written as linear sum of the immediately precoding samples.

5. Which of the following LPC uses code book?
a) Multiple excited LPC
b) Residual excited LPC
c) LPC Vocoders
d) Code excited LPC

View Answer

Answer: d [Reason:] Code excited LPC uses code book. In this method, the coder and decoder have a predetermined code book of stochastic (zero mean white Gaussian) excitation signals.

6. How many past samples are used by linear predictive coders to estimate present sample?
a) 100-150
b) 10-15
c) 1
d) 1000-1100

View Answer

Answer: b [Reason:] LPCs uses weighted sum of past p samples to estimate the present samples. The number of past samples used by linear predictive coders ranges from 10 to 15.

7. Which of the non-linear transform is generally used to improve the coding of reflection coefficient?
a) Long area ratio transform
b) Mutual information
c) Least square
d) Interpolation

View Answer

Answer: a [Reason:] Long area ratio (LAR) transform is generally used to improve the coding of reflection coefficient. This non linear transformation reduces the sensitivity of reflection coefficients to quantization errors. LAR performs an inverse hyperbolic tangent mapping of reflection coefficients.

8. Which of the following LPC uses two sources at the receiver?
a) Multiple excited LPC
b) Residual excited LPC
c) LPC Vocoders
d) Code excited LPC

View Answer

Answer: c [Reason:] LPC vocoder uses two sources at the receiver, one of white noise and the other with a series of pulses at the current pitch rate. The selection of either of these excitation methods is based on voiced/unvoiced decision made at the transmitter.

9. Which of the following LPC produces a buzzy twang in the synthesized speech?
a) Multiple excited LPC
b) Residual excited LPC
c) LPC Vocoders
d) Code excited LPC

View Answer

Answer: c [Reason:] LPC vocoder requires that the transmitter extract pitch frequency information which is often very difficult. Moreover, the phase coherence between the harmonic components of the excitation pulse tends to produce a buzzy twang in the synthesized speech.

10. The problem of buzzy twang in synthesized speech is mitigated by multipulse excited LPC or code excited LPC. State whether True or False.
a) True
b) False

View Answer

Answer: a [Reason:] LPC vocoder produces buzzy twang in the synthesized speech due to phase coherence between the harmonic components of the excitation pulses. This problem is mitigated by multipulse excited or code excited LPC.

11. Multipulse excited LPC requires pitch detection.
a) True
b) False

View Answer

Answer: b [Reason:] Multipulse excited LPC does not require pitch detection and the prediction residual is better approximated by several pulses per pitch period. This is the reason for better speech quality.

Communications MCQ Set 5

1. Which UE category supports 64 QAM on the uplink?
a) Only category 5
b) Only category 4
c) Only category 3
d) Category 3,4 and 5

View Answer

Answer: a [Reason:] Category information is used to allow the eNB to communicate effectively with all the UEs connected to it. The UE-category defines a combined uplink and downlink capability. Only UE category 5 supports 64 QAM on the uplink.

2. What type of handovers is supported by LTE?
a) Hard handover only
b) Soft handover only
c) Hard and soft handover
d) Hard, soft and softest handover

View Answer

Answer: a [Reason:] LTE supports only hard handover. It does not receives data from two frequencies at the same time because switching between different carrier frequencies is very fast so soft handover is not required.

3. What is the minimum amount of RF spectrum needed for an FDD LTE radio channel?
a) 1.4 MHz
b) 2.8 MHz
c) 5 MHz
d) 20 MHz

View Answer

Answer: b [Reason:] In telecommunication, Long-Term Evolution (LTE) is a standard for high-speed wireless communication for mobile devices and data terminals, based on the GSM/EDGE and UMTS/HSPA technologies. The minimum amount of RF spectrum needed for an FDD LTE radio channel is 2.8 MHz.

4. Which organization is responsible for developing LTE standards?
a) UMTS
b) 3GPP
c) 3GPP2
d) ISO

View Answer

Answer: b [Reason:] The 3rd Generation Partnership Project (3GPP) is collaboration between groups of telecommunications standards associations, known as the Organizational Partners. LTE (Long Term Evolution) introduced in 3GPP R8, is the access part of the Evolved Packet System (EPS).

5. Which channel indicates the number of symbols used by the PDCCH?
a) PHICH
b) PDCCH
c) PBCH
d) PCFICH

View Answer

Answer: d [Reason:] PCFIH channel indicates the number of symbols used by the PDCCH. The actual number of OFDM symbols occupied in any given subframe is indicated in the PCFICH (Physical Control Format Indicator Channel), which is located in the first OFDM symbol of each subframe.

6. How often can resources be allocated to the UE?
a) Every symbol
b) Every slot
c) Every subframe
d) Every frame

View Answer

Answer: c [Reason:] Resources can be located to the UE every subframe. CCE Index is the CCE number at which the control channel data (PDCCH) is allocated. Normally this index changes for each subframe, i.e. even the same PDCCH data (e.g, a PDCCH for the same UE) allocated in each subframe changes subframe by subframe.

7. What is the largest channel bandwidth a UE is required to support in LTE?
a) 10 MHz
b) 20 MHz
c) 1.4 MHz
d) 5 MHz

View Answer

Answer: b [Reason:] The LTE format was first proposed by NTT DoCoMo of Japan and has been adopted as the international standard. LTE-Advanced accommodates the geographically available spectrum for channels above 20 MHz.

8. In LTE, what is the benefit of PAPR reduction in the uplink?
a) Improved uplink coverage
b) Lower UE power consumption
c) Reduced equalizer complexity
d) Improved uplink coverage, lower UE power consumption and reduced equalizer

View Answer

Answer: d [Reason:] PAPR is the relation between the maximum power of a sample in a given OFDM transmit symbol divided by the average power of that OFDM symbol. PAPR reduction in the uplink leads to improved uplink coverage, lower UE power consumption and reduced equalizer complexity.

9. Which RLC mode adds the least amount of delay to user traffic?
a) Unacknowledged mode (UM)
b) Acknowledged mode (AM)
c) Low latency mode (LM)
d) Transparent mode (TM)

View Answer

Answer: d [Reason:] The transparent mode entity in RLC does not add any overhead to the upper layer SDUs. The entity just transmits the SDUs coming from upper layer to MAC.

10. How much bandwidth is required to transmit the primary and secondary synchronization signals?
a) 1.08 MHz
b) 1.4 MHz
c) 930 kHz
d) 20 MHz

View Answer

Answer: a [Reason:] Cell synchronization is the very first step when UE wants to camp on any cell. 1.08 MHZ is required to transmit the primary and secondary synchronization signals.