Communications MCQ Set 1
1. Which of the following is not a service offered by SS7?
a) Touchstar
b) 800 services
c) Alternate billing services
d) 400 services
Answer
Answer: d [Reason:] There are three main types of services offered by the SS7 network. They are the Touchstar, 800 services, and alternate billing services.
2. Which of the following service is called CLASS?
a) Touchstar
b) 800 services
c) Alternate billing services
d) 400 services
Answer
Answer: a [Reason:] Touchstar is also known as CLASS. It is a group of switch controlled services that provide its users with certain call management capabilities.
3. Which of the following service of SS& provides call return, call forwarding?
a) Touchstar
b) 800 services
c) Alternate billing services
d) 400 services
Answer
Answer: a [Reason:] Touchstar is one of the three main type of services offered by SS7. It provides services such as call return, call forwarding, repeat dialling, call block, call tracing and caller ID.
4. Alternating billing service uses CCS network.
a) True
b) False
Answer
Answer: a [Reason:] Alternate billing service and line information database use the CCS network to enable the calling party to bill a call to a personal number (third party number, calling card, or collect etc.) from any number.
5. SS7 was first developed by _______
a) ITU
b) Ericsson
c) CCITT
d) Motorola
Answer
Answer: c [Reason:] SS7 is an outgrowth of the out of band signalling first developed by the CCITT under common channel signalling standard, CCS no. 6.
6. Which of the layer of OSI is associated with NSP of SS&?
a) All layers
b) Upper three
c) Lowest three
d) Middle two
Answer
Answer: c [Reason:] The lowest three layers of the OSI model are handled in SS7 by the network service part (NSP) of the protocol. NSP is made up of three message transfer parts (MTPs) and the signalling connection control part (SCCP) of the SS7 protocol.
7. The function of ____ is to transfer and deliver signalling network.
a) MTP
b) CCS
c) ARDIS
d) CDPD
Answer
Answer: a [Reason:] The function of the MTP is to ensure that signalling traffic can be transferred and delivered reliably between the end users and the network. MTP is provided at three levels.
8. Signalling data functions are provided by ____
a) MTP level 1
b) MTP level 2
c) MTP level 3
d) TCAP
Answer
Answer: a [Reason:] Signalling data link functions (MTP level 1) provide an interface to the actual physical channel over which communication takes place. Physical channels may include copper wire, twisted pair, fibre, mobile radio and are transparent to the higher layers.
9. Signalling data functions are provided by ____
a) MTP level 1
b) MTP level 2
c) MTP level 3
d) TCAP
Answer
Answer: b [Reason:] Signalling data link functions are provided by MTP level 2. They correspond to the second layer in the OSI reference model and provide a reliable link for the transfer of traffic between two directly connected signalling points.
10. ______ provide procedures that transfer messages between signalling nodes.
a) MTP level 1
b) MTP level 2
c) MTP level 3
d) TCAP
Answer
Answer: c [Reason:] Signalling network functions (MTP level 3) provide procedures that ransfer messages between signalling nodes. As in ISDN, there are two types of MTP level 3 functions: signalling message handling and signalling network management.
11. Which of the following does not include in SS7 user part?
a) MTP
b) ISUP
c) TCAP
d) OMAP
Answer
Answer: a [Reason:] The SS7 user part includes the ISDN user part (ISUP), the transaction capabilities application part (TCAP) and operations maintenance and administration part (OMAP). The telephone user part (TUP) and the data user part (DUP) are included in the ISUP.
12. The TCAP part in SS7 refers to _____ layer of OSI.
a) Physical
b) Network
c) Data Link
d) Application
Answer
Answer: d [Reason:] The transaction capabilities application part in SS7 refers to the application layer which invokes the services of the SCCP and the MTP in a hierarchical format. TCAP is concerned with remote operations.
Communications MCQ Set 2
1. Which of the following is not a small scale multipath measurement technique?
a) Indirect pulse measurement
b) Direct pulse measurement
c) Spread spectrum sliding correlator measurement
d) Swept frequency measurement
Answer
Answer: a [Reason:] Because of the importance of multipath structure in determining the small scale fading effects, a number of wideband channel sounding techniques have been developed. These are direct pulse measurement, spread spectrum sliding correlator and swept frequency measurement.
2. Direct RF pulse systems are complex. (True/False)
a) True
b) False
Answer
Answer: b [Reason:] Direct RF pulse systems have a very attractive system. They lack complexity because of the use of off the shelf equipment. This system can also provide a local average power delay profile.
3. What is the main disadvantage of RF pulse system?
a) Complexity
b) Not real time
c) Interference and noise
d) Simplicity
Answer
Answer: c [Reason:] The main problem with RF pulse system is that it is subject to interference and noise. This is due to wide passband filter required for multipath time resolution.
4. Why is the phase of individual multipath components are not received in RF Pulse system?
a) Due to use of duplexer
b) Due to use of ADC
c) Due to use of flip flops
d) Due to use of envelope detector
Answer
Answer: d [Reason:] Another disadvantage of RF pulse system is that phase of individual multipath components are not received due to use of an envelope detector. However, use of a coherent detector permits measurement of the multipath phase using this technique.
5. Spread spectrum sliding correlator has better ________ in comparison to RF pulse system.
a) Dynamic range
b) Frequency
c) Power density
d) Structure
Answer
Answer: a [Reason:] While the probing signal may be wideband, it is possible to detect the transmitted signal using a narrowband receiver preceded by a wideband mixer. Thus, it improves the dynamic range of the system in comparison to RF pulse system.
6. If a PN sequence has chip duration of Tc, then chip rate is given by ____-
a) Tc
b) 1/ Tc
c) 2 Tc
d) Tc2
Answer
Answer: b [Reason:] In a spread spectrum channel sounder, a carrier signal is spread over a large bandwidth. It is done by mixing it with a binary pseudo number (PN) sequence having a chip duration of Tc and chip rate of 1/ Tc.
7. The maximal length of PN sequence, the sequence length for n number of shift register is ______
a) 2n
b) 1/2n
c) 2n -1
d) 2n +1
Answer
Answer: c [Reason:] For a maximal length pseudo number (PN) sequence, the sequence length is 2n -1. Here, n is the number of shift registers in the sequence register.
8. The ratio between transmitter chip clock rate and the difference between the transmitter chip clock and difference between transmitter and receiver chip clock rates is called ___________
a) Slide factor
b) Chip factor
c) Reuse factor
d) Shape factor
Answer
Answer: a [Reason:] The slide factor is defined as ratio between transmitter chip clock rate and the difference between the transmitter chip clock and difference between transmitter and receiver chip clock rates. It is unit less quantity.
9. Which of the following is not an advantage of spread spectrum channel sounding system?
a) Rejection of passband noise
b) Real time
c) Coverage range improvement
d) Less transmitter power required
Answer
Answer: b [Reason:] One of the main advantage of spread spectrum channel sounding system is ability to reject passsband noise. Thus, it improves the coverage range for a given transmitter power. Also, required transmitter powers can be considerably lower compared to direct pulse systems due to inherent processing gain.
10. Which of the following is a disadvantage of spread spectrum channel sounding system?
a) Passband noise
b) Less coverage area
c) Large transmitter power required
d) Not in real time
Answer
Answer: d [Reason:] A disadvantage of spread spectrum system as compared to direct RF pulse system, is that measurements are not made in real time. But they are compiled as the PN codes slide past one another.
11. There is no relationship between time domain and frequency domain techniques. State whether True or False.
a) True
b) False
Answer
Answer: a [Reason:] There is a dual relationship between time domain and frequency domain techniques. Therefore, it is possible to measure the channel impulse response in the frequency domain. Frequency domain channel sounder is used for measuring channel impulse response.
12. Transmissivity is a _________ domain response.
a) Time
b) Frequency
c) Time-frequency
d) Spatial
Answer
Answer: b [Reason:] Transmissivity response is a frequency domain response. It is frequency domain representation of channel impulse response. This response is then converted to the time domain using inverse discrete Fourier transform.
13. Frequency domain channel sounding technique do not require hard wired synchronization between transmitter and receiver.(True/False)
a) True
b) False
Answer
Answer: b [Reason:] Frequency domain channel sounding system requires careful calibration and hardwired synchronization between transmitter and receiver. However, this technique works well and indirectly provides amplitude and phase information in the time domain.
14. Frequency domain channel sounding system is _______ in nature.
a) Real time
b) Digital
c) Non real time
d) Analog
Answer
Answer: c [Reason:] Frequency domain channel sounding system makes measurements of non real time nature. For time varying channels, the channel frequency response can change rapidly giving an erroneous impulse response measurement.
15. Faster sweep time can be accomplished by ________
a) Increasing frequency steps
b) Increasing time slots
c) Reducing time slots
d) Reducing frequency steps
Answer
Answer: d [Reason:] A faster sweep time can be accomplished by reducing the number of frequency steps. But this sacrifices the time resolution and excess delay range in the time domain.
Communications MCQ Set 3
1. The choice of speech coder does not depend on cell size used.
a) True
b) False
Answer
Answer: b [Reason:] The choice of speech coder depends on the cell size used. When the cell size is sufficiently small such that high spectral efficiency is achieved through frequency reuse, it may be sufficient to use a simple high rate speech codec.
2. Which of the following is an important factor in determining spectral efficiency of the system?
a) Multiple access technique
b) Cell size
c) Modulation technique
d) Vocoder
Answer
Answer: a [Reason:] The type of multiple access technique used is an important factor in determining the spectral efficiency of the system. It strongly influences the choice of speech codec.
3. Type of modulation does not affect the choice of speech codec?
a) True
b) False
Answer
Answer: b [Reason:] The type of modulation employed has a considerable impact on the choice of speech codec. Using bandwidth efficient modulation scheme can lower the bit rate reduction requirements on the speech codec and vice versa.
4. Which of the following is the name of original speech coder used in the pan European digital cellular standard GSM?
a) Multipulse excited codec
b) Residual excited codec
c) Regular pulse excited long term prediction
d) Code excited codec
Answer
Answer: c [Reason:] The original speech coder used in the pan European digital cellular standard GSM goes by a rather grandiose name of regular pulse excited long term prediction (RPE-LTP) codec. This codec has a bit rate of 13 kbps.
5. Which of the following is true for baseband RELP codec?
a) Good quality of speech, low complexity
b) Good quality of speech, high complexity
c) Bad quality of speech, low complexity
d) Bad quality of speech, high complexity
Answer
Answer: a [Reason:] The advantage of baseband RELP codec is that it provides good quality speech at low complexity. The speech quality is sometimes limited due to tonal noise introduced by the process of high frequency generation.
6. Which of the following is true for MPE-LTP codec?
a) Good quality of speech, low complexity
b) Good quality of speech, high complexity
c) Bad quality of speech, low complexity
d) Bad quality of speech, high complexity
Answer
Answer: b [Reason:] The MPE-LTP technique produces excellent speech quality at high complexity. It is not much affected by bit errors present in the channel.
7. RPE-LTP codec combines the advantage of RELP codec and CELP codec. State whether True or False.
a) True
b) False
Answer
Answer: b [Reason:] The RPE-LTP codec combines the advantages of the earlier French proposed RELP codec with those of the multipulse excited long term prediction (MPE-LTP) codec proposed by Germany.
8. Which of the following codec is used by IS-136?
a) Residual Excited Linear Predictive Coders
b) Multipulse Excited LPC
c) LPC Vocoders
d) Vector sum excited LPC
Answer
Answer: d [Reason:] The US digital cellular system, IS-136 uses a vector sum excited linear predictive coder (VSELP). This coder operates at a raw data rate of 7950 bits/s and a total data rate of 13 kbps after channel coding.
9. VSELP speech coder is a variant of ______
a) CELP
b) MPE_LTP
c) RELP
d) RPE-LTP
Answer
Answer: a [Reason:] The VSELP speech coder is a variant of the CELP type vocoders. The code books in the VSELP encoder are organised with a predefined structure such that a brute-force search is avoided.
10. Which of the following is true for VSELP?
a) Low speech quality, modest computational complexity, robust to channel errors
b) Highest speech quality, low computational complexity, channel errors
c) Highest speech quality, high computational complexity, robust to channel errors
d) Highest speech quality, modest computational complexity, robust to channel errors
Answer
Answer: d [Reason:] VSELP speech coder is designed to accomplish the three goals of highest speech quality, modest computational complexity and robustness to channel errors. The code books used by VSELP impart high speech quality and increased robustness to channel errors.
11. What is DAM in speech coding system?
a) Diagnostic Acceptability Measure
b) Digital Acceptability Measure
c) Diagnostic Accessibility Measure
d) Digital Accessibility Measure
Answer
Answer: a [Reason:] The diagnostic acceptability measure is used in speech coding system. It is used for evaluation of acceptability of speech coding systems.
12. ________ exaggerates the bit errors originally received at the base station.
a) Non linear transformation
b) Tandem signalling
c) Large cell size
d) Complex vocoders
Answer
Answer: b [Reason:] Tandem signalling tends to exaggerate the bit errors originally received at the base station. Tandem signalling is difficult to protect against but is an important evaluation criterion in the evaluation of speech coders.
Communications MCQ Set 4
1. The transmission bandwidth of spread spectrum techniques is equal to the minimum required signal bandwidth.
a) True
b) False
Answer
Answer: b [Reason:] Spread spectrum techniques employ a transmission bandwidth that is several orders of magnitude greater than the minimum required signal bandwidth. On the other hand, primary objective of all the modulation schemes is to minimize the required transmission bandwidth.
2. Why spread spectrum technique is inefficient for a single user?
a) Large transmission bandwidth
b) Small transmission bandwidth
c) Fixed transmission bandwidth
d) Fixed null bandwidth
Answer
Answer: a [Reason:] Spread spectrum systems are bandwidth inefficient for single users. But in spread spectrum systems, many users can simultaneously use the same bandwidth without significantly interfering with one another. It is one of the advantages of spread spectrum.
3. Which of the following is not a property of spread spectrum techniques?
a) Interference rejection capability
b) Multipath fading
c) Frequency planning elimination
d) Multiple user, multiple access interface
Answer
Answer: b [Reason:] Resistance to multipath fading is one of the fundamental reasons for considering spread spectrum systems for wireless communication. Since spread spectrum signals have uniform energy over a very large bandwidth, at any given time only a small portion of the spectrum will undergo fading.
4. Which of the following is not a characteristic of PN sequence?
a) Nearly equal number of 0s and1s
b) Low correlation between shifted version of sequence
c) Non deterministic
d) Low cross-correlation between any two sequences
Answer
Answer: c [Reason:] Pseudo-Noise sequences are deterministic in nature. Certain characteristics of PN sequence are nearly equal number of 0s and 1s, very low correlation between shifted versions of the sequence, very low cross correlation between any two sequences.
5. PN sequence can be generated using sequential logic circuits.
a) True
b) False
Answer
Answer: True [Reason:] PN sequence is usually generated using sequential logic circuits. When the feedback logic consists of exclusive OR gates, the shift register is called a linear PN sequence generator.
6. The period of a PN sequence produced by a linear m stage shift register cannot exceed _____ symbols.
a) 2m
b) m
c) 2m
d) 2m-1
Answer
Answer: d [Reason:] There are exactly 2m-1 non-zero states for an m stage feedback shift register. Thus, the period of a PN sequence produced by a linear m stage shift register cannot exceed 2m-1.
7. DSSS system spreads the baseband signal by ________ the baseband pulses with a pseudo noise sequence.
a) Adding
b) Subtracting
c) Multiplying
d) Dividing
Answer
Answer: c [Reason:] A direct sequence spread spectrum (DS-SS) system spreads the baseband data by directly multiplying the baseband data pulses with a pseudo noise sequence. And the pseudo noise sequence is produced by a pseudo noise code generator.
8. Frequency hopping involves a periodic change of transmission _______
a) Signal
b) Frequency
c) Phase
d) Amplitude
Answer
Answer: b [Reason:] Frequency hopping involves a periodic change of transmission frequency. A frequency hopping signal is regarded as a sequence of modulated data bursts with time varying, pseudo random carrier frequencies.
9. What is the set of possible carrier frequencies in FH-SS?
a) Hopset
b) Hop
c) Chips
d) Symbols
Answer
Answer: a [Reason:] The set of possible carrier frequencies in FH-SS is called hopset. Hopping occurs of a frequency band that includes a number of channels. Each channel is defined as a spectral region with central frequency in the hopset.
10. The bandwidth of the channel used in the hopset is called _________
a) Hopping bandwidth
b) Total hopping bandwidth
c) Instantaneous bandwidth
d) 3 dB bandwidth
Answer
Answer: c [Reason:] The bandwidth of a channel used in the hopset is called the instantaneous bandwidth. And the bandwidth of the spectrum over which the hopping occurs is called total hopping bandwidth.
11. The processing gain of FH systems is given by ratio of _______
a) Hopping bandwidth and hopping period
b) Instantaneous bandwidth and hopping duration
c) 3 dB bandwidth and bit rate
d) Total hopping bandwidth and instantaneous bandwidth
Answer
Answer: d [Reason:] The processing gain of frequency hopping (FH) systems is given by Bss/B. Here, Bss and B denote the total hopping bandwidth and instantaneous bandwidth respectively.
12. FH systems do not have collisions. State whether True or False.
a) True
b) False
Answer
Answer: b [Reason:] It is possible to have collisions in an FH system where an undesired user transmits in the same channel at the same time as the desired user. Whenever an undesired signal occupies a particular hopping channel in FH, the noise and interference in the channel are translated in frequency.
13. In fast frequency hopping, hopping rate is less than the information symbol rate. State whether True or False.
a) True
b) False
Answer
Answer: b [Reason:] Fast frequency hopping occurs if there is more than one frequency hop during each transmitted symbol. Thus, in fast frequency hopping the hopping rate equals or exceeds the information symbol rate.
Communications MCQ Set 5
1. SSMA uses signals which have a transmission bandwidth that is smaller than the minimum required RF bandwidth.
a) True
b) False
Answer
Answer: b [Reason:] Spread spectrum multiple access (SSMA) uses signals which have a transmission bandwidth that is several orders of magnitude greater than the minimum required RF bandwidth.
2. PN sequence converts _______ signal to ______ signal.
a) Narrowband, wideband
b) Wideband, narrowband
c) Unmodulated, modulated
d) Low frequency, high frequency
Answer
Answer: a [Reason:] A pseudo-noise (PN) sequence converts a narrowband signal to a wideband noise like signal before transmission. SSMA provides immunity to multipath interference and robust multiple access capability.
3. SSMA is bandwidth efficient when used with a single user. State whether True or False.
a) True
b) False
Answer
Answer: b [Reason:] SSMA is not very bandwidth efficient when used by a single user. However, since many users can share the same spread spectrum bandwidth without interfering with one another, spread spectrum systems become bandwidth efficient in a multiple user environment.
4. ______ is a digital multiple access system in which carrier frequencies are varied in pseudorandom order.
a) CDMA
b) FCDMA
c) FHMA
d) SDMA
Answer
Answer: c [Reason:] Frequency hopped multiple access (FHMA) is a digital multiple access system in which the carrier frequencies of the individual users are varied in a pseudorandom fashion within a wideband channel.
5. If the rate of change of the carrier frequency is greater than the symbol rate, then the system is referred as ______
a) Fast frequency hopping system
b) Slow frequency hopping system
c) Time division frequency hopping system
d) Code division multiple access system
Answer
Answer: a [Reason:] If the rate of change of the carrier frequency is greater than the symbol rate, then the system is referred as a fast frequency hopping system. If the channel changes at a rate less than or equal to the symbol rate, it is called slow frequency hopping.
6. A frequency hopped system does not provide security. State whether True or False.
a) True
b) False
Answer
Answer: b [Reason:] A frequency hopped system provides a level of security, especially when a large number of channels are used. Since, an unintended receiver that does not know the pseudorandom sequence of frequency slots must retune rapidly to search for the signal it wishes to intercept.
7. All users in CDMA system uses _____- carrier frequency.
a) Different
b) Two
c) Ten
d) Same
Answer
Answer: d [Reason:] All users in CDMA system use the same carrier frequency and may transmit simultaneously. Each user has its own pseudorandom codeword which is approximately orthogonal to all other codewords.
8. ______ problem occurs when many mobile users share the same channel.
a) Near-far
b) Activation
c) Line of sight
d) Windowing
Answer
Answer: a [Reason:] The near far problem occurs when many mobile users share the same channel. In general, the strongest received mobile signal will capture the demodulator at a base station.
9. In CDMA, symbol duration is ______ than channel delay spread.
a) Equal
b) Greater
c) Slightly greater
d) Much less
Answer
Answer: d [Reason:] In CDMA, symbol (chip) duration is very short and usually much less than the channel delay spread. Thus, channel data rates are very high in CDMA systems.
10. _______ is used to improve reception by collecting time delayed versions of the required signal.
a) RAKE receiver
b) Equalizer
c) Frequency modulator
d) High pass filter
Answer
Answer: a [Reason:] A RAKE receiver can be used to improve reception by collecting time delayed versions of the required signals. Since PN sequences have low autocorrelation, multipath which is delayed by more than a chip appear as noise.
11. _______ arises from the fact that the spreading sequences of different users are not orthogonal.
a) Near-far problem
b) Line of sight
c) Windowing
d) Self jamming
Answer
Answer: d [Reason:] Self jamming arises from the fact that the spreading sequences of different users are not orthogonal. Hence, in the despreading of a particular PN code, non zero contributions to the receiver decision statistic for a desired user arise from the transmissions of other users in the system.
