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## Communications MCQ Set 1

1. Frequency division multiple access (FDMA) assigns ______ channels to _______ users.
a) Individual, individual
b) Many, individual
c) Individual, many
d) Many, many

Answer: a [Reason:] Frequency division multiple access (FDMA) assigns individual channels to individual users. Each user is allocated a unique frequency band or channel. These channels are assigned on demand to users who request service.

2. During the period of call, other users can share the same channel in FDMA. State whether True or False.
a) True
b) False

Answer: b [Reason:] In FDMA systems, no other user can share the same channel during the period of call. In FDD systems, the users are assigned a channel as a pair of frequencies; one is used for the forward channel while the other frequency is used for the reverse channel.

3. The FDMA channel carries ____ phone circuit at a time.
a) Ten
b) Two
c) One
d) Several

Answer: c [Reason:] The FDMA channel carries one phone circuit at a time. Each individual band or channel is wide enough to accommodate the signal spectra of the transmissions to be propagated.

4. If the FDMA channel is not in use, it can be used by other users. State whether True or False.
a) True
b) False

Answer: b [Reason:] If an FDMA channel is not in use, the it sits idle and cannot be used by other users to increase or share capacity. It is essentially a wasted resource.

5. The bandwidth of FDMA channel is ______
a) Wide
b) Narrow
c) Large
d) Zero

Answer: b [Reason:] The bandwidth of FDMA channels is relatively narrow as each channel supports only one circuit per carrier. That is, FDMA is usually implemented in narrow band systems.

6. The symbol time in FDMA systems is _________ thus intersymbol interference is ______
a) Large, high
b) Small, low
c) Small, high
d) Large, low

Answer: d [Reason:] The symbol time of a narrowband signal is large as compared to the average delay spread. This implies that the amount of intersymbol interference is low and, thus, little or no equalization is required in FDMA narrowband systems.

7. Due to _________ transmission scheme, ______ bits are needed for overhead in FDMA systems.
a) Continuous, few
b) Discontinuous, few
c) Continuous, many
d) Discontinuous, many

Answer: a [Reason:] Since FDMA is a continuous transmission scheme, fewer bits are needed for overhead purposes (such as synchronization and framing bits) as compared to TDMA.

8. Which of the following is not true for FDMA systems as compared to TDMA systems?
a) Low complexity
b) Lower cell site system cost
c) Tight RF filtering
d) Narrow bandwidth

Answer: b [Reason:] FDMA systems have higher cell site system costs as compared to TDMA systems. It is due to single channel per carrier design, and the need to use costly bandpass filters to eliminate spurious radiation at the base station.

9. __________ is undesired RF radiation.
a) Intermodulation frequency
b) Intermediate frequency
c) Instantaneous frequency
d) Instrumental frequency

Answer: a [Reason:] Intermodulation (IM) frequency is undesired RF radiation which can interfere with other channels in the FDMA systems. The nonlinearities cause signal spreading in the frequency domain and generate IM frequency.

10. ______ is based on FDMA/FDD.
a) GSM
b) W-CDMA
c) Cordless telephone
d) AMPS

Answer: d [Reason:] The first US analog cellular system, the Advanced Mobile Phone System (AMPS) is based on FDMA/FDD. A single user occupies a single channel while the call is in progress.

11. In US AMPS, 416 channels are allocated to various operators with 10 kHz guard band and channel between them is 30 kHz. What is the spectrum allocation given to each operator?
a) 12.5 kHz
b) 30 kHz
c) 12.5 MHz
d) 30 MHz

Answer: c [Reason:] Spectrum allocated to each cellular operator is 12.5 MHz. As Bt=NBc+2Bguard; which is equal to 416*30*103+2(10*103) =12.5 MHz.

## Communications MCQ Set 2

1. Cellular concept replaces many low power transmitters to a single high power transmitter. (True or False)
a) True
b) False

Answer: b [Reason:] Cellular concept is a system level idea that replaces a single high power transmitter to many low power transmitters. High power transmitters lead to large cell, and thus it was impossible to use the same frequencies throughout the systems. But, it is possible with low power transmitter.

2. Why neighbouring stations are assigned different group of channels in cellular system?
a) To minimize interference
b) To minimize area
c) To maximize throughput
d) To maximize capacity of each cell

Answer: a [Reason:] Neighbouring base stations are assigned different group of channels. It minimizes the interference between base stations and the users under their control.

3. What is a cell in cellular system?
a) A group of cells
b) A group of subscribers
c) A small geographical area
d) A large group of mobile systems

Answer: a [Reason:] Cell is a small geographic area in a cellular system. Each cellular base station within a cell is allocated a group of radio channels that could be used in another cell.

4. What is frequency reuse?
a) Process of selecting and allocating channels
b) Process of selection of mobile users
c) Process of selecting frequency of mobile equipment
d) Process of selection of number of cells

Answer: a [Reason:] Frequency reuse is the process of using the same radio frequencies on radio transmitter sites within a geographic area. They are separated by sufficient distance to cause minimal interference with each other.

5. Which of the following is a universally adopted shape of cell?
a) Square
b) Circle
c) Triangle
d) Hexagon

Answer: d [Reason:] Hexagonal cell shape is a simplistic model of radio coverage for each base station. It has been universally adopted since the hexagon permits easy and manageable analysis of a cellular system.

6. Actual radio coverage of a cell is called _____
a) Fingerprint
b) Footprint
c) Imprint
d) Matrix

Answer: b [Reason:] Actual radio coverage of a cell is known as the footprint. It is determined from field measurements or propagation prediction models. Although the real footprint is amorphous in nature, a regular cell shape is needed for systematic system design.

7. Why the shape of cell is not circle?
a) Omni directionality
b) Small area
c) Overlapping regions or gaps are left
d) Complex design

Answer: c [Reason:] Circle is the first natural choice to represent the coverage area of a base station. But while adopting this shape, adjacent cells cannot be overlaid upon a map without leaving gaps or creating overlapping regions.

8. What is the main reason to adopt hexagon shape in comparison to square and triangle?
a) Largest area
b) Simple design
c) Small area
d) Single directional

Answer: a [Reason:] For a given distance between the center of a polygon and its farthest perimeter points, the hexagon has the largest area. Thus, by using the hexagon geometry, the fewest number of cells can cover a geographic region.

9. Which type of antenna is used for center excited cells?
a) Dipole antenna
b) Grid antenna
c) Sectored antenna
d) Omnidirectional antenna

Answer: d [Reason:] For center excited cells, base station transmitters are used at the center of cell. To cover the whole cell, omnidirectional antenna is the best choice for base station transmitters.

10. Which type of antenna is used for edge excited cells?
a) Omnidirectional antenna
b) Grid antenna
c) Sectored directional antenna
d) Dipole antenna

Answer: c [Reason:] For edge excited cell, mostly base station transmitters are placed on three of the six cell vertices. To cover the assigned portion of a cell, sectored directional antenna is the best choice.

11. For a cellular system, if there are N cells and each cell is allocated k channel. What is the total number of available radio channels, S?
a) S=k*N
b) S=k/N
c) S=N/k
d) S=kN

Answer: a [Reason:] If there is a cellular system with total of S duplex channels. Each cell is allocated a group of k channels and there are total N cells in the system, S channels are divide among N cells into unique and disjoint channel groups. Therefore, total number of radio channel is the product of total number of cells in the system (N) and number of channel allocated to each cell (k).

12. What is a cluster in a cellular system?
a) Group of frequencies
b) Group of cells
c) Group of subscribers
d) Group of mobile systems

Answer: b [Reason:] Cluster is group of N cells. These cells use the complete set of frequency available for the cellular system at that location.

13. What is a frequency reuse factor for N number of cells in a system?
a) N
b) N2
c) 2*N
d) 1/N

Answer: d [Reason:] The frequency reuse factor is defined as 1 over the number of cells in the cluster of the system (N). It is given by 1/N since each cell within a cluster is only assigned 1/N of the total available channels in the system.

14. Capacity of a cellular system is directly proportional to _____
a) Number of cells
b) Number of times a cluster is replicated
c) Number of Base stations
d) Number of users

Answer: b [Reason:] The capacity of a cellular system is directly proportional to the number of times a cluster is replicated in a fixed area. If the cluster size N is reduced while the cell size is kept constant, more clusters are required to cover a given area, and hence more capacity is achieved.

15. A spectrum of 30 MHz is allocated to a cellular system which uses two 25 KHz simplex channels to provide full duplex voice channels. What is the number of channels available per cell for 4 cell reuse factor?
a) 150 channels
b) 600 channels
c) 50 channels
d) 85 channels

Answer: a [Reason:] Total bandwidth is 30 MHz. And the channel bandwidth is 50 KHz/duplex channel (25KHz*2). Therefore, total available channels are 600 channels (30,000/50). For 4 cell reuse factor, total number of channels available per cell will be 150 channels (600/4).

## Communications MCQ Set 3

1. Which of the following is not an open standard?
a) Bluetooth
b) WWW
c) HTML
d) VPN

Answer: d [Reason:] An open standard is a standard that is publicly available. It has various rights to use associated with it. Bluetooth is an open standard that has been embraced by over thousand manufacturers of electronic appliances. VPN (Virtual Private Network) is a private network.

2. What is the nominal range of Bluetooth?
a) 1 Km
b) 10 m
c) 1 m
d) 10 Km

Answer: b [Reason:] The power of the transmitter governs the range over which a Bluetooth device can operate. Generally, Bluetooth devices are said to fall into one of three classes. The most common kind is class 2, and it operates in 10 m range.

3. Bluetooth standard is named after ______
a) King Ronaldo Bluetooth
b) Pope Vincent Bluetooth
c) King Herald Bluetooth
d) Pope Francis Bluetooth

Answer: c [Reason:] Bluetooth standard is named after King Harald Bluetooth. He was the 10th century Viking who united Denmark and Norway. The Bluetooth standard aims to unify the connectivity chores of appliances within the personal workspace of an individual.

4. Bluetooth operates in which band?
a) Ka Band
b) L Band
c) Ku Band
d) 2.4 GHz ISM Band

Answer: d [Reason:] Bluetooth is a radio technology operating in 2.4 GHz frequency band. Bluetooth is best suited for low-bandwidth applications like transferring sound data with telephones (i.e. with a Bluetooth headset) or byte data with hand-held computers (transferring files) or keyboard.

5. Which of the following scheme is used by Bluetooth?
a) Frequency hopping TDD scheme
b) Frequency hopping FDD scheme
c) DSSS TDD scheme
d) DSSS FDD scheme

Answer: a [Reason:] Frequency hopping TDD (Time Division Duplexing) scheme is used by Bluetooth. Frequency hopping provides a form of multiple access among co-located devices in different piconets.

6. What is the range of time slot in Bluetooth?
a) 120 milliseconds
b) 625 microseconds
c) 577 microseconds
d) 5.7 seconds

Answer: b [Reason:] Bluetooth uses a timeslot of 625 microseconds. A data channel hops randomly 1600 times per second between 79 RF channels. Thus, each channel is divided into time slots 625 microseconds.

7. Which modulation scheme is used by Bluetooth?
a) DQPSK
b) MSK
c) GFSK
d) BPSK

Answer: c [Reason:] Bluetooth uses GFSK (Gaussian Frequency Shift Keying). When GFSK is used for Bluetooth modulation, the frequency of the carrier is shifted to carry the modulation. By doing this the Bluetooth modulation achieves a bandwidth of 1 MHz with stringent filter requirements to prevent interference on other channels.

8. What is the channel symbol rate in Bluetooth for each user?
a) 270.833 Kbps
b) 1 Gbps
c) 100 Mbps
d) 1 Mbps

Answer: d [Reason:] Each user in Bluetooth uses a radio channel symbol rate of 1 Mbps using GFSK modulation. The frequency hopping scheme of each Bluetooth user is determined from a cyclic code with a length of 1027-1.

9. What is the raw channel bit error rate of Bluetooth?
a) 10-3
b) 10-10
c) 103
d) 10-1

Answer: a [Reason:] Bluetooth has a bit error rate (BER) of 10-3. Bluetooth uses a number of forward error control (FEC) coding and automatic repeat request (ARR) schemes to achieve this bit rate.

10. Which of the following standard committee specifies Bluetooth and other Personal Area Networks (PAN)?
a) IEEE 802.11b
b) IEEE 802.15
c) IEEE 802.11g
d) IEEE 802.16

Answer: b [Reason:] IEEE 802.15 standards committee is formed to provide an international forum for developing Bluetooth and other PANs. PANs are used to interconnect pocket PCs, personal digital assistants (PDAs), cell phones, light projectors and other appliances.

## Communications MCQ Set 4

1. Which of the following is not used to improve received signal quality over small scale times and distance?
a) Modulation
b) Equalization
c) Diversity
d) Channel coding

Answer: a [Reason:] Equalization, diversity and channel coding are the three techniques which are used to improve received signal quality and link performance over small scale times and distance. But, the approach, cost, complexity and effectiveness varies in wireless communication system.

2. Equalization is used to compensate __________
a) Peak signal to noise ratio
b) Intersymbol interference
d) Noises present in the signal

Answer: b [Reason:] Equalization compensate the intersymbol interference (ISI) created by multipath within time dispersive channels. An equalizer within a receiver compensates for the average range of the expected channel amplitude and delay characteristics.

3. Training and tracking are the operating modes of _________
a) Diversity techniques
b) Channel coding techniques
c) Equalization techniques
d) Demodulation techniques

Answer: c [Reason:] General operating modes of an adaptive equalizer includes training and tracking. A known fixed length training sequence is sent by the transmitter so that the receiver’s equalizer may adapt to a proper setting for minimum bit error rate detection.

4. An equalizer is said to be converged when it is properly _______
a) Trained
b) Tracked
c) Installed
d) Used

Answer: a [Reason:] When an equalizer has been properly trained, it is said to have converged. Equalizers are commonly used in digital communication systems where user data is segmented into short time blocks or time slots.

5. Time for convergence of an equalizer is not a function of _______
a) Equalizer algorithm
b) Equalizer structure
c) Time rate of change of multipath radio channel
d) Transmitter characteristics

Answer: d [Reason:] The timespan over which an equalizer converges is a function of the equalizer algorithm, the equalizer structure and the time rate of change of the multipath radio channel. Equalizers require proper retraining in order to maintain effective ISI cancellation.

6. Equalizer is usually implemented in __________
a) Transmitter
b) Baseband or at IF in a receiver
d) Modulator stage

Answer: b [Reason:] An equalizer is implemented at baseband or at IF in a receiver. Since, the baseband complex envelope expression can be used to represent bandpass waveform, the channel response, the demodulated signal and adaptive equalizer algorithm are usually simulated and implemented at baseband.

7. Equalizer is ________ of the channel.
a) Opposite
b) Same characteristics
c) Inverse filter

Answer: c [Reason:] An equalizer is actually an inverse filter of the channel. The goal of equalization is to satisfy that the combination of the transmitter, channel and receiver be an all pass channel.

8. ______ controls the adaptive algorithm in an equalizer.
a) Error signal
b) Transmitted signal
d) Channel impulse response

Answer: a [Reason:] The adaptive algorithm is controlled by the error signal. The error signal is derived by comparing the output of the equalizer and some signal which is either an exact scaled replica of the transmitted signal or represents a property of transmitted signal.

9. The adaptive algorithms in equalizer that do not require training sequence are called ________
b) Blind algorithms

Answer: b [Reason:] Blind algorithms exploit the characteristics of the transmitted signal and do not require training sequence. These type of algorithm are able to acquire equalization through property restoral techniques of transmitted signal.

10. Which of the following is a blind algorithm?
b) Constant modulus algorithm

Answer: b [Reason:] Blind algorithm technique uses algorithms such as the constant modulus algorithm (CMA) and the spectral coherence restoral algorithm (SCORE). CMA is used for constant envelope modulation and forces the equalizer weights to maintain a constant envelope on the received signal.

## Communications MCQ Set 5

1. Which of the following is the world’s first cellular system to specify digital modulation and network level architecture?
a) GSM
b) AMPS
c) CDMA
d) IS-54

Answer: a [Reason:] GSM was the world’s first cellular system to specify digital modulation and level architectures and services. It is the world’s most popular 2G technology. It was developed to solve the fragmentation problems of the first cellular systems in Europe.

2. Previously in 1980s, GSM stands for ______
a) Global system for mobile
b) Groupe special mobile
c) Global special mobile
d) Groupe system mobile

Answer: b [Reason:] In the mid-1980s GSM was called by the name Groupe special mobile. In 1992, GSM changed its name to Global System for Mobile Communication for marketing reasons.

3. Who set the standards of GSM?
a) ITU
b) AT & T
c) ETSI
d) USDC

Answer: c [Reason:] The setting of standards for GSM is under the aegis of the European Technical Standards Institute (ETSI). GSM task was to specify common mobile communication system for Europe in the 900 MHZ band.

4. Which of the following does not come under the teleservices of GSM?
a) Standard mobile telephony
b) Mobile originated traffic
c) Base originated traffic
d) Packet switched traffic

Answer: d [Reason:] GSM services follow ISDN guidelines and are classified as either teleservices or data services. Teleservices include standard mobile telephony and mobile originated or base originated traffic.

5. Which of the following comes under supplementary ISDN services?
a) Emergency calling
b) Packet switched protocols
c) Call diversion
d) Standard mobile telephony

Answer: c [Reason:] Supplementary ISDN services are digital in nature. They include call diversion, closed user groups, and caller identification, and are not available in analog mobile networks. Supplementary services also include short messaging service (SMS).

6. Which of the following memory device stores information such as subscriber’s identification number in GSM?
a) Register
b) Flip flop
c) SIM
d) SMS

Answer: c [Reason:] SIM (subscriber identity module) is a memory device that stores information such as the subscriber’s identification number, the networks and countries where the subscriber is entitled to service, privacy keys, and other user specific information.

7. Which of the following feature makes impossible to eavesdrop on GSM radio transmission?
a) SIM
b) On the air privacy
c) SMS
d) Packet switched traffic

Answer: b [Reason:] The on the air privacy feature of GSM makes impossible to eavesdrop on a GSM radio transmission. The privacy is made possible by encrypting the digital bit stream sent by a GSM transmitter, according to a specific secret cryptographic key that is known only to the cellular carrier.

8. Which of the following does not come under subsystem of GSM architecture?
a) BSS
b) NSS
c) OSS
d) Channel

Answer: d [Reason:] The GSM architecture consists of three major interconnected subsystems that interact between themselves and with the users through certain network interfaces. The subsystems are BSS (Base Station Subsystem), NSS (Network and Switching Subsystem) and OSS (Operation Support Subsystem).

9. Which of the following subsystem provides radio transmission between mobile station and MSC?
a) BSS
b) NSS
c) OSS
d) BSC

Answer: a [Reason:] The BSS provides and manages radio transmission paths between the mobile stations and the Mobile Switching Center (MSC). It also manages the radio interface between the mobile stations and all other subsystems of GSM.

10. ____ manages the switching function in GSM.
a) BSS
b) NSS
c) OSS
d) MSC