Category: Wireless & Mobile Communications Objective Questions

Basic Wireless & Mobile Communications Questions and Answers on “Small- Scale Multipath Propagation”.

1. Small scale fading describes the _________ fluctuations of the amplitude, phases of a signal.
a) Rapid
b) Slow
c) Instantaneous
d) Different
Answer: a
Clarification: Small scale fading or simply fading, is used to describe the rapid fluctuations of amplitudes, phases, or multipath delays of a radio signal over a short period of time or travel distance. It ignores the large scale path loss.

2. Fading is caused by interference.
a) True
b) False
Answer: a
Clarification: Fading is caused by interference. It is caused by interference between two or more versions of the transmitted signal which arrive at the receiver at slightly different times.

3. Which of the following is not an effect caused by multipath in radio channel?
a) Rapid changes in signal strength
b) Random frequency modulation
c) Power of base station
d) Time dispersion
Answer: c
Clarification: Rapid changes in signal strength over a small travel distance are caused due to multipath. It causes random frequency modulation due to varying Doppler shifts on different multipath signals. Time dispersion is also caused by multipath propagation delays.

4. In urban areas, fading occurs due to height of mobile antenna ________ than height of surrounding structure.
a) Same
b) Smaller
c) Greater
d) Very larger
Answer: b
Clarification: In urban areas, fading occurs because height of the mobile antenna is below the height of surrounding structures. Therefore, there is no single line of sight path to the base station.

5. Fading does not occur when mobile receiver is stationary.
a) True
b) False
Answer: b
Clarification: The received signal may fade even when the mobile receiver is stationary. It is due to the movement of surrounding objects in the radio channel. The multipath components combine vectorially at the receiver antenna and cause signal to distort or fade.

6. Apparent shift in frequency in multipath wave is caused due to relative motion between________
a) Base station and MSC
b) Mobile and surrounding objects
c) Mobile and MSC
d) Mobile and base station
Answer: d
Clarification: Due to relative motion between mobile and base station, each multipath wave experiences an apparent shift in frequency. This shift in received signal frequency due to motion called Doppler shift.

7. Doppler shift is directly proportional to __________
a) Velocity
b) Height of antenna
c) Power of receiving antenna
d) Power of transmitter
Answer: a
Clarification: The shift in received signal frequency due to motion is called Doppler shift. It is directly proportional to the velocity and direction of motion of mobile with respect to the direction of arrival of the received multipath wave.

8. Which of the following factor does not influence small scale fading?
a) Multipath propagation
b) Power density of base station
c) Speed of mobile
d) Speed of surrounding objects
Answer: b
Clarification: Many physical factors in radio channel influence small scale fading. Multipath propagation, speed of mobile, speed of surrounding objects, transmission bandwidth of the signal influences small scale fading in a large way.

9. Signal will distort if transmitted signal bandwidth is greater than bandwidth of __________
a) Receiver
b) Radio channel
c) Multipath channel
d) Transceiver
Answer: c
Clarification: Received signal will be distorted if transmitted signal bandwidth is greater than bandwidth of multipath channel. But received signal strength will not fade much over a local area.

10. What is a measure of the maximum frequency difference for which signals are strongly correlated in amplitude?
a) Coherence bandwidth
b) Narrow bandwidth
c) Incoherent bandwidth
d) Wide bandwidth
Answer: a
Clarification: The bandwidth of the channel can be quantified by the coherence bandwidth. It is related to the specific multipath structure of channel. It is a measure of maximum frequency difference for which signals are strongly correlated in amplitude.

11. The Doppler shift for mobile moving with constant velocity, v is given by _______
a) (v*cos θ)/λ
b) v/λ
c) v*cos θ
d) v*λ
Answer: a
Clarification: Doppler shift is given by (v*cos θ)/λ. This formula relates the Doppler shift to the mobile velocity and spatial angle between the direction of motion of mobile and the direction of arrival of the wave.

12. Doppler shift is positive if mobile is moving away from direction of arrival of the wave.
a) True
b) False
Answer: b
Clarification: Doppler shift is positive if the mobile is moving toward the direction of arrival of the wave, as the apparent received frequency is increased. And if the mobile is moving away from the direction of arrival of the wave, Doppler shift is negative.

basic questions and answers on all areas of Wireless & Mobile Communications, .

Wireless & Mobile Communications Multiple Choice Questions on “Spread Spectrum Modulation Techniques”.

1. The transmission bandwidth of spread spectrum techniques is equal to the minimum required signal bandwidth.
a) True
b) False
Answer: b
Clarification: 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: a
Clarification: 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: b
Clarification: 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: c
Clarification: 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: a
Clarification: 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) 2^m
d) 2^m-1
Answer: d
Clarification: There are exactly 2^m-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 2^m-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: c
Clarification: 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: b
Clarification: 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: a
Clarification: 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: c
Clarification: 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: d
Clarification: The processing gain of frequency hopping (FH) systems is given by B_ss/B. Here, B_ss and B denote the total hopping bandwidth and instantaneous bandwidth respectively.

12. FH systems do not have collisions.
a) True
b) False
Answer: b
Clarification: 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.
a) True
b) False
Answer: b
Clarification: 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.

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Wireless & Mobile Communications Multiple Choice Questions on “Frequency Division Multiple Access (FDMA)”.

1. Frequency division multiple access (FDMA) assigns ______ channels to _______ users.
a) Individual, individual
b) Many, individual
c) Individual, many
d) Many, many
Answer: a
Clarification: 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.
a) True
b) False
Answer: b
Clarification: 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
Clarification: 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.
a) True
b) False
Answer: b
Clarification: If an FDMA channel is not in use, 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: Spectrum allocated to each cellular operator is 12.5 MHz. As B_t = NB_c + 2B_guard; which is equal to 416*30*10³+2(10*10³) = 12.5 MHz.

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Wireless & Mobile Communications Multiple Choice Questions on “CDMA Digital Cellular Standard (IS-95)”.

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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
c) RAKE receiver
d) Frequency divider
Answer: c
Clarification: 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
Answer: b
Clarification: CT2 was the second generation of cordless telephones introduced in Great Britain in 1989. It is used to provide telepoint services which allow a subscriber to use CT2 handsets at a public telepoint.

10. CT2 is analog version of first generation cordless telephones.
a) True
b) False
Answer: b
Clarification: CT2 is a digital version of the first generation, analog, cordless telephones. When compared with analog cordless phones, CT2 offers good speech quality and is more resistant to interference.

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Wireless & Mobile Communications Multiple Choice Questions & Answers (MCQs) on “Wireless Local Area Networks (WLANs)”.

1. What is the full form of WLAN?
a) Wide Local Area Network
b) Wireless Local Area Network
c) Wireless Land Access Network
d) Wireless Local Area Node
Answer: b
Clarification: WLAN stands for Wireless Local Area Network. Wireless networks is increasingly used as a replacement for wires within homes, buildings, and office settings through the deployment of wireless local area networks (WLANs).

2. WLANs use high power levels and generally require a license for spectrum use.
a) True
b) False
Answer: b
Clarification: WLANs use low power and generally do not require a license for spectrum. They provide ad hoc high data transmission rate connections deployed by individuals. In the late 1980s, FCC provided licence free bands for low power spread spectrum devices in ISM band, which is used by WLAN.

3. What is the name of 300 MHz of unlicensed spectrum allocated by FCC in ISM band?
a) UNII
b) Unlicensed PCS
c) Millimetre wave
d) Bluetooth
Answer: a
Clarification: FCC allocated 300 MHz of unlicensed spectrum in the ISM bands. This allocation is called the Unlicensed National Information Infrastructure (UNII) band. It was allocated for the express purpose of supporting low power license free spread spectrum data communication.

4. Which of the following specifies a set of media access control (MAC) and physical layer specifications for implementing WLANs?
a) IEEE 802.16
b) IEEE 802.3
c) IEEE 802.11
d) IEEE 802.15
Answer: c
Clarification: IEEE 802.11 is a set of media access control and physical layer specification for implementing WLAN computer communication. It was founded in 1987 to begin standardization of spread spectrum WLANs for use in the ISM bands.

5. Which of the following is not a standard of WLAN?
a) HIPER-LAN
b) HIPERLAN/2
c) IEEE 802.11b
d) AMPS
Answer: d
Clarification: AMPS is a standard of first generation network. HIPERLAN is a WLAN standard developed in Europe in mid 1990s. HIPERLAN/2 is also developed in Europe that provides upto 54 Mbps of user data.

6. Which of the following is the 802.11 High Rate Standard?
a) IEEE 802.15
b) IEEE 802.15.4
c) IEEE 802.11g
d) IEEE 802.11b
Answer: d
Clarification: IEEE 802.11b was a high rate standard approved in 1999. It provided new data rate capabilities of 11 Mbps, 5.5 Mbps in addition to the original 2 Mbps and 1 Mbps user rates of IEEE 802.11.

7. Which of the following spread spectrum techniques were used in the original IEEE 802.11 standard?
a) FHSS and DSSS
b) THSS and FHSS
c) THSS and DSSS
d) Hybrid technique
Answer: a
Clarification: Original IEEE 802.11 used both the approaches of FHSS (Frequency Hopping Spread Spectrum) and DSSS (Direct Sequence Spread Spectrum). But from late 2001s, only DSSS modems are used within IEEE 802.11.

8. Which of the following WLAN standard has been named Wi-Fi?
a) IEEE 802.6
b) IEEE 802.15.4
c) DSSS IEEE 802.11b
d) IEEE 802.11g
Answer: c
Clarification: The DSSS IEEE 802.11b standard has been named Wi-Fi by the Wireless Ethernet Compatibility Alliance. It is a group that promotes adoption of 802.11 DSSS WLAN.

9. Which of the following is developing CCK-OFDM?
a) IEEE 802.11a
b) IEEE 802.11b
c) IEEE 802.15.4
d) IEEE 802.11g
Answer: d
Clarification: IEEE 802.11g is developing CCK-OFDM (Complementary Code Keying Orthogonal Frequency Division Multiplexing) standards. It will support roaming capabilities and dual band use for public WLAN networks. It also has backward compatibility with 802.11b technologies.

10. What is the data rate of HomeRF 2.0?
a) 10 Mbps
b) 54 Mbps
c) 200 Mbps
d) 1 Mbps
Answer: a
Clarification: HomeRF 2.0 has data rate of the order of 10 Mbps. The FHSS proponents of IEEE 802.11 have formed the HomeRF standard that supports the frequency hopping equipment. In 2001, HomeRF developed a 10 Mbps FHSS standard called HomeRF 2.0.

11. HIPER-LAN stands for ____________
a) High Precision Radio Local Area Network
b) High Performance Radio Local Area Network
c) High Precision Radio Land Area Network
d) Huge Performance Radio Link Access Node
Answer: b
Clarification: HIPER-LAN stands for High Performance Radio Local Area Network. It was developed in Europe in mid 1990s. It was intended to provide individual wireless LANs for computer communication.

12. What is the range of asynchronous user data rates provided by HIPER-LAN?
a) 1-100 Mbps
b) 50-100 Mbps
c) 1-20 Mbps
d) 500 Mbps to 1 Gbps
Answer: c
Clarification: HIPER-LAN provides asynchronous user data rates of between 1 to 20 Mbps, as well as time bounded messaging of rates of 64 kbps to 2.048 Mbps. It uses 5.2 GHz and 17.1 GHz frequency bands.

13. What is the name of the European WLAN standard that provides user data rate upto 54 Mbps?
a) UNII
b) WISP
c) MMAC
d) HIPERLAN/2
Answer: d
Clarification: HIPERLAN/2 has emerged as the next generation European WLAN standard. It provides upto 54 Mbps of user data to a variety of networks. The networks includes the ATM backbone, IP based networks and the UMTS network.

14. What is WISP?
a) Wideband Internet Service Protocol
b) Wireless Internet Service Provider
c) Wireless Instantaneous Source Provider
d) Wideband Internet Source Protocol
Answer: b
Clarification: WISP is wireless Internet Service Provider used to explore public LANs (publican). It builds a nationwide infrastructure of WLAN access points in selected hotels, restaurants or airports. It then charges a monthly subscription fee to users who wish to have always on Internet access in those selected locations.

15. The price of WLAN hardware is more than 3G telephones and fixed wireless equipment.
a) True
b) False
Answer: b
Clarification: As, WLAN could be used to provide access for the last 100 meters into homes and businesses. Therefore, the price of WLAN hardware is far below 3G telephones and fixed wireless equipment.

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