Category: Wireless & Mobile Communications Objective Questions

Wireless & Mobile Communications Multiple Choice Questions on “Outdoor Propagation Models”.

1. Which of the following is not an outdoor propagation model?
a) Longley-Rice model
b) Ericson Multiple Breakpoint Model
c) Hata model
d) Okumura model
Answer: b
Clarification: Ericson multiple breakpoint model is an indoor propagation model. Longley-Rice, Hata and Okumura model are outdoor propagation models. Most of these models are based on a systematic interpretation of measurement data obtained in the service area.

2. Longley –Rice model is applicable to _________
a) Point to point communication
b) All to all communication
c) Point to multipoint communication
d) Multipoint microwave distribution sstem
Answer: a
Clarification: The Longley-Rice model is applicable to point-to-point communication systems in the frequency range from 40 MHz to 100 GHz. They are applicable for different kinds of terrain. Terrain profile may vary from a simple curved Earth profile to a highly mountainous profile.

3. Longley-Rice prediction model is also referred as _________
a) Okumura model
b) Hata model
c) ITS irregular terrain model
d) Bertoni model
Answer: c
Clarification: The Longley Rice prediction model is also referred to as ITS irregular terrain model. The model is based on electromagnetic theory and on statistical analyses of both terrain features and radio measurements. It predicts the median attenuation of a radio signal as a function of distance and the variability of the signal in time and in space.

4. The extra term for additional attenuation due to urban clutter near the receiving antenna is called __________
a) Power factor
b) Urban gain
c) Clutter factor
d) Urban factor
Answer: d
Clarification: The urban factor (UF) is derived by comparing the predictions by the original Longley –Rice model with those obtained by Okumura. It deals with radio propagation in urban areas and is relevant to mobile radio.

5. Longley Rice model’s merit is to provide corrections due to environmental factors.
a) True
b) False
Answer: b
Clarification: One shortcoming of the Longley –Rice model is that it does not provide a way of determining corrections due to environmental factors. It does not consider correction factors to account for the effects of buildings and foliage. Multipath is also not considered.

6. Which method is used by Edwards and Durkin algorithm to calculate the loss associated with diffraction edges?
a) Epstein and Peterson method
b) Interpolation method
c) Knife edge diffraction method
d) Fresnel- Kirchoff method
Answer: a
Clarification: The Edwards and Durkin algorithm uses Epstein and Peterson method to calculate the loss associated with two diffraction edges. It is the sum of two attenuations. First is loss at second diffraction edge caused by first diffraction edge. And second is the loss at receiver caused by second diffraction edge.

7. Durkin’s model can read digital elevation map.
a) True
b) False
Answer: a
Clarification: Durkin’s model is very attractive because it can read in a digital elevation map and perform a site specific propagation computation on the elevation data. It can produce a signal strength contour that is reported to be good within a few dB.

8. Which of the most widely used model for signal prediction in urban areas?
a) Ericsson Multiple Breakpoint Model
b) Log distance path loss model
c) Okumura model
d) Attenuation factor model
Answer: c
Clarification: Okumura’s model is one of the most widely used models for signal prediction in urban areas. This model is applicable for frequencies in the range 150 MHz to 1920 MHz (Extrapolated upto 3000 MHz).

9. Okumura model is applicable for distances of _________
a) 1 m to 10 m
b) 1 km to 100 km
c) 100 km to 1000 km
d) 10 km to 10000 km
Answer: b
Clarification: Okumura’s model is applicable for distances of 1 km to 100 km. It can be used for base station antenna heights ranging from 30 m to 1000 m. Okumura developed a set of curves giving the median attenuation relative to free space in an urban area.

10. Okumura model is considered to be complex in predicting path loss.
a) True
b) False
Answer: b
Clarification: Okumura’s model is considered to be among the simplest and best in terms of accuracy in path loss prediction for mature cellular and land mobile radio system. It is very practical and has become a standard for system planning in modern land mobile system in Japan.

11. Which of the following is the major disadvantage of the Okumura model?
a) Complex
b) Inaccurate
c) Not practical
d) Slow response to rapid change in terrain
Answer: d
Clarification: The major disadvantage with the model is its slow response to rapid changes in terrain. Therefore the model is fairly good in urban and suburban areas, but not as good in rural areas. Common standard deviations between predicted and measured path loss values are 10 dB to 14 dB.

12. The Hata model is empirical formulation of which model?
a) Okumura model
b) Longley- Rice model
c) Durkin’s model
d) Walfisch and Bertoni model
Answer: a
Clarification: The Hata model is an empirical formulation of the graphical path loss data provided by Okumura. It is valid from 150 MHz to 1500 MHz. Hata presented the urban area propagation loss as a standard formulation. It supplied correct Equations for application to other situations.

13. Hata model is well suited for _________
a) Personal communication system
b) Large cell mobile radio system
c) Small cell mobile radio system
d) Every mobile radio system
Answer: b
Clarification: Hata model is well suited for large cell mobile radio systems. But it is not well suited for personal communication system (PCS) which have cells on the order of 1 km radius. Hata model does not have any path specific corrections which are available in Okumura model.

14. Which of the following considers the impact of rooftops and building?
a) Okumura model
b) Hata model
c) Walfisch and Bertoni model
d) Longley- Rice model
Answer: c
Clarification: The impact of rooftops and building height is considered by Walfisch and Bertoni model. It uses diffraction to predict average signal strength at street level. It considers path loss to be a product of three factors.

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Wireless & Mobile Communications Multiple Choice Questions on “Constant Envelope Modulation”.

1. In non-linear modulation, the amplitude of the carrier varies with the variation of modulating signal.
a) True
b) False
Answer: b
Clarification: In non-linear modulation, the amplitude of the carrier is constant regardless of the variation in the modulating signal. Many practical mobile radio communication systems use these types of nonlinear modulation methods.

2. In constant envelope family of modulation, class C amplifiers introduces degradation in spectrum occupancy.
a) True
b) False
Answer: b
Clarification: The constant envelope family of modulation has an advantage of satisfying various conditions. In this, power efficient Class C amplifiers can be used without introducing degradation in the spectrum occupancy of the transmitted signal.

3. Constant envelope modulation techniques occupy ______ bandwidth than linear modulation schemes.
a) Larger
b) Smaller
c) Same
d) Twice
Answer: a
Clarification: Constant envelope modulation technique occupies a larger bandwidth than linear modulation technique. It is one of the disadvantage of constant envelope modulation. It is not well suited where bandwidth efficiency is more important than power efficiency.

4. In BFSK __________ of constant amplitude carrier signal is switched between two values.
a) Phase
b) Angle
c) Frequency
d) Amplitude
Answer: c
Clarification: In BFSK, the frequency of a constant amplitude carrier signal is switched between two values according to the two possible message states. These states are called high and low tunes, which corresponds to binary 1 or 0.

5. MSK stands for ________
a) Maximum shift keying
b) Minimum shift keying
c) Minimum space keying
d) Maximum space keying
Answer: b
Clarification: MSK stands for minimum shift keying. It is a special type of continuous phase shift keying. It is form of digital modulation technique that was developed in 1950s.

6. What is the modulation index of MSK?
a) 0.1
b) 1
c) 0.5
d) 0
Answer: c
Clarification: Minimum shift keying is a special type of CPFSK. Its peak frequency deviation is equal to ¼ the bit rate. In other words, MSK is continuous phase FSK with a modulation index of 0.5.

7. The modulation index of an FSK signal is similar to modulation index of ________
a) Amplitude modulation
b) Phase modulation
c) QPSK
d) Frequency modulation
Answer: d
Clarification: The modulation index of an FSK signal is similar to FM modulation index. It is defined by (2∆F)/Rb. Here ∆F is the peak RF frequency and Rb is the bit rate.

8. The name minimum phase shift keying implies minimum _________
a) Frequency separation
b) Amplitude separation
c) Phase change
d) Amplitude deviation
Answer: a
Clarification: The name minimum phase shift keying implies minimum frequency separation, i.e. the bandwidth that allows orthogonal detection. A modulation index of 0.5 corresponds to the minimum frequency spacing that allows two FSK signals to be coherently orthogonal.

9. MSK is sometimes also referred as _________
a) Slow FSK
b) Fast FSK
c) Slow PSK
d) Fast PSK
Answer: b
Clarification: Minimum shift keying is sometimes also referred as fast FSK. It is so called because frequency spacing used is only half as much as that used in conventional noncoherent frequency shift keying.

10. Which of the following is not a property of MSK?
a) Variable envelope
b) Spectral efficiency
c) Good BER performance
d) Self synchronizing capability
Answer: a
Clarification: MSK has a constant envelope. It is a spectrally efficient scheme. It possesses properties such as constant envelope, spectral efficiency, good BER performance and self-synchronizing capability.

11. MSK is a special form of OQPSK.
a) True
b) False
Answer: a
Clarification: Yes, MSK can be thought of as a special form of offset quadrature phase shift keying. The condition is that baseband rectangular pulses are replaced with half sinusoidal pulses.

12. GMSK is a ________ of MSK.
a) Integral
b) Opposite
c) Derivative
d) Similar
Answer: c
Clarification: Gaussian minimum phase shift keying is a simple binary modulation scheme. It is viewed as a derivative of MSK. GMSK considerably reduces the sidelobe levels in the transmitted spectrum.

13. Which of the following holds true for GMSK?
a) Minimum ISI
b) Minimum error rate
c) Good spectral efficiency
d) Variable envelope property
Answer: c
Clarification: GMSK sacrifices the irreducible error rate caused by partial response signalling in exchange for extremely good spectral efficiency and constant envelope properties. And the premodulation Gaussian filtering introduces ISI in the transmitted signal.

14. MSK has complex demodulation and synchronization circuits.
a) True
b) False
Answer: b
Clarification: MSK has simple demodulation and synchronization circuits. It has various other advantages like continuous phase property makes it highly desirable for highly reactive loads. Due to these advantages, MSK is a popular modulation scheme for mobile radio communication.

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Wireless & Mobile Communications Multiple Choice Questions & Answers (MCQs) on “Speech Codecs”.

1. The choice of speech coder does not depend on cell size used.
a) True
b) False
Answer: b
Clarification: 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: a
Clarification: 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. The type of modulation does not affect the choice of speech codec.
a) True
b) False
Answer: b
Clarification: 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: c
Clarification: 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: a
Clarification: 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: b
Clarification: 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.
a) True
b) False
Answer: b
Clarification: 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: d
Clarification: 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: a
Clarification: 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: d
Clarification: 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: a
Clarification: 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: b
Clarification: 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.

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Wireless & Mobile Communications Multiple Choice Questions on “Global System for Mobile (GSM)”.

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
Clarification: 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
Clarification: 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 sets the standards of GSM?
a) ITU
b) AT & T
c) ETSI
d) USDC
Answer: c
Clarification: The setting of standards for GSM is under the aegis of the European Technical Standards Institute (ETSI). GSM task was to specify a 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Answer: b
Clarification: NSS (Network and Switching Subsystem) manages the switching functions of the system. It allows the MSCs to communicate with other networks such as PSTN and ISDN.

11. __________ supports the operation and maintenance of GSM.
a) BSS
b) NSS
c) OSS
d) MSC
Answer: c
Clarification: The OSS (Operation Support Subsystem) supports the operation and maintenance of GSM. It allows system engineers to monitor, diagnose, and troubleshoot all aspects of GSM.

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Wireless & Mobile Communications Multiple Choice Questions on “3G W-CDMA (UMTS)”.

1. What is the full form of UMTS?
a) Universal Mobile Telephone System
b) Ubiquitous Mobile Telephone System
c) Ubiquitous Mobile Telemetry System
d) Universal Machine Telemedicine System
Answer: a
Clarification: UMTS (Universal Mobile Telephone System) is a visionary air interface standard that was introduced in 1996. European carriers, manufacturers, and government regulators collectively developed the early version of UMTS as an open air interface standard for third generation wireless telecommunication.

3. UMTS use which multiple access technique?
a) CDMA
b) TDMA
c) FDMA
d) SDMA
Answer: a
Clarification: Although UMTS is designed to operate on evolved GSM core networks, it uses code division multiple access (CDMA) for its air interface. The majority of the 3G systems in operation employ CDMA, while the rest use TDMA. CDMA allows various users to share a channel at the same time, while TDMA allows users to share the same channel by chopping it into different time slots.

3. UMTS does not has backward compatibility with ____________
a) GSM
b) IS-136
c) IS-95
d) GPRS
Answer: c
Clarification: UMTS assures backward compatibility with the second generation GSM, IS-136 and PDC TDMA technologies. It is also compatible with all 2.5G TDMA techniques like GPRS and EDGE. But it does not provide compatibility to CDMA technologies of 2G and 2.5 G. IS-95 is a CDMA standard of 2G.

4. UMTS is also known as _____________
a) IS-95
b) GPRS
c) CdmaOne
d) W-CDMA
Answer: d
Clarification: UMTS uses Wideband CDMA (W-CDMA) to carry the radio transmissions. Therefore, it is also referred as W-CDMA. W-CDMA offers greater spectral efficiency and bandwidth to mobile network operators.

5. What is the chip rate of W-CDMA?
a) 1.2288 Mcps
b) 3.84 Mcps
c) 270.833 Ksps
d) 100 Mcps
Answer: b
Clarification: W-CDMA uses a chip rate of 3.84 Mcps. Chip rate is the product of symbol rate and spreading factor. If the symbol rate is 960 Kbps and spreading factor is 4 for W-CDMA, then the chip rate is 3.84 Mcps. The chip rate for Cdma2000 and GSM are 1.2288 Mcps and 27.0833 Ksps respectively.

6. W-CDMA works in FDD mode only.
a) True
b) False
Answer: b
Clarification: W-CDMA works in both FDD and TDD mode. W-CDMA developed for wide area cellular coverage uses FDD. And TDD is used by W-CDMA for indoor cordless type applications.

7. How much packet data rate per user is supported by W-CDMA if the user is stationary?
a) 2.048 Kbps
b) 100 Mbps
c) 2.048 Mbps
d) 1 Gbps
Answer: c
Clarification: If the user is stationary, W-CDMA supports packet data rates upto 2.048Mbps per user. Thus, it allows high quality data, multimedia, streaming audio video and broadcast type services to consumers. Future version of WCDMA will support stationary user data rates in excess of 8Mbps.

8. What is the minimum spectrum allocation required by W-CDMA?
a) 5 MHz
b) 20MHz
c) 1.25 MHz
d) 200 KHz
Answer: a
Clarification: W-CDMA/UMTS requires a minimum spectrum allocation of 5 MHz. Using this bandwidth, it has the capacity to carry over 100 simultaneous voice calls. It is able to carry data at speeds up to 2 Mbps in its original format. 20 MHz is the bandwidth defined for LTE. CdmaOne uses a bandwidth of 1.25 MHz. GSM’s bandwidth is 200 KHz.

9. W-CDMA requires a complete change of RF equipment at each base station.
a) True
b) False
Answer: a
Clarification: W-CDMA is designed to provide backward compatibility and interoperability for all GSM, IS-136/PDC, GPRS and EDGE equipment. But due to a wider air interface bandwidth of W-CDMA, it requires a complete change of RF-equipment at each base station.

10. How much increase in spectral efficiency is provided by W-CDMA in comparison to GSM?
a) Two times
b) Three times
c) No increase
d) Six times
Answer: d
Clarification: W-CDMA can provide at least six times an increase in spectral efficiency over GSM at system level. Such a wider bandwidth is chosen to higher data rates as low as 8 kbps to as high as 2 Mbps on a single 5 MHz W-CDMA radio channel.

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Wireless & Mobile Communications Multiple Choice Questions on “Indoor Propagation Models”.

1. The variability of the environment is slower for a smaller range of T-R separation distances in indoor models.
a) True
b) False
Answer: b
Clarification: The indoor radio channel differs from the traditional mobile radio channel. The distances covered are much smaller and variability of the environment is much greater for a much smaller range of T-R separation distances.

2. Propagation within building is not influenced by _________
a) Layout of the building
b) Construction materials
c) Building type
d) Trees outside the building
Answer: d
Clarification: It has been observed that propagation within buildings is strongly influenced by specific features. These features are layout of the building, the construction materials, and the building type.

3. Smaller propagation distances make it more difficult to insure far-field radiation for all receiver location and types of antenna.
a) True
b) False
Answer: a
Clarification: Smaller propagation distances make it more difficult to insure far field radiation for all receiver location and types of antenna. The condition is very variable for smaller propagation distances.

4. What is hard partition?
a) Partition as part of the building
b) Partition that can be moved
c) Partition not touching ceiling
d) Partition between different floors
Answer: a
Clarification: Partitions that are formed as part of the building structure are called hard partitions. Partitions vary widely in their electrical and physical characteristics. Thus, it makes difficult in applying general models to specific indoor installation.

5. Partitions that can be moved are called _______
a) Soft partitions
b) Hard partitions
c) Disk partition
d) Dynamic partition
Answer: a
Clarification: Partitions that may be moved are called soft partitions. They do not span to the ceiling. Office buildings often have large open areas which are constructed by using moveable office partitions. Thus, space can be reconfigured easily.

6. Losses between the floors of the building can be determined using ________
a) Internal dimensions
b) Material used to create antenna
c) External dimension
d) Line of sight path
Answer: c
Clarification: The losses between floors of a building are determined by external dimensions as well as materials of the building. It is also determined using type of construction used to create the floors and external surroundings.

7. Technique of drawing a single ray between the transmitter and receiver is called ______
a) Secondary ray tracing
b) Primary ray tracing
c) Line of sight
d) Straight line tracing
Answer: b
Clarification: PAF (path attenuation factor) represents a specific obstruction encountered by a ray drawn between the transmitter and receiver in 3-D. This technique of drawing a single ray between transmitter and receiver is called primary ray tracing.

8. ________ is a process of converting plain text into cipher text.
a) Authentication
b) Decryption
c) Encryption
d) Compression
Answer: c
Clarification: Encryption is the most effective way to achieve data security. It is the process of encoding a message in such a way that only authorized parties can access it. Encryption does not itself prevent interference but denies the intelligible content to a would-be interceptor.

9. _______ reduces the cell size to increase capacity.
a) Intelligent cell approach
b) Microcell approach
c) Top down approach
d) Bottom up approach
Answer: b
Clarification: The microcell zone concept is used in cellular systems to specifically increase the capacity and coverage in cellular systems. A handoff is not required at MSC when the mobile travels between zones within a cell or when a mobile travels from one zone to another within the cell.

10. _________ configuration describes a desktop in an office.
a) Mobile and wired
b) Fixed and wired
c) Fixed and wireless
d) Mobile and wireless
Answer: b
Clarification: Fixed and wired configuration describes a desktop in an office. The device use fixed networks for performance reasons. Neither weight nor power consumption of the devices allows for mobile usage.

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