250+ TOP MCQs on Diffraction and Answers

Wireless & Mobile Communications Multiple Choice Questions on “Diffraction”.

1. Diffraction occurs when radio path between Tx. And Rx. Is obstructed by ____________
a) Surface having sharp irregularities
b) Smooth irregularities
c) Rough surface
d) All types of surfaces
Answer: a
Clarification: Diffraction occurs when radio path between transmitter and receiver is obstructed by a surface that has sharp irregularities (edges). The secondary waves resulting from the obstructing surface are present throughout the space and even behind the obstacle.

2. At high frequencies, diffraction does not depends on ___________
a) Geometry of the object
b) Distance between Tx and Rx
c) Amplitude of incident wave
d) Polarization of incident wave
Answer: b
Clarification: At high frequency, diffraction depends on the geometry of the object, as well as the amplitude, phase, and polarization of the incident wave at the point of diffraction. It gives rise to a bending of waves even when line of sight does not exist between transmitter and receiver.

3. Diffraction allows radio signals to propagate around ________
a) Continuous surface
b) Smooth surface
c) Curved surface of Earth
d) Does not allow propagation
Answer: c
Clarification: Diffraction allows radio signals to propagate around the curved surface of the Earth. Signals can propagate beyond the horizon and to propagate behind obstruction. It is the slight bending of light as it passes around the edge of an object.

4. Which principle explains the phenomenon of diffraction?
a) Principle of Simultaneity
b) Pascal’s Principle
c) Archimedes’ Principle
d) Huygen’s principle
Answer: d
Clarification: The phenomenon of diffraction can be explained by Huygen’s principle. It states that all points on a wavefront can be considered as point sources for the production of secondary wavelets. And these wavelets combine to produce a new wavefront in direction of propagation.

5. Diffraction is caused by propagation of secondary wavelets into _______
a) Bright region
b) Shadowed region
c) Smooth region
d) Large region
Answer: b
Clarification: Diffraction is caused due to propagation of secondary wavelets into a shadowed region. The field strength in the shadowed region is the vector sum of the electric field components of all the secondary wavelets in the space around the obstacle.

6. Difference between the direct path and the diffracted path is called _______
a) Average length
b) Radio path length
c) Excess path length
d) Wavelength
Answer: c
Clarification: Excess path length denoted by ∆, is the difference between the direct path and the diffracted path. It is calculated with the help of Fresnel zone geometry.

7. The phase difference between a direct line of sight path and diffracted path is function of _______
a) Height and position of obstruction
b) Only height
c) Operating frequency
d) Polarization
Answer: a
Clarification: The phase difference between a direct line of sight path and diffracted path is a function of height and position of the diffraction. It is also a function of transmitter and receiver location.

8. Which of the following explains the concept of diffraction loss?
a) Principle of Simultaneity
b) Pascal’s Principle
c) Fresnel zone
d) Archimedes’ Principle
Answer: c
Clarification: The concept of diffraction loss is a function of the path difference around an obstruction. It can be explained by Fresnel zones. Fresnel zones represent successive regions where secondary waves have a path length from Tx to Rx which are nλ/2 greater than total path length.

9. In mobile communication system, diffraction loss occurs due to ______
a) Dielectric medium
b) Obstruction
c) Electric field
d) Operating frequency
Answer: b
Clarification: Diffraction loss occurs from the blockage of secondary waves such that only a portion of the energy is diffracted around an obstacle. An obstruction causes a blockage of energy from source some of the Fresnel zones, allowing only some of the transmitted energy to reach the receiver.

10. For predicting the field strength in a given service area, it is essential to estimate ______
a) Polarization
b) Magnetic field
c) Height of transmitter
d) Signal attenuation
Answer: d
Clarification: Estimating the signal attenuation caused by diffraction of radio waves over hills and buildings is essential in predicting the field strength in a given service area. In practice, prediction is a process of theoretical approximation modified by necessary empirical corrections.

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250+ TOP MCQs on Digital Modulation and Answers

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

1. Modern mobile communication systems use analog modulation techniques.
a) True
b) False

Answer: b
Clarification: Modern mobile communication systems use digital modulation techniques. Advancements in VLSI and digital signal processing technology have made digital modulation more cost effective than analog transmission systems.

2. Which of the following is not an advantage of digital modulation?
a) Greater noise immunity
b) Greater security
c) Easier multiplexing
d) Less bandwidth requirement

Answer: d
Clarification: Digital modulation offer many advantages over analog modulation. Some advantages include greater noise immunity and robustness. They provide easier multiplexing of various forms of information and greater security.

3. A desirable modulation scheme provides _________ bit error rates at __________ received signal to noise ratios.
a) Low, low
b) Low, high
c) High, high
d) High, low

Answer: a
Clarification: A desirable modulation scheme provides low bit error rates at low received signal to noise ratios. They perform well in multipath and fading conditions, occupies a minimum bandwidth and is easy and cost effective to implement.

4. The performance of modulation scheme is not measured in terms of __________
a) Power efficiency
b) Bandwidth efficiency
c) Cost and complexity
d) Transmitted power

Answer: d
Clarification: The performance of modulation scheme is often measured in terms of its power efficiency and bandwidth efficiency. Other factors also affect the choice of modulation scheme, such as cost and complexity of the subscriber receiver and modulation which is simple to detect.

5. In digital communication system, in order to increase noise immunity, it is necessary to increase _________
a) Signal power
b) Signal amplitude
c) Signal frequency
d) Signal magnitude

Answer: a
Clarification: In digital communication system, in order to increase noise immunity, it is necessary to increase signal power. However, the amount by which the signal power should be increased to obtain a certain level of fidelity depends on the particular type of modulation employed.

6. Which of the following is the ratio of signal energy per bit to noise power spectral density?
a) Bandwidth efficiency
b) Spectral density
c) Power efficiency
d) Power density

Answer: c
Clarification: Power efficiency is often expressed as the ratio of signal energy per bit to noise power spectral density required at the receiver input for a certain probability of error. Power efficiency is a measure of how favourably the trade-off between fidelity and signal power is made.

7. Increasing the data rate implies the increase in pulse width of digital symbol.
a) True
b) False

Answer: b
Clarification: There is an unavoidable relationship between data rate and bandwidth occupancy. Increasing the data rate implies decreasing the pulse width of a digital symbol, which increases the bandwidth of the signal.

8. Which of the following is the ratio of the throughput data rate per Hertz?
a) Bandwidth efficiency
b) Spectral density
c) Power efficiency
d) Power density

Answer: a
Clarification: Bandwidth efficiency reflects how efficiently the allocated bandwidth is utilized. It is defined as the ratio of throughput data rate per Hertz in a given bandwidth. It describes the ability of a modulation scheme to accommodate data within a limited bandwidth.

9. Which of the following is defined as the range of frequencies over which the signal has a non zero power spectral density?
a) Null to null bandwidth
b) Half power bandwidth
c) 3 dB bandwidth
d) Absolute bandwidth

Answer: d
Clarification: The absolute bandwidth is defined as the range of frequencies over which the signal has a non-zero power spectral density. For symbols represented as rectangular baseband pulses, the PSD profile extends over an infinite range of frequencies, and has an absolute bandwidth of infinity.

10. _______ is equal to width of main spectral lobe.
a) Null to null bandwidth
b) Half power bandwidth
c) 3 dB bandwidth
d) Absolute bandwidth

Answer: a
Clarification: Null to null bandwidth is a simpler and more widely accepted measure of bandwidth. It is equal to the width of main spectral lobe.

11. Half power bandwidth is also called ______
a) Absolute bandwidth
b) Null to null bandwidth
c) 3 dB bandwidth
d) Zero dB bandwidth

Answer: c
Clarification: Half power bandwidth is also called the 3 dB bandwidth. It is defined as the interval between frequencies at which the PSD has dropped to half power, or 3 dB below the peak value.

250+ TOP MCQs on Frequency Domain Coding of Speech and Answers

Wireless & Mobile Communications Multiple Choice Questions & Answers on “Frequency Domain Coding of Speech”.

1. Frequency domain coders divides the speech signal into __________
a) A set of frequency components
b) A set of different amplitudes
c) A set of time delays
d) A set of phase components
Answer: a
Clarification: In the class of frequency domain coders, the speech signal is divided into a set of frequency components. Each frequency component is quantized and encoded separately.

2. In frequency domain coding of speech, the number of bits used to encode each frequency component is constant.
a) True
b) False
Answer: b
Clarification: Frequency domain coders have an advantage that number of bits used to encode each frequency component can be dynamically varied. They can also be shared among different bands.

3. Quantization is a ________ process.
a) Linear
b) Direct
c) Non-linear
d) Indirect
Answer: c
Clarification: Quantization is a non-linear process. It produces distortion products that are typically broad in spectrum.

4. Sub band coding codes the short time transform of a windowed signal.
a) True
b) False
Answer: b
Clarification: It is function of block transform coding. However, a sub band coder divides the speech signal into many smaller sub bands and encodes each sub band separately according to some perceptual criterion.

5. Which of the following is one of the most frequently used transform in speech coding?
a) Fourier transform
b) Wavelet transform
c) Shearlet transform
d) Discrete cosine transform
Answer: d
Clarification: DCT (discrete cosine transform) is one of the most attractive and frequently used transforms for speech coding. Fast algorithms developed for computing the DCT in a computationally efficient manner are used.

6. What does ATC stands for in speech coders?
a) Automatic transform code
b) Air traffic controller
c) Active thermal convection
d) Adaptive transform coding
Answer: d
Clarification: In speech coding, ATC stands for adaptive transform coding. It is form of frequency domain coder that encodes the speech at bit rates in the range of 9.6 kbps and 20 kbps.

7. Waveform coders and Vocoders are the types of ____________
a) Speech coders
b) Modulation technique
c) Frequency translation methods
d) Channel allocation for transmission
Answer: a
Clarification: Speech coders can be classified into waveform coders and Vocoders. Waveform coders convert the analog signal into digital signal. Vocoders exploit the special properties of speech signal to reduce the bit rate.

8. The type of frequency domain coding that divides the speech signal into sub bands is _____
a) Waveform coding
b) Vocoders
c) Block transform coding
d) Sub-band coding
Answer: d
Clarification: Sub band coding (SBC) is a method where the speech signal is subdivided into several frequency bands and each band is digitally encoded separately. The audible frequency spectrum 20Hz-20 KHz is divided into frequency sub-bands using a bank of finite impulse response (FIR) filter and output of each filter is sampled and encoded.

9. Speech coders are categorized on the basis of __________
a) Signal compression techniques
b) Frequency of signal
c) Bandwidth of the signal
d) Phase of the signal
Answer: a
Clarification: Speech coders are categorised on the basis of signal compression techniques. Speech coding is an art of compressing and then encoding speech signals.

10. The speech coding technique that is dependent on the prior knowledge of the signal is __________
a) Waveform coders
b) Vocoders
c) Sub band coding
d) Block transform
Answer: b
Clarification: Vocoders are dependent on the prior knowledge of the signals. They capture the characteristic elements of audio signal and then uses this characteristic signal to affect other audio signals.

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250+ TOP MCQs on Personal Communication Services/Networks (PCS/PCNs) and Answers

Wireless & Mobile Communications online test on “Personal Communication Services/Networks (PCS/PCNs)”.

1. PCS/PCN provides only wired communication.
a) True
b) False
Answer: b
Clarification: The objective of personal communication systems (PCS) or personal communication networks (PCNs) is to provide ubiquitous wireless communications coverage, enabling users to access telephone network and Internet.

2. Concept of PCS/PCN is based on _________
a) Advanced intelligent network
b) Artificial intelligent network
c) CDPD
d) SS7
Answer: a
Clarification: The concept of PCS/PCN is based on an advanced intelligent network (AIN). The mobile and fixed networks will be integrated to provide universal access to the network and its databases.

3. AIN has different telephone numbers for both wireline and wireless services.
a) True
b) False
Answer: b
Clarification: AIN (advanced intelligent network) allow its users to have a single telephone number to be used for both wireless and wireline services.

4. Circuit switching has more advantage than packet switching for PCS.
a) True
b) False
Answer: b
Clarification: Packet switching technology will have more advantages for PCS/PCN than circuit switching. PCS is required to serve a wide range of services including voice, data, e-mail and digital video.

5. ______ is used for transmission of packets in the cellular switched architecture.
a) Packet switching techniques
b) Circuit switching techniques
c) Packet and circuit switched technique
d) Datagram technique
Answer: a
Clarification: Packet switching techniques are used for transmission of packets in the cellular switched architecture. Packet switching is attractive for wireless networks because the addresses and other information in packet headers make it possible for dispersed network elements to respond to a mobile user without intervention central controllers.

6. Which of the following has the function to accept information from PSTN?
a) TIU
b) WIU
c) BIU
d) CIU
Answer: a
Clarification: The function of TIU (Trunk interface unit) is to accept information from the PSTN. TIU acts as the physical layer and transforms the standard format of the PSTN into the wireless access physical layer.

7. PTI is the address of ____________
a) TIU
b) WIU
c) BIU
d) CIU
Answer: a
Clarification: PTI (Permanent terminal identifier) is the address of TIU. It is the address from where he call is originated.

8. _______ is the information contained in the packet header of TIU.
a) PTI
b) VCI
c) PRMA
d) CDPD
Answer: b
Clarification: VCI (virtual circuit identifier) is the information contained in the packet header of TIU. It is used to identify the route through which the transmission will take place.

9. Transmission protocol, PRMA stands for _________
a) Pocket reservation multiple access
b) Packet register multiple access
c) Pocket register multiple access
d) Packet reservation multiple access
Answer: d
Clarification: Transmission protocol, PRMA stands for packet reservation multiple access. PRMA is a time division multiplex (TDM) based multiple access protocol that allows a group of spatially dispersed terminals to transmit packet voice and low bit rate data over a common channel.

10. UMTS stands for ____________
a) Universal mobile telecommunication system
b) Universal mobile telephone system
c) United multiplex telecommunication system
d) Universal mobile telecommunication system
Answer: d
Clarification: UMTS stands for Universal mobile telecommunication system. It is a system that is capable of providing a variety of mobile services to a wide range of global mobile communication standards.

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250+ TOP MCQs on 4G Network Architecture and Answers

Wireless & Mobile Communications Multiple Choice Questions on “4G Network Architecture”.

1. Which type of cell provides the best level of service for average subscribers?
a) Acceptance cell
b) Barred cell
c) Reserved cell
d) Suitable cell
Answer: d
Clarification: A suitable cell is a cell on which the UE may camp on to obtain normal service. The UE shall have a valid USIM and such a cell shall fulfil all the following requirements. It provides the best level of service for average subscribers.

2. With the normal cyclic prefix, how many symbols are contained in 1 frame?
a) 7
b) 140
c) 12
d) 40
Answer: b
Clarification: There are two different type of Cyclic Prefix. One is normal Cyclic Prefix and the other is ‘Extended Cyclic Prefix’ which is longer than the Normal Cyclic Prefix. Normal cyclic prefix contains 140 symbols in 1 frame.

3. What is the PBCH scrambled with?
a) Current frame number
b) Physical cell ID
c) UE’s CRNTI
d) Not scrambled
Answer: b
Clarification: The PBCH is scrambled prior to modulation with a cell-specific sequence that depends on the cells’ identity. In contrast to the synchronization signals, the PBCH is transmitted on the 72 reserved subcarriers, which are QPSK-modulated.

4. What is the length of the shortest possible PDCCH in bits?
a) 144
b) 288
c) 72
d) 576
Answer: c
Clarification: PDCCH is a physical channel that carries downlink control information (DCI). Shortest possible PDCCH is 72 bits.

5. What is the average uploading speed of 4G LTE network?
a) 1-3 Gbps
b) 2-5 Gbps
c) 1-3 Mbps
d) 2-5 Mbps
Answer: d
Clarification: Verizon 4G LTE wireless broadband is 10 times faster than 3G able to handle download speeds between 5 and 12 Mbps (Megabits per second) and upload speeds between 2 and 5 Mbps.

6. Which of the following is not a part of the characteristic of 4G network?
a) Multirate management
b) Fully converged services
c) Software dependency
d) Diverse user devices
Answer: a
Clarification: 4G is the fourth generation of broadband cellular network technology, succeeding 3G. Its characteristics include fully converged services, software dependency and diverse user devices.

7. What does SGSN stands for?
a) Serving GPRS Support Node
b) Supporting GGSN Support Node
c) Supporting GPRS Support Node
d) Supporting Gateway Support Node
Answer: a
Clarification: The Serving GPRS Support Node (SGSN) is a main component of the GPRS network, which handles all packet switched data within the network, e.g. the mobility management and authentication of the users. The SGSN performs the same functions as the MSC for voice traffic.

8. What location management feature is supported by 4G?
a) Concatenated Location Registration
b) Concurrent Location Register
c) Concatenated Management
d) Collated Location Registration
Answer: a
Clarification: 4G supports concatenated location registration. Concatenated location registration reports to the network that they are concatenated to a common object.

9. In 2007 ____________ announced its plan to transmit its network to 4G standard LTE with joint efforts of Vodafone group.
a) Verizon Wireless
b) AirTouch
c) Netflix
d) V Cast
Answer: a
Clarification: In 2007, Verizon announced plans to develop and deploy its fourth generation mobile broadband network using LTE, the technology developed within the Third Generation Partnership Project (3GPP) standards organization.

10. Hybrid ARQ is part of the ____________ layer.
a) PDCP
b) RLC
c) MAC
d) PHY
Answer: c
Clarification: Hybrid automatic repeat request (hybrid ARQ or HARQ) is a combination of high-rate forward error-correcting coding and ARQ error-control. It is part of the MAC layer.

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250+ TOP MCQs on Second Generation (2G) Cellular Networks and Answers

Wireless & Mobile Communications Interview Questions and Answers on “Second Generation (2G) Cellular Networks”.

1. Which of the following multiple access techniques are used by second generation cellular systems?
a) FDMA/FDD and TDMA/FDD
b) TDMA/FDD and CDMA/FDD
c) FDMA/FDD and CDMA/FDD
d) FDMA/FDD only
Answer: b
Clarification: First generation cellular system used FDMA/FDD techniques. Second generation standards uses TDMA/FDD and CDMA/FDD multiple access techniques. 2G networks are digital.

2. Which one is not a TDMA standard of second generation networks?
a) GSM
b) IS-136
c) AMPS
d) PDC
Answer: c
Clarification: GSM (Global System Mobile), IS-136 (Interim Standard 136) and PDC (Pacific Digital Cellular) are the three most popular TDMA standards of second generation. AMPS is a first generation standard.

3. Which of the following is a CDMA standard of second generation network?
a) IS-95
b) IS-136
c) ETACS
d) EDGE
Answer: a
Clarification: Interim Standard 95 (IS-95) is the most popular CDMA standard of second generation networks. IS-136 is a TDMA standard of 2G. EDGE is a standard of 2.5G and ETACS is a 1G standard.

4. Popular 2G CDMA standard IS-95 is also known as ______________
a) CdmaOne
b) CdmaTwo
c) IS-136
d) IS-95B
Answer: a
Clarification: The popular 2G CDMA standard, Interim Standard (IS-95) is also known as CdmaOne. The 2.5G CDMA standard, IS-95B is called CdmaTwo. And IS-136 is a TDMA standard for 2G.

5. How many users or voice channels are supported for each 200 KHz channel in GSM?
a) Eight
b) Three
c) Sixty four
d) Twelve
Answer: a
Clarification: GSM is a circuit switched system that divides each 200 KHz channel into eight 25 KHz time slots, i.e. each radio channel is divided into eight voice channels.

6. How many voice channels are supported for each 30 KHz radio channel in IS-136?
a) Eight
b) Thirty
c) Three
d) Sixteen
Answer: c
Clarification: Interim Standard 136 (IS-136) was popularly known as North American Digital Cellular (NADC) system. It divides each 30 KHz radio channel into three time slots, each of 10 KHz.

7. How many users are supported in IS-95 for each 1.25 MHz?
a) Eight
b) Sixty four
c) Sixteen
d) Twenty five
Answer: b
Clarification: IS-95 supports upto 64 users which are orthogonally coded and simultaneously transmitted on each 1.25 MHz. The services of IS-95 standard are short messaging service, slotted paging, over-the-air activation, enhanced mobile station identities etc.

8. Which modulation technique is used by GSM?
a) GMSK
b) BPSK
c) QPSK
d) GFSK
Answer: a
Clarification: GSM uses a form of modulation known as GMSK (Gaussian Minimum Shift Keying). It is a form of modulation with no phase discontinuities and provides data transmission with efficient spectrum usage.

9. IS-95 uses which modulation technique?
a) GMSK
b) BPSK
c) QAM
d) AFSK
Answer: b
Clarification: IS- 95 uses BPSK (Binary Phase Shift Keying) with quadrature spreading. It is regarded as one of the most robust digital modulation technique and is used for long distance wireless communication.

10. IS-136 uses which modulation technique?
a) π/4 DQPSK
b) BPSK
c) GMSK
d) AFSK
Answer: a
Clarification: IS-136 uses π/4 DQPSK modulation technique. This technique allows a bit rate of 48.6 Kbit/s with 30 KHz channel spacing which gives a bandwidth efficiency of 1.62 bit/s/Hz.

11. Which is one of the disadvantages of 2G standards?
a) Short Messaging Service (SMS)
b) Digital modulation
c) Limited capacity
d) Limited Internet Browsing
Answer: d
Clarification: 2G technologies use circuit switched data modems that limits data users to a single circuit switched voice channel. The advantages of 2G network are that they are digital in nature and supports SMS service.

12. GSM (Global System for Mobile) was earlier also known as _____________
a) Group System Mobile
b) Global Special Meaning
c) Group Special Mobile
d) Global Special Mobile
Answer: c
Clarification: GSM was earlier known as Group Special Mobile. As it became more global, the meaning of acronym was changed to Global System for Mobile.

13. 2G CDMA standard, IS-95, was proposed by which company?
a) Nippon Telephone and Telegraph (NTT)
b) Qualcomm
c) Bellcore and Motorola
d) AT&T Bell Laboratories
Answer: b
Clarification: IS-95 was proposed by Qualcomm in early 1990s. Later it was adopted as a standard by Telecommunications Industry Association in TIA/EIA/IS-95 release published in 1995.

14. Which one of the following 2G standard is used in Japan?
a) IS-136
b) GSM
c) PDC
d) AMPS
Answer: c
Clarification: PDC (Personal Digital Cellular) was standardized by Japanese Ministry of Posts and Telecommunication in 1991. It is similar to IS- 136, but with 25 KHz voice channels to be compatible with the Japanese analog channels.

15. The 2G GSM technology uses a carrier separation of _____________
a) 1.25 MHz
b) 200 KHz
c) 30 KHz
d) 300 KHz
Answer: b
Clarification: The Global System for Mobile (GSM) uses a carrier separation of 200 KHz, each channel supporting upto eight users.

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