Category: Wireless & Mobile Communications Objective Questions

Wireless & Mobile Communications Multiple Choice Questions on “Long Term Evolution (LTE)”.

1. Which UE category supports 64 QAM on the uplink?
a) Only category 5
b) Only category 4
c) Only category 3
d) Category 3,4 and 5
Answer: a
Clarification: Category information is used to allow the eNB to communicate effectively with all the UEs connected to it. The UE-category defines a combined uplink and downlink capability. Only UE category 5 supports 64 QAM on the uplink.

2. What type of handovers is supported by LTE?
a) Hard handover only
b) Soft handover only
c) Hard and soft handover
d) Hard, soft and softest handover
Answer: a
Clarification: LTE supports only hard handover. It does not receive data from two frequencies at the same time because switching between different carrier frequencies is very fast so soft handover is not required.

3. What is the minimum amount of RF spectrum needed for an FDD LTE radio channel?
a) 1.4 MHz
b) 2.8 MHz
c) 5 MHz
d) 20 MHz
Answer: b
Clarification: In telecommunication, Long-Term Evolution (LTE) is a standard for high-speed wireless communication for mobile devices and data terminals, based on the GSM/EDGE and UMTS/HSPA technologies. The minimum amount of RF spectrum needed for an FDD LTE radio channel is 2.8 MHz.

4. Which organization is responsible for developing LTE standards?
a) UMTS
b) 3GPP
c) 3GPP2
d) ISO
Answer: b
Clarification: The 3rd Generation Partnership Project (3GPP) is a collaboration between groups of telecommunications standards associations, known as the Organizational Partners. LTE (Long Term Evolution) introduced in 3GPP R8, is the access part of the Evolved Packet System (EPS).

5. Which channel indicates the number of symbols used by the PDCCH?
a) PHICH
b) PDCCH
c) PBCH
d) PCFICH
Answer: d
Clarification: PCFIH channel indicates the number of symbols used by the PDCCH. The actual number of OFDM symbols occupied in any given subframe is indicated in the PCFICH (Physical Control Format Indicator Channel), which is located in the first OFDM symbol of each subframe.

6. How often can resources be allocated to the UE?
a) Every symbol
b) Every slot
c) Every subframe
d) Every frame
Answer: c
Clarification: Resources can be located to the UE every subframe. CCE Index is the CCE number at which the control channel data (PDCCH) is allocated. Normally this index changes for each subframe, i.e. even the same PDCCH data (e.g, a PDCCH for the same UE) allocated in each subframe changes subframe by subframe.

7. What is the largest channel bandwidth a UE is required to support in LTE?
a) 10 MHz
b) 20 MHz
c) 1.4 MHz
d) 5 MHz
Answer: b
Clarification: The LTE format was first proposed by NTT DoCoMo of Japan and has been adopted as the international standard. LTE-Advanced accommodates the geographically available spectrum for channels above 20 MHz.

8. In LTE, what is the benefit of PAPR reduction in the uplink?
a) Improved uplink coverage
b) Lower UE power consumption
c) Reduced equalizer complexity
d) Improved uplink coverage, lower UE power consumption and reduced equalizer
Answer: d
Clarification: PAPR is the relation between the maximum power of a sample in a given OFDM transmit symbol divided by the average power of that OFDM symbol. PAPR reduction in the uplink leads to improved uplink coverage, lower UE power consumption and reduced equalizer complexity.

9. Which RLC mode adds the least amount of delay to user traffic?
a) Unacknowledged mode (UM)
b) Acknowledged mode (AM)
c) Low latency mode (LM)
d) Transparent mode (TM)
Answer: d
Clarification: The transparent mode entity in RLC does not add any overhead to the upper layer SDUs. The entity just transmits the SDUs coming from upper layer to MAC.

10. How much bandwidth is required to transmit the primary and secondary synchronization signals?
a) 1.08 MHz
b) 1.4 MHz
c) 930 kHz
d) 20 MHz
Answer: a
Clarification: Cell synchronization is the very first step when UE wants to camp on any cell. 1.08 MHZ is required to transmit the primary and secondary synchronization signals.

.

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.

.

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.

all areas of Wireless & Mobile Communications, .

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.

all areas of Wireless & Mobile Communications for online tests, .