Wireless & Mobile Communications Questions & Answers – Frequency Division Multiple Access (FDMA)

This set of Wireless & Mobile Communications Multiple Choice Questions & Answers (MCQs) focuses 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
View Answer

Answer: a
Explanation: 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
View Answer

Answer: b
Explanation: 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
View Answer

Answer: c
Explanation: 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.
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4. If the FDMA channel is not in use, it can be used by other users.
a) True
b) False
View Answer

Answer: b
Explanation: 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
View Answer

Answer: b
Explanation: 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
View Answer

Answer: d
Explanation: 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
View Answer

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

Answer: b
Explanation: 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
View Answer

Answer: a
Explanation: 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.
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10. __________ is based on FDMA/FDD.
a) GSM
b) W-CDMA
c) Cordless telephone
d) AMPS
View Answer

Answer: d
Explanation: 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
View Answer

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

Sanfoundry Global Education & Learning Series – Wireless & Mobile Communications.

To practice all areas of Wireless & Mobile Communications, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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