Wireless & Mobile Communications Questions & Answers – Time Division Multiple Access (TDMA)

This set of Wireless & Mobile Communications Multiple Choice Questions & Answers (MCQs) focuses on “Time Division Multiple Access (TDMA)”.

1. TDMA systems transmit in a continuous way.
a) True
b) False
View Answer

Answer: b
Explanation: TDMA systems transmit data in a buffer and burst method. Thus, the transmission for any user is not continuous.

2. Preamble contains __________
a) Address
b) Data
c) Guard bits
d) Trail bits
View Answer

Answer: a
Explanation: TDMA frame is made up of a preamble, an information message and the trail bits. In a TDMA frame, the preamble contains the address and synchronization information that both the base station and the subscribers use to identify each other.

3. __________ are utilized to allow synchronization of the receivers between different slots and frames.
a) Preamble
b) Data
c) Guard bits
d) Trail bits
View Answer

Answer: c
Explanation: Guard times are utilized to allow synchronization of the receivers between different slots and frames. TDMA/FDD systems intentionally induce several time slots of delay between the forward and reverse time slots for a particular user.
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4. Which of the following is not true for TDMA?
a) Single carrier frequency for single user
b) Discontinuous data transmission
c) No requirement of duplexers
d) High transmission rates
View Answer

Answer: a
Explanation: TDMA share a single carrier frequency with several users, where each user makes use of non-overlapping time slots. The number of time slots per frame depends on several factors, such as modulation technique, available bandwidth etc.

5. Because of _______ transmissions in TDMA, the handoff process in __________
a) Continuous, complex
b) Continuous, simple
c) Discontinuous, complex
d) Discontinuous, simple
View Answer

Answer: d
Explanation: Because of discontinuous transmissions in TDMA, the handoff process is much simpler for a subscriber unit, since it is able to listen for other base stations during idle time slots.
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6. __________ synchronization overhead is required in TDMA due to _______ transmission.
a) High, burst
b) High, continuous
c) Low, burst
d) No, burst
View Answer

Answer: a
Explanation: High synchronization overhead is required in TDMA systems because of burst transmissions. TDMA transmissions are slotted, and this requires the receivers to be synchronized for each data burst.

7. TDMA allocates a single time per frame to different users.
a) True
b) False
View Answer

Answer: b
Explanation: TDMA has an advantage that it can allocate different numbers of time slots per frame to different users. Thus, bandwidth can be supplied on demand to different users by concatenating or reassigning time slots based on priority.
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8. ___________ of TDMA system is a measure of the percentage of transmitted data that contains information as opposed to providing overhead for the access scheme.
a) Efficiency
b) Figure of merit
c) Signal to noise ratio
d) Mean
View Answer

Answer: a
Explanation: Efficiency of TDMA system is a measure of the percentage of transmitted data that contains information as opposed to providing overhead for the access scheme. The frame efficiency is the percentage of bits per frame which contain transmitted data.

9. A TDMA system uses 25 MHz for the forward link, which is broken into radio channels of 200 kHz. If 8 speech channels are supported on a single radio channel, how many simultaneous users can be accommodated?
a) 25
b) 200
c) 1600
d) 1000
View Answer

Answer: d
Explanation: For a TDMA system that uses 25 MHz for the forward link, which is broken into radio channels of 200 kHz. If 8 speech channels are supported on a single radio channel, 1000 simultaneous users can be accommodated as N = (25 MHz)/(200 kHz/8).
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10. What is the time duration of a bit if data is transmitted at 270.833 kbps in the channel?
a) 270.833 s
b) 3 μs
c) 3.692 μs
d) 3.692 s
View Answer

Answer: c
Explanation: If data is transmitted at 270.833 kbps in the channel, the time duration of a bit will be 3.692 μs, as Tb = (1/270.833 kbps) = 3.692 μs.

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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