Lect 5

Question 1

Radio band

Answer 1

Small contiguous section of the frequency spectrum.

Question 2

Multiplexing/multiple access

Answer 2

Method of sharing a channel to many users.

Question 3

Channel

Answer 3

System resource allocated to a mobile user enabling the user to establish communication with the network.

Question 4

Classes of multiple access techniques

Answer 4

1. Reservation based
2. Random

Question 5

Types of reservation based

Answer 5

1. FDMA
2. TDMA
3. CDMA

Question 6

Types of Random

Answer 6

1. CSMA
2. ALOHA
3. ISMA

Question 7

Freq Division Duplexing channels

Answer 7

1. Uplink
2. Downlink

Question 8

What’s a Narroband system

Answer 8

System relating bandwidth of a single channel to the entire bandwidth.

Question 9

Narrowband FDMA

Answer 9

User is assigned a channel which isn’t shared by others.

Question 10

Narrowband TDMA

Answer 10

Users share a channel but allocates unique timeslot to each user in a cyclical fashion.

Question 11

CDMA

Answer 11

Allows all transmiters to access the channel at the same time.

Question 12

Guard band

Answer 12

Narrow frequency band between adjacent frequency channels to avoid interference employed in FDMA systems.

Question 13

Features of FDMA

Answer 13

1. BS and MS transmit simultaneously and continuously.
2. Uses duplexers
3. Implemented in narrowband systems.
4. Less complex.
5. Higher cell site system costs.

Question 14

FDMA disadvantages

Answer 14

1. Low capacity in terms of channel availability.
   2.Frequency wastage due to guardbands.

Question 15

Number of channels in FDMA syst

Answer 15

N = (Bt - 2Bguard)/Bc

Where Bt = total bandwidth
Bc = channel bandwidth
Bguard = guard bandwidth

Question 16

TDMA Co-channel interference

Answer 16

When 2 users overlap in time.

Question 17

Components of TDMA frame

Answer 17

1. Preamble
2. Info message
3. Tail bits

Question 18

TDMA

Answer 18

whole channel available to users but each user uses it for a fraction of time.

Question 19

TDMA features

Answer 19

1. Transmission is not continuous
2. Handoff is simpler
3. Duplexers not required
4. High sync overhead is required
5. Larger overheads

Question 20

TDMA advantages

Answer 20

1. Single freq can be used by many users.
2. No interference due to time gap.
3. No guard band needed so all freqs are used.

Question 21

TDMA disavantages

Answer 21

1.User needs precise synchronization.
2. Reduced data rate.

Question 22

Efficiency of TDMA

Answer 22

Measure of percentage of transmitted data that contains info as opposed to providing overhead.

Question 23

Frame efficiency

Answer 23

Percentage of bits per frame which contain transmitted data.

Question 24

No. Of overhead bits per frame

Answer 24

Boh = NrBr + NrBg + NtBp + NtBg

Where,
Nr = no. of reference burst/frame
Nt = no. of traffic burst/frame
Br = no. of bits/reference burst
Bp = no. of bits/preamble
Bg = no. of bits/guard time

Question 25

Total no. of bits per frame

Answer 25

bT = Tf*R Tf = frame duratiion R = channel biy rate

Question 26

Efficiency of TDMA frame

Answer 26

= (1 - bOH/bT) * 100%

Question 27

No. Of channels in TDMA system

Answer 27

N = m(Bt - 2Bguard)/Bc m = max users per channel

Question 28

CDMA

Answer 28

The narrowband message signal is multiplied by very large bandwidth signal called spreading signal.

Question 29

Orthogonality

Answer 29

Property that allows multiple info to be transmitted over a common channel with succesful detection.

Question 30

Advantages of CDMA

Answer 30

1. Improvement in network capacity 2. Attractive for 3G high speed internet use 3. Easy for a terminal to communicate with two base stations

Question 31

Disadvantages

Answer 31

1. Can't offer roaming. 2. Perfomance degrades with increase in number of users. 3.CDMA network is not matured.

Question 32

Application of CDMA

Answer 32

1. Adopted by 3G cellular telecom systems. 2. Merged with GSM to give high speed 4G internet services.

Question 33

Spectral efficiency depends on.

Answer 33

1. Channel spacing. 2. Freq reuse factor. 3. Cell area. 4. Modulation techniques. 5. Multiple access techniques.

Question 34

Spectral efficiency of modulation

Answer 34

= 1/(Bc*N*Ac) Channels/MHz/Km sqr Bc = channel spacing in MHz N = freq reuse factor Ac = area covered by a cell in km sqr

Question 35

Multiple access Spectral efficiency

Answer 35

Ratio of total time or freq dedicated for traffic transmission to total time or freq available to the system.

Question 36

FDMA multiple access spectr effi.

Answer 36

na = (Bc*Nt)/Bw na = multiple access spec. effi. Nt = total no. of traffic channels Bc = channel spacing Bw = sys bandwidth

Question 37

TDMA spec efficiency

Answer 37

na = t*Mt/Tf t = duration of timeslot that carries data Tf = frame duration Mt = no. of time slots/frame

Question 38

Overall Spectral efficiency

Answer 38

n = nm * na