FDMA - IDMA - CDMA (Code Division Multiple Access) Problems, Exercises of Mobile Communication Systems

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prepared by

Dr : Adly Dr:Reham

Eng : Ibrahem Garrah

Sec

Mobile Communication

SFE

Notes:

 Channelization is a multiple-access method in which the available bandwidth of a link is shared in time, frequency, or through code, between different stations.

Frequency-Division Multiple Access (FDMA)

 In FDMA, the available bandwidth is divided into frequency bands.  Each station is allocated a band to send its data. Each band is reserved for a specific station, and it belongs to the station all the time.  To prevent station interferences, the allocated bands are separated from one another by small guard bands.  In FDMA, the bandwidth is divided into channels. Time-Division Multiple Access (TDMA)

 In TDMA, the stations share the bandwidth of the channel in time.  Each station is allocated a time slot during which it can send data.  Each station transmits its data in its assigned time slot.  In TDMA, bandwidth is just one channel that is timeshared.  The main problem with TDMA lies in achieving synchronization between the different stations.  Each station needs to know the beginning of its slot and the location of its slot.  This may be difficult because of propagation delays introduced in the system if the stations are spread over a large area.  To compensate for the delays, we can insert guard times.  Synchronization is normally accomplished by having some synchronization bits (normally referred to as preamble bits) at the beginning of each slot. Code-Division Multiple Access

 CDMA differs from FDMA because only one channel occupies the entire bandwidth of the link.  It differs from TDMA because all stations can send data simultaneously; there is no timesharing.  In CDMA, one channel carries all transmissions simultaneously.

Sheet 4

1. The following Figure depicts a simplified scheme for CDMA encoding and decoding. There are seven logical channels, all using DSSS with a spreading code of 7 bits. Assume that all sources are synchronized. If all seven sources transmit a data bit, in the form of a 7-bit sequence, the signals from all sources combine at the receiver so that two positive or two negative values reinforce and a positive and negative value cancel. To decode a given channel, the receiver multiplies the incoming composite signal by the spreading code for that channel, sums the result, and assigns binary 1 for a positive value and binary 0 for a negative value.

a. What are the spreading codes for the seven channels? b. Determine the receiver output measurement for channel 1 and the bit valueassigned. c. Repeat part (b) for channel 2

Solution:

a. • Channel 0: ( 1 , 1 , 1 ,- 1 ,- 1 , 1 ,- 1 )

• Channel 1 : (- 1 , 1 , 1 , 1 ,- 1 ,- 1 , 1 )
• Channel 2 : ( 1 ,- 1 , 1 , 1 , 1 ,- 1 ,- 1 )•
• Channel 3 : (- 1 , 1 ,- 1 , 1 , 1 , 1 ,- 1 )
• Channel 4 : (- 1 ,- 1 , 1 ,- 1 , 1 , 1 , 1 )
• Channel 5 : ( 1 ,- 1 ,- 1 , 1 ,- 1 , 1 , 1 )
• Channel 6 : ( 1 , 1 ,- 1 ,- 1 , 1 ,- 1 , 1 )

b. Multiplying the composite signal with channel 1’s code, we have (- 5 , 1 , 1 ,- 3 ,- 1 , 3 ,- 3 ), the receiver output measurement is the sum of the elements:- 7 , and the bit value is 0. c. Multiplying the composite signal with channel 2’s code, we have ( 5 ,- 1 , 1 ,- 3 , 1 , 3 , 3 ), the receiver output measurement is the sum of the elements: 9 , and the bit value is 1.

2. Find the chips for a network with a. Two stations b. Four stations

Solution:

We can use the rows of W 2 and W 4 in above figure: a. For a two-station network, we have: [+1 +1] and [+1 -1] b. For a four-station network we have: [+1 +1 +1 +1] and [+1 -1 +1 -1] and [+1 +1 -1 -1] and [+1 -1 -1 +1]

3. Create the W 2 and W 4 tables using W 1 = [-1].

Ans. Like question 2 but with reverse sign

4. Check to see if the following set of chips can belong to an orthogonal system. [+1,+1] and [+1, -1]

Ans. Because if I multiply each code with the other it gets 0 and if we

multiply each code by itself it gets N (number of stations)

5. In CDMA system ,how station 3 can detect the data sent by station 2 if both stations 1 and 3 are silent, station 2 sends bit 0 and station 4 sends bit 1,show also the corresponding signals for each station and the signal that is on the common channel.

Ans. Step 1: get warsh table