1. (c)CDMA- Manzur Ashraf1 Code Division Multiple Access (CDMA) © Manzur Ashraf

2. (c)CDMA- Manzur Ashraf2 Preface CDMA allows each station to transmit over the entire frequency spectrum all the time. Multiple simultaneous transmissions are separated using coding theory. CDMA also relaxes the assumption that colliding frames are totally garbled. Instead it assumes that multiple signals add linearly. GSM uses a combination of ALOHA, TDM & FDM. CDMA is another method for allocating a wireless channel.

3. (c)CDMA- Manzur Ashraf3 The key to CDMA is to able to extract the desired signal while rejecting everything else as random noise. Each bit-time is subdivided into m short intervals called ‘chips’. Typically there are 64 or 128 chips/bit but we will use 8 chips/bit in the following example. Each station is assigned a unique m-bit code or chip sequence. To transmit a 1 bit, a station sends its chip sequence. Thus, for m=8, if station A is assigned by 00011011, it sends a ‘1’ bit by sending 00011011 & a ‘0’ bit by sending 11100100 (1’s complement of chip sequence)

4. (c)CDMA- Manzur Ashraf4 Increasing the amount of information to be sent from b bits/sec to mb chips/sec can be done if bandwidth available is increased by a factor of m, making CDMA a form of spread spectrum communication. If we have a 1-MHz band available for 100 stations, with FDM each one would have 10 KHz and could send 10 kbps (let 1 bit/Hz). With CDMA, each station uses full 1 MHz, so chip rate is 1 Megachip/sec. With fewer than 100 chips/bit, the effective B/W per station is higher for CDMA than FDMA.

5. (c)CDMA- Manzur Ashraf5 Channel allocation Let we assume, bipolar notation as binary 0 being -1 and binary 1 being +1. So ‘1’ bit of station now becomes (-1-1-1+1+1-1+1+1). Each station has a unique chip sequence. Let ‘S’ denotes m-chip vector for station ‘S’ and ~S for its negation. All chip sequences are pair-wise orthogonal, which means normalized inner product of any two distinct chip sequences, S & T (S.T) is ‘zero’. S.T= 1/m∑ 1<=i<=m S i T i = 0

6. (c)CDMA- Manzur Ashraf6 If S.T=0 then S.~T=0 also. Again, S.S= 1/m∑ 1<=i<=m (±1) 2 = 1 S.~S= -1 (Try to calculate) During each bit time, a station can tranbsmit a 1 by sending its chip sequence, it can transmit a 0 by sending negative of its chip sequence or it can be silent. We assume all stations are synchronous is time, so all chip sequences begin at the same time. When two or more stations transmit simultaneously their bipolar signals add linearly.

7. (c)CDMA- Manzur Ashraf7 For example, if in one chip period 3 stations output +1 and one station outputs -1, the result is +2. One can think of it as adding three stations outputting +1 volts and one station outputting -1 volt gives 2 volts.

8. (c)CDMA- Manzur Ashraf8 Let ‘chip sequence’ of four stations are: A:0 0 0 1 1 0 1 1 A:(-1-1-1+1+1-1+1+1) B:0 0 1 0 1 1 1 0 A:(-1-1+1-1+1+1+1-1) A:0 1 0 1 1 1 0 0 A:(-1+1-1+1+1+1-1-1) A:0 1 0 0 0 0 1 0 A:(-1+1-1-1-1-1+1-1) Six examples: --1- C S1=(-1+1-1+1+1+1-1-1) -11- B+C S2=(-2 0 0 0 +2+2 0 -2) 10-- A+ ~B S3=(0 0 -2 +2 0 -2 0 -2) 101- A+ ~B +C S4=(-1+1-3+3-1-1-1+1) 1111 A+B+C+D S5=(-4 0 -2 0 +2 0 +2 -2) 1101 A+B+ ~C+D S6=(-2-2 0 -2 0 -2 +4 0) S1.C=(1+1+1+1+1+1+1+1)/8=1 (YES) S2.C=(2+0+0+0+2+2+0+2)/8=1 (YES) S3.C=(0+0+2+2+0-2+0-2)/8=0 (NO) S4.C=(1+1+3+3+1-1+1-1)/8=1 (YES) S5.C=(4+0+2+0+2+0-2+2)/8=1 (YES) S6.C=(2-2+0-2+0-2-4+0)/8=-1 (Complemented) [ Note: YES=‘1’; No=absent Complemented=‘0’ ]