CSE 4215/5431: Mobile Communications Winter 2011

1. CSE 4215/5431:Mobile Communications Winter 2011 Suprakash Datta datta@cs.yorku.ca Office: CSEB 3043 Phone: 416-736-2100 ext 77875 Course page: http://www.cs.yorku.ca/course/4215 Some slides are adapted from the book website CSE 4215, Winter 2011

2. Spread Spectrum • What? • Why? • How? CSE 4215, Winter 2011

3. Spread spectrum technology Problem of radio transmission: frequency dependent fading can wipe out narrow band signals for duration of the interference Solution: spread the narrow band signal into a broad band signal using a special code protection against narrow band interference Side effects: coexistence of several signals without dynamic coordination tap-proof Alternatives: Direct Sequence, Frequency Hopping signal power interference power spread signal spread interference detection at receiver f f CSE 4215, Winter 2011

4. Effects of spreading and interference dP/df dP/df user signal broadband interference narrowband interference i) ii) f f sender dP/df dP/df dP/df iii) iv) v) f f f receiver CSE 4215, Winter 2011

5. Spreading and frequency selective fading channelquality 2 2 2 2 2 1 frequency spreadspectrum channelquality 2 1 5 6 narrowband channels 3 4 frequency narrow bandsignal guard space spread spectrum channels CSE 4215, Winter 2011

6. Spread Spectrum CSE 4215, Winter 2011

7. Spread Spectrum - sender • Input is fed into a channel encoder • Produces analog signal with narrow bandwidth • Signal is further modulated using sequence of digits • Spreading code or spreading sequence • Generated by pseudonoise, or pseudo-random number generator • Effect of modulation is to increase bandwidth of signal to be transmitted CSE 4215, Winter 2011

8. Spread Spectrum - receiver • At the receiving end, digit sequence is used to demodulate the spread spectrum signal • Signal is fed into a channel decoder to recover data CSE 4215, Winter 2011

9. Frequency Hoping Spread Spectrum (FHSS) • Signal is broadcast over seemingly random series of radio frequencies • A number of channels allocated for the FH signal • Width of each channel corresponds to bandwidth of input signal • Signal hops from frequency to frequency at fixed intervals • Transmitter operates in one channel at a time • Bits are transmitted using some encoding scheme • At each successive interval, a new carrier frequency is selected CSE 4215, Winter 2011

10. FHSS - contd • Channel sequence dictated by spreading code • Receiver, hopping between frequencies in synchronization with transmitter, picks up message • Advantages • Eavesdroppers hear only unintelligible blips • Attempts to jam signal on one frequency succeed only at knocking out a few bits CSE 4215, Winter 2011

11. FHSS - illustration CSE 4215, Winter 2011

12. FHSS details - 1 Discrete changes of carrier frequency sequence of frequency changes determined via pseudo random number sequence Two versions Fast Hopping: several frequencies per user bit Slow Hopping: several user bits per frequency Advantages frequency selective fading and interference limited to short period simple implementation uses only small portion of spectrum at any time Disadvantages not as robust as DSSS simpler to detect CSE 4215, Winter 2011

13. FHSS - iIIustration tb user data 0 1 0 1 1 t f td f3 slow hopping (3 bits/hop) f2 f1 t td f f3 fast hopping (3 hops/bit) f2 f1 t tb: bit period td: dwell time CSE 4215, Winter 2011

14. FHSS Performance Considerations • Large number of frequencies used • Results in a system that is quite resistant to jamming • Jammer must jam all frequencies • With fixed power, this reduces the jamming power in any one frequency band CSE 4215, Winter 2011

15. FHSS and Retransmissions • What happens when a packet is corrupt and has to be retransmitted? • IEEE 802.11: max time of each hop: 400ms, max packet length: 30 ms. CSE 4215, Winter 2011

16. FHSS and WLAN access points • IEEE 802.11 FHSS WLAN specifies 78 hopping channels separated by 1 MHz in 3 groups • (0,3,6,9,…, 75), (1,4,7,…, 76), (2,5,8,…,77) • Allows installation of 3 AP’s in the same area. CSE 4215, Winter 2011

17. Direct Sequence Spread Spectrum (DSSS) • Each bit in original signal is represented by multiple bits in the transmitted signal • Spreading code spreads signal across a wider frequency band • Spread is in direct proportion to number of bits used • One technique combines digital information stream with the spreading code bit stream using exclusive-OR CSE 4215, Winter 2011

18. DSSS illustration

19. DSSS Using BPSK

20. Spectral view of DSSS CSE 4215, Winter 2011

21. Code-Division Multiple Access (CDMA) • Basic Principles of CDMA • D = rate of data signal • Break each bit into kchips • Chips are a user-specific fixed pattern • Chip data rate of new channel = kD CSE 4215, Winter 2011

22. CDMA Example • If k=6 and code is a sequence of 1s and -1s • For a ‘1’ bit, A sends code as chip pattern • <c1, c2, c3, c4, c5, c6> • For a ‘0’ bit, A sends complement of code • <-c1, -c2, -c3, -c4, -c5, -c6> • Receiver knows sender’s code and performs electronic decode function • <d1, d2, d3, d4, d5, d6> = received chip pattern • <c1, c2, c3, c4, c5, c6> = sender’s code CSE 4215, Winter 2011

23. CDMA Example • User A code = <1, –1, –1, 1, –1, 1> • To send a 1 bit = <1, –1, –1, 1, –1, 1> • To send a 0 bit = <–1, 1, 1, –1, 1, –1> • User B code = <1, 1, –1, – 1, 1, 1> • To send a 1 bit = <1, 1, –1, –1, 1, 1> • Receiver receiving with A’s code • (A’s code) x (received chip pattern) • User A ‘1’ bit: 6 -> 1 • User A ‘0’ bit: -6 -> 0 • User B ‘1’ bit: 0 -> unwanted signal ignored CSE 4215, Winter 2011

24. CDMA for DSSS

25. CDMA in theory Sender A sends Ad = 1, key Ak = 010011 (assign: “0”= -1, “1”= +1) sending signal As = Ad * Ak = (-1, +1, -1, -1, +1, +1) Sender B sends Bd = 0, key Bk = 110101 (assign: “0”= -1, “1”= +1) sending signal Bs = Bd * Bk = (-1, -1, +1, -1, +1, -1) Both signals superimpose in space interference neglected (noise etc.) As + Bs = (-2, 0, 0, -2, +2, 0) Receiver wants to receive signal from sender A apply key Ak bitwise (inner product) Ae = (-2, 0, 0, -2, +2, 0)  Ak = 2 + 0 + 0 + 2 + 2 + 0 = 6 result greater than 0, therefore, original bit was “1” receiving B Be = (-2, 0, 0, -2, +2, 0)  Bk = -2 + 0 + 0 - 2 - 2 + 0 = -6, i.e. “0” CSE 4215, Winter 2011