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DIGITAL SPREAD SPECTRUM SYSTEMS

DIGITAL SPREAD SPECTRUM SYSTEMS. ENG-737. Wright State University James P. Stephens. GOLD CODE IMPLEMENTATION. Gold Codes are used by GPS and are constructed by the linear combination of two m-sequences of length n=10 There are 1023 possible codes possible for n=10

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DIGITAL SPREAD SPECTRUM SYSTEMS

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  1. DIGITAL SPREAD SPECTRUM SYSTEMS ENG-737 Wright State University James P. Stephens

  2. GOLD CODE IMPLEMENTATION • Gold Codes are used by GPS and are constructed by the linear combination of two m-sequences of length n=10 • There are 1023 possible codes possible for n=10 • Each different code is generated by inputting a different initial fill into the G2 Coder • Each GPS satellite is assigned a different code

  3. GPS C/A CODER

  4. KASAMI CODES • Kasami sequences are one of the most important types of binary sequence sets because of their very low cross-correlation and their large number of available sets • There are two different sets of Kasami sequences, Kasami sequences of the ‘small set’ and sequences of the ‘large set’ • A procedure similar to that used for generating Gold sequences will generate the ‘small set’ of Kasami sequences with M = 2n/2 binary sequences of period N = 2n/2 + 1 • In this procedure, we begin with an m-sequence ‘a’ and we form the sequence a’ by decimating ‘a’ by 2n/2 + 1 • It can be verified that the resulting sequence a’ is an m-sequence with period 2n/2 - 1

  5. b = 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 Repeats 1 2 3 4 5 KASAMI CODE IMPLEMENTATION X4 + X + 1 q = 2n/2 + 1 = 5 m = 2n/2 - 1 = 3 Where, q = decimation value m = period of a’ a a = 1 1 1 1 0 1 0 1 1 0 0 1 0 0 0 a’ = 1 1 0 • a xor b = 0 0 1 0 1 1 1 0 1 1 1 1 1 1 0 • Kasami codes are generated by cyclically shifting ‘a’ 2n/2 -2 = 2 times • Including a and b there are 2n/2 = 4 sequences

  6. KASAMI CODE IMPLEMENTATION Example: • Let n=10, therefore, N=2n - 1 = 1023 (length of ‘a’) • The decimation value is 2n/2 + 1 = 33 which is used to create a’ • 1023/33 = 31 which will be the length of a’ • If we observe 1023 bits of sequence a’, we will see 33 repetitions of the 31-bit sequence which we will call sequence ‘b’ • Now taking 1023 bits of sequence ‘a’ and ‘b’ we form a new set of sequences by adding (modulo-2 addition) the bits from ‘a’ and the bits from ‘b’ and all 2n/2 – 2 cyclic shifts of the bits from ‘b’ • By including ‘a’ in the set, we obtain a set of 2n/2 = 32 binary sequences of length 1023 • All the elements of a ‘small set’ of Kasami sequences can be generated in this manner

  7. KASAMI CODE IMPLEMENTATION • The autocorrelation and cross-correlation functions provide excellent properties, as good or better, than Gold Codes • The ‘large set’ of Kasami sequences is generated in a similar manner with the addition of another register • The two registers are a preferred pair as in Gold Code and therefore when combined with the decimated sequence, produce all the associated Gold Codes and the Kasami sequences for an even larger let of sequences

  8. FACTORS FOR DETERMINING SIGNALING FORMAT • Signal spectrum • Synchronization • Interference and noise immunity • Error detection capability • Cost and complexity Before we begin a more in-depth discussion of direct sequence spread spectrum, it will be helpful to compare various encoding and / or signaling techniques used in digital communications

  9. DIGITAL SIGNAL ENCODING FORMATS • Biphase-Space Always a transition at beginning of interval • 1 = no transition in middle of interval • 0 = transition in middle of interval • Differential Manchester Always a transition at middle of interval • 1 = no transition at beginning of interval • 0 = transition at beginning of interval • Delay Modulation (Miller) • 1 = transition in middle of interval • 0 = no transition if followed by 1, transition at end of interval if followed by 1 • Bipolar • 1 = pulse in first half of bit interval, alternating polarity from pulse to pulse • 0 = no pulse • Nonreturn to zero-level (NRZ-L) • 1 = high level • 0 = low level • Nonreturn to zero-mark (NRZ-M) • 1 = transition at beginning of interval • 0 = no transition • Nonreturn to zero-space (NRZ-S) • 1 = no transition • 0 = transition at beginning of interval • Return to zero (RZ) • 1 = pulse in first half of bit interval • 0 = no pulse • Biphase-Level (Manchester) • 1 = transition from hi to lo in middle of interval • 0 = transition from lo to hi in middle of interval • Biphase-Mark Always a transition at beginning of interval • 1 = transition in middle of interval • 0 = no transition in middle of interval

  10. DIGITAL SIGNAL ENCODING FORMATS

  11. DIFFERENTIALLY ENCODING AND DECODING Encoder ‘1’s are converted to 180o phase shifts, ‘0’s are unchanged Decoder ‘1’s are inserted for every transition, ‘0’s if no transition

  12. NRZ-M NRZ-S Input Input + + + + + + + + + + + + + + + + + + + + DIFFERENTIAL ENCODING 1 0 1 1 0 0 0 1 1 0 1 1 0 1 1 0 0 0 1 1 0 1 0 0 1 0 0 0 0 1 0 0 1 1 0 0 0 1 0 1 1 1 0 0

  13. DIGITAL SIGNAL ENCODING FORMATS • Phase-encoding schemes are used in magnetic recording systems, optical communication, and in some satellite telemetry links • Schemes with transitions during each interval are self-clocking • Schemes that transition in the middle are naturally shorter pulses and require greater bandwidth • Differential encoding supports non-coherent detection

  14. DIRECT SEQUENCE SYSTEMS • DSSS is the most common commercially • Sometimes called PN spread spectrum • Used in CDMA Cellular systems, GPS, some earlier cordless telephones, and 802.11(b) • DSSS directly modulates a carrier with a high rate code that is combined with data • DSSS usually employs PSK and the code is often combined with data by mod-2 addition, i.e. code inversion keying • Predominantly, in practice, a DSSS transmitted signal is either: • BPSK (Binary Phase Shift Keyed) • QPSK (Quadrature Phase Shifted Keyed) • MSK (Minimum Shifted Keyed)

  15. X Data + Code Output Carrier BINARY SHIFT KEYING • This technique is implemented with a ‘Balanced Modulator’ • Two basic types of modulators are: • Single balanced • Double balanced • Three port devices in which  1’s on the code + data input cause 180 degree phase shifts of the carrier

  16. BINARY PHASE SHIFT KEYING (BPSK) Data + Code Typically more than one cycle per chip 1800 Phase Shifts BPSK Carrier

  17. BPSK POWER SPECTRAL DENSITY Suppressed Carrier Discrete spectral lines

  18. SUPPRESSED CARRIER Reasons that make suppressed carrier desirable are: • More difficult for adversary to detect signal • Power not wasted on carrier • Signal has constant envelop level so that power efficiency is maximized for the bandwidth used • Bi-phase modulators are simple, stable, low cost devices

  19. BPSK CIRCUIT IMPLEMENTATION

  20. 00 1800 PHASOR REPRESENTATION • BPSK is called antipodal • Antipodal means that two symbols meet the following criteria: s1 = -s2 • BPSK other than 1800 is not antipodal

  21. 900 00 1800 2700 QUADRATURE PSK OR QPSK • QPSK does not degrade as seriously as BPSK when passed through non-linearity simultaneous with interference • Bandwidth is one-half required by BPSK at same data rate (or twice the data rate in the same bandwidth)

  22. x Σ x x x ~ 900 QPSK BLOCK DIAGRAM m1(t)cos(2πft) Code 1 cos(2πft) Data 1 Carrier SQPSK sin(2πft) Code 2 m2(t)sin(2πft) m2(t) Data 2 SQPSK(t) = m1cos(2πft) + m2sin(2πft) m1(t) Twice the data same BW

  23. x Σ x x ~ 900 ALTERNATIVE IMPLEMENTATION OF QPSK 2-bit serial to parallel Code Data QPSK Half the BW same data rate

  24. BALANCED MODULATORS

  25. BPSK MODULATION

  26. CARRIER SUPPRESSION • Carrier suppression may be expressed in dB in accordance with the following expression: V = 10 log [ B/ (A sin  + A’ sin )] Where, B = amplitude of the correct output signal (i.e. when A=A’ and  =  = 00) A = 00 signal amplitude A’ = 1800 signal amplitude  = A phase offset  = A’ phase offset

  27. CARRIER SUPPRESSION Example: Compute the carrier suppression of biphase-balanced modulation in dB if the amplitude of the correct signal (B) is 10, the zero and 1800 signal amplitude is 5 (A and A’), and the phase offsets ( and ) are: (a) 90 degrees V = 10 log [10/(5 sin 900 + 5 sin 900)] = 0 dB (no carrier suppression) (b) 60 degrees V = 10 log [10/(5 sin 600 + 5 sin 600)] = 0.62 dB (c) 1 degree V = 10 long [10/5 sin 10 + 5 sin 10)] = 17.58 dB

  28. QPSK MODULATOR

  29. BPSK AND QPSK SPECTRA 66

  30. MQPSK / OQPSK / SQPSK • Modified QPSK such that by shifting the I and Q clocks, no phase transition will occur larger than 900 QPSK SQPSK • 00 = 00 • 01 = 900 • = 1800 • = 2700

  31. MINIMUM SHIFT KEYING

  32. DSSS MODULATION COMPARISON

  33. WHICH MODULATION WOULD YOU CHOOSE ? Depends upon: • Effects on synchronization • Sidelobe energy / bandwidth • Complexity of modulator / demodulator • Effects of jamming in interference • Impact on size, weight, power, and reliability

  34. DETERMINING THE NUMBER OF SIMULTANEOUS USERS • Many DSSS users can transmit messages simultaneously over the same channel bandwidth provided each user has his own PN code sequence • Digital communications in which each transmitter/receive user pair has its own distinct signature code is called Code Division Multiple Access (CDMA) • In cellular systems, a base station transmits signals to Nu mobile receivers using Nu orthogonal PN sequences, one for each receiver • These Nu signals are perfectly synchronized so that they can arrive in synchronism due to the orthogonality of the codes • However, this synchronization cannot always be achieved, particularly in the uplink (mobile to base station)

  35. DETERMINING THE NUMBER OF SIMULTANEOUS USERS • In demodulation of each DSSS signal at base station, the signals from the other simultaneous users of the channel appear as additive interference • Assume equal power of all simultaneous users at the base station (achieved via adaptive power control), the desired SNR is: • In determining the maximum number of users, we assumed that the codes are orthogonal and the interference from the other users adds on a power basis and dominates the noise term • This is why the design of a large set of PN sequences with good correlation properties is important

  36. DETERMINING THE NUMBER OF SIMULTANEOUS USERS Example: Desired level of performance for a user in a CDMA system requires Eb/Jo = 10 dB Determine the maximum number of simultaneous users that can be accommodated in a CDMA system if the bandwidth-to-bit ratio is 100 (PG) and the coding gain is 6 dB:

  37. NEAR FAR PROBLEM FOR DSSS • There are many ways in which the received powers can be unequal for a DSSS multiple access system • For the analysis that follows, assume that all users transmit with equal power, but are different distances from the ‘jth’ receiver • Then the received power from the ‘ith’ transmitter may be represented as: Pi = Po / di Where, Po = received power at unit distance di = distance from the ith transmitter to the jth receiver  = propagation law • The parameter  is the propagation law and depends upon the medium in which the transmission takes place • In free space  = 2, at UHF over an ideal earth  tends to change between 3 and 4 (determined experimentally)

  38. NEAR FAR PROBLEM FOR DSSS • It is possible to represent the ratio of the power received from the ith transmitter to that received from the jth transmitter, which is the desired signal • This is shown by: • The SNR at the output of the jth receiver may now be written as:

  39. NEAR FAR PROBLEM FOR DSSS • Solving for the term that is related to the distances gives: • The term subtracts off for the intended signal • To find the capacity of a CDMA system where all powers are equal and the distances are the same, let

  40. NEAR FAR PROBLEM FOR DSSS Therefore, for all users U, • Note that the number of users has been reduced by a factor of 3 simply by one transmitter being 2.5 times closer than all of the others • The system would fail completely as a multiple access system if, dj/di> 2.78 since only one user could be supported and none of the other users would be received with the desired SNRo Subtracts off closer user and intended user

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