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Quiz Determine the minimum number of shift register stages required to create a maximal length PN sequence which has a repetition time greater than 10 seconds and a chip rate of 1.23 Mc/s. 12.3 x 106 2 N 1 N log 2 (12.3 x 106 1) 23.5 24 stages Multiple Access Systems FDMA 30 kHz FM channels – 900 Mhz band #users NT = Bsys/Bch TDMA 3:1 compression of Old FDMA channels CDMA 2/4 PSK 1.23 Mhz Spreading codes Transmit Power control to equalize incoming power levels Cellular Protocol Number of cells in cluster Reuse factor Kr = a2 + ab + b2 dI Kr = 173 N cell r 3 Bsys 2 3 Bch K r 2 1 5 7 d 0.3 I r 5 7 6 3 6 Kr 1 4 Cell Radius r For Co-channel Separation distance dI 4 2 2 C d 0.17 I I0 r 2 4 4 1 Cellular CDMA Capacity Since all cells use the same bandwidth, and mobile unit power is controlled, it is only the presence of other users that limit capacity. NCDMA G 1 E / I 1 v b 0 For Eb/I0 = 7 dB, NCDMA = 25 G = Bs/Rv Bs = Spreading Chip Rate ~ 1.23 Mb/s Rv = Voice Bit Rate ~ 9.6 kb/s = 0.5 Neighboring cell interference = 0.85 Power Control Factor v = 0.6 Voice Activity factor CDMA is much less susceptible to fading than narrowband FM because of broad bandwidth used. Cellular Radio Propagation Path Loss Doppler (for texting while driving) Slow log-normal fading Fast multipath fading 30-40 dB variations over fractional wavelengths Time dispersion Burst errors Rayleigh statistics if LOS is not dominant 1 2 Pe e 2 =C N Rayleigh multipath + Dominant LOS > Rician fading statistics Pe 1 2 k Note that Rice e 2 k Rayleigh as k k= 0 LOS MP Digital Transmission n bits/symbol M = 2n channel states (phasors) Channel must switch between states every Ts = 1/fs seconds, Thus fi = nfs Bandwidth efficiency fi B unitless (Bits/sec per hz) Since there are n bits per symbol, Es = n Eb and fs Es = fi Eb For M-PSK, phase noise required for an error is 1 2 N 2 M M And the noise voltage to signal voltage for an error would be Vn Vˆs sin N sin N Vˆs Vˆs Probability of Error Vˆs 2Vs , RMS is the magnitude of the symbol phasor. VˆS VN , RMS 2VS , RMS VN , RMS 2 S , RMS 2 N , RMS 2V V 2 Es N 0 1 2 The probability of symbol error is given by 2 E 12 s PE ( M ) 2Erfc sin N 0 The probability of bit error is PE ( M ) Pb n Erfc(x) is the complimentary error function for Gaussian Distributions: Erfc( x) e x u2 2 du e x2 2 x 2 Probability of Error (cont) For M-FSK, M B Mf s Ts The probability of symbol error is M 1 Es 2 N0 2n 1 Es 2 N0 PE M e e 2 2 PE M Pb n Viterbi sez . . . 2n 1 Pb n PE 2 1