Differential Phase Shift Keying (DPSK)

Transcription

1 Dfferental Phase Shft Keyng (DPSK) BPSK need to synchronze the carrer. DPSK no such need. Key dea: transmt the dfference between adjacent messages, not messages themselves. Implementaton: b = b m m = 1 b = b k k 1 k k k k 1 m = 0 b = b k k k 1 (7.1) where s bnary (mod-) addton and ( ) s bnary negaton. Snce BPSK (RF) modulaton s used: - same spectrum ( f etc.), - same rate/spectral effcency. Lecture 8 4-Nov-15 1(6)

2 Demodulator (sub-optmal): Pe ( ) 3dB loss Q γ (7.) Demodulator: (optmal) BMF = bandpass matched flter, ( ) Optmal probablty of error (BER): Pe h t = p( t)cos ω t. 1 = e γ (7.3) Lecture 8 4-Nov-15 (6)

3 Q.: fnd expressons for sgnals at each pont of demodulator y t = x t. Compare to the suboptmal ( ) assumng no nose ( ) ( ) demodulator. BER: 1dB loss to BPSK. 1 BER n AWGN channel OOK BPSK DPSK DPSK-A SNR: Eb/N0 [db] Lecture 8 4-Nov-15 3(6)

4 Sgnal Constellaton In-phase/quadrature representaton 1 : Complex form: ( ) = cos( ω + ϕ) x t A t = Acos ϕcosωt Asn ϕsn ωt = AI cosωt AQ sn ωt I ( ) ( ) ( ) { j ω t+ϕ } x t = Acos ω t + ϕ = Re Ae = Re { jωt A } ce Q (7.4) (7.5) j where Ac = Ae ϕ = complex ampltude, ω =carrer frequency, A = carrer ampltude (real). A = A + ja = Acos ϕ jasn ϕ (7.6) c I Q c I Q A = A = A + A (7.7) 1 ( Ac ) tan ( AQ AI ) ϕ = arg = (7.8) 1 For smplcty, assume p( t) ( t T ) = Π and consder 1 pulse only. Lecture 8 4-Nov-15 4(6)

5 A = A t ϕ = ϕ c ( ) ( t) ( ) A A t = c modulaton BPSK constellaton: ( ) ( ) x t = s t cos ωt; ( ) s t = ± 1, A = ± 1, A = 0; A = ± 1, ϕ = 0,180 I Q c T.S. Rappaport, Wreless Communcatons, Prentce Hall, 00 Lecture 8 4-Nov-15 5(6)

6 Quadrature phase shft keyng. QPSK BPSK: ϕ = 0 or π QPSK: ϕ = 0, π, 3π or π 4 ( bts nstead of 1). forms: I π x( t) = Acos ( ω t + θ ), θ =, = 0,1,,3 π π or θ = + 4 Q form: Constellaton: ( ) = cosω sn ω, x t A t A t A I I Q A A = ±, AQ = ± (7.9) (7.10) Lecture 8 4-Nov-15 6(6)

7 ( b ): combnaton of I and Q BPSK: ( ) x t = a cosωt b sn ωt ( ) a, b = ± 1 I and Q data (7.11) T.S. Rappaport, Wreless Communcatons, Prentce Hall, 00 Lecture 8 4-Nov-15 7(6)

8 QPSK: Propertes BPSK: 1 bt/symbol (snc) QPSK: bt/symbol (snc) twce SE! (same f ) bt SE = η = Rb Ts QPSK f = 1 T = s ( ) (7.1) 1btT BPSK: η = s = 1 (7.13) 1T s I/Q data sequences: constructed n the same way as for BPSK, (7.14) ( ) = ( ), ( ) = ( ) m t a p t T m t b p t T I Q.e. separate baseband BPSK over I and Q channels. Bandpass modulated sgnal: (7.15) ( ) = ( ) cos( ω + θ ) x t A p t T t θ = encodes data, e.g. θ1 θ 00, 01, 10 θ, 11 θ 3 4 Lecture 8 4-Nov-15 8(6)

9 QPSK = BPSK ELG4179: Wreless Communcaton Fundamentals S.Loyka QPSK Modulator (Tx) BBM = baseband BPSK modulator, DS = data spltter. ( ) = ( ), ( ) = ( ) m t a p t T m t b p t T I Q { } m a, b, a, b = ± 1 ( ), ( ) m t m t = baseband BPSK-modulated sgnals. I Q (7.16) Lecture 8 4-Nov-15 9(6)

10 QPSK Demodulator (Rx) ( ) to p t, MF = baseband matched flter ( ) DC = data combner. Probablty of bt error (BER): where b ( ) ( ) b P = Q γ = Q γ (7.17) E E γ γ = = = = =. b s b SNR/bt; Es Eb N0 N0 Lecture 8 4-Nov-15 10(6)

11 Bandwdth of QPSK p( t) ( RC ) = : 1+ α Rb 1+ α f = ; frf = Rb (7.18) R b η = 1+ α = Rs = = bt rate [ bts s] (7.19) T s [ bts s Hz] QPSK=xBPSK (7.0) Key parameters (for any modulaton): data rate R b BER (error probablty) P e bandwdth f ( f RF ) or spectral effcency (SE) η Lecture 8 4-Nov-15 11(6)

12 QPSK Constellaton n Nose: SNR = 0 db QPSK Constellaton Nosy QPSK N=1000 symbols transmtted. Lecture 8 4-Nov-15 1(6)

13 QPSK Constellaton n Nose: SNR = 10 db QPSK Constellaton Nosy QPSK N=1000 symbols transmtted. Lecture 8 4-Nov-15 13(6)

14 QPSK Constellaton n Nose: SNR = 0 db QPSK Constellaton Nosy QPSK N=1000 symbols transmtted. Lecture 8 4-Nov-15 14(6)

15 Quadrature Ampltude Modulaton (QAM) QAM key dea: use n-phase ( cosω t) and quadrature ( sn ω t) for PAM smultaneously. x rate of 1 channel ndependent channels as M-PAM: T cos ωt sn ω tdt = 0 (7.1) 0 ( ) ( ), 1,..., s t = A p t = M (7.) M-QAM: M - QAM = M PAM M PAM I Q RF sgnal of M-QAM: use M PAM on I and Q: ( ) ( ) cos ( ) ( ) ( ) ( ) ( ) x t = Am t ω t Am t sn ω t I c Q c m t = a p t, m t = b p t, 0 t T I Q a, b : represent I and Q bts = + 1, = L,..., L 1, L = M (7.3) Lecture 8 4-Nov-15 15(6)

16 Bass functons of QAM: ( ) ( ) ( ) ( ) ψ t = p t cos ω t, ψ t = p t sn ω t (7.4) 1 c Note the orthogonalty property: provded that S p ( f ) S ( ) ct f p( t ) and cos ω (or sn ω c t Q.: Prove ths. c t T ψ1 ( t) ψ ( t) dt = 0 (7.5) 0 cos 0, ω =.e. the spectrum of ) do not overlap. c T.S. Rappaport, Wreless Communcatons, Prentce Hall, 00 Lecture 8 4-Nov-15 16(6)

17 Alternatve form of RF QAM sgnal: x = a ψ t + b ψ t (7.6) ( t) 1( ) ( ) Mnmum symbol energy E mn : E mn = A p T 0 E p ( ) energy of ( ) E = p t dt = p t (7.7) Probablty of symbol error (symbol error rate - SER) P s: P s 1 E 4 1 Q mn M N 0 1 3E = 4 1 Q av M N ( M 1) 0 (7.8) where E av The BER P b : = average symbol energy, E av ( M ) 1 = Emn (7.9) 3 1 log Ps Pb Ps M, 1 Pb log Ps (7.30) M Lecture 8 4-Nov-15 17(6)

18 1 BER of M-QAM n AWGN channel BPSK M=4 M=8 M=16 M=3 M= SNR: Eb/N0 [db] Q.: reproduce the graph. Q.: how much extra SNR do you need to add 1 extra bt at the same BER? Adaptve modulaton: keep BER (almost) constant. Lecture 8 4-Nov-15 18(6)

19 4G systems: Optmzed for hgh-speed data servce (Internet), VoIP. Two major standards: LTE (Long Term Evoluton) and WMax (Worldwde Interoperablty for Mcrowave Access). LTE Standard Modulaton: OFDM + QPSK/16QAM/64QAM, up to 0MHz bandwdth. Rates: see below. 3GPP Long Term Evoluton: System Overvew, Product Development, and Test Challenges. Applcaton Note, Aglent. Note: MIMO = multple-nput multple-output, or mult-antenna system. SISO = sngle-nput sngle-output, or sngle-antenna system. Lecture 8 4-Nov-15 19(6)

20 IEEE 80.11n WF standard 80.11n Prmer, Whtepaper, ArMagnet, August 05, 008. Lecture 8 4-Nov-15 0(6)

21 Baseband/RF bandwdth; spectral effcency of M-QAM: 1+ α 1+ α R f = Rs = log R η = b = f RF log M 1+ α b M [ bt s Hz] (7.31) Complex form of QAM sgnal: ( ) ( ) { jω } ct ( ) I ( ) Q ( ) x t = Re m t e, m t = m t + jm t (7.3) Sgnal constellaton: va a + jb M = BPSK, M = 4 QPSK ( ) ( ) M -QAM = M PAM M PAM Demodulaton: va I Q ( ) ( ) ( ) ( ) { cos c } { sn } m t = LPF x t ω t m t = LPF x t ω t + baseband demodulaton of ( ), ( ) M PAM ) c m t m t (separately, as I Q (7.33) Lecture 8 4-Nov-15 1(6)

22 QAM Modulator BM = baseband modulator m DS = data spltter, { a, b } BPF = bandpass flter BM = PAM modulator for a p( t ), or s ( t) = a p( t T ); s ( t) = b p( t T ). I Q The RF QAM sgnal s (sngle pulse): ( ) ( ) ( ) x t = a p t cos ωt b p t sn ωt, 0 t T (7.34) Lecture 8 4-Nov-15 (6)

23 QAM Demodulator Demodulator: down-converson, MF + detecton. ( ) MF = matched flter for p( t ). BM = bt mappng, (, ) ( ) D = detector, a b. ( sˆ, sˆ ) ( a, b ). I Q I, Q = n-phase and quadrature channels. QAM demodulator = I-PAM + Q-PAM demod. In practce: M = 8,16,64,...,104. Lecture 8 4-Nov-15 3(6)

24 Baseband (BB): snc: ELG4179: Wreless Communcaton Fundamentals S.Loyka Bandwdth and Spectral Effcency 1 f = ; RC: T s 1+ α f = ; rect: T s RC = rased-cosne pulse.; rect = rectangular pulse. 1 f = ; (7.35) T s Passband or RF: If DSB s employed: snc: 1 f = ; RC: T s 1+ α f = ; rect: T s f = ; (7.36) T.e. f = f. If SSB, take 1/ of DSB bandwdth. RF BB s Spectral effcency (SE): R b SE: [ bts s Hz].e. how many b/s per unt bandwdth (Hz). η = (7.37) f Lecture 8 4-Nov-15 4(6)

25 Spectral Effcency Assume RC pulse everywhere; f not, adjust accordng to (7.35). General relatonshp: Baseband (BB): M-PAM: log M Rb = Rs log M = (7.38) T Rb log = M η = [ b s Hz], (7.39) f 1 + α s Passband or RF: M-PAM (-PAM = BPSK): DSB: M-QAM (4-QAM = QPSK): f MQAM DSB: log M η = 1+ α, SSB: log M η = 1+ α = M,.e. Q.: whch s better? (7.40) log MQAM log = M η = 1+ α 1+ α, (7.41) QAM = M PAM M PAM I Q Lecture 8 4-Nov-15 5(6)

26 Summary DPSK. QPSK. QAM. Sgnal constellaton. Modulators/demodulators. Bandwdth and spectral effcency. BER, SER. Readng: Rappaport, Ch. 6 ( ). L.W. Couch II, Dgtal and Analog Communcaton Systems, 7th Edton, Prentce Hall, 007. (other edtons are OK as well) Other books (see the reference lst). Note: Do not forget to do end-of-chapter problems. Remember the learnng effcency pyramd! Lecture 8 4-Nov-15 6(6)