Consider a 2-D constellation, suppose that basis signals =cosine and sine. Each constellation symbol corresponds to a vector with two real components

Transcription

1 TUTORIAL ON DIGITAL MODULATIONS Part 3: 4-PSK [2--26] Roberto Garello, Politecnico di Torino Free download (for personal use only) at:

2 Quadrature modulation Consider a 2-D constellation, suppose that basis signals =cosine and sine b ( t) = p( t)cos(2 π f t) b ( t) = p( t)sin(2 π f t) 2 Each constellation symbol corresponds to a vector with two real components b ( ( t ) 2 M = { s = ( α, β )} i i i β i s i α i b ( t) b ( t) R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 2

3 Quadrature modulation Binary information sequence Symbol sequence v [ n] H T st [ n] M R DURATION T k 2 DURATION T α[ n] R β[ n] R DURATION T Transmitted signal = + = + s( t) α[ n] b ( t nt ) β[ n] b ( t nt ) a( t) b( t) n 2 n R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 3

4 Quadrature modulation Spectrum of a(t): a( t) = α[ n] b ( t nt ) = α[ n] p( t nt ) cos 2π ft n n ( ) 2 2 Ga = x P( f f) + P( f f) x R when p(t) = ideal low pass filter f f R T = R R T = R R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 4

5 Quadrature modulation Spectrum of b(t): b( t) = β[ n] b ( t nt) = β[ n] p( t nt ) sin 2π ft n n 2 2 Gb = y P( f f) + P( f f) y R when p(t) = ideal low pass filter ( ) f f R R T = R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 5 R T = R

6 Quadrature modulation s( t) = a( t) + b( t) It can be proved that for these two basis signals G ( f ) = G ( f ) + G ( f ) s a b R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 6

7 Quadrature modulation s( t) = a( t) + b( t) G = G + G s a b 2 2 Ga = x P( f f) + P( f f) x R 2 2 G b = y P( f f) + P( f f) y R 2 2 Gs = z P( f f) + P( f f) z R G a and G b have the same shape and live on the same frequencies This is also the case for G s The spectrum of s(t) only depends on P(f) 2 R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 7

8 Quadrature modulation Example when p(t) = ideal low pass filter 2 2 Gs = z P( f f) + P( f f) z R f f R R T = R T = R R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 8

9 I/Q component b( t) β i s i Given a quadrature modulation, let us consider its transmitted waveform α i b ( t) s( t) = a( t) + b( t) = = α[ n] p( t nt ) cos 2 π f t + β[ n] p( t nt) sin 2π f t n n ( ) ( ) i( t ) q( t) I component (in phase) Q component (in quadrature) R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 9

10 Complex envelope [ ] ( π ) [ ] ( π ) s( t) = i( t) cos 2 f t + q( t) sin 2 f t Complex envelope s% ( t) = i( t) jq( t) i( t) = α[ n] p( t nt ) q( t) = β[ n] p( t nt) n n Complex symbol γ[ n] = α[ n] jβ[ n] s% ( t) = γ[ n] p( t nt ) n R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26]

11 Complex envelope s% ( t) = γ[ n] p( t nt ) n γ[ n] = α[ n] jβ[ n] b( t) β i s i γ = α jβ i i i α i b ( t) Quadrature constellation as a set of complex numbers M m { γ i αi jβ i} i = = = R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26]

12 Analytic signal [ ] ( π ) [ ] ( π ) s( t) = i( t) cos 2 f t + q( t) sin 2 f t s% ( t) = i( t) jq( t) s t s% t e s& t [ ] j2π ft ( ) = Re ( ) = Re ( ) 2 ( ) = s( t) e j π f t s& t % Analytic signal s& ( t) = s% ( t) e = γ[ n] p( t nt ) e n j2π f t j2π f t R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 2

13 4-PSK: : characteristics. Band-pass modulation 2. 2D signal set 3. Basis signals p(t)cos(2πf t) e p(t)sin(2πf t) 4. Costellation = 4 signals, equidistant on a circle 5. Information associated to the carrier phase R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 3

14 4-PSK: : constellation SIGNAL SET M = { s ( t) = Ap( t)cos(2 π f t), s ( t) = Ap( t)sin(2 π f t) 2 s ( t) = Ap( t)cos(2 π f t), s ( t) = Ap( t)sin(2 π f t) } 3 4 If we write M s ( t ) = Ap ( t )cos(2 π f t ), π s2( t) = Ap( t)sin(2 π ft) = Ap( t)cos 2 π ft, 2 = s3( t) = Ap( t)cos(2 π ft) = Ap( t)cos( 2 π ft π ), 3π s4( t) = Ap( t)sin(2 π ft) = Ap( t)cos 2π ft 2 Information associated to the carrier phase R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 4

15 4-PSK: : constellation SIGNAL SET M = { s ( t) = Ap( t)cos(2 π f t ϕ ) } π ϕi = ( i ) 2 4 i i i= Basis signals b ( t) = p( t)cos(2 π f t) b ( t) = p( t)sin(2 π f t) 2 VECTOR SET M = { s = ( A,), s = (, A), s = ( A,), s = (, A) } R R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 5

16 4-PSK: : constellation VECTOR SET M = { s = ( A,), s = (, A), s = ( A,), s = (, A) } R s 2 s 3 s s 4 R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 6

17 4-PSK: : constellation SIGNAL SET (with arbitary starting phase) M = { s ( t) = Ap( t)cos(2 π f t ϕ ) } 4 i i i= π ϕi = Φ + ( i ) 2 s ( t) = ( Acos ϕ ) p( t)cos(2 π f t) + ( Asin ϕ ) p( t)sin(2 π f t) i i i Versors b ( t) = p( t)cos(2 π f t) b ( t) = p( t)sin(2 π f t) 2 Vector set M = { s = ( α, β ) } R α = 4 2 i i i i= Acosϕ Asinϕ π ϕi = Φ + ( i ) 2 7 R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 7 i β = i i i

18 4-PSK: : constellation Example: Φ = M = { s = ( A,), s = (, A), s = ( A,), s = (, A)} R M = { s = ( α, β ) } R α = i β = i 4 2 i i i i= Acosϕ Asinϕ i i π π 3π ϕi = ( i ),, π, R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 8

19 4-PSK: : constellation Example: Φ = π 4 M = { s = ( α, α), s = ( + α, α), s = ( + α, + α), s = ( α, + α)} R M = { s = ( α, β ) } R α = i β = i 4 2 i i i i= Acosϕ Asinϕ i i π π π 3π 5π 7π ϕi = + ( i ),,, α = A 2 R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 9

20 4-PSK: : binary labeling Example of Gray labeling e : H 2 M e() = s e() = s e() = s e() = s 2 3 R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 2

21 4-PSK: : transmitted waveform m = 4 k = 2 T = 2T b Each symbol has duration T Each symbol component (α and β) lasts for T second R = R b 2 Transmitted waveform s( t) = α[ n] p( t nt ) cos(2 π ft) + β[ n] p( t nt ) sin(2 π ft) n n i( t ) q( t) I component (in phase) Q component (in quadrature) R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 2

22 4-PSK: : transmitted waveform example for p( t) = PT ( t) f = 2R b T α = T t/t R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 22

23 4-PSK: : analytic signal s( t) = α[ n] p( t nt ) cos(2 π ft) + β[ n] p( t nt ) sin(2 π ft) n n i( t ) q( t) [ ] 2 s( t) = Re s& ( t) = Re s% ( t) e j π f t s% ( t) = i( t) jq( t) = γ[ n] p( t nt) n γ[ n] = α[ n] jβ[ n] R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 23

24 4-PSK: : analytic signal s% ( t) = γ[ n] p( t nt ) n γ[ n] = α[ n] jβ[ n] { ( ), ( ), ( ), ( ),} M = s = a ja s = a ja s = a + ja s = a + ja R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 24

25 4-PSK: : bandwidth and spectral efficiency Transmitted waveform s( t) = α[ n] p( t nt ) cos(2 π ft) + β[ n] p( t nt ) sin(2 π ft) n n 2 2 Gs ( f ) = z P( f f) + P( f + f) z R Each symbol α[n] and β[n] has time duration T = 2T b R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 25

26 4-PSK: : bandwidth and spectral efficiency Case : p(t) = ideal low pass filter f f R R /T /T Total bandwidth (ideal case) B id = R = R 2 b Spectral efficiency (ideal case) Rb η id = = 2 bps / Hz B id R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 26

27 4-PSK: : bandwidth and spectral efficiency Case 2: p(t) = RRC filter with roll off α f f Total bandwidth R b B = R( + α ) = ( + α) 2 T T R ( + α ) R ( + α ) ( + α ) ( +α ) Spectral efficiency R b η = = B 2 ( + α) bps / Hz R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 27

28 Exercize Given a bandpass channel with bandwidth B = 4 Hz, centred around f =2 GHz, compute the maximum bit rate R b we can transmit over it with a 4-PSK constellation in the two cases: Ideal low pass filter RRC filter with α=.25 R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 28

29 Important It can be used for baseband channels, too Note the penalty with respect to the 2-PAM is recovered! The 2-PAM is a base-band modulation with η=2 The 2-PSK is a band-pass modulation with η= (due to the left part of the spectrum which is now on the positive frequencies) The 4-PSK is a band-pass modulation with η=2 (it uses both cosine and sine) R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 29

30 4-PSK: : modulator u [ ] T i vt [ n] k e ( ) α[ n] β[ n] cos 2 p( t) ( π f t) i ( t) 9 cos sin s( t) p( t) q( t) R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 3

31 4-PSK: : demodulator p( t) t + nt ρ [ n] 9 cos 2 sin 2 ( π f t) ( π f t) SYMBOL SYNCHRONIZATION R ML CRITERION s [ ] R n v [ n ] u [ i] e - R k R CARRIER SYNCHRONIZATION cos 2 ( π f t) p( t) t + nt ρ2[ n] R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 3

32 4-PSK: : Eye diagram 4-PSK constellation with RRC filter (α=.5) Canale I Canale Q R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 32

33 4-PSK: : intepretation The 4-PSK vector set can be viewed as the Cartesian product of two 2-PSK constellations ( A, A) ( A ) ( A) R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 33

34 4-PSK: : intepretation This is also true for the binary Gray labeling (first bit = I component, second bit = Q component) R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 34

35 4-PSK: : intepretation The AWGN channel adds two Gaussian components which are statistically independent R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 35

36 4-PSK: : intepretation The Voronoi regions of 4-PSK signals are the Cartesian product of the Voronoi regions of the constituent 2-PSK constellations R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 36

37 4-PSK: : intepretation The Voronoi regions of 4-PSK signals are the Cartesian product of the Voronoi regions of the constituent 2-PSK constellations Given the received vector (ρ [n], ρ 2 [n]) The sign of the first component ρ [n] determines the first received bit The sign of the second component ρ 2 [n] determines the second received bit R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 37

38 4-PSK: : intepretation The 4-PSK modulation can be viewed as the Cartesian product of two 2-PSK constellations transmitted over two independent channels R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 38

39 4-PSK: : modulator u [ ] T i R b S/P R b / 2 / 2 R b e e( ) cos 2 ( π f t) e e( ) p( t) p( t) i ( t) q( t) 9 cos sin s( t) R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 39

40 4-PSK: : demodulator cos 2 p( t) ( π f t) t + nt SYMBOL SYNCHRONIZATION 9 9 sin 2 ( π f t) R ρ [ n] ML CRITERION e - P/S u [ ] R i CARRIER SYNCHRONIZATION cos 2 ( π f t) p( t) t + nt ρ2[ n] ML CRITERION e - R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 4

41 4-PSK: : intepretation The Cartesian product interpretation clarifies why a 4-PSK constellation. Has the same BER performance of a 2-PSK 2. Has double spectral efficiency (two sequences with half bit-rate transmitted on the same frequencies) R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 4

42 4-PSK: : error probability BER.. E-3 E-4 E-5 E-6 E-7 E-8 E-9 E- E- E-2 E-3 E-4 E b BER = 2 erfc N BER Eb/N [db] R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 42

43 4-PSK: : applications Probably the most used digital modulation Satellite links Terrestrial radio links (with low spectral efficiency) GPS/Galileo UMTS The simplest QAM constellation (4-PSK = 4-QAM) used in many OFDM applications including ADSL, WiFi (82.n), WiMAX (82.6m) R. Garello. Tutorial on digital modulations - Part b 4-PSK [2--26] 43