Digital Transmission Via Carrier Modulation Carrier Phase Modulation BPSK(Binary PSK):M=2 QPSK(Quadrature PSK):M=4 MPSK(M-ary PSK):M>2
Digital Transmission Via Carrier Modulation Carrier Phase Modulation BPSK(Binary PSK) : M=2 QPSK(Quadrature PSK): M=4 MPSK(M-ary PSK): M > 2
Carrier Phase Modulation Carrier phase is used to transmit digital information via digital phase modulation M-ary phase modulation (M=2k) 2πm m=0,1,M-1 M Binary phase modulation(M=2) ■0=0,0=元 M carrier phase modulated signal waveform 0=g,0c2x+2m=0L-1 Transmitting filter pulse shape Amplitude
Carrier Phase Modulation ◼ Carrier phase is used to transmit digital information via digital phase modulation ◼ M-ary phase modulation (M=2k ) ◼ ◼ Binary phase modulation (M=2) ◼ ◼ M carrier phase modulated signal waveform ◼ 2 m , 0,1,., 1 m m M M = = − 0 0 = = 0, 2 m T c ( ) ( )cos(2 ), 0,1,., 1 m u t Ag t f t m M M = + = − Transmitting filter pulse shape Amplitude
PSK(Phase Shift Keying) Digital phase modulation is called PSK PSK signals have equal energy See Figure 6.7 page 289 ■5n=」o)dt=于」gudh=5,for all m Normalized power case(Rectangular g(t)) ·80=27,0≤1≤T今)g0d=1→A= Constant envelope
PSK(Phase Shift Keying) ◼ Digital phase modulation is called PSK ◼ PSK signals have equal energy ◼ See Figure 6.7 page 289 ◼ ◼ Normalized power case (Rectangular gT (t)) ◼ → → ◼ 2 2 2 ( ) ( ) , 2 m m T s A u t dt g t dt for all m − − = = 1 2 ( ) 1 2 g t dt T − = A = s g t T t T T ( ) 2 , 0 = 2 2 ( ) cos(2 ), 0,1,., 1 s m c m u t f t m M T M = + = − Constant envelope
PSK with 2 orthogonal basis Changing representation of waveform 2πm u (t)=Agr(t)cos(2ft+ M -5(tco2fo)-in(2r)sin -cot(codim-gx()in2x M0 45(t) ms Two orthogonal basis:v(),v() PSK can be represented geometrically as two- dimensional with components SDe,5s
PSK with 2 orthogonal basis ◼ Changing representation of waveform ◼ Two orthogonal basis: ◼ PSK can be represented geometrically as twodimensional with components 2 ( ) ( ) cos(2 ) 2 2 ( ){cos(2 ) cos( ) sin(2 )sin( )} 2 2 cos( ){ ( ) cos(2 )} sin( ){ ( )sin(2 )} m T c s T c c s T c s T c m u t Ag t f t M m m g t f t f t M M m m g t f t g t f t M M = + = − = + − mc s ms s 1 ( )t 2 ( )t 1 2 ( ), ( ) t t , mc ms s s
Constellation of PSK Two dimensional =6)-(W5cs2W5m20 M=2 M=4 M=8 001 000 √50了 (5,0) 10 100 +10 Same as binary PAM with antipodal signals Why Gray code is used
Constellation of PSK ◼ Two dimensional ◼ 2 2 ( , ) ( cos , sin ) m mc ms s s m m s s s M M = = ( ,0) s ( ,0) − s M=2 Same as binary PAM with antipodal signals s M=4 01 00 11 10 s M=8 011 001 111 101 000 100 110 010 Why Gray code is used ?
Phase demodulation and detection Received signal through AWGN channel r(t)=um(t)+n(t) =Smc8(t)cos(2πft)-Sms8r(t)sin(2πft) +n.(t)cos(2πfct)-n,(t)sin(2πft) Optimal Receiver Two correlator (or Matched Filter) Two dimensional Signal Optimum Detector
Phase demodulation and detection ◼ Received signal through AWGN channel ◼ ◼ Optimal Receiver ◼ Two correlator (or Matched Filter) ◼ Two dimensional Signal ◼ Optimum Detector ( ) ( ) ( ) ( ) cos(2 ) ( )sin(2 ) ( ) cos(2 ) ( )sin(2 ) m mc T c ms T c c c s c r t u t n t s g t f t s g t f t n t f t n t f t = + = − + −
Optimum Receiver for PSK Two Correlator cs2W+80n0h=瓦 2 m ne PSK in Sample r(t)= & Hold u,(t)+n(t) cos(2πft) Osc. Phase Comparator sin(2πft) Sample & Hold 5=瓦m2+上s00h=瓦sm2W+m 2πm M
Optimum Receiver for PSK ◼ Two Correlator Sample & Hold PSK in cos(2 ) c f t 0 ( ) t d Sample & Hold sin(2 ) c f t 0 ( ) t d Osc. ( ) ( ) ( ) m r t u t n t = + Phase Comparator 2 1 2 cos ( ) ( ) cos 2 c s T c s c m m r g t n t dt n M M − = + = + 2 1 2 sin ( ) ( ) sin 2 s s T s s s m m r g t n t dt n M M − = + = +
Noise component ■ Because n(t)and n(t)are uncorrelated zero mean Gaussian process Zero mean:EIne]=E[n,]=EInn,]=0 With variance:N 2 Example of M=4 r=收(m2同m+） M h=(V下，+n,n.) 5=(n,√5+n) What do you think is 5=(-V下，+n,n) optimal detector ? 5=(n。,-V,+n)
Noise component ◼ Because nc (t) and ns (t) are uncorrelated zero mean Gaussian process ◼ Zero mean: ◼ With variance: ◼ Example of M=4 E n E n E n n [ ] [ ] [ ] 0 c s c s = = = 2 2 0 [ ] [ ] 2 c s N E n E n = = 2 2 ( , ) ( cos , sin ) c s s c s c m m r r r n n M M = = + + 0 1 2 3 ( , ) ( , ) ( , ) ( , ) s c c c s c s c c c s c r n n r n n r n n r n n = + = + = − + = − + What do you think is optimal detector ??
Optimal Detector ■Example of M=4 Phase of received signal:0=tan-()= πm M 3π π 4 4 0,=π =0 3π 0,2 5π 7π 0= 0 4 4
Optimal Detector ◼ Example of M=4 ◼ Phase of received signal: 1 2 tan ( ) s r c r m r M − = = r = 0 2 r = r = 3 2 r = 4 T = 3 4 T = 5 4 T = 7 4 T =
Probability of Error Hard to get closed-form expression ■M=2case Same as binary PAM:P.= 25) N ■Where 3。is energy per bit M>4 case 49ax-a臣m3-8m Equivalent bit-error probability for M-ary PSK Approximalion: See figure 6.10 page 293,for Probability of Symbol error for M-ary PSK
Probability of Error ◼ Hard to get closed-form expression ◼ M=2 case ◼ Same as binary PAM : ◼ Where is energy per bit ◼ M>4 case ◼ Approximation: ◼ Equivalent bit-error probability for M-ary PSK ◼ Approximation: ◼ See figure 6.10 page 293, for Probability of Symbol error for M-ary PSK 2 0 2 ( ) b P Q N = b 0 0 2 2 2 ( sin ) 2 ( sin ) s b M k P Q Q N M N M M b P P k