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Digital Baseband Modulation

Digital Baseband Modulation
1 Lecture 7. Digital Communications Part II. Digital Modulation • Digital Baseband Modulation • Digital Bandpass Modulation Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 72 Digital Communications Analog Signal Bit sequence t 0001101110…… Digital Digital AD Source Baseband Bandpass Conversion SOURCE Modulation Modulation t t Baseband Bandpass Channel Channel Digital Digital DA User Baseband Bandpass Conversion Demodulation Demodulation Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 73 Digital Modulation t Digital Baseband Baseband Channel Modulation Bit sequence Modulated signal 0001101110…… Digital Bandpass Bandpass Channel Modulation t • How to choose proper digital waveforms to “carry” the digits Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 74 Digital Modulation Digital Bit sequence Modulated Baseband/Bandpass Baseband/Bandpass Signal Channel 0001101110…… Modulation • Bit Rate: number of bits transmitted in unit time • Required channel bandwidth: determined by the bandwidth of the modulated signal. • Bandwidth Efficiency: Information Bit Rate R b  Required Channel Bandwidth B h Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 75 Digital Baseband Modulation • Pulse Amplitude Modulation (PAM) • Pulse Shaping Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 76 Digital Baseband Modulation • Choose baseband signals to carry the digits. – Each baseband signal can carry multiple bits. • Each baseband signal carries 1 bit. Binary • Bit Rate: R1/ b • Totally 2 baseband signals are required. • Each baseband signal carries a symbol (with log M bits). 2 Mary •Symbol Rate: RM log / R1/ Bit Rate: b 2 s • Totally M baseband signals are required. Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 77 Digital Baseband Modulation • Focus on “amplitude modulation” – The baseband signals have the same shape, but different amplitudes. – Timedomain representation of the modulated signal:  s()tZv(tn)  n n  Z is a discrete random variable with where PrZaM  1/ , 1i ,...,M, n ni v(t) is a unit baseband signal. – Power spectrum of the modulated signal: Read the 2  supplemental  1 2 m  2 Z Gf() V()f  f material for  sZ     m   details. Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 78 Pulse Amplitude Modulation (PAM) • Binary PAM • Binary OnOff Keying (OOK) • 4ary PAM Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 79 Binary PAM a negative rectangular pulse 1: a positive rectangular pulse 0: with amplitude A and width  with amplitude A and width  A  110 100 11… s()tZv(tn)  n s(t) n  PrZ 1 1/ 2  n At , 0   vt ()  0, otherwise  Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 710 Power Spectrum of Binary PAM 2   1 m 2  2 Z Gf() V()f  f sZ    m   22 With Binary PAM: Vf()Asinc(f) Gf()Asinc(f) BPAM 2  0,1 Z Z Gf() BPAM 2 A f   0 See Textbook (Sec. 3.2) or Reference Proakis Salehi (Sec. 8.2) for more details. Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 711 Effective Bandwidth of Binary PAM 22 G (f) Afsinc ( ) BPAM 2 A f  0  90 bandwidth: 1/ 90 power 95 bandwidth: 2/ 95 power • Suppose 90 of signal power must pass through the channel (90 inband power): B1/ Required Channel Bandwidth: h90 B R hb 90 R1/ Bit rate: b • Suppose 95 of signal power must pass through the channel (95 inband power): B 2/ 2R Required Channel Bandwidth: h95 b Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 712 Bandwidth Efficiency of Binary PAM Information Bit Rate R b  • Bandwidth Efficiency : Required Channel Bandwidth B h • Bandwidth Efficiency of Binary PAM: R1/ b 1 with 90 inband power BPAM B1/ h90  0.5 with 95 inband power BPAM B 2/ h95 What if the two pulses have unsymmetrical amplitudes Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 713 Binary OnOff Keying (OOK) 1: a positive rectangular pulse 0: nothing (can be regarded as a with amplitude A and width  pulse with amplitude 0) A  110 100 11… s()tZv(tn)  n s(t) n  PrZZ1 Pr 01/ 2  nn At , 0   vt ()  0, otherwise  Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 714 Power Spectrum of Binary OOK 2   1 m 2  2 Z Gf() V()f  f sZ    m   1 2 Gf() A sinc(f )  BOOK With Binary OOK: Vf()Asinc(f)  2  1/ 2,1/ 4  11 m  Z Z  f    44   m  Gf() BOOK 1 2 A 4 … … f   0 Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 715 Bandwidth Efficiency of Binary OOK Information Bit Rate R b • Bandwidth Efficiency :  Required Channel Bandwidth B h • Bandwidth Efficiency of Binary OOK: R1/ b  1 with 90 inband power BOOK B1/ h90  0.5 with 95 inband power BOOK B 2/ h95 Can we improve the bandwidth efficiency without sacrificing the inband power Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 716 4ary PAM Each waveform carries 2bit information. • 4ary PAM: 11: 01: 00: 10:  s()tZv(tn)  n 1 1 0 1 0 0 1 0 0 0 … n  s(t)  PrZZ1 Pr 1/ 3 nn  PrZZ 1 Pr1/ 3 nn 1/ 4 At , 0   vt ()  0, otherwise  Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 717 Power Spectrum of 4ary PAM 2   1 m 2  2 Z Gf() V()f  f sZ    m   22 5 Gf()Asinc(f) With 4ary PAM: Vf()Asinc(f) 4PAM 9 2  0, 5/9 Z Z G (f) 4PAM 2 5 A 9 f   0 B1/ • Required channel bandwidth with 90 inband power: h90 B 2/ • Required channel bandwidth with 95 inband power: h95 Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 718 Bandwidth Efficiency of 4ary PAM •Symbol rate: R1/ S RR22 / •Bit rate: bS • Require channel bandwidth: 1 with 90 inband power: B1/  R R h90 S b 2 with 95 inband power:  R B 2/ 2R b h95 S  2 with 90 inband power 4PAM  1 with 95 inband power 4PAM 4ary PAM achieves higher bandwidth efficiency than binary PAM Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 719 Bandwidth Efficiency of Mary PAM • Suppose there are totally M distinct amplitude (power) levels. • How many bits are carried by each symbol k M 2 kM log 2 • What is the relationship between symbol rate R and bit rate R S b RRk / R kR or Sb bS • What is the required channel bandwidth with 90 inband power B RRk / hS 90 b Tradeoff between bandwidth efficiency and fidelity performance • Bandwidth Efficiency of Mary PAM kM log with 90 inband power MPAM 2 • A larger M also leads to a smaller minimal amplitude difference – higher error probability (to be discussed). Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 720 Pulse Shaping • InterSymbol Interference (ISI) • SincShaped Pulse and RaisedCosine Pulse Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 721 Transmission over Bandlimited Channel • Frequency domain Baseband Channel PAM signal H(f) 2 Gf()G ()f H(f) YPAM G (f) PAM The signal distortion B 0 B f h h incurred by channel is f 0 always nonzero • Time domain  y()ts ()t h()t Zx(tn)  n PAM signal Baseband Channel n   h(t) xt ()vt ()h() t st () Z v(tn)  n n  Intersymbol Sample y(t) at m, m=1,2,…, we have  Interference y() m Z xm(n)Zx(0) Zxm(n)  nm n (ISI) nn  m Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 722 ISI and Eye Diagram • An eye diagram is constructed by plotting overlapping ksymbol segments of a baseband signal. • An eye diagram can be displayed on an oscillo scope by triggering the time sweep of the oscilloscope. See Reference Ziemer Tranter (Sec. 4.6) for more details about eye diagram. • ISI is caused by insufficient channel bandwidth. • Any better choice than rectangular pulse SincShaped pulse Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 723 SincShaped Pulse v(t)  Asinc( f ) V(f) A A f /2 t 0/2 1/ 2/ 1/ 2/  Rectangular Pulse V(f) v(t)=Asinc(t/) A A t 1/(2)1/(2)  0 f  SincShaped Pulse Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 724 Binary SincShapedPulse Modulated Signal a negative sincshaped pulse 1: a positive sincshaped pulse 0: with amplitude A and first with amplitude A and first  crossingzero point crossingzero point  A 0 t  0 A t  s(t) 111 0 0 1 1 … … t … …  PrZ 1 1/ 2   n s()tZv(tn)  n vt () Asinc(t/)  n  Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 725 Power Spectrum of SincShapedPulse Modulated Signal 2   1 m 2  2 Z Gf() V()f  f sZ    m   2 1 2 Gf ()A , f  BSSP 2 With Binary SincShaped 0,1 Z Z 1 Pulse Modulated Signal: Vf ()A , f  2 G (f) BSSP 2 A 1/(2)1/(2) 0 f R1/  2 Bit Rate: b BSSP (with 100 inband power)  R /2 B1/(2) Required channel bandwidth: b h Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 726 SincShapedPulse Modulated Signal over Bandlimited Channel • Frequency domain Baseband Channel Binary SincShapedPulse signal 2 H(f) Gf()G ()f H(f) YBSSP G (f) BSSP G (f) Y 2 A 2 A B 0 B f h h f 1/(2)1/(2) 0 f 1/(2)1/(2) 0 • Time domain Binary SincShapedPulse signal Baseband Channel yt ()s() th() t  h(t) st () Z v(tn)  n n  Zero ISI at t=m y(t) 111 0 0 11 … … t … …  Are there any other (better) choices to achieve zero ISI Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 727 Nyquist PulseShaping Criterion for Zero ISI Nyquist pulseshaping criterion for zero ISI A necessary and sufficient condition for x(t) to satisfy 1, n 0  xn()  0, n 0   m Xf(). is that its Fourier transform X(f) satisfies   m  Suppose that there is a baseband channel with the channel bandwidth W. To pass a modulated signal with symbol rate 1/ through the channel: •If 1/WW, there is no way to satisfy the Nyquist pulseshaping criterion for zero ISI.  m Xf()    m  …… …… f 2/ 1/ 0 W 1/ 1/W 2/ Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 728 Nyquist PulseShaping Criterion for Zero ISI According to Nyquist pulseshaping criterion for zero ISI:  If the symbol rate 1/2W, there is no way that we can design a system with zero ISI. , f W   If the symbol rate 1/=2W, we must have Xf()  0, otherwise  • The maximum symbol rate for zero ISI is 2W. • In the binary case, the highest bandwidth efficiency for zeroISI is 2, which is achieved by the binary sincshapedpulse modulated signal.  If the symbol rate 1/2W, we have numerous choices. One of them is called RaisedCosine Pulse. See Reference Proakis Salehi (Sec. 8.3.1) for more details. Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 729 RaisedCosine Pulse: Tradeoff between Bandwidth Efficiency and Robustness V(f) v(t)=Asinc(t/) SincShaped Pulse: A A 1 1 t    0 f 2 2 • Strong ISI at tn. 1 • Perfect synchronization is required at the receiver side. 0 2 1   cos(2 t) 2 vt ( ) Asinc(t / ) RaisedCosine Pulse:  2 V(f) 1(4t)  A A 1 1 f  t  0  2 2 More robust •Larger  Larger bandwidth Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 730 Summary I: Digital Baseband Modulation Complexity Bandwidth Efficiency Low 1 (90 inband power) Binary PAM PAM 2 (90 inband power) Low 4ary PAM High Binary Sinc 2 (100 inband power) (Susceptible to ShapedPulse timing jitter) Modulation R b Binary Raised 12 1 R 2 b Moderate CosinePulse (100 inband power) Modulation Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 731 Digital Bandpass Modulation • Binary ASK • Binary FSK • Binary PSK • Quaternary PSK Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 732 Digital Bandpass Modulation • How to transmit a baseband signal over a bandpass channel 1 01 0 Amplitude Binary Amplitude Shift Keying modulation (BASK) Carrier Signal t cos(2f t) c Frequency Binary Frequency Shift Keying modulation (BFSK) t t Phase Binary Phase Shift Keying modulation (BPSK) t Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 733 Binary Amplitude Shift Keying (ASK) • Generate a binary ASK signal: – Send the carrier signal if the information bit is “1”; – Send 0 volts if the information bit is “0”. 10 1 0 s () t  s ()tf cos(2t) BASK BOOK c t Binary OnOff Keyingst () BOOK t x cos(2f t) c t Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 734 Power Spectrum of BASK • Power spectrum of Binary OOK:  11 1 m 2  Gf() A sinc(f ) f   BOOK    44   m  Read the • Power spectrum of Binary ASK: supplemental material for 1 Gf() G (ff)G (ff) details. BASK BOOK c BOOK c 4 G () f BASK … ……… f f f f f f 0 f c c c c c c 2/ 2R b Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 735 Bandwidth Efficiency of BASK G () f BASK fR fR fR fR f f 0 f c b c b c b c b c c 2R b The bandwidth of BASK signal is twice of that of its baseband signal (binary OnOff Keying) • The required channel bandwidth for 90 inband power: BR 2 hb 90 • Bandwidth Efficiency of BASK: with 90 inband power  0.5 BASK  0.25 with 95 inband power BASK Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 736 Binary Frequency Shift Keying (BFSK) • Generate a binary FSK signal: Frequency offset Acos(2 (ftf ) ) – Send the signal if the information bit is “1”; c – Send the signal if the information bit is “0”. Acos(2 (ftf ) ) c s () t s () t s () t cos(2 ( f ft ) ) cos(2 ( fft ) ) b1,BFSK bB 2, AFSK BFSK c c 0 b1  A b 1 i  i st ()  st () b2,BFSK  bB 1, FSK A b 0 0 b 0 i i  Binary OnOff Keying Binary OnOff Keying 01 0 1 t Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 737 Bandwidth Efficiency of BFSK 1 G(f) Gf( ()f f) Gf( ()f f) BFSK 4 b1,BFSK c b1,BFSK c 1  ( Gf(ff))Gf((ff)) 4 bB 2, FSK c bB 2, FSK c G () f BFSK f () f f () f f ff ff 0 c c c c 2f 22f R b • The required channel bandwidth for 90 inband power: Bf22 R hb 90 1 • Bandwidth efficiency of BFSK:   0.5  0.5 BFSK BASK 1/ fR (with 90 inband power) b The bandwidth efficiency of BFSK signal is lower than that of BASK signal Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 738 Binary Phase Shift Keying (BPSK) • Generate a binary PSK signal: Acos(2 ft) – Send the signal if the information bit is “1”; c – Send the signal if the information bit is “0”. Af cos(2t ) c Acos(2 ft) c 10 1 0 s () t  s ()tf cos(2t) BPSK BPAM c t st () Binary PAM BPAM t x cos(2f t) c t Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 739 Bandwidth Efficiency of BPSK 1 G () f G (ff ) G (ff ) BPSK 4 BPAM c BPAM c G () f BPSK … ……… f f 0 f c c 2R b • The required channel bandwidth for 90 inband power: BR 2 hb 90 • Bandwidth Efficiency of BPSK: with 90 inband power  0.5 BPSK  0.25 with 95 inband power BPSK The bandwidth efficiency of BPSK signal is the same as that of BASK signal Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 740 Mary PSK • Mary PSK: transmitting pulses with M possible different carrier phases, and allowing each pulse to represent log M bits. 2 st ()Acos(2ft)  Binary PSK: “1” 1 c st ()Acos(2ft) “0” 2 c  Quaternary PSK: “11” st ()Acos(2ft ( /4)) 1 c (QPSK) “10” st ()Acos(2ft /4) 2 c st ()Acos(2ft3/ 4) “00” 3 c st ()Acos(2ft5/ 4) “01” 4 c Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 741 QPSK AA st () Acos(2ft /4) cos(2 ft) sin(2 ft) “1 1” cc 1 c 22 AA st ( ) Acos(2 ft / 4) cos(2ft) sin(2ft) “1 0” 2 cc c 22 AA st () Acos(2 ft 3 /4) cos(2ft) sin(2ft) “0 0” 3 cc c 22 AA st () Acos(2 ft 5 /4) cos(2ft) sin(2ft) “0 1” 4 cc c 22 A QPSK signal can be decomposed into the sum of two PSK signals: an inphase component and a quadrature component. AA st ( ) d cos(2 ft)d sin(2ft) QPSK I c Q c 22 1 if 1 b  1 if 1 b  2i 21 i d  d Q  I 1 if 0 b 1if 0 b 2i   21 i Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 742 QPSK Modulator AA st ( ) d cos(2 ft)d sin(2ft) QPSK I c Q c 22 1 if 1 b  1 if 1 b  2i 21 i d d Q  I 1 if 0 b 1if 0 b  2i 21 i  A cos(2 f t) c 2 • Modulator Mapping b d 2i1 I 1 if 1 b  21 i d x  I s 1if 0 b Bit series QPSK  21 i Serial to Parallel + b i b 2i d Mapping Q x 1 if 1 b  2i d  Q 1 if 0 b A  2i sin(2 f t) c 2 Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 743 Bandwidth Efficiency of QPSK G () f QPSK … ……… f f f f f f 0 f c c c c c c 2/ • Symbol rate: R1/ • Bit rate:RR22/ SQ , PSK bQ,, PSK S QPSK BR 2 • Required Channel Bandwidth: R hS 90 ,QPSK bQ , PSK BR 4  2R hS 95 ,QPSK bQ , PSK • Bandwidth Efficiency:  1 with 90 inband power QPSK with 95 inband power  0.5 QPSK QPSK achieves higher bandwidth efficiency than BPSK Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 744 Summary II: Digital Bandpass Modulation Bandwidth Efficiency (90 inband power) Binary ASK 0.5 1 0.5 Binary FSK 1/ f R b Binary PSK 0.5 1 QPSK Lin Dai (City University of Hong Kong) EE3008 Principles of Communications Lecture 7
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