通信原理英文版教學(xué)課件：Lecture 7 Digital Bandpass Transmission System

1、1Communication Theory2Communication TheoryLecture 7 Digital Bandpass Transmission System3IntroductionDigital modulationDigital band pass transmission systemTwo methods of digital modulationsUsing analog modulation methodsKeying: amplitude shift keying, frequency shift keying, phase shift keyingBinar

2、y modulation and M-ary modulationamplitude shift keying frequency shift keying phase shift keyingLecture 7 Digital bandpass transmission system47.1 Binary digital modulation7.1.1 Binary amplitude shift keying (2ASK)On-off shift keying (OOK)WaveformLecture 7 Digital bandpass transmission system5Gener

3、al expression of 2ASK Ts symbol duration g(t) baseband pulse waveform with duration Ts . an The level of the N-th symbol valueThe corresponding 2ASK of signal is OOK signal.Lecture 7 Digital bandpass transmission system62ASK signal generationAnalog modulationKeyingLecture 7 Digital bandpass transmis

4、sion system72ASK signal demodulationNon-coherent demodulation (envelope detector)Coherent demodulation (synchronous detection)Lecture 7 Digital bandpass transmission system8The time domain waveform of the non-coherent demodulation Lecture 7 Digital bandpass transmission system9PSD 2ASK signal can be

5、 expressed as where s(t) is the binary unipolar random rectangular pulse sequenceAssume: Ps (f) the PSD of s(t) P2ASK (f) the PSD of 2ASK signalthe PSD of 2ASK signal is the linear combination of baseband signal power spectrum Ps (f)Lecture 7 Digital bandpass transmission system10The general express

6、ion for the unipolar random pulse sequences power spectrum iswhere fs = 1/Ts G(f) is the frequency spectrum of g (t)for all integers that m 0Lecture 7 Digital bandpass transmission system11When P =1/2,the PSD of 2ASK signal isLecture 7 Digital bandpass transmission system12The PSD of 2ASKLecture 7 D

7、igital bandpass transmission system13The bandwidth of 2ASK signal is twice the bandwidth of the baseband signal, where fs = 1/Ts The transmission bandwidth of 2ASK signal is twice the symbol rate.Lecture 7 Digital bandpass transmission system147.1.2 Binary Frequency Shift Keying (2FSK)Principle: For

8、 2FSK, the carrier frequency changes with a binary baseband signal, whose expression isLecture 7 Digital bandpass transmission system15Waveforms2FSK signals (a) can be decomposed into (b) and (c). Lecture 7 Digital bandpass transmission system16 g(t) rectangular pulse; Ts symbol duration; n and n ar

9、e the initial phase of the nth signal. If n=0 and n=0, the expression of 2FSKsignal can be simplified as Lecture 7 Digital bandpass transmission system172FSK signal generationAnalog FMKeying, the phases between adjacent symbols are not continuous.Lecture 7 Digital bandpass transmission system18Demod

10、ulationNon-coherent demodulationLecture 7 Digital bandpass transmission system19Coherent demodulationLecture 7 Digital bandpass transmission system20Frequency discriminator, differential detection Zero crossing detectionLecture 7 Digital bandpass transmission system21PSD Discontinuous phase 2FSK sig

11、nal can be viewed as the superposition of two 2ASK signals with different carrier frequencies The PSD is When P = , we haveLecture 7 Digital bandpass transmission system22 Lecture 7 Digital bandpass transmission system23If | f1 f2 | fs, the continuous spectrum has two peaks.The 2FSK signals bandwidt

12、h is approximated aswhere fs = 1/Ts is the bandwidth of baseband signal.Lecture 7 Digital bandpass transmission system247.1.3 Binary Phase Shift Keying (2PSK) 2PSK signal expressionThe time domain expression of the 2PSK signal is where n denotes the phase of the nth symbolThe expression above can be

13、 rewritten as:Lecture 7 Digital bandpass transmission system25Two symbols are with the same waveform and opposite polarity. wherewhere g(t) is rectangular pulse with width Ts andIt is also called a binary absolute phase-shift keying.Lecture 7 Digital bandpass transmission system26Typical waveformLec

14、ture 7 Digital bandpass transmission system272PSK signal generationAnalog modulationKeying Lecture 7 Digital bandpass transmission system282PSK demodulationLecture 7 Digital bandpass transmission system29Lecture 7 Digital bandpass transmission system“Inversed phenomenon”Long continuous cosine wavefo

15、rmDifferential phase shift keying (DPSK)30PSDCompared with 2ASK,2ASK：2PSK：the expression is almost the same. Only difference is s (t) (an is different). 2ASK is unipolar while 2PSK is bipolar. The PSD of the 2PSK isLecture 7 Digital bandpass transmission system31 The PSD of bipolar rectangular rando

16、m pulse sequence isSubstituting it into the PSD of 2PSK, we obtainWhen P =1/2, We obtainLecture 7 Digital bandpass transmission system32PSD The PSDs of 2PSK and 2ASK signals are similar. No carrier component.Lecture 7 Digital bandpass transmission system337.1.4 Binary differential phase shift keying

17、 (2DPSK)Principle 2DPSK uses the relative phase changes of adjacent symbols to transmit signal.Assume is the difference between the adjacent symbols, Lecture 7 Digital bandpass transmission system34Corresponding 2DPSK signal waveform：The information symbol is determined by the difference between the

18、 phase changes of the adjacent symbols.Lecture 7 Digital bandpass transmission system35The relationship between digital information and can also be defined asVector expression of 2DPSKIn method B, current symbol change its phase of /2 compared to the previous one. There must be a phase transition be

19、tween adjacent symbol. (a) Method A(b) Method BLecture 7 Digital bandpass transmission system362DPSK signal generation Coherent differential code follows:where represents mode 2 plus, bn-1 is the previous code of bn. The inverse form of the above procedure is decoding.Lecture 7 Digital bandpass tran

20、smission system372DPSK demodulationCoherent demodulation + decoding Lecture 7 Digital bandpass transmission system38Differential demodulation (phase compare)Lecture 7 Digital bandpass transmission system39PSDThe PSD of 2DPSK and 2PSK are the same. The signal bandwidth is:which is the same as 2ASK.Le

21、cture 7 Digital bandpass transmission system407.2 Anti-noise performance of binary digital modulation systemIntroductionBit-error-rateConditions: Constant channel Additive white Gaussian noise (AWGN)Lecture 7 Digital bandpass transmission system417.2.1 Anti-noise performance of 2ASK systemPerformanc

22、e of coherent demodulationAnalytical modelLecture 7 Digital bandpass transmission system42Assume in a symbol duration Ts, the waveform send from the transmitter can be expressed aswhereThe input waveform of the receiver iswhere ui(t) is the waveform of uT(t) going through the channel and ni(t) is AW

23、GN with 0 mean.Lecture 7 Digital bandpass transmission system43Assume that the transmitting coefficient is K. Let a =AK, Assume that the band-pass filter at the receiver is ideal. where n(t) is the output noise through ideal band-pass filter. Lecture 7 Digital bandpass transmission system44From Chap

24、ter 3, n(t) is narrowband Gaussian noise, soMultiply y(t) with coherent carrier 2cos ct. After a low pass filter, the input waveform of the sampling decision isLecture 7 Digital bandpass transmission system45where a is the signal component, nc(t) is Gaussian with mean 0 variance n2. x(t) is a Gaussi

25、an random process with mean a (when sending “1”) or 0 (when sending “0”) and variance n2. Assuming that the sampling time of the kth symbol is kTs, then sample of x(t) at kTsis a Gaussian random variable whose PDF is Lecture 7 Digital bandpass transmission system46when sending “1” andwhen sending “0

26、”.If take Decision threshold as b, the decision rule would be: when x b take “1” when x b take “0” Lecture 7 Digital bandpass transmission system47 when x b take “1” when x b take “0” When sending “1”, the probability of receive “0” iswhereWhen sending “0”, the probability of receive “1” isLecture 7

27、 Digital bandpass transmission system48Assume sending “1” with probability P(1), sending “0” with probability P(0). The BER of 2ASK using coherent demodulation isWhen P(1) , P(0) and f1(x) , f0(x) are fixed, the BER Pe is closely relative with decision threshold b.Lecture 7 Digital bandpass transmis

28、sion system49Optimal threshold The BER Pe is equal to the shaded area. The shaded area is minimized when b take the b* -the intersection point of P(1)f1(x) and P(0)f0(x). Lecture 7 Digital bandpass transmission system50The optimal decision threshold is derived by letting the partial derivative Pe ov

29、er b be equal to 0we get so We obtainAfter simplification,Lecture 7 Digital bandpass transmission system51 If the probabilities of sending “1” and “0” are equal, the best decision threshold is b* = a / 2The BER of 2ASK is where is the SNR of the demodulator input. When r 1, Lecture 7 Digital bandpas

30、s transmission system52Performance of envelope detectorAnalytical model： When sending “1”, the output is When sending “0”, the output is Lecture 7 Digital bandpass transmission system53From Chapter 3, their PDFs arewhere n2 is the variance of narrowband Gaussian noise n(t).Lecture 7 Digital bandpass

31、 transmission system54Let the decision threshold be b, when sample V b take “1” when sample V b take “0” When sending “1”, the probability of mistaking it as “0” isThe integration can be calculated using Marcum Q function, Lecture 7 Digital bandpass transmission system55Letwhere r = a2 / n2 is the S

32、NR b0 =b /n is the normalized thresholdLecture 7 Digital bandpass transmission system56When sending “0”, the probability of mistaking it as “0” isThe BER of the system is:When P(1) = P(0), we have:Lecture 7 Digital bandpass transmission system57The BER Pe is half of the shaded area in the figure bel

33、ow. When b0 is at the intersection point of f1(V) and f0(V), the BER Pe is minimized. Lecture 7 Digital bandpass transmission system58Optimal thresholdBy lettingwe obtainWhen P(1) = P(0) From f1(V) and f0(V) and the formula above, we obtainLecture 7 Digital bandpass transmission system59The optimal

34、threshold is approximated as We have The normalized threshold b0* is To arbitrary SNR r， b0*lies between 21/2 and (r/2)1/2Lecture 7 Digital bandpass transmission system60 Practically, the system always work at high SNR. The optimal threshold should be taken The BER is when r the lower bound of the f

35、ormula above is Lecture 7 Digital bandpass transmission system61【Ex】A 2ASK system with the symbol rate RB = 4.8 106 Baud, sends “1” and “0” with the same probability. The receiver employ both coherent demodulation and envelope detector. The input signal amplitude is a = 1 mV. The AWGN is with PSD n0

36、 = 2 10-15 W/Hz.(1) Obtain the BER of coherent demodulation; (2) Obtain the BER of envelope detector . 【Sol】(1) The bandwidth of received bandpass filter is The average noise power of the bandpass filter is The SNR isLecture 7 Digital bandpass transmission system62The BER of coherent demodulation is

37、The BER of envelope detector isFor large SNR, the BER performance of envelope detector is close to that of coherent demodulation.Lecture 7 Digital bandpass transmission system637.2.2 Anti-noise performance of 2FSK systemPerformance of coherent demodulationAnalytical model Lecture 7 Digital bandpass

38、transmission system64Symbol “1” - f1(1) Symbol “0” - f2(1) In a symbol duration Ts, the 2FSK signal iswhereLecture 7 Digital bandpass transmission system65The input of the receiver iswhere ni (t) is AWGN with zero mean.Lecture 7 Digital bandpass transmission system66 The output of the bandpass filte

39、r at the receiver is: where n1(t) and n2(t) are the bandpass filtered narrowband Gaussian noise with zero mean and variance n2. The only difference is the central frequency.Lecture 7 Digital bandpass transmission system67Assume that the symbol “1” is sent within the time (0, Ts). The outputs of the

40、two bandpass filters areAfter coherent demodulation, Upper branch Lower branchwhere a is signal component, n1c(t) and n2c(t) are low-pass filtered Gaussian noise with zero mean and variance n2 . Lecture 7 Digital bandpass transmission system68The PDFs of x1(t) and x2(t) areThe sampled values of x1(t

41、) and x2(t) are x1 and x2. where z = x1 x2. z is Gauss random variables, with the mean a and variance z2 = 2 n2 .Lecture 7 Digital bandpass transmission system69Let the PDF of z be f(z). Similarly, The BER of coherent demodulation isAt high SNR, the above equation is approximated asLecture 7 Digital

42、 bandpass transmission system70Performance of envelope detectorAnalysis modelLecture 7 Digital bandpass transmission system71The output of the two-way envelope detector Upper branch Lower branch The sampled values of V1(t) and V2(t) are V1 and V2. V1 is generalized Rayleigh distribution and V2 is Ra

43、yleigh distribution, whose PDFs are Lecture 7 Digital bandpass transmission system72The probability of error isLetand substitute into the above equation.Lecture 7 Digital bandpass transmission system73From the Marcum Q function,we have Similarly, The BER of envelope detector isLecture 7 Digital band

44、pass transmission system74 【Ex 7.2.2】The equivalent bandwidth of 2FSK signal is 2400Hz. The frequencies of 2FSK signal are f1 = 980 Hz, f2 = 1580 Hz. The symbol rate RB = 300 B. The input SNR at the receiver is 6dB. Obtain(1) 2FSK signal bandwidth;(2) The BER of envelope detector;(3) The BER of cohe

45、rent demodulation【Sol】(1) (2) The bandwidth of the band-pass filter in the upper and lower branch is approximatelyLecture 7 Digital bandpass transmission system75Chapter 7:Digital band pass Transfer system It is only of the equivalent bandwidth of the channel (2400Hz). The noise power is reduced by

46、1/4. The output SNR after the bandpass filter is The BER of envelope detector is (3) Similarly, the BER of coherent demodulation is76Chapter 7:Digital band pass Transfer system7.2.3 Anti-noise performance of 2PSK and 2DPSK systemsThe 2PSK and 2DPSK signals with a symbol duration Ts arewheresT(t) rep

47、resents 2PSK signals of the original digital information (absolute code) and sT(t) represents 2DPSK signals of the relative code.77Performance of coherent demodulation for 2PSK systemAnalysis modelThe bandpass filter output at the receiver isAfter coherent demodulation, the sampled output isLecture

48、7 Digital bandpass transmission system78nc(t) is Gaussian noise with zero mean and variance n2. The PDF of x(t) is When P(1)=P(0), the optimal decision threshold is b* = 0. Similarly, Lecture 7 Digital bandpass transmission system79The BER of 2PSK system At high SNR, the above equation is approximat

49、ed asLecture 7 Digital bandpass transmission system80Performance of coherent demodulation for 2DPSK systemAnalysis model: coherent demodulation+decoderLecture 7 Digital bandpass transmission system81The simplified model is shown as follows: Decoding error（ Error-free ） （ A wrong code ） （ 2 consecuti

50、ve wrong codes ） （ Continuous n wrong code ） Lecture 7 Digital bandpass transmission system82BER Let Pe be the BER of bn and Pe be the BER of bn, where Pn is the probability of continuous n error in bn. The Pn is equal to the probability “n error with no error at their two ends”, Lecture 7 Digital b

51、andpass transmission system83Since the BER is always less than 1, If Pe is small, then Pe / Pe 2 If Pe is large, Pe 1/2, then Pe / Pe 1 This means that Pe is always larger than Pe . Lecture 7 Digital bandpass transmission system84Performance differential demodulation for 2DPSK systemAnalysis model L

52、ecture 7 Digital bandpass transmission system85Assume the current code is “1”, and the former code is “1” two signal y1(t) and y2(t) is expressed as where a is the signal amplitude, n1(t) and n2(t) are narrowband Gauss noise. n1(t) and n2(t) are independent. After the lowpass filter, The sampled out

53、put isLecture 7 Digital bandpass transmission system86Decision rule If x 0, “1” - correct If x larger bandwidthNormally, symbol duration = n carrier periodLecture 7 Digital bandpass transmission system111QPSK ModulationMultiplication circuit-sin0tcarrierMultiplyMultiply/2Phase Serial / parallelTrans

54、formAddcos0tA(t)s(t)abLecture 7 Digital bandpass transmission system112Serial / parallel transform012345(a) Input baseband symbolt024(b) Parallel brancht135(c) Parallel branchtLecture 7 Digital bandpass transmission system113Vector diagram Binary code “1” bipolar pulse of “+1” Binary code “0” bipola

55、r pulse of “-1”01110010a(1)a(0)b(1)b(0)Figure 7-39 QPSK vector diagramLecture 7 Digital bandpass transmission system114Selection methodSerial / parallelconvertPhaseselectionBandpassfilter4-phase carriergenerator1432abLecture 7 Digital bandpass transmission system115QPSK DemodulationBlock diagramCarr

56、ierMultiplyLowpassfiltersample/2MultiplyLowpassfiltersampleParallel/serialA(t)s(t)abcos0t-sin0tTimingExtractLecture 7 Digital bandpass transmission system116Offset QPSK(OQPSK)Disadvantages of QPSK: maximum phase difference is 180 = larger bandwidthOQPSK: maximum phase difference = 90 abk009001011270

57、10180Lecture 7 Digital bandpass transmission system117Comparison of OQPSK and QPSK waveformsa1a3a5a7a2a6a4a8a2a4a1a3a5a7a6a8Lecture 7 Digital bandpass transmission system118/4 Phase shift QPSK Advantages: maximum phase shift is 135.451110(a) (b)010011010010Lecture 7 Digital bandpass transmission sys

58、tem1197.4.4 M-ary differential phase shift keying(MDPSK)abkMethod AMethod B0090135010451127031510180225Lecture 7 Digital bandpass transmission system120MDPSK GenerationFirst methodabcdEncoderAdds(t)A(t)Serial / parallelConvert-/4CarrierMultiplyMultiply/4Lecture 7 Digital bandpass transmission system

59、121Relative phase shiftPrevious symbol and phase Current symbol and the phaseak bkkck-1 dk-1k-1ck dkk0 0900 00 11 11 00901802700 11 11 00 090180 27000 100 00 11 11 00901802700 00 11 11 00901802701 12700 00 11 11 00901802701 00 00 11 1 2700901801 01800 00 11 11 00901802701 11 00 00 1180270 0 90Lectur

60、e 7 Digital bandpass transmission system122The encoderThe second method:Selection methodRead-onlyMemoryTTakbkckdkdk-1ck-1Lecture 7 Digital bandpass transmission system123Coherent demodulation + decoderbacdA(t)-/4MultiplyMultiply/4s(t)Low passLow passsamplesampleParallel / serialConvertDecodertimingC