EE4512 Analog and Digital Communications Chapter 4. Chapter 4 Receiver Design

Transcription

1 Chapter 4 Receiver Design

2 Chapter 4 Receiver Design Probability of Bit Error Pages

3 Probability of Bit Error The low pass filtered and sampled PAM signal results in an expression for the probability of bit error P b (S&M p ). 127). A is the amplitude at the sampling point and γ is the attenuation of the channel (0 γ 1) P{ ith bit in error } = P(b i = 0) P{ n o [ (i-1) 1)T b + T b /2 ] < γa A } + P(b i = 1) P{ n o [ (i-1) 1)T b + T b /2 ] γa A }

4 Review of Probability and Stochastic Processes (S&M p ) 132) Probability distribution function F X (a) = P{ X = a } Probability density function f X (x) = d F X (x) / dx Mean (or expected value) µ X = x f X (x) dx Variance σ X2 = (x µ X ) f X (x) dx E { (X( µ X ) 2 }

5 Review of Probability and Stochastic Processes (S&M p ) 132) Joint probability distribution function Joint probability density function F X,Y (a (a,, b) = P{ X = a and Y = b } f X,Y (x, y)) = 2 F X,Y (x, y) / x y Conditional probabilities P { X > a and event Z } = P { event Z } P{ X > a event Z }

6 Chapter 4 Receiver Design Examining Thermal Noise Pages

7 Johnson Nyquist noise or thermal noise is the electronic noise generated by the thermal agitation of the charge carriers (usually the electrons) ) inside an electrical conductor at equilibrium. This thermal noise was first measured by John B. Johnson at Bell Labs in He described his findings to Harry Nyquist,, also at Bell Labs, who was able to explain the results. Harry Nyquist

8 Thermal (or Gaussian) noise is approximately white, meaning that the power spectral density is equal throughout the frequency spectrum.. Additionally, the amplitude of the signal has very nearly a Gaussian probability density function with mean µ n = 0. S&M Figure 4-34 µ n = 0 σ n = 1

9 Since thermal noise has a Gaussian probability density function the probability that a noise voltage n(t) ) at time t o will be less than or equal to a threshold γa A is (S&M Eq. 4.27): γa 1 (x -µ 2 ) ( n) P{ n(to γa } = FX γa) = exp dx 2 2π σ 2σ n n and the probability that a noise voltage n(t) ) at time t o will be greater than a threshold γa is (S&M Eq. 4.28): 2 1 (x -µ ) > ( n) P{ n(to γa } = 1 FX γa) = exp 2 dx γa 2π σn 2σn

10 The probabilistic properties of thermal noise do not change with time (stationarity( stationarity). Thermal noise is an insidious property of communication systems that limits the speed of reliable data transmission and the detection of weak signals.

11 A MATLAB and Simulink simulation verifies the spectral characteristics of thermal noise and the performance of low-pass filtered f cutoff = khz thermal noise. MS Figure 1.11

12 Thermal noise PSD = N o, f MS Figure 1.12 N o N o (single-sided sided spectrum) N o Thermal noise LPF PSD = N o, f < khz N o khz

13 Thermal noise PSD = N o, f MS Figure 1.12 uncorrelated Thermal noise autocorrelation MS Figure 1.14 uncorrelated

14 Thermal noise LPF PSD = N o, f < khz uncorrelated LPF thermal noise autocorrelation MS Figure 1.14 correlated

15 Simulink Histogram block from the Statistics, Signal Processing Blockset is analogous to the pdf. The Vector Scope block from the Sinks, Signal Processing Blockset, displays the histogram. Simulink blocks to calculate the mean µ and variance σ 2 are in Statistics. NoiseHistogram.mdl

16 Thermal noise histogram display µ = 10 σ 2 = 1 Gaussian pdf LPF f cutoff = 1 khz thermal noise histogram display µ = σ o2 = 0.64 pdf remains Gaussian after LPF

17 LPF thermal noise has an average normalized power: σ 2 o = N o f cutoff (S&M p. 135) The probability that the ith-bit is received in error is: P(b = 0) P{ n(t ) < γa } + P(b = 1) P{ n(t ) γa } = i o i o γa 2 1 (x -µ o ) P(b i = 0) exp 2 dx + 2π σo 2σo 2 1 (x -µ o ) P(b i = 1) exp 2 dx γa 2π σo 2σo

18 If and only if the binary thresholds are symmetrical ( γa( and γa), the P(b i = 0) + P(b i = 1) = 1 and the Gaussian normal pdfs are symmetrical the probability that the ith-bit is received in error becomes (S&M p. 136): P(b = 0) P{ n(t ) < γa } + P(b = 1) P{ n(t ) γa } = i o i o γa 1 2π σ o (x -µ o ) exp 2 2σo 2 dx -γa γa

19 Chapter 4 Receiver Design Gaussian Probability Density Function, Probability of Bit Error Pages

20 Gaussian (normal) probability density function (pdf)

21 Gaussian (normal) probability distribution function

22 Gaussian (Normal) Probability Distribution Abraham de Moivre was a French mathematician famous for de Moivre's formula,, which links complex numbers and trigonometry,, and for his work on the normal distribution and probability theory in He wrote a book on probability theory entitled The Doctrine of Chances which was said to be highly prized by gamblers Gauss rigorously justified and extended the work in 1809.

23 Gaussian (Normal) Probability Distribution Johann Carl Friedrich Gauss was a German mathematician and scientist who contributed significantly to many fields, including number theory, geometry, electrostatics, astronomy and optics

24 Gaussian pdfs µ = 0, σ = 1 S&M Figure 4-6a4 µ = 1.6, σ = 1 S&M Figure 4-6b4

25 Gaussian pdfs µ = 0, σ = 2 S&M Figure 4-6c4 µ = 1, σ = 2 S&M Figure 4-6d4

26 The probability of bit error is the area under the Gaussian pdf from the threshold a to which could be tabulated. However, the probability of bit error is determined by three independent variables (a,( µ and σ) ) and this would be an unwieldy table. Q-function S&M Figure 4-74

27 Rather, construct a single table with µ =0 and σ = 1 for the probability of bit error as the area under the Gaussian pdf as a function of the threshold a only which is known as the complementary error or Q-function. Q-function S&M Figure 4-74

28 The Q-function for µ =0 and σ = 1 as a function of the threshold a is listed in Table 4-14 (S&M p. 141) and Appendix B (MS p )

29 Reading the Q-function table correctly What is Q(1.82)? What is Q(2.63)? What is Q(3.18)?

30 Reading the Q-function table correctly Q(1.82) = Q(2.63) = Q(3.18) =

31 What if µ 0? It can be shown that the Q-function table remains valid if the threshold variable in the table is changed from a to a µ.. Note that the areas under the Gaussian pdfs are the same. S&M Figure 4-8a4 Figure 4-8b4 a a µ

32 What if µ 0 and σ 1? It can also be shown that the Q-function table remains valid if the threshold variable in the table is first changed from a to a µ and a S&M Figure 4-9a4 Figure 4-9b4 a µ

33 then the threshold variable in the Q-function table is changed from a µ to (a( µ) ) / σ. Note that argument on the x-axis is compressed by σ and the y-axis is expanded by σ. a µ S&M Figure 4-9b4 Figure 4-9c4 (a µ) / σ

34 For simple baseband PAM the probability of bit error P b is expressed by the Q-function: Q noise margin of sampled value P b = Q average normalized noise power at the input to the single point sampler sampling at T b /2 binary rectangular PAM data with AWGN

35 Chapter 4 Receiver Design Optimal Receiver: The Matched Filter or Correlation Receiver Pages

36 The simple baseband PAM receiver structure is: But is this the best that there is?? What about sampling an odd number of times (like 3) during each bit time T b? T b

37 Although sampling can be increased to a very large, odd number of samples during T b, there is an optimal way: Since LPF in PAM improved performance, assume that the optimal processing is a linear filter H(f) ) (S&M p. 150)

38 After development (S&M p ) 153) the optimal processing is a linear filter H o (f) The optimal linear filter H o (f) ) has an impulse response h o (t) ) and is known as a matched filter since the processing is matched to input signal s(t): H o(f) = k S (f) exp( j 2π f i Tb ) -1-1 h o(t) = F { H o(f) } = k F { S (f) exp( j 2π f i Tb ) } h (t) = k s(it t) o b impulse response matched to s(t)

39 The impulse response of the optimum filter h o (t) ) is a scaled (by k), time delayed (by iti b ) and time reversed (function of iti b t): h (t) = k s(it t) o b

40 When optimum processing is used the argument inside the Q-function Q is maximized (S&M p ) 154) and the probability of bit error P b is: P P P b b b r(i T b) h(i Tb) = Q maximum σ p r(i Tb) * h(i Tb) = Q maximum 2 σp 2 E b = Q S&M Eq No where E b is the energy per bit of the received signal. 2

41 The optimum filter H o (f) ) is equivalent to the correlation receiver (S&M p ). 156). Optimum Filter h (t) = k s(i t) o T b Correlation Receiver

42 Since the optimum filter H o (f) ) and the correlation receiver are equivalent, with s 1 (t) = s(t) ) for a matched filter and r(t) ) = γ s(t) ) where γ is the communication channel attenuation, the energy-per per-bit E b is (S&M p. 156, Eq 4.62): E b it b (i-1) T itb 2 2 = γ s(t) γ s(t) dt = γ s(t) dt b (i-1) T b

43 The expected or mean value a i (it b ) is the output of the correlation receiver when r(t) ) = γ s i (t) ) and n(t) ) = 0 where γ is the communication channel attenuation. it b a(i T )= γ s (t) s (t) dt S&M Eq i b i 1 (i-1) T b

44 P b = Q( (2 E b / N o ) ) and the ratio E b / N o can be expressed in db: 10 log 10 (E b / N o ). The resulting plot of P b verses E b / N o in db is a characteristic of binary symmetric PAM with AWGN. S&M Figure 4-144

45 Symmetric binary PAM implies that the two transmitted signals for binary 1 and binary 0 s(t) and the resulting outputs a(it b ) from the correlation receiver are equal in magnitude but opposite in sign: Threshold = 0 s bi=1 =1 (t) = s bi=0 (t) S&M Figure 4-154

46 The probability of bit error for equally-likely likely binary symmetric PAM is the sum of the error regions shown. P(b i =0) = P(b i =1) = 0.5 S&M Figure 4-164

47 The binary PAM signals are symmetrical and the threshold is 0 (equidistant( from the means or expected values ± a(it b ) ).). The error regions are equal in area. S&M Figure 4-164

48 The probability of bit error does not minimize if the correlation receiver threshold is misadjusted (τ 0). With a misadjusted threshold the apriori probabilities are now important since the area of the error regions are no longer equal. S&M Figure 4-174

49 The correlation receiver is also known as the integrate- and-dump dump which describes the process. matched filter or correlation receiver MS Figure 2.29

50 EE4512 Analog and Digital Communications Chapter 3 Chapter 2 Baseband Modulation and Demodulation Optimum Binary Baseband Receiver: The Correlation Receiver Pages

51 The correlation receiver can be simulated in Simulink: MS Figure 2.29

52 The Simulink Integrate and Dump block is in the Comm Filters, Communications Blockset

53 The parameters of the Simulink Integrate and Dump block are the integration period and offset in simulation samples. f simulation T simulation simulation = 50 khz simulation = 1/f simulation 0.02 msec T b = 1 msec simulation = Integration period = T b /T simulation = 1/ 0.02 = 50 samples Offset = 0 samples

54 The complete binary symmetrical rectangular PAM digital communication system with BER analysis and optimum receiver. MS Figure 2.30 Transmitter Receiver BER

55 Observed BER as a function of SNR for binary rectangular PAM in a LPF simple receiver (LPF( LPF) ) and the optimum correlation receiver (CR( CR) ) with normalized signal power = 25 W (MS Table 2.3 p. 27 and Table 2.6 p. 43). SNR db AWGN σ 2 V 2 BER (LPF( LPF) BER (CR( CR)

56 The energy per bit E b = V 2 -sec (S&M Eq. 4.62, p. 156) for rectangular ± 5 V PAM with the channel attenuation γ = 1: E b it b (i-1) T itb 2 2 = γ s(t) γ s(t) dt = γ s(t) dt b (i-1) T b The observed bit error rate (BER) can be compared to the theoretical probability of bit error P b (S&M Eq. 4.58, p. 154) to validate the basic simulation. P b 2 Eb = Q No

57 The ratio of the energy per bit E b to the noise power spectral density N o (E b / N o ) rather than the SNR is the conventional metric used for BER performance. E b it b (i-1) T itb 2 2 = γ s(t) γ s(t) dt = γ s(t) dt b P b 2 Eb = Q No (i-1) T b S&M Eq S&M Eq. 4.58

58 The BER and P b comparison (MS Table 2.7, p. 44): Table 2.7 Observed BER and Theoretical P b as a Function of E b / N o in a Binary Symmetrical Rectangular PAM Digital Communication System with Optimum Receiver E b / N o db BER P b

59 Chapter 4 Receiver Design Correlation Receiver for Asymmetric PAM, Optimum Thresholds, Synchronization Pages

60 Asymmetric PAM signals do not have equal means or expected values of the output of the correlation receiver: a 2 (it b ) a 1 (it b ) S&M Figure 4-194

61 The optimum threshold τ opt is again equidistant between the means or expected values of the output of the correlation receiver: τ opt a(i T )+a(i T ) = 2 b 1 2 b S&M Eq S&M Figure Here a 2 (it b ) + τ opt = a 1 (it b ) τopt and the threshold is equidistant from the means or expected values.

62 If the apriori probabilities are not equal then the optimum threshold τ opt is not equidistant from the means or expected value of output of the correlation receiver. An asymmetric binary PAM signal is shown: a 2 (it b ) a 1 (it b ) S&M Figure 4-234

63 The optimum threshold τ opt where the apriori probabilities are (1 M) ) and M (which sums to 1) is: τ opt = 2 σ ln ( M / (1 M) ) + a (i T ) a (i T ) o 2 b 1 b [ T T ] 2 a (i ) a (i ) 2 b 1 b S&M Eq if M = 0.5 then: τ opt a(i T )+a(i T ) = 2 b 1 2 b S&M Eq. 4.71

64 Asymmetric PAM with an optimum threshold τ opt has a probability of bit error P b : S&M Figure P P P b b b =Q a 1(i Tb) a 2(i Tb) 2 σo 2 [ a (i T ) a (i T ) ] N =Q σ = 1 b 2 b 2 o 2 o 4 σ o 2 S&M Eq and Eq itb E { ( )} 2 d = Q where E = d γ s 1(t) s 2(t) dt 2 No (i-1) Tb

65 The optimum correlation receiver for asymmetric binary PAM uses the difference signal s 1 (t) s 2 (t) as the reference: S&M Figure 4-204

66 The optimum correlation receiver can be reconfigured as an alternate but universal structure which can be used for both asymmetric or symmetric binary PAM signals: S&M Figure 4-214

67 If the apriori probabilities are not equal then the optimum threshold τ opt is not equidistant from the means or expected value of output of the correlation receiver. An asymmetric binary PAM signal is shown: a 2 (it b ) a 1 (it b ) S&M Figure 4-234

68 The optimum threshold τ opt where the apriori probabilities are (1 M) ) and M (which sums to 1) is: τ opt = 2 σ ln ( M / (1 M) ) + a (i T ) a (i T ) o 2 b 1 b [ T T ] 2 a (i ) a (i ) 2 b 1 b S&M Eq if M = 0.5 then: τ opt a(i T )+a(i T ) = 2 b 1 2 b S&M Eq. 4.71

69 The probability of bit error P b then is: S&M Eq a(i 1 Tb) τopt τ opt a(i 2 Tb) Pb = M Q + (1 M) Q σo σo ( ) 2 ( ) 2 a(i 1 Tb) τopt τ opt a(i 2 Tb) P = Q b M + (1 M) Q 2 No 2 No

70 If the apriori probabilities are equal (M( = 0.5): S&M Figure a(i T = 2 b)+a(i T 1 b) τopt 2 and P b becomes: P P b b ( ) 2 ( ) 2 a(i 1 Tb) τopt τ opt a(i 2 Tb) = 0.5 Q Q 2 No 2 No 2 ( ) a(i 1 Tb) - a(i 2 Tb) = Q Ed = Q S&M p N o 2 No

71 The BER and P b comparison (MS Table 2.8, p. 49): Table 2.8 Observed BER and Theoretical P b as a Function of E d / N o in an Asymmetrical Binary Rectangular PAM Digital Communication System with Optimum Receiver. E d / N o db BER P b

72 Since for binary rectangular PAM E d = 4 E b (S&M p. 168) the BER performance for asymmetric PAM is comparable to symmetric PAM if a E d / N o is reduced by 6 db (10 log (4) = 6) : E d / N o db BER Asymmetric PAM (MS Table 2.8, p. 49) P b E b / N o db BER P b Symmetric PAM (MS Table 2.7, p. 44)

73 Chapter 2 Baseband Modulation and Demodulation The Correlation Receiver for Baseband Asymmetrical Signals Pages

74 The optimum threshold τ opt requires the additive Gaussian noise variance σ 2 as processed by the correlation receiver or σ o2 : τ opt = 2 σ ln ( M / (1 M) ) + a (i T ) a (i T ) o 2 b 1 b [ T T ] 2 a (i ) a (i ) 2 b 1 b σ o MS Figure 2.31

75 The optimum threshold τ opt also requires the apriori probabilities P 1 = M and P 0 = M 1: τ opt = 2 σ ln ( M / (1 M ) ) + a (i T ) a (i T ) o 2 b 1 b [ T T ] 2 a (i ) a (i ) 2 b 1 b Here P 1 = = M and P 0 = 1 M = Mean MS Figure 2.32

76 The Mean and Variance blocks are in the Statistics, Signal Processing Blockset

77 The parameter of the Mean block can be set for running mean or an average over the simulation period. The result can be shown by a Display block from the Simulink Blockset. MS Figure 2.32

78 The parameter of the Variance block can be set for running variance or an average over the simulation period. The result can be shown by a Display block from the Simulink Blockset.

79 The Simulink simulation for binary asymmetrical PAM with BER analysis has a threshold adjustment based on the estimate of τ opt from measurements of σ o and M. τ opt = 2 σ ln ( M / (1 M) ) + a (i T ) a (i T ) o 2 b 1 b [ T T ] 2 a (i ) a (i ) 2 b 1 b Reference MS Figure 2.33 Threshold

80 The Simulink simulation for the alternative and universal structure for the correlation receiver: MS Figure 2.34 S&M Figure 4-214

81 Synchronization (not considered here) provides timing recovery or the exact beginning and end of a bit time T b : S&M Figure start of T b?

82 Synchronization of both symbol time and carrier frequency and phase is one of the advanced topics in EE4542 Telecommunications Engineering.. The effect is called jitter.

83 Chapter 4 Receiver Design Multi-level level PAM (M-ary PAM) Pages

84 Multi-level level (M-ary) PAM is another means to minimize the bandwidth required for a data transmission rate r b b/sec. Rather than transmitting a binary signal in a bit time T b, send a multi-level level (usually a power-of of-2) signal during the same period called the symbol time T S. A multi-state comparator determines the received symbol which is then decoded to the received bits. S&M Figure M = 4

85 The three optimum thresholds if the apriori probabilities are equally likely (P i = 0.25) are: τ τ τ opt1 opt2 opt3 = = = a(i T )+a(i T ) 1 S 2 2 a(i T )+a(i T ) 2 S 3 2 a(i T )+a(i T ) 3 S 4 2 S S S S&M Figure 4.49 S&M Eq

86 The probability of symbol error P s, where M = 2 n is the number of levels, can be shown to be: where: E d,symbol i T { } j k = γ s(t)-s (t) dt s (i-1) T ( ) E 2M 1 P= s Q M 2N s d,symbol o 2 M = 4 S&M Eq S&M Eq

87 There are 2 (M 1) error regions due to only M = 4 adjacent regions being misinterpreted with M = 2 n equally probable symbols. The probability of occurrence P j for a misinterpreted symbol is also equally likely and is: 1 P= j 2M 1 SVU Eq ( )

88 The amplitudes of the M-ary M (M = 4) rectangular PAM signal is ± 3A and ± A. The energy difference for a symbol E d,symbol = 4γ4 2 A 2 T S (S&M Eq ): E d,symbol i T { } j k = γ s(t)-s (t) dt s (i-1) T s 2 The average received energy per symbol E avg,symbol 5γ 2 A 2 T S (S&M Eq ) avg,symbol = S&M Figure modified M =

89 Then E d,symbol = 0.4 E avg,symbol and substitute (M = 4): 3 E P= s Q 2 2N d,symbol o S&M Eq b 3 P= s Q E N avg,symbol o S&M Eq a S&M Figure modified M =

90 The average energy per bit E b = 0.5 E avg,symbol since M =4 and there are two bits per symbol: 3 P= s Q E N avg,symbol o S&M Eq a 3 P= s Q N o E b 10 S&M Eq b S&M Figure modified M =

91 The symbols transmitted can have 1 bit in error 4 / 6 of the time and 2 bits in error 2 / 6 of the time: Transmitted Di-Bit Received Di-Bit Bits In Error Pb,4-level = P(1 of 2 bits in error) + P(2 of 2 bits in error) Pb,4-level = Ps + Ps = P S&M Eq s

92 The probability of bit error P b,4-symbol = 2 P s / 3 but can that be improved? Change the assignment of symbols to di- bits as a Gray code and there is only 1 bit in error for each of the six error regions and P b,4-symbol = P s / 2 Transmitted Di-Bit Received Di-Bit Bits In Error Pb,4-level = P(1 of 2 bits in error) Pb,4-level = Ps = Ps 62 2

93 The simple change to a Gray coded symbol improves the probability of bit error P b : P P b b E b 0.8 Eb = Q = Q 32 No No E b Eb = Q = Q 22 No 4 No Straight binary coded Gray coded S&M Figure

94 Frank Gray was a researcher at Bell Labs s who made numerous innovations in television and is remembered for the Gray code.. The Gray code appeared in his 1953 patent and is a binary system often used in electronics.. Gray also conducted pioneering research on the development of television.. He proposed an early form of the flying spot scanning system for TV cameras in 1927, and helped develop a two-way way mechanically scanned TV system in Frank Gray

95 Chapter 2 Baseband Modulation and Demodulation Multilevel (M-ary) Pulse Amplitude Modulation Pages 49-55

96 The Simulink simulation of a 4-level 4 straight binary (not( Gray coded) ) rectangular PAM system with BER analysis: Transmitter MS Figure 2.35 Receiver BER bit to symbol converter symbol to bit converter

97 Although Simulink provides a Bit to Integer block and an Integer to Bit block from the Utility Blocks, Communication Blockset, these blocks use a vector of bits rather than from and to a serial bitstream.

98 A bit to symbol converter Simulink subsystem can be configured from Downsampler blocks and a Delay block from the Signal Operations, Signal Processing Blockset and the Gain and Sum blocks from the Simulink, Math Operations to produce the 4-level 4 PAM signal (0, 1, 2 or 3) from the input di-bit (00, 01, 10, or 11) with the most significant bit (MSB) first. MS Figure 2.36

99 The Simulink bit to symbol converter can be easily reconfigured to produce an 8-level 8 PAM signal (0, 1, 2, 7) from the input tri-bit (000, 001, 111) with the most significant bit (MSB) first. MS Figure 2.37

100 A symbol to bit converter Simulink subsystem can be configured from Constant blocks, Pulse Generator blocks and a Multiport Switch block from the Signal Routing, Simulink Blockset to produce the di-bit (MSB first) from the 4-level PAM signal. MS Figure 2.40

101 A symbol to bit converter Simulink subsystem can be easily reconfigured to produce the tri-bit (MSB first) from an 8-level 8 PAM signal. MS Figure 2.41

102 Binary data to 4-level 4 PAM converter MS Figure 2.36

103 Transmit binary data, r b = 1 kb/sec 01 MSB LSB Transmit 4-level 4 PAM signal, r S = 500 symbols/sec 1

104 M-ary PAM transmits n bits per symbol (M = 2 n ) but has the same rectangular pulse shape as binary PAM. The normalized power spectral density for M-ary M PAM PSD M has the same sinc shape as that for binary PAM PSD B but uses the symbol time T S rather than the bit time T b : M 2 avg ( TS ) ( T ) PSD (f) = A T sinc π f 2 PSD B(f) = A Tb sinc π b f S

105 Observed BER and theoretical P b as a function of E b / N o in a 4-level4 rectangular straight-binary coded PAM digital communication system with optimum receiver (MS Table 2.11 p. 58). E b / N o db BER P b

106 The M-ary M PAM PSD uses the average amplitude A avg where P j is the apriori probability of occurrence of the amplitude A j. 2 M = 2 A A P avg j j j=1 avg,

107 The M-ary M (M = 4) PAM PSD can be determined by a Simulink model Fig239.mdl MS Figure 2.39

108 The Simulink simulation of a 4-level 4 straight binary (not( Gray coded) ) rectangular PAM system measures the BER performance. MS Figure 2.35 Transmitter Receiver BER

109 The rectangular M-ary M pulse width is the entire symbol time T S and is optimum in the bandwidth sense. The bandwidth is similar to that for binary PAM but with T b = T S (cf. S&M Table p. 86). Table Bandwidth of a Binary Rectangular M-ary M PAM Signal as a Percentage of the Total Power (MS p. 53) Bandwidth (Hz) Percentage of Total Power 1/T S 90% 1.5/T S 93% 2/T S 95% 3/T S 96.5% 4/T S 97.5% 5/T S 98%

110 PSD M M = 4, r b = 1 kb/sec, first null bandwidth = 500 Hz 500 Hz PSD B r b = 1 kb/sec, first null bandwidth = 1 khz 1 khz

111 Chapter 2 Baseband Modulation and Demodulation Performance of M-ary M Pulse Amplitude Modulation Gray Encoded Data Pages 57-60

112 The Simulink simulation of a 4-level 4 straight binary Gray encoded rectangular PAM system with BER analysis: Transmitter MS Figure 2.42 Receiver bit to Gray encoded symbol converter BER Gray encoded symbol to bit converter

113 The M = 4 level symbol (0, 1, 2 and 3) is Gray encoded by a Lookup Table Block from the Simulink Blockset. MS Figure 2.43 M = 4 symbol bit to Gray encoded symbol converter

114 The M = 4 level symbol (0, 1, 2 and 3) is Gray encoded by a Lookup Table Block from the Simulink Blockset.

115 The M = 4 level symbol (0, 1, 2, and 3) is Gray encoded by a Lookup Table Block by mapping [0, 1, 2, 3] to [0, 1, 3, 2] Gray coding

116 The M = 4 level Gray encoded symbol is decoded as di-bits by a Lookup Table Block from the Simulink Blockset. M = 4 symbol MS Figure 2.44 Gray encoded symbol to bit converter

117 The M = 4 level symbol (0, 1, 2, and 3) is Gray decoded by a Lookup Table Block by mapping [0, 1, 2, 3] to [0, 1, 3, 2] Gray coding

118 The PSD of a Gray coded rectangular PAM signal is the same as a straight-binary rectangular PAM signal. The coding does not change the PSD. The PSD is only affected by the pulse shape and data rate. PSD M M = 4, r b = 1 kb/sec, first null bandwidth = 500 Hz 500 Hz

119 The Simulink simulation of a 4-level 4 straight binary Gray encoded rectangular PAM system measures the BER performance. MS Figure 2.42 Transmitter Receiver BER

120 Observed BER and theoretical P b as a function of E b / N o in a 4-level4 rectangular Gray coded PAM digital communication system with optimum receiver (MS Table 2.12 p. 60). E b / N o db BER P b

121 Observed BER as a function of E b / N o for binary rectangular PAM for straight-binary coded (SBC( SBC) ) and Gray coded (GC) data (MS Table 2.11 p. 58 and MS Table 2.12 p. 60). E b / N o db BER (SBC( SBC) BER (GC( GC) Although a slight improvement, Gray coding improves BER performance without an increase in E b.

122 End of Chapter 4 Receiver Design