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1、1Communication Theory2Communication TheoryLecture 6 Digital Baseband Transmission System3Lecture 6 Digital baseband transmission systemlIntroductionnDigital baseband signal: unmodulated digital signal, it occupies the spectrum from the zero-frequency or low frequency.nDigital baseband transmission s
2、ystemnDigital bandpass transmission systemnWhy digital baseband transmission systemuWidely used in short-range data communication systemuContains many of the basic problems of the bandpass transmission uAny linear modulation of the bandpass transmission system is equivalent to a baseband transmissio
3、n system.4l6.1 Digital baseband signal and its spectral characteristicsn6.1.1 Digital baseband signaluSeveral basic baseband signal waveformLecture 6 Digital baseband transmission system5pUnipolar waveform: single polarity, easy to be generated using TTL, CMOS circuit. Disadvantage: with DC componen
4、t, short-distance transmissionpBipolar waveform: no DC component. The decision threshold is zero value, which does not subject to changes of channel characteristics. Lecture 6 Digital baseband transmission system6pUnipolar return-to-zero (RZ) waveform: RZ waveform is with duty cycle 50% . The timing
5、 information is extracted from the unipolar RZ waveform.pBipolar RZ waveform: the receiver easily identifies the beginning and ending of each symbol time, which facilitates synchronization.Lecture 6 Digital baseband transmission system7pDifferential waveform:pMulti-level waveform :Lecture 6 Digital
6、baseband transmission system8uExpression of the digital baseband signal: an the corresponding level value of the n-th symbol Ts symbol duration g(t) pulse waveform Lecture 6 Digital baseband transmission system9n6.1.2 Spectral characteristics of the baseband signaluSpectral characteristics pThe digi
7、tal baseband signal is a random pulse sequences. No deterministic frequency spectrum. Only power spectral density.uThe expression of the random pulse sequences Let a binary random pulse sequence as shown below:Lecture 6 Digital baseband transmission system10 Ts symbol durationg1(t), g2(t) waveforms
8、represent “0” and “1”pThe probabilities of g1(t) and g2(t) are P and (1-P), respectively. We have Lecture 6 Digital baseband transmission system11ps(t) can be decomposed into a steady-state waveform v(t) and the alternating waveform u(t). v(t) is the statistical averaging of random sequence s(t), v(
9、t) is periodic deterministic signal with period Ts. Lecture 6 Digital baseband transmission system12Alternating waveform u(t) is the difference between s(t) and v(t). Lecture 6 Digital baseband transmission system13uv(t) has the PSD Pv(f)v(t) is with period Tswhose Fourier series are in the interval
10、 （-Ts/2，Ts/2） Lecture 6 Digital baseband transmission system14Because only has value in（-Ts/2，Ts/2）, the limits of integration can be modified as (- , ). According to the relationship between the PSD of periodic signal and the Fourier series, we obtainLecture 6 Digital baseband transmission system15
11、uu(t) has the PSD Pu(f) Because u(t) is a random pulse sequences, of the power type, the PSD of u(t) can be founded by UT (f) the PSD of u(t)s truncated function uT(t) E Statistical averaging T truncated time, which is equal to（2N+1） symbol duration T = (2N+1) Lecture 6 Digital baseband transmission
12、 system16We first obtain the frequency spectrum of uT(t).Lecture 6 Digital baseband transmission system17The Statistical averaging isWhen m = nLecture 6 Digital baseband transmission system18When m nThus, the Statistical averaging only exists when m = n Lecture 6 Digital baseband transmission system
13、19The PSD of u (t) is The PSD of alternating waveform, Pu (f), is continuous.Lecture 6 Digital baseband transmission system20us(t) has the PSD Ps(f)Because s(t) = u(t) + v(t), we have which is double side PSD. The single side PSD isLecture 6 Digital baseband transmission system21wherefs = 1/Ts symbo
14、l rateTs symbol duration G1(f) and G2(f) are the Fourier transforms of g1(t) and g2(t)Lecture 6 Digital baseband transmission system22The PSD of binary random pulse sequences Ps(f) may contain continuous PSD and the discrete PSD.The continuous PSD is present when g1(t) and g2(t) are not identical. T
15、he PSD depends on frequency spectrum of g1(t) and g2(t) as well as the probability of occurrence P.The existence of discrete PSD depends on g1(t), g2(t) and P. Normally, it is always present. However, when g1(t) = - g2(t) with probability P=1/2, there is NO discrete PSD. Discrete PSD determines whet
16、her there is a DC component and timing components.Lecture 6 Digital baseband transmission system23uEx6.1 Obtain the PSDs of unipolar NRZ and RZ rectangular pulse sequenceSol For unipolar waveform, assuming g1(t) = 0, g2(t) = g(t) We obtain When P=1/2,Lecture 6 Digital baseband transmission system24p
17、Discussion：If the waveform represents “1” g2(t) = g(t) is NRZ Rectangular pulse its PSD iswhen f = mfs, if m = 0, G(0) = Ts Sa(0) 0, the PSD Ps(f) contains DC component. If m is nonzero integer, , no timing component.Lecture 6 Digital baseband transmission system25Thus,is reduced toLecture 6 Digital
18、 baseband transmission system26If the waveform represents “1”g2(t) = g(t) is half-duty zero rectangular pulse with pulse width = Ts /2, the PSD of waveform is when f = mfs , if m = 0, G(0) = Ts Sa(0)/2 0, so the PSD Ps(f) have DC component. if m is odd, there is timing componentif m is even, there i
19、s no discrete spectrum. The PSD Ps(f) isLecture 6 Digital baseband transmission system27pPSDs of unipolar NRZ and RZ rectangular pulse sequenceLecture 6 Digital baseband transmission system28u【Ex6-2】Obtain the PSD of the bipolar NRZ and RZ rectangular pulse sequence 【Sol】For bipolar waveform, assumi
20、ng g1(t) = - g2(t) = g(t), we havewhen P = Lecture 6 Digital baseband transmission system29pDiscussion:If g(t) is NRZ rectangular pulse with height 1.If g(t) is half-duty RZ rectangular pulse with height 1Lecture 6 Digital baseband transmission system30pPSDs of the bipolar NRZ and RZ rectangular pul
21、se sequenceLecture 6 Digital baseband transmission system31pThe bandwidth of binary baseband signal depends on G1(f) and G2(f) .The smaller the duty ratio of the time the larger band width of the waveform occupied. pWhether there is a discrete PSD in unipolar baseband signal depends on the duty cycl
22、e of the rectangular pulse. Lecture 6 Digital baseband transmission system32l6.2 The common code for the baseband transmission.nMain requirementsucode: suitable for transmissionuelectrical waveform: suitable for the transmissionTwo independent but related problems Lecture 6 Digital baseband transmis
23、sion system33n6.2.1 Code selection principleuNon-DC, low-frequency component is as little as possibleuRich timing informationuMain lobe of the PSD is narrowuNot sensitive to the variance of information source statistics characteristicsuInherent error detection capabilitiesuSimple Codec Lecture 6 Dig
24、ital baseband transmission system34n6.2.2 Several codesuAMI: Alternate mark inversion codepEncoding rules: message “1” is alternately converted to “+1” and “-1”, and 0 (space) remains unchanged.pE.g.message： 0 1 1 0 0 0 0 0 0 0 1 1 0 0 1 1 AMI： 0 -1 +1 0 0 0 0 0 0 0 1 +1 0 0 1 +1 pThree levels: posi
25、tive, negative, zeroLecture 6 Digital baseband transmission system35pAdvantage: No DC component, the encoding and decoding circuit is simple, error detection capabilities by alternating polarity. After a full-wave rectification, AMI-RZ waveform is converted to a unipolar RZ waveformpDisadvantage: Fo
26、r a long code string of “0”, the timing information is difficult to extract. Lecture 6 Digital baseband transmission system36uHDB3: 3-order high-density bipolar codepModification of the AMI code. Assure that the number of continuous “0” code is no more than 3.pEncoding Rules: (1) Check the number of
27、 “0” in the message code. When the number of continuous 0 is less than or equal to 3, HDB3 code is the same as the AMI code(2) When the number of continuous “0” is more than 3, 4 continuous “0” is defined as B00V. (3) V has the same polarity with the adjacent non- “0” pulse before. The neighboring V
28、 should have opposite polarity.Lecture 6 Digital baseband transmission system37(4) The value of B is optional 0, +1 or -1, so that the V simultaneously satisfy the two requirements of (3);(5) The V code should have opposite polarity with the non- “0” pulse behind.pE.g.message： 1 0 0 0 0 1 0 0 0 0 1
29、1 0 0 0 0 0 0 0 0 l 1 AMI ： -1 0 0 0 0 +1 0 0 0 0 -1 +1 0 0 0 0 0 0 0 0 -1 +1 HDB ： -1 0 0 0 V +1 0 0 0 +V -1 +1-B 0 0 V +B 0 0 +V -l +1 Lecture 6 Digital baseband transmission system38pDecoding of HDB3：From the received symbol sequence, we can easily find the destruction point B00V. Lecture 6 Digit
30、al baseband transmission system39uManchester code (biphase code)pE.g.message: 1 1 0 0 1 0 1Manchester code: 10 10 01 01 10 01 10pAdvantage and disadvantageThe Manchester code is a bipolar NRZ waveform, only two opposite polarity level. Transitions are present it in the center of each symbol interval
31、, thus rich timing information. No DC component. The encoding is simple. The disadvantage is the reduced bandwidth efficiency.Lecture 6 Digital baseband transmission system40p(a) Manchester code p(b) Miller codep(c) CMI codeLecture 6 Digital baseband transmission system41uBlock encodingpnBmB code: n
32、-bit binary code of the original information stream is replaced with a new m-bit binary code. E.g. In 4B5B encoding, for 4-bit encoding, there are 24 = 16 different combinations. For 5-bit encoding, there are 25 = 32different combinationsLecture 6 Digital baseband transmission system42pnBmT code: n-
33、bit binary code of the original information stream is replaced with a new m three ternary code. E.g. In 4B3T encoding, four binary code is converted into the three ternary code. In the same symbol rate, the 4B3T code information rate is greater than 1B1T, improving the bandwidth efficiency.Lecture 6
34、 Digital baseband transmission system43l6.3 Digital baseband signal transmission and intersymbol interference (ISI)n6.3.1 Digital baseband signal transmission systemLecture 6 Digital baseband transmission system44uBaseband waveform diagraminputcodingTransmitterchannelreceiversynchronizationRecovery
35、errorLecture 6 Digital baseband transmission system45uISIuCaused by multipath propagation or the inherent non-linear frequency response of a channel causing successive symbols to blur together. Lecture 6 Digital baseband transmission system46n6.3.2 Quantitative analysis of the digital baseband signa
36、l transmissionuDigital baseband signal transmission model an The transmitting filter input sequence of symbols, 0, 1 or -1, +1 d (t) Corresponding baseband signalDecLecture 6 Digital baseband transmission system47uTransmitter outputgT (t) The impulse response of the transmission filterLet the transf
37、er function of transmission filter be GT ()uTotal transmission characteristicsLet the channel transfer function be C(), the transfer function of receiving filter be GR (). The total transmission characteristics of the base-band transmission system is whose impulse response isLecture 6 Digital baseba
38、nd transmission system48uThe output signalwhere nR(t) is additive noise n(t) passing the receiving filter .uSampling:pE.g. in order to obtain the estimation of kth symbol ak , we sample r(t) at t = kTs + t0. where ak h(t0) is the sampled value of the kth received symbol, which determines ak; the sec
39、ond term is sum of the symbols other than the kth symbol (ISI). Lecture 6 Digital baseband transmission system49The third term nR(kTS + t0) is the output noise in the sampling instant. We make decision as followswhen r (kTs + t0 ) Vd , ak is “1”when r (kTs + t0 ) Vd , ak is “0”。When the ISI and nois
40、e is small enough to ensure our decision correct.Lecture 6 Digital baseband transmission system50l6.4 Baseband transmission characteristics without ISI n6.4.1 The basic idea to eliminate ISI To eliminate the ISI, Because an is randomized, to eliminate ISI through canceling them out is not feasible.
41、Thus we consider the requirement of h(t).Lecture 6 Digital baseband transmission system51 If we let h (k-n)Ts +t0=0 at Ts+ t0 , 2Ts +t0 , the ISI can be eliminated as the figure shown below Lecture 6 Digital baseband transmission system52n6.4.2 No ISI conditionuTime-domain If we sample h(t) at discr
42、ete time, we obtain non-zero value at the time t = 0, and zero at all the other sampling points.Lecture 6 Digital baseband transmission system53uFrequency domainFrom the relations between h (t) and H()When t = kTs ,Rewritten asLecture 6 Digital baseband transmission system54Using variable substituti
43、onwe have d = d, = +2i/Ts . When = (2i1)/Ts, = /Ts . Thus,Interchanging the order of summation and integral, Lecture 6 Digital baseband transmission system55 From Fourier transform, if F() has period 2/Ts, we can use Fourier series representationCompare it with h(kTs), we know that is the Fourier se
44、ries of h(kTs).Lecture 6 Digital baseband transmission system56From NO ISI condition in time domain, we obtain NO ISI condition in frequency domainorcalled Nyquist first condition.Lecture 6 Digital baseband transmission system57Lecture 6 Digital baseband transmission system58n6.4.3 No ISI designuIde
45、al low-pass characteristicsLecture 6 Digital baseband transmission system59Channel impulse response h(t) has periodically zero at t = kTs (k 0). As long as the receiver samples at t = kTs, the transmission can be achieved without ISI.Lecture 6 Digital baseband transmission system60The bandwidth ispT
46、ransmission rate RB = 1/Ts , no ISIpTransmission rate is larger than RB = 1/Ts, there exists ISI. RB is called the Nyquist rate. B=1/2Ts is Nyquist bandwidth.Bandwidth efficiency: However, it is not practical.Lecture 6 Digital baseband transmission system61uCosine roll-off characteristicsp“roll-off”
47、: Make the edge of an ideal low-pass filter slowly declinepcosine roll-off characteristics, as shown : Cosine roll-off characteristicsLecture 6 Digital baseband transmission system62p Transfer function of cosine roll-off characteristics Corresponding h(t) is Roll-off factor is defined asLecture 6 Di
48、gital baseband transmission system63where fN Nyquist bandwidth ， f the amount of bandwidth beyond the Nyquist bandwidthThe bandwidth increases asMaximum bandwidth efficiency isLecture 6 Digital baseband transmission system64pWhen =0, ideal low-pass systempWhen =1, whose channel impulse response is L
49、ecture 6 Digital baseband transmission system65l6.5 Anti-noise performance of the baseband transmission systemnAnalysis model n(t) Additive white Gaussian noise (AWGN), two-side PSD n0 /2. nR (t) PSD whose power is抽样判决Lecture 6 Digital baseband transmission system66 The PDF of nR (t) V - the instant
50、aneous value of the noise nR (kTs) Lecture 6 Digital baseband transmission system67n6.5.1 Binary bipolar baseband system Binary bipolar signal is + A or -A (corresponding to “1” or “0”). The sampled waveform x(t) at the receiver A+ nR(kTs) has PDF-A+ nR(kTs) has PDFLecture 6 Digital baseband transmi
51、ssion system68 Select an appropriate level Vd for decision thresholdLecture 6 Digital baseband transmission system69uMistake “1” as “0” P(0/1)uMistake “0” as “1” P(1/0)=Lecture 6 Digital baseband transmission system70Lecture 6 Digital baseband transmission system71u“1” probability P (1)u“0” probabil
52、ity P (0)uBit error rate (BER) Lecture 6 Digital baseband transmission system72If P(1) = P(0) = 1/2. BERThe larger A/ n , the smaller Pe. Lecture 6 Digital baseband transmission system73n6.5.2 Binary unipolar baseband system For unipolar signal, we only need to move the following figure f0(x) curve
53、distribution center from A to 0.Lecture 6 Digital baseband transmission system74When P(1) = P(0) = ,Vd* = A/2Lecture 6 Digital baseband transmission system75l6.6 Eye patternnEye pattern is an effective experimental method.nObserve the baseband signal waveform using an oscilloscope.nConnect the input
54、 of the sampler with an oscilloscope, and adjust the horizontal scanning period of the oscilloscope so that the received symbols synchronized.nThe oscilloscope display the graphic like a human eye.Lecture 6 Digital baseband transmission system76nEye patternLecture 6 Digital baseband transmission sys
55、tem77nEye patternLecture 6 Digital baseband transmission system78nEye pattern photosLecture 6 Digital baseband transmission system79l6.7 Partial response and time domain equalizern6.7.1 Partial response systemuArtificially introduction of ISIuBaseband system using partial response waveform transmiss
56、ion is known as a partial response system.Lecture 6 Digital baseband transmission system80nFirst kind partial response waveformLecture 6 Digital baseband transmission system81uThe expression of the synthesized waveform ispFrom the figure, g(t)=1 at adjacent sampling time t =Ts/2. Lecture 6 Digital b
57、aseband transmission system82uFrequency Spectrum of g(t) uBandwidth B = 1/2Ts (Hz), the same as the ideal rectangular filteruThe bandwidth efficiency is Lecture 6 Digital baseband transmission system83uBinary symbol sequence ak uak takes the values of +1 or -1 uThe sampled output Ck isCk = ak + ak-1
58、 or ak = Ck - ak-1Lecture 6 Digital baseband transmission system84uProblems existpHigher bandwidth efficiency, greater tail attenuationpError propagation problemLecture 6 Digital baseband transmission system85 input symbol 1 0 1 1 0 0 0 1 0 1 ak Transmitter +1 1 +1 +1 1 1 1 +1 1 +1 +1CkTransmitter 0
59、 0 +2 0 2 2 0 0 0 +2 Ck Receiver 0 0 +2 0 2 0 0 0 0 +2 ak Receiver +1 1 +1 +1 1 1 +1 1 +1 1 +3 Lecture 6 Digital baseband transmission system86uReason of error propagation: The introduction of ISI cause the independent symbols becomes correlated symbols Ck = ak + ak-1which is called related coding.
60、Lecture 6 Digital baseband transmission system87uPrecodingpprecoding rule: bk = ak bk-1 ak = bk bk-1 uRelated coding Ck = bk + bk-1uModulo 2 decision Ckmod2 = bk + bk-1mod2 = bk bk-1 = ak ak = Ckmod2 We can obtain ak without the knowledge of ak-1.Lecture 6 Digital baseband transmission system88uExam
61、ple: ak and bk is binary bipolar code, which have the value +1 and -1 (corresponding to “1” and “0”) ak 1 0 1 1 0 0 0 1 0 1 1 bk-1 0 1 1 0 1 1 1 1 0 0 1 bk 1 1 0 1 1 1 1 0 0 1 0 Ck 0 +2 0 0 +2 +2 +2 0 2 0 0 Ck 0 +2 0 0 +2 +2 +2 0 0 0 0 ak 1 0 1 1 0 0 0 1 1 1 1 Decision rule:Lecture 6 Digital baseban
62、d transmission system89uBlock diagram of first kind partial response systempFig (a) - Principle block diagrampFig (b) - Actual system block diagram Lecture 6 Digital baseband transmission system90nThe general form of the partial response where R1, R2, , Rn are weighting coefficients. When R1 =1, R2
63、=1, and others are equal to 0, it is first kind partial response waveform. Frequency spectrum of g(t)Lecture 6 Digital baseband transmission system91uRelated code: assuming that the input data sequence ak, the related code isThe number of level Ck is depend on the value of ak and Rm. Lecture 6 Digit
64、al baseband transmission system92uPrecoding - Related coding - Modulo decisionuPrecoding where ak and bk are assumed to be L-ary, and “+” is “moldulo L sum”.uRelated coding bk uModulo decision ak = Ckmod LLecture 6 Digital baseband transmission system93nFive types commonly used partial response wave
65、formsLecture 6 Digital baseband transmission system94nAdvantages and disadvantages of partial response systemuAchieve 2B/Hz bandwidth efficiency.uWhen the input data is L-ary, the level number of the partial response system is more than L. Worse anti-noise performance. Lecture 6 Digital baseband tra
66、nsmission system95n6.7.2 Time domain equalizationuWhat is an equalizer? reduce ISIuThe type of equalizerspFrequency domain equalizerpThe time domain equalizerLecture 6 Digital baseband transmission system96uTime domain equalization Channel impulse response h(t), whose FT is H(). We will prove: if a
67、filter whose impulse response is is inserted between the filter and sampled decision, we can theoretically eliminate ISI. 【Proof】Let the FT of hT(t) be T(). IfsatisfyISI is eliminated. Lecture 6 Digital baseband transmission system97Substituting intoWe obtainIf T() is periodic function with period 2
68、/Ts,we haveThe condition of eliminating ISI.Lecture 6 Digital baseband transmission system98 Since T() is periodic function with period 2/Ts, T() can be represented by Fourier series Lecture 6 Digital baseband transmission system99whose inverse Fourier transform isLecture 6 Digital baseband transmis
69、sion system100The filter depends on the tap coefficients Cn.Since Cn is adjustable, it is able to adapt to channel variation.Theoretically, filter with infinite taps completely eliminates ISI. Actually, it is not practical.Lecture 6 Digital baseband transmission system101pA filter with 2N +1 taps ha
70、s the impulse response e(t),Lecture 6 Digital baseband transmission system102Let the input be x(t), without additional noise. After equalization, the output y(t) is The sampling time t = kTsThereforeLecture 6 Digital baseband transmission system103u【Ex 6-3】 A three-tap filter, with C-1= -1/4, C0 = 1
71、, C+1 = -1/2. The input x(t) is x-1 = , x0 = 1, x+1 = . Find the equalizer output y(t) at the sampling points. 【Sol】 Since When k = 0,When k = 1,When k = -1,We can also obtain y-2 = -1/16，y+2 = -1/4, with others all 0Lecture 6 Digital baseband transmission system104uThe equalization criterion Peak d
72、istortion criterion and mean square distortion criterion pPeak distortionpMean square distortion definitionLecture 6 Digital baseband transmission system105pMinimum peak method - Zero ForcingThe peak distortion before equalization (initial distortion) is expressed asIf xk is normalized, x0 = 1, then
73、 the above formula becomesLet yk be normalized, y0 = 1D0 =Lecture 6 Digital baseband transmission system106orandSubstituting We obtainC0x0 += 1C0 = 1 -yk =Lecture 6 Digital baseband transmission system107Substituting into peak distortion definition,we obtain Visible, in the input sequence of xk give
74、n below the circumstance, peak distortion D is the tap coefficient Ci ( except C0 ) function. Obviously, the D solving minimum Ci is what we care.yk =|Lecture 6 Digital baseband transmission system108Lucky has proved: if the initial distortion D01, then the coefficient Ci isThus we have 2N+1 linear
75、equationsLecture 6 Digital baseband transmission system109In matrix form, Lecture 6 Digital baseband transmission system110【Ex6-4】 Design a 3-tap zero forcing equalizer, to reduce ISI. Assume x-2 = 0, x-1 = 0.1, x0 = 1, x1 = -0.2, x2 = 0.1. Obtain 3-tap coefficients and calculate the peak distortion
76、.【Sol】Based on the equation, we have the following equationLecture 6 Digital baseband transmission system111The solution isFromwe obtainInput peak distortionOutput peak distortionPeak distortion decreases 4.6 times.Lecture 6 Digital baseband transmission system112pMinimum mean square distortion equa
77、lizerLecture 6 Digital baseband transmission system113pMinimum mean square distortion equalizerLet the transmitting sequence be ak, the equalizer input be x(t), and the sampled value of equalizer output be yk. The error is Mean square error (MSE) is defined asSubstituting we obtainLecture 6 Digital
78、baseband transmission system114We minimize the MSE by obtaining the partial derivatives of Ci,where represents an error value. Here error is caused by ISI and noise.Lecture 6 Digital baseband transmission system115Fromwe know that to have Eek xk-i=0, the error ek is uncorrelated with the equalizer input xk-i (|i| N). Lecture 6 Digital baseband transmission system116Block diagram of a 3-tap adaptive equalizerLecture 6 Digital baseband transmission system117l6.8 ConclusionLecture 6 Digital baseband transmission system
