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UCSB ECE 594 - Jitter in Digital Communication Systems

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IntroductionBackgroundReceiver DecisionsThe One or Zero DecisionTiming of the Sampling InstantPredicting the BER Caused by JitterProbability of Error at Each Sampling InstantSampling Time ProbabilityBER due to JitterConclusions6hfan404.doc 03/05/02Application Note:HFAN-4.0.4Rev 0; 03/02Jitter in Digital Communication Systems, Part 2MAXIM High-Frequency/Fiber Communications Group Maxim Integrated ProductsApplication Note HFAN-4.0.4 (Rev. 0, 03/02)Maxim Integrated ProductsPage 2 of 7Jitter in Digital Communication Systems, Part 21 IntroductionA previous application note on jitter, HFAN-4.0.3"Jitter in Digital Communication Systems, Part 1,"defined jitter and its various sub-components. Thepurpose of this application note is to answer thequestion, "So now that we know what jitter is, whyshould I care?" To answer this question, we willexplore some of the ways that jitter causes bit errorsin digital communication systems.2 BackgroundA basic characteristic of digital communicationssystems is the need for synchronization between thebinary encoded data (the bit stream) and the variouscircuit elements in the transmitter and receiver. Bitsynchronization information is generally conveyedseparately in the transmitter and receiver by the bitclock, which is a square wave signal that has afrequency (in Hz) equal to the data rate (in bits persecond). The relationship between an NRZ encodedbit stream and the bit clock is illustrated in Figure 1.A fundamental problem is how to get the bitsynchronization information from the transmitter tothe receiver. In general, digital communicationsystems transmit only the bit stream and thenregenerate the bit clock at the receiver through useof a clock and data recovery (CDR) circuit such asMaxim's MAX3873, MAX3875, or MAX3877.Distortions and noise in the received bit stream aswell as imperfections in receiver bit clockregeneration result in mistiming (jitter) between thereceived bit stream and the regenerated bit clock thatcan cause bit errors.3 Receiver DecisionsThe receiver in a digital communication system(illustrated in Figure 2) is tasked with accuratelymaking two decisions: (1) when to sample thereceived bit stream, and (2) whether the sampledvalue represents a binary one or zero. The bit clockcontrols the timing of the first decision. Jitterbetween the bit clock and the bit stream may causethe receiver to sample the bit stream at the wrongtime, which can result in bit errors.To better understand the relationship between jitterand the resulting bit errors, it is necessary tounderstand the details of the two decisions made bythe receiver about each bit. We will first discuss thesecond decision (one or zero?) and then come backto the first decision (when to sample?).3.1 The One or Zero DecisionThe decision circuit in a basic receiver simplycompares the sampled voltage, v(t), to a referencevalue, γ, called the decision threshold. If v(t) isgreater than γ, it indicates that a binary one was sent,whereas if v(t) is less than γ, it indicates that abinary zero was sent. Assuming perfectsynchronization between the bit stream and the bitclock, the major obstacle to making the correctdecision is noise in the received data. 1 0 0 1 0 1 1 1 1 0 0 1 1V1t1t2t3t4t5t6t7t8t9t10tn-2tn-1tntn – tn-1 = 1 Bit Period (Tb) = 1Unit Interval (UI) = 1/datarateFigure 1. NRZ encoded bit streamBit Clockt0NRZ Bit Stream……Figure 2. Block-diagram of a fiber-optic receiverVCC+–+–DATACLKDecision CircuitDQPLLApplication Note HFAN-4.0.4 (Rev. 0, 03/02)Maxim Integrated ProductsPage 3 of 7If we assume that additive white Gaussian noise(AWGN) is the dominant cause of erroneousdecisions, then we can calculate the statisticalprobability of making such a decision. Theprobability density function for v(t) with AWGN canbe written mathematically as:2)(21221]),([−−=xvtvxxsetvPROBσπσσ(1)where vS is the voltage sent by the transmitter (themean value of the density function), v(t) is thesampled voltage value in the receiver at time, t, andσ is the standard deviation of the noise. Equation (1)is illustrated in Figure 3.In binary signaling, vS can take on one of twovoltage levels, which we will call vS0 and vS1, andthe probability of making an erroneous decision inthe receiver is:P[ε] = P[v(t) > γ | vS = vS0] P[vS0] + P[v(t) < γ | vS = vS1] P[vS1] (2)where P[ε] is the probability of error and P[x | y]represents the probability of x given y. If we assumean equal probability of sending vS0 versus vS1 (50%mark density), then P[vS0] = P[vS1] = 0.5. Also, inorder to simplify the example, we will assume thatthe same noise effects either voltage level (i.e., σ0 =σ1), which means that P[v(t) > γ | vS = vS0] = P[v(t) <γ | vS = vS1]. Using these assumptions, equation (2)can be reduced to: P[ε] = P[v(t) > γ | vS = vS0] × 0.5 + P[v(t) < γ | vS = vS1] × 0.5 = ∫∞−γσdttvPROB ]),([211 + ∫∞γσdttvPROB ]),([210 (3)where PROB[v(t),σx] is defined in equation (1).This result is illustrated in Figure 4.vSv(t)PROB[v(t)]Figure 3. AWGN probability density functionσFigure 4. Probability of error for binary signalingvS0vS1γP[v(t) | vS = vS1]P[v(t) | vS = vS0]v(t)PROB[v(t)]v(t) =Signal+ Noiseσ1σ0P[v(t) > γ | vS = vS1]P[v(t) < γ | vS = vS0]γApplication Note HFAN-4.0.4 (Rev. 0, 03/02)Maxim Integrated ProductsPage 4 of 7From Figure 4 and equations (2) and (3) we canconclude that the probability of error is equal to thearea under the tails of the density functions thatextend beyond the threshold, γ. This area, and thusthe bit error ratio (BER), is determined by twofactors: (1) the standard deviations of the noise (σ0and σ1) and (2) the voltage difference between vS0and vS1 (i.e., the signal-to-noise ratio).It is important to note that for the special case whenσ0 = σ1, the threshold is halfway between the oneand zero levels (i.e., γ = (vS1−vS0)/2). For the moregeneral case when σ0 ≠ σ1, the optimum thresholdfor minimum BER will be higher or lower than(vS1−vS0)/2. For optimum performance, then, thedecision circuit include an adjustable threshold level,as in Maxim's MAX3877 and MAX3878.To simplify computation of the probability of biterror we can rewrite equation (3) in terms of theerror function, Er(x), which is defined as1: ∫∞−=xxdxxEre2221)(π(4)for a standard normal distribution (i.e., mean = 0,


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