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Chapter 4 Digital Transmission

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Chapter 4 Digital Transmission Chapter 4 Digital Transmission 4.1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4-1 DIGITAL-TO-DIGITAL CONVERSION In this section, we see how we can represent digital data by using digital signals. The conversion involves three techniques: line coding, block coding, and scrambling. Line coding is always needed; block coding and scrambling may or may not be needed. Topics discussed in this section: Line Coding Line Coding Schemes Block Coding Scrambling 4.2 Line Coding Line coding is the process of converting digital data to digital signals At the sender, data elements are encoded into signal elements At the receiver, signal elements are decoded into data elements 4.3 Figure 4.1 Line coding and decoding 4.4 Characteristics of line coding Data Element vs. Signal Element Data Rate vs. Signal Rate Bandwidth Baseline Wandering DC Components Self-Synchronization Built-in Error Detection Immunity to noise and interference Complexity 4.5 Signal Elements vs. Data Elements A data element is the smallest entity that can represent a piece of information (a bit) A signal element is the shortest unit in time of a digital signal (a baud) Data elements are what we need to send Signals elements are what we can send The ratio, r, is defined as the number of data elements carried by each signal element 4.6 Figure 4.2 Signal element versus data element 4.7 Data Rate vs. Signal Rate The data rate (or bit rate), N, is the number of data elements (bits) sent in one second The signal rate (or baud rate or pulse rate or modulation rate), S, is the number of signal elements (bauds) sent in one second The goal is to increase the data rate (and hence the speed of transmission) while decreasing the signal rate (and hence the required bandwidth) Where c is the case factor that depends on the case 4.8 Example 4.1 A signal is carrying data in which one data element is encoded as one signal element (r = 1). If the bit rate is 100 kbps, what is the average value of the baud rate if c is between 0 and 1? Solution We assume that the average value of c is 1/2 . The baud rate is then 4.9 The Bandwidth As learned previously, a digital signal that carries information is nonperiodic and its bandwidth is continuous and infinite Also, many of its components have very small amplitude that can be ignored Therefore, the “effective bandwidth” of the real- life digital signal is finite The bandwidth is proportional to the signal rate 1 Given N: B = c × N × min r 1 Given B: N = × B × r max c 4.10 Example 4.2 The maximum data rate of a channel (see Chapter 3) is Nmax = 2 × B × log2 L (defined by the Nyquist formula). Does this agree with the previous formula for Nmax? Solution A signal with L levels actually can carry log2L bits per level. If each level corresponds to one signal element and we assume the average case (c = 1/2), then we have 4.11 Note Although the actual bandwidth of a digital signal is infinite, the effective bandwidth is finite. 4.12 Baseline Wandering While decoding the received signal, the receiver, calculates the running average of the received signal power, called the “baseline” The incoming signal power is evaluated against the baseline in order to determine the value of the data element Long strings of 0’s or 1’s can cause a drift of the baseline, called “baseline wandering”, which makes it hard for the receiver to correctly decode the signal A good line coding scheme should prevent baseline wandering 4.13 DC Components Long constant voltage levels (DC) in digital signals create low-frequency components DC components are mostly filtered out in systems that can’t pass low frequencies A good line coding scheme should have no DC components 4.14 Self-synchronization The sender and receiver must be exactly synchronized in order for the received signals to be interpreted correctly Bit duration, the start, and the end of the bits should be exactly identified by the receiver If the sender and the receiver are running at different clock rates, the bit intervals will not match and the receiver may misinterpret the signals A good “self-synchronized” line coding scheme should keep the receiver well-synchronized 4.15 Figure 4.3 Effect of lack of synchronization 4.16 Example 4.3 In a digital transmission, the receiver clock is 0.1 percent faster than the sender clock. How many extra bits per second does the receiver receive if the data rate is 1 kbps? How many if the data rate is 1 Mbps? Solution At 1 kbps, the receiver receives 1001 bps instead of 1000 bps. At 1 Mbps, the receiver receives 1,001,000 bps instead of 1,000,000 bps. 4.17 Figure 4.4 Line coding schemes 4.18 Figure 4.5 Unipolar NRZ (Non-Return to Zero) scheme Unipolar: All signal levels at one side of the time axis NRZ since the signal does not return to zero in the middle of the bit Relatively very costly scheme in terms of normalized power Hence, not a popular scheme nowadays 4.19 Figure 4.6 Polar NRZ-L (NRZ-Level) and NRZ-I (NRZ-Invert) schemes Polar: The signal levels are on both sides of the time axis NRZ-L: the voltage level determines the value of the bit NRZ-I: the change or lack of change determines the value Baseline Wandering and Synchronization problems: in both, but twice as severe in NRZ-L Both have the DC Component problem 4.20 Note NRZ-L and NRZ-I both have an average signal rate of N/2 Bd. 4.21 Note NRZ-L and NRZ-I both have a DC component problem. 4.22 Example 4.4 A system is using NRZ-I to transfer 1 Mbps data. What are the average signal rate and minimum bandwidth? Solution The average signal rate is S = N/2 = 500 kbaud. The minimum bandwidth for this average baud rate is Bmin = S = 500 kHz. 4.23 Figure 4.7 Polar RZ (Return-to-Zero) scheme RZ: the signal changes at the beginning and in middle of bit Solves the synchronization problem in NRZ schemes No DC Component problem Needs two signal levels to encode a bit more bandwidth Uses 3 signal levels more complex Not a popular scheme 4.24 Figure 4.8 Polar biphase: Manchester and differential Manchester schemes Overcome the problems of NRZ-L & NRZ-I The cost is doubling the signal rate and hence the minimum bandwidth 4.25 Note In Manchester and differential Manchester encoding, the transition at the middle of the bit is used for synchronization. 4.26 Note The minimum bandwidth of Manchester and differential Manchester is 2 times that of NRZ. 4.27 Note In bipolar encoding, we use three levels: positive, zero, and negative. 4.28 Figure 4.9 Bipolar schemes: AMI and pseudoternary AMI (Alternate Mark Inversion) = Alternate 1 Inversion AMI: 0 = zero volt, 1 = alternating positive and negative Pseudoternary: 1 = zero volt, 0 = alter. pos. and neg. No DC component. Why? What about synchronization? 4.29 Multi-Level Line Encoding Schemes (mBnL) The goal is to increase the number of bits per baud by encoding a pattern of m DE’s into a pattern of n SE’s m Binary data elements 2 data patterns n L signal levels L signal patterns m n If 2 = L each data pattern has a signal pattern m n If 2 < L data patterns form a subset of the signal patterns Extra signal patterns can be used for better synch. and error detection m n If 2 > L not enough signal patterns (not allowed) mBnL Binary patterns of length m, signal patterns of length n with L signal levels L can be replaced by: B (Binary) L=2 (e.g.; 2B2B) T (Ternary) L=3 (e.g.; 8B6T) Q (Quaternary) L=4 (e.g.; 2B1Q) 4.30 Note In mBnL schemes, a pattern of m data elements is encoded as a pattern of n signal elements in which 2m ≤ Ln. 4.31 Figure 4.10 Multilevel: 2B1Q (2 Binary 1 Quaternary) scheme 2 Faster data rate. Why? 2 1 L = 4 complex receiver & 2 = 4 no extra signal patterns What about the DC Comp.? Not balanced (scrambling used to help) What about synchronization and baseline wandering? (same) 4.32 Figure 4.11 Multilevel: 8B6T (Eight Binary 6 Ternary) scheme 6 8 3 - 2 = 473 extra signal patterns good synchronization and error detection Each pattern has a weight of either 0 or +1 DC values If two consecutive patterns with +1 weight, the sender inverts the second pattern. Why? How does the receiver recognize the inverted pattern? Savg = ½ x N x 6/8 = 3N/8 4.33 Figure 4.12 Multilevel: 4D-PAM5 scheme 4-dimentional five-level pulse amplitude modulation Level-0 is only used for error-correction Equivalent to 8B4Q scheme with Smax = N x 4/8 = N/2 Four signal are sent simultaneously over four different wires Smax = N/8 Used in Gigabit LANs over 4 copper cables 4.34 Figure 4.13 Multi-transition: MLT-3 scheme If next bit is 0 no transition (problematic long string of 0’s) If next bit is 1 and current level is not zero next level is 0 If next bit is 1 and current level is zero alternate the nonzero level The worst case is a periodic signal with period of 4/N Smax = N/4 4.35 Tab le 4.1 Summary of line coding schemes 4.36 Block Coding The process of stuffing the bit stream with redundant bits in order to: Ensure synchronization Detect errors The bit stream is divided into groups of m bits (called blocks) Each group is substituted with a different (usually larger) group of n bits (a code) This is referred to as mB/nB coding 4.37 Note Block coding is normally referred to as mB/nB coding; it replaces each m-bit group with an n-bit group.
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