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## Chapter 4 Digital Transmission

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**Chapter 4**Digital Transmission**Data**Analog: Continuous value data (sound, light, temperature) Digital: Discrete value (text, integers, symbols) Signal Analog: Continuously varying electromagnetic wave Digital: Series of voltage pulses (square wave) Components of Data Communication**Analog data to analog signal**Inexpensive, easy conversion (eg telephone) Used in traditional analog telephony Analog data to digital signal Requires a codec (encoder/decoder) Allows use of digital telephony, voice mail Analog Data-->Signal Options**Digital data to analog signal**Requires modem (modulator/demodulator) Necessary when analog transmission is used Digital data to digital signal Less expensive when large amounts of data are involved More reliable because no conversion is involved Digital Data-->Signal Options**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 SchemesBlock Coding Scrambling**Figure 4.2 Signal element versus data element**r = number of data elements / number of signal elements**Data Rate Vs. Signal Rate**• Data rate: the number of data elements (bits) sent in 1s (bps). It’s also called the bit rate • Signal rate: the number of signal elements sent in 1s (baud). It’s also called the pulse rate, the modulation rate, or the baud rate. • We wish to: • 1. increase the data rate (increase the speed of transmission) • 2. decrease the signal rate (decrease the bandwidth requirement) • Worst case, best case, and average case of r • S = c * N / r baud**Baseline wandering**Baseline: running average of the received signal power DC Components Constant digital signal creates low frequencies Self-synchronization Receiver Setting the clock matching the sender’s**Most common, easiest method is different voltage levels for**the two binary digits Typically, negative=1 and positive=0 Known as NRZ-L, or nonreturn-to-zero level, because signal never returns to zero, and the voltage during a bit transmission is level Digital Encodingof Digital Data**Differential version is NRZI (NRZ, invert on ones)**Change=1, no change=0 Advantage of differential encoding is that it is more reliable to detect a change in polarity than it is to accurately detect a specific level Differential NRZ**Difficult to determine where one bit ends and the next**begins In NRZ-L, long strings of ones and zeroes would appear as constant voltage pulses Timing is critical, because any drift results in lack of synchronization and incorrect bit values being transmitted Problems With NRZ**Transition in the middle of each bit period**Transition provides clocking and data Low-to-high=1 , high-to-low=0 Used in Ethernet Manchester Code**Midbit transition is only for clocking**Transition at beginning of bit period=0 Transition absent at beginning=1 Has added advantage of differential encoding Used in token-ring Differential Manchester**Figure 4.8 Polar biphase: Manchester and differential**Manchester schemes**High=0, Low=1**• No change at begin=0, Change at begin=1 • H-to-L=0, L-to-H=1 • Change at begin=0, No change at begin=1**Bipolar schemes: AMI (Alternate Mark Inversion) and**pseudoternary**Multilevel Schemes**• In mBnL schemes, a pattern of m data elements is encoded as a pattern of n signal elements in which 2m ≤ Ln • m: the length of the binary pattern • B: binary data • n: the length of the signal pattern • L: number of levels in the signaling • B for l=2 binary • T for l=3 ternary • Q for l=4 quaternary**Figure 4.10 Multilevel: 2B1Q scheme**Used in DSL**Block Coding**• Redundancy is needed to ensure synchronization and to provide error detecting • Block coding is normally referred to as mB/nB coding • it replaces each m-bit group with an n-bit group • m < n**Figure 4.15 Using block coding 4B/5B with NRZ-I line coding**scheme**Scrambling**• It modifies the bipolar AMI encoding (no DC component, but having the problem of synchronization) • It does not increase the number of bits • It provides synchronization • It uses some specific form of bits to replace a sequence of 0s**Figure 4.19 Two cases of B8ZS scrambling technique**B8ZS substitutes eight consecutive zeros with 000VB0VB**Figure 4.20 Different situations in HDB3 scrambling**technique HDB3 substitutes four consecutive zeros with 000V or B00V depending on the number of nonzero pulses after the last substitution.**4-2 ANALOG-TO-DIGITAL CONVERSION**The tendency today is to change an analog signal to digital data. In this section we describe two techniques, pulse code modulationanddelta modulation.**According to the Nyquist theorem, the sampling rate must be**at least 2 times the highest frequency contained in the signal. What can we get from this: 1. we can sample a signal only if the signal is band-limited 2. the sampling rate must be at least 2 times the highest frequency, not the bandwidth**Figure 4.23 Nyquist sampling rate for low-pass and bandpass**signals**Figure 4.24 Recovery of a sampled sine wave for different**sampling rates**Example**An example related is the seemingly backward rotation of the wheels of a forward-moving car in a movie. This can be explained by under-sampling. A movie is filmed at 24 frames per second. If a wheel is rotating more than 12 times per second, the under-sampling creates the impression of a backward rotation.**Example**A complex low-pass signal has a bandwidth of 200 kHz. What is the minimum sampling rate for this signal? Solution The bandwidth of a low-pass signal is between 0 and f, where f is the maximum frequency in the signal. Therefore, we can sample this signal at 2 times the highest frequency (200 kHz). The sampling rate is therefore 400,000 samples per second.**Example**A complex bandpass signal has a bandwidth of 200 kHz. What is the minimum sampling rate for this signal? Solution We cannot find the minimum sampling rate in this case because we do not know where the bandwidth starts or ends. We do not know the maximum frequency in the signal.**Contribution of the quantization error to SNRdb**SNRdb= 6.02nb + 1.76 dB nb: bits per sample (related to the number of level L) What is the SNRdB in the example of Figure 4.26? Solution We have eight levels and 3 bits per sample, so SNRdB = 6.02 x 3 + 1.76 = 19.82 dB Increasing the number of levels increases the SNR.**Example**A telephone subscriber line must have an SNRdB above 40. What is the minimum number of bits per sample? Solution We can calculate the number of bits as Telephone companies usually assign 7 or 8 bits per sample.**The minimum bandwidth of the digital signal is nb times**greater than the bandwidth of the analog signal. Bmin= nb x Banalog We have a low-pass analog signal of 4 kHz. If we send the analog signal, we need a channel with a minimum bandwidth of 4 kHz. If we digitize the signal and send 8 bits per sample, we need a channel with a minimum bandwidth of 8 × 4 kHz = 32 kHz.