4.1

Chapter 4

Digital Transmission

4.2

Components of Data Communication

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)

4.3

Analog Data-->Signal Options

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

4.4

Digital 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

4.5

4-1 DIGITAL-TO-DIGITAL CONVERSION4-1 DIGITAL-TO-DIGITAL CONVERSION

In this section, we see how we can represent digital In this section, we see how we can represent digital data by using digital signals. The conversion involves data by using digital signals. The conversion involves three techniques: three techniques: line codingline coding, , block codingblock coding, and , and scramblingscrambling. Line coding is always needed; block . Line coding is always needed; block coding and scrambling may or may not be needed.coding and scrambling may or may not be needed.

Line CodingLine Coding SchemesBlock CodingScrambling

Topics discussed in this section:Topics discussed in this section:

4.6

Figure 4.1 Line coding and decoding

4.7

Figure 4.2 Signal element versus data element

r = number of data elements / number of signal elements

4.8

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) 3. Worst case, best case, and average case of r4. S = c * N / r baud

4.9

Baseline wanderingBaseline: running average of the

received signal power

DC ComponentsConstant digital signal creates low

frequencies

Self-synchronizationReceiver Setting the clock matching the

sender’s

4.10

Figure 4.3 Effect of lack of synchronization

4.11

Figure 4.4 Line coding schemes

4.12

Figure 4.5 Unipolar NRZ scheme

4.13

Digital Encodingof Digital Data

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

4.14

Differential NRZ

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

4.15

Problems With 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

4.16

Figure 4.6 Polar NRZ-L and NRZ-I schemes

4.17

Figure 4.7 Polar RZ scheme

4.18

Manchester Code

Transition in the middle of each bit period

Transition provides clocking and data Low-to-high=1 , high-to-low=0 Used in Ethernet

4.19

Differential Manchester

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

4.20

Figure 4.8 Polar biphase: Manchester and differential Manchester schemes

4.21

• 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

4.22

Bipolar schemes: AMI (Alternate Mark Inversion) and pseudoternary

4.23

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

4.24

Figure 4.10 Multilevel: 2B1Q scheme

Used in DSL

4.25

Figure 4.11 Multilevel: 8B6T scheme

4.26

Figure 4.13 Multitransition: MLT-3 scheme

4.27

Table 4.1 Summary of line coding schemes

Polar

4.28

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

4.29

Figure 4.14 Block coding concept

4.30

Figure 4.15 Using block coding 4B/5B with NRZ-I line coding scheme

4.31

Table 4.2 4B/5B mapping codes

4.32

Figure 4.16 Substitution in 4B/5B block coding

4.33

Figure 4.17 8B/10B block encoding

4.34

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

4.35

Figure 4.19 Two cases of B8ZS scrambling technique

B8ZS substitutes eight consecutive zeros with 000VB0VB

4.36

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.

4.37

4-2 ANALOG-TO-DIGITAL CONVERSION4-2 ANALOG-TO-DIGITAL CONVERSION

The tendency today is to change an analog signal to The tendency today is to change an analog signal to digital data. digital data.

In this section we describe two techniques, In this section we describe two techniques, pulse code modulationpulse code modulation andand delta modulationdelta modulation..

4.38

Figure 4.21 Components of PCM encoder

4.39

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

4.40

Figure 4.23 Nyquist sampling rate for low-pass and bandpass signals

4.41

Figure 4.24 Recovery of a sampled sine wave for different sampling rates

4.42

Figure 4.25 Sampling of a clock with only one hand

4.43

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

4.44

A complex low-pass signal has a bandwidth of 200 kHz. What is the minimum sampling rate for this signal?

SolutionThe 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

4.45

A complex bandpass signal has a bandwidth of 200 kHz. What is the minimum sampling rate for this signal?

SolutionWe 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.

Example

4.46

Figure 4.26 Quantization and encoding of a sampled signal

4.47

What is the SNRdB in the example of Figure 4.26?SolutionWe 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.

Contribution of the quantization error to SNRdb

SNRdb= 6.02nb + 1.76 dBnb: bits per sample (related to the number of level L)

4.48

A telephone subscriber line must have an SNRdB above 40. What is the minimum number of bits per sample?

SolutionWe can calculate the number of bits as

Example

Telephone companies usually assign 7 or 8 bits per sample.

4.49

PCM decoder: recovers the original signal

4.50

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.

The minimum bandwidth of the digital signal is nb times greater than the bandwidth of the analog signal.

Bmin= nb x Banalog

4.51

DM (delta modulation) finds the change from the previous sampleNext bit is 1, if amplitude of the analog signal is largerNext bit is 0, if amplitude of the analog signal is smaller

4.52

Figure 4.29 Delta modulation components

4.53

Figure 4.30 Delta demodulation components

4.54

4-3 TRANSMISSION MODES4-3 TRANSMISSION MODES

1. The transmission of binary data across a link can 1. The transmission of binary data across a link can be accomplished in either parallel or serial mode. be accomplished in either parallel or serial mode. 2. In parallel mode, multiple bits are sent with each 2. In parallel mode, multiple bits are sent with each clock tick. clock tick. 3. In serial mode, 1 bit is sent with each clock tick. 3. In serial mode, 1 bit is sent with each clock tick. 4. there are three subclasses of serial transmission: 4. there are three subclasses of serial transmission: asynchronous, synchronous, and isochronous.asynchronous, synchronous, and isochronous.

4.55

Figure 4.31 Data transmission and modes

4.56

Figure 4.32 Parallel transmission

4.57

Figure 4.33 Serial transmission

4.58

Asynchronous transmission1. We send 1 start bit (0) at the beginning and 1 or more stop bits (1s) at the end of each byte. 2. There may be a gap between each byte.3. Extra bits and gaps are used to alert the receiver, and allow it to synchronize with the data stream.4. Asynchronous here means “asynchronous at the byte level,”but the bits are still synchronized, their durations are the same.

4.59

Synchronous transmissionIn synchronous transmission, we send bits one after another without start or stop bits or gaps. It is the responsibility of the receiver to group the bits.