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Chapter 4 Digital Transmission. 4-1 DIGITAL-TO-DIGITAL CONVERSION. line coding , block coding , and scrambling . Line coding is always needed; block coding and scrambling may or may not be needed. Figure 4.2 Signal element versus data element.

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slide1

Chapter 4

Digital Transmission

slide2

4-1 DIGITAL-TO-DIGITAL CONVERSION

line coding, block coding, and scrambling. Line coding is always needed; block coding and scrambling may or may not be needed.

4.#

slide3

Figure 4.2 Signal element versus data element

r = number of data elements / number of signal elements

slide4

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

slide6

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
slide8

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
slide11

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
slide13

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
slide14

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.

slide16

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

slide18

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.

slide19

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.

slide20

DM (delta modulation) finds the change from the previous sample

Next bit is 1, if amplitude of the analog signal is larger

Next bit is 0, if amplitude of the analog signal is smaller

slide22

Chapter 5

Analog Transmission

slide25

1. Data element vs. signal element

2. Bit rate is the number of bits per second.

2. Baud rate is the number of signal elements per second. 3. In the analog transmission of digital data, the baud rate is less than or equal to the bit rate.

S = N x 1/r baud r = log2L

slide26

Figure 5.3 Binary amplitude shift keying

B = (1+d) x S = (1+d) x N x 1/r

qam quadrature amplitude modulation
QAM – Quadrature Amplitude Modulation
  • Modulation technique used in the cable/video networking world
  • Instead of a single signal change representing only 1 bps – multiple bits can be represented by a single signal change
  • Combination of phase shifting and amplitude shifting (8 phases, 2 amplitudes)
slide34

Figure 5.16 Amplitude modulation

The total bandwidth required for AM can be determined from the bandwidth of the audio signal: BAM = 2B.

slide36

Figure 5.20 Phase modulation

The total bandwidth required for PM can be determined from the bandwidth and maximum amplitude of the modulating signal:BPM = 2(1 + β)B.

slide37

Chapter 6

Bandwidth Utilization:

Multiplexing and Spreading

slide40

Figure 6.4 FDM process

FDM is an analog multiplexing technique that combines analog signals.

slide43

Figure 6.10 Wavelength-division multiplexing

WDM is an analog multiplexing technique to combine optical signals.

slide44

Figure 6.12 TDM

  • TDM is a digital multiplexing technique for combining several low-rate channels into one high-rate one.
  • Two types: synchronous and statistical
slide45

Figure 6.13 Synchronous time-division multiplexing

  • In synchronous TDM, each input connection has an allotment in the output even if it is not sending data.
  • In synchronous TDM, the data rate of the link is n times faster, and the unit duration is n times shorter.
slide46

Figure 6.17 Example 6.9

Solution

Figure 6.17 shows the output for four arbitrary inputs. The link carries 50,000 frames per second. The frame duration is therefore 1/50,000 s or 20 μs. The frame rate is 50,000 frames per second, and each frame carries 8 bits; the bit rate is 50,000 × 8 = 400,000 bits or 400 kbps. The bit duration is 1/400,000 s, or 2.5 μs.

slide47

Figure 6.18 Empty slots

Synchronous TDM is not always efficient

slide53

Figure 6.27 Spread spectrum

Bss >> B

1 Wrap message in a protective envelope for a more secure transmission.

2 the expanding must be done independently

3 two types: frequency hopping spread spectrum (FHSS) and direct sequence spread spectrum (DSSS)

slide56

Figure 6.32 DSSS

Direct sequence spread spectrum

Replace each data bit with n bits using a spreading code

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