Digital Transmission

1
Digital Transmission

• Chapter 4

2
Digital To Digital Conversion

• Data element vs signal element, r.
• a digital data element is a bit.
• a signal element carries the data element from
  the source to the destination.
• If people are data elements, cars and busses are
  the signal elements.

3
Data rate vs signal rate

• Data rate is measured in bps.
• Signal rate is the number of signals sent in one
  second. (a signal may contain more than one bit).
  Signal rate is the baud rate, aka
• Modulation rate
• Pulse rate

4
Data rate vs Signal rate

• S c (N/r)
• S baud rate
• c case factor 0 lt c lt 1
• The average is c .5
• N data rate in bps
• r ratio of data elements to signal elements.

5
Figure 4.2 Signal element versus data element
6
Baud Rate

• Buad rate determines the bandwidth based on the
  ratio of data elements to signal elements.
• Traffic congestion is not determined by the
  number of people on the road, but the number of
  vehicles.
• Using busses to reduce cars on the road reduces
  overall congestion.

7

• Let N max bit rate
• B bandwidth
• r data elements / signal element
• N ½ B r

8
Example

• Let r 1 (one data element / one signal element)
• Let c the average is ½
• B 100kbps
• Find the baud rate S
• S ½ 100kbps 1 50kbaud

9
Digital Signal Problems

• DC Components ( baseline wandering )
• Synchronization

10
DC Components

• A constant output generates a DC signal.
• Some systems cannot pass low frequencies easily.
  
• Example A telephone line cannot pass frequencies
  below 200Hz.

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Synchronization

• The intervals between bits must match between
  sender and receiver. If the intervals are
  different, then the data will not be translated
  correctly.
• Example sender requires 3 time units per bit,
  the receiver uses 2 time units between bits. If
  two 1s are sent, the receiver will translate them
  as 3 bits.

12
Figure 4.3 Effect of lack of synchronization
13
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.
14
Line Coding

• Unipolar
• Polar
• Bipolar
• Multilevel
• Multitransitional

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Figure 4.5 Unipolar NRZ scheme
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NRZ

• Non-return-to zero (NRZ) implies that the signal
  does not return to zero during the transmission
  of the bit.

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Figure 4.6 Polar NRZ-L and NRZ-I schemes
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Polar Line Coding

• NRZ-Level the voltage level determines the bit
  value.
• NRZ-Invert the change in voltage level at the
  bit boundary determines the bit value.
• No change relative to the last bit is a zero
• change relative to the last bit is a one.

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Polar RZ

• 3 values positive, zero, and negative
• Signal changes (returns to zero, RZ) during the
  bit transmission.
• Positive implies a one
• Negative implies a zero
• Return to zero during the bit transmission to
  synchronize the sender and receiver.

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Figure 4.7 Polar RZ scheme
21
Polar RZ

• Requires twice the bandwidth compared to NRZ
  methods because of the change during the bit
  transmission.

22
Figure 4.8 Polar biphase Manchester and
differential Manchester schemes
23
Manchester Line Coding (polar)

• No synchronization problems due to the transition
  of each bit.
• Requires twice the bandwidth of NRZ.
• Not suited for long distances (LANs only)
• Differential Manchester is used by Ethernet.

24
Bipolar Line Coding

• Voltage levels are positive, zero, or negative.
• One alternates between positive and negative
  voltage.
• Zero is no voltage.

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Bipolar DC

• Zero voltage has zero amplitude. Therefore there
  is no DC component.
• Commonly used for long distance communication.
• Bipolar line coding is subject to synchronization
  problems.

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AMI Pseudoternary

• AMI Alternate mark inversion
• Pseudoternary

27
Figure 4.9 Bipolar schemes AMI and pseudoternary
28
Multilevel Line Coding mBnL

• m data elements
• n signal elements
• B is binary, 2m
• L different levels T three, Q 4.
• Note 2m lt Ln, otherwise there are not enough
  signal elements to represent each data element.

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2B1Q

• Q is for quaternary (4), note that 22 41
• Uses a translation table for each possible bit
  pattern
• Used for DSL

30
Figure 4.10 Multilevel 2B1Q scheme
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2B1Q

• In North America, 2B1Q operates at 40kHz
• What is the bit rate N for 2B1Q?
• N 2flog2(L)
• What is L for 2B1Q?

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2B1Q

• Answer N 160kbps

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Figure 4.11 Multilevel 8B6T scheme
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8B6T

• 8 binary, 6 ternary
• 3 levels (ternary)
• 256 28 lt 36 729
• Each 8 bit data pattern is converted into a 6 bit
  pattern using one of 3 voltage levels (-,0,).
• Used by 100Mbps Ethernet (Fast Ethernet)

35
4D-PAM5

• 4 dimensional, 5-level pulse amplitude
  modulation.
• 4 wires carrying simultaneous signals.
• Five voltage levels -2, -1, 0, 1, 2.
• Used by gigabit ethernet

36
Figure 4.12 Multilevel 4D-PAM5 scheme
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Table 4.1 Summary of line coding schemes
38
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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Block Coding 4B/5B

• Replace each 4 bit group (a nibble) with a 5 bit
  group.
• Division divide the data into 4 bit nibbles.
• Substitution substitute the 4 bit nibble with a
  5 bit code so there is no more than one leading 0
  on the left of the 5 bit word.
• (8B/10B is also available)

40
Figure 4.16 Substitution in 4B/5B block coding
41
Table 4.2 4B/5B mapping codes
42

• What impact does 4B/5B block coding have on
  bandwidth?

43
Scrambling B8ZS HDB3

• B8ZS 8 consecutive zero bits are substituted
  with 000VB0VB
• V and B designate a pattern of polarized ones.
  This will reduce the DC problem for long strings
  of 0.

44
Figure 4.19 Two cases of B8ZS scrambling
technique
45
PCM Pulse Code Modulation

• PCM converts analog to digital
• Sample the analog signal
• Quantize the sample
• Encode

46
PCM sampling

• Sampled at a regular interval, T, where f 1/T.
• The highest frequency signal is determined by the
  Nyquist theoem. The sample rate is twice the
  maximum frequency.

47
Figure 4.24 Recovery of a sampled sine wave for
different sampling rates
48
Figure 4.22 Three different sampling methods for
PCM
49
PCM Quantization

• Signal has a maximum and minimum amplitude.
• Divide the range of amplitudes into L zones
  each of equal height.
• Delta is (Vmax-Vmin)/L
• Assign a value to each midpoint of each zone
• See figure 4.26 page 125 in text.

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Figure 4.26 Quantization and encoding of a
sampled signal
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Encoding

• Normalized PAM is the signal value/delta
• The quantization code is then converted to its
  binary equivalent.

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PCM SNR(db)

• SNR(db) 6.02 n 1.76
• Where n bit depth
• If 40db is required for phone audio, what is the
  bit depth required (the next power of two).
  Answer n 6.4 -gt n 8.

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PCM Sample Rate

• What bit rate is required for voice and what is
  the bit rate if the bandwidth is 0-4000Hz?
• Sample rate 2fmax (due to Nyquist)
• Bit rate Sample rate bit depth

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Delta Modulation

• Measures the change from the previous sample.
• DM requires oversampling beyond the Nyquist rate
  to achieve the same SN compared to PCM.

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Transmission Modes

• Parallel Transmission
• Serial Transmission
• Asynchronous
• Synchronous
• Isosynchronous

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Figure 4.31 Data transmission and modes
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Parallel

• Use n-wires to transmit n bits simultaneously.
• Usually limited to short distances due to the
  cost of multiple strands of wire bundled together.

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Figure 4.32 Parallel transmission
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Figure 4.33 Serial transmission
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Serial Asynch (byte level)

• Each byte is sent with a start bit (0) and one
  stop bit (1) at the end of each byte.
• The start bit alerts the receiver that data will
  arrive.
• The stop bit can be used as an error check.
• Idle time is allowed between the transmission of
  bytes. This idle time allows for synchronization.
• The sender and receiver are synchronized during
  the duration of the byte transfer (10 bits in
  all).

61
Serial Synch (more than 1 byte)

• A continuous stream of bits are sent from source
  to receiver. The two must be synchronized during
  the entire data frame transfer. Gaps are allowed
  between frames.
• This requires synchronization but also results in
  the fastest data transfer (no start or stop bits,
  no gaps within a frame).

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Serial isosynchronous

• Frames arrive at regular intervals audio and
  video transmission are examples.