Chap 4

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Chapter 4
Digital
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Chapter 4: Outline
4.1 DIGITAL-TO-DIGITAL CONVERSION
4.2 ANALOG-TO-DIGITAL CONVERSION
4.1 TRANSMISSION MODES
4-1 DIGITAL-TO-DIGITAL CONVERSION
In this section, we explore how digital data is
represented using digital signals.
4.3
4-1 DIGITAL-TO-DIGITAL CONVERSION
The conversion involves three techniques:
• line coding,
• block coding, and
• scrambling.
4.4
4-1 DIGITAL-TO-DIGITAL CONVERSION
Line coding is always needed;
block coding and scrambling may or may not be
needed.
4.5
4.4.1 Line Coding
Line coding converts a sequence of bits to a digital
signal.
• At the sender, digital data are encoded into a
digital signal
• At the receiver, the digital data are recreated by
decoding the digital signal.
4.6
Figure 4.1: Line coding and decoding
4.7
4.4.1 Line Coding
Define the ratio of data elements to signal elements
with r = (data elements) / (signal elements).
4.8
Figure 4.2: Signal elements versus data elements
4.9
4.4.1 Line Coding
Let S be the number of signals per second. This is
baud rate.
The relationship between bandwidth, N-bps, and baud
rate, S-signals/sec, is:
S = c * N * 1/r
The value c is a unit-less value: 0 < c <=1
4.10
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 ½ (the average value between 0 and 1)?
4.11
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 ½ ?
Solution
The baud rate is then
4.12
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?
4.13
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?
S = ½ N / r solve for N,
N=2Sr
4.14
Digital Transmission
Factors impacting Digital Transmission
coding:
1. Synchronization
2. Baseline wandering
3. DC Components
Synchronization
If the clocks between the sender and
receiver are not perfectly synchronized,
there will be errors (too many or too few
bits at the receiver).
We need a way to synchronize the signal
between sender and receiver.
Baseline Wandering
The average power is computed at the
receiver and compared to ground. This
difference is used to measure the relative
voltage of incoming signals.
Large changes in average power will result
in errors when attempting to detect the
incoming voltage.
DC Components
Not all channels can carry low frequency
information. DC components will lead to
errors when converting the signal
elements back into data elements.
Figure 4.3: Synchronization
4.19
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?
4.20
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?
0.1% is .001
Total bits received = (1 + .001)*1000 bits/sec
Total bits received = 1001 (one extra bit).
4.21
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?
Solution
At 1 kbps, the receiver receives 1001 bps instead of 1000
bps.
4.22
4.4.2 Line Coding Schemes
We can roughly divide line coding schemes into five
broad categories, as shown in Figure 4.4.
.
4.23
Figure 4.4: Line coding scheme
4.24
Figure 4.5: Unipolar NRZ scheme
4.25
Unipolar
Not a good choice because if long streams
of 1s occur, we get DC components and
baseline wandering.
Figure 4.6: Polar schemes (NRZ-L and NRZ-I)
4.27
Unipolar: NRZ-L
Long streams of zeros or ones will cause DC
components and baseline wandering.
Unipolar: NRZ-I
Long streams of ones are OK, but streams
of zeros introduce DC components and
baseline wandering.
Example 4.4
A system is using NRZ-I to transfer 10-Mbps data. What
are the signal rate and bandwidth? Case factor c = ½
4.30
Example 4.4
A system is using NRZ-I to transfer 10-Mbps data. What
are the signal rate and bandwidth? Case factor c = ½
NRZ-I has r = 1 (1 bit / signal)
S=cN/r
4.31
Example 4.4
A system is using NRZ-I to transfer 1-Mbps data. What are
the signal rate and bandwidth? Case factor c = ½
NRZ-I has r = 1 (1 bit / signal)
S=cN/r
S = ½ (1Mbps) / 1 = .5 Mbaud = 500 Kbaud
4.32
Example 4.4
A system is using NRZ-I to transfer 1-Mbps data. What are
the signal rate and bandwidth?
r = log2(L)
N = 2 B log2( L) solve for B
B=½N/r
B = 1Mbps/2 = .5 MHz = 500 KHz
4.33
Figure 4.7: Polar schemes (RZ)
4.34
Polar Line Coding
No DC components, no baseline wandering.
These require 2 signal elements per data
element (more bandwidth requirement)
Figure 4.8: Polar biphase
4.36
Bipolar
Bipolar – positive and negative amplitudes
above and below the time axis.
• AMI
• Pseudoternary
AMI
Alternate Mark Inversion
Every other 1-bit has a 180 degree phase
change
The 0-bit has zero amplitude
Used for some long distance
communications.
Pseudoternary
“three” states:
0-bits alternate phase
1-bits have zero amplitude
Figure 4.9: Bipolar schemes: AMI and pseudoternary
4.40
Multilevel Schemes
The goal is to send more bits per signal.
We designate the different schemes by this
notation: mBnL
m = number of bits per signal element
B = two possible data elements (0 or 1)
binary
L = number of levels
Multilevel Schemes
The goal is to send more bits per signal.
We designate the different schemes by this
notation: mBnL
B^m <= L^n
mBnL Schemes
Example: 2B1Q (used for DSL)
2=2
B=2
n=1
L=4 (Q is for quad)
Figure 4.10: Multilevel: 2B1Q (the diagram is wrong!)
4.44
8B6T
2^8 <= 3^6,
256 <= 729
this is the early line-code implementation of
fast-Ethernet.
8B6T
Bit patterns are weighted as 0 (neutral) or
+1.
If there are two consecutive +1 weighted bit
patterns, the second one is inverted.
This maintains DC balance.
Figure 4.11: Multilevel: 8B6T (early version of fast Ethernet)
4.47
4D-PAM5
A different category of multilevel line
coding:
4-dimensional 5-level pulse amplitude
modulation.
4D-PAM5
Four wires transmit in parallel using 8B4Q
over each wire. This is used for Gigabit
Ethernet.
Figure 4.12: Multilevel: 4D-PAMS scheme
4.50
Table 4.1: Summary of line coding schemes
4.51
4.4.3 Block Coding
We need redundancy to ensure synchronization and to
provide some kind of inherent error detecting. Block
coding can give us this redundancy and improve the
performance of line coding. In general, block coding
changes a block of m bits into a block of n bits, where
n is larger than m. Block coding is referred to as an
mB/nB encoding technique.
4.52
Figure 4.14: Block coding concept
4.53
Figure 4.15: Using block coding 4B/5B with NRZ-I line coding scheme
4.54
Table 4.2: 4B/5B mapping codes
4.55
Block Coding
nB/mB
Coding 2^n bits as 2^m bits, m > n.
Figure 4.16: Substitution in 4B/5B block coding
4.57
Example 4.5
We need to send data at a 1-Mbps rate. What is the
minimum required bandwidth, using
1. a combination of 4B/5B and NRZ-I; or
2. Manchester coding?
C = ½ for both
4.58
Example 4.5
We need to send data at a 1-Mbps rate. What is the
minimum required bandwidth, using
1. a combination of 4B/5B and NRZ-I; or
2. Manchester coding
C = ½ for both cases
There is a 25% increase in bits due to the overhead in
case-1
The link must therefore support a 1.25 Mbps rate to move
1-Mbps data.
4.59
Example 4.5
We need to send data at a 1-Mbps rate. What is the
minimum required bandwidth, using
1. a combination of 4B/5B and NRZ-I; or,
2. Manchester coding
NRZ-I encodes one bit per signal, r = 1.
Manchester encoding encodes one bit per two signals: r =
½
4.60
Example 4.5
We need to send data at a 1-Mbps rate. What is the
minimum required bandwidth, using
1. a combination of 4B/5B and NRZ-I; or,
2. Manchester coding?
C = ½ for both cases.
B=cN/r
NRZ-I : B = ½ (1.25Mbps) / 1 = 626KHz
Man : B = ½ ( 1 Mbps) / .5 = 1. MHz
4.61
Figure 4.17: 8B/10B block encoding
4.62
4.4.4 Scrambling
We modify line and block coding to include
scrambling.
4.63
4.4.4 Scrambling
The system inserts the required pulses based on the
defined scrambling rules. Two common scrambling
techniques are B8ZS and HDB3.
4.64
Figure 4.18: AMI used with scrambling
4.65
Figure 4.19: Two cases of B8ZS scrambling technique
4.66
4-2 ANALOG-TO-DIGITAL CONVERSION
The techniques described in Section 4.1 convert
digital data to digital signals. Sometimes, however,
we have an analog signal such as one created by a
microphone or camera. We have seen in Chapter 3
that a digital signal is superior to an analog signal.
The tendency today is to change an analog signal to
digital data. In this section we describe two
techniques, pulse code modulation and delta
modulation.
4.67
4.2.1 Pulse Code Modulation (PCM)
The most common technique to change an analog
signal to digital data (digitization) is called pulse code
modulation (PCM). A PCM encoder has three
processes, as shown in Figure 4.24.
4.68
Figure 4.21: Components of PCM encoder
4.69
Figure 4.22: Three different sampling methods for PCM
4.70
Figure 4.23: Nyquist sampling rate for low-pass and bandpass signals
4.71
Figure 4.24: Recovery of a sine wave with different sampling rates.
4.72
Example 4.9
Telephone companies digitize voice by assuming a
maximum frequency of 4000 Hz. The sampling rate
therefore is 8000 samples per second.
4.73
Example 4.10
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.
4.74
Example 4.11
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.
4.75
Figure 4.26: Quantization and encoding of a sampled signal
4.76
SNR based on sample size
SNR(dB) = 6.02 n + 1.76.
4.77
Example 4.12
What is the SNRdB in the example of Figure 4.26?
Solution
We can use the formula to find the quantization. We have
eight
levels
and
3
bits
per
sample,
so
SNRdB = 6.02(3) + 4.76 = 19.82 dB. Increasing the number
of levels increases the SNR.
4.78
Example 4.13
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.
4.79
Example 4.14
We want to digitize the human voice. What is the bit rate,
assuming 8 bits per sample?
Solution
The human voice normally contains frequencies from 0 to
4000 Hz. So the sampling rate and bit rate are calculated
as follows:
4.80
Example 4.15
Transmitting a digital signal using n-bit samples requires:
B(transmission) = sample size in bits *Bandwidth of signal
What bandwidth is required to transmit a digital phone
call? (4KHz, 8bits per sample).
4.81
4-3 TRANSMISSION MODES
Of primary concern when we are considering the
transmission of data from one device to another is
the wiring, and of primary concern when we are
considering the wiring is the data stream. Do we
send 1 bit at a time; or do we group bits into larger
groups and, if so, how? The transmission of binary
data across a link can be accomplished in either
parallel or serial mode. and isochronous (see Figure
4.31).
4.82
Figure 4.31: Data transmission modes
4.83
4.3.1 Parallel Transmission
Line coding is the process of converting digital data to
digital signals. We assume that data, in the form of
text, numbers, graphical images, audio, or video, are
stored in computer memory as sequences of bits (see
Chapter 1). Line coding converts a sequence of bits to
a digital signal. At the sender, digital data are encoded
into a digital signal; at the receiver, the digital data are
recreated by decoding the digital signal. Figure 4.1
shows the process.
4.84
Figure 4.32: Parallel transmission
4.85
4.3.2 Serial Transmission
In serial transmission one bit follows another, so we
need only one communication channel rather than n to
transmit data between two communicating devices
(see Figure 4.33)..
4.86
Figure 4.33: Serial transmission
4.87
Figure 4.34: Asynchronous transmission
4.88
Figure 4.35: Synchronous transmission
Direction of flow
11110111
4.89
Frame
11111011 11110110
Frame
•••
Frame
11110111
11110011

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