Chapter 4
Digital
Transmission
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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 the case factor: 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
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 = 2 S r , by substitution,
2 B log2 ( L ) = 2 S r
r = log2( L )
4.15
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?
2 B log2 ( L ) = 2 S r
r = log2( L )
B in cycles /sec is the same as S signals/sec, where a cycle is
a signal for this case.
4.16
Figure 4.3: Synchronization
4.17
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?
4.18
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).
4.19
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?
Solution
At 1 kbps, the receiver receives 1001 bps instead of 1000
bps.
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 Mbps?
At 1 Mbps, the receiver receives 1,001,000 bps instead of
1,000,000 bps.
4.21
Figure 4.3: Effect of lack of synchronization
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
Figure 4.6: Polar schemes (NRZ-L and NRZ-I)
4.26
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
4.27
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
4.28
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
4.29
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
4.30
Figure 4.7: Polar schemes (RZ)
4.31
Figure 4.8: Polar biphase
4.32
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
4.36
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
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!)
4.40
8B6T
2^8 <= 3^6,
256 <= 729
this is the early line-code implementation of
fast Ethernet. It does not use Manchester
coding.
Figure 4.11: Multilevel: 8B6T (early version of fast Ethernet)
4.42
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
4.45
Figure 4.13: Multi-transition MLT-3 scheme
4.46
MLT-3
see page 108 for the rules
Table 4.1: Summary of line coding schemes
4.48
4.4.3 Block Coding
Block coding provides redundancy
synchronization and error detecting.
4.49
to
ensure
4.4.3 Block 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.
4.50
Figure 4.14: Block coding concept
4.51
Figure 4.15: Using block coding 4B/5B with NRZ-I line coding scheme
4.52
Table 4.2: 4B/5B mapping codes
4.53
Block Coding
nB/mB
Coding 2^n bits as 2^m bits, m > n.
Figure 4.16: Substitution in 4B/5B block coding
4.55
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
4.56
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 1Mbps data.
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
NRZ-I encodes one bit per signal, r = 1.
Manchester encoding encodes one bit per two signals: r = ½
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.
B=cN/r
NRZ-I : B = ½ (1.25Mbps) / 1 = 626KHz
Man : B = ½ ( 1 Mbps) / .5 = 1. MHz
4.59
Figure 4.17: 8B/10B block encoding
4.60
4.4.4 Scrambling
Scrambling is used with long distance bipolar AMI
line coding.
4.61
4.4.4 Scrambling
Scrambling helps synchronize when the data stream
contains a sequence of 8 consecutive zeros.
4.62
4.4.4 Scrambling
B8ZS is the designation of AMI with scrambling.
8 consecutive zeros is coded as 000VB0VB.
4.63
Figure 4.19: Two cases of B8ZS scrambling technique
4.64
4-2 ANALOG-TO-DIGITAL CONVERSION
The tendency today is to change an analog signal to
digital data.
Example:
Voice is converted to a digital stream by a phone
then converted to an analog signal representing
digital data.
4.65
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).
4.66
4.2.1 Pulse Code Modulation (PCM)
A PCM encoder has three steps:
Sampling
• Encoding
• Quantizing
•
4.67
Figure 4.21: Components of PCM encoder
4.68
Sampling
Sampling – a snapshot of the analog signal
is recorded at regular time intervals.
If the time interval is short, the digital signal is
a good estimate of the original analog signal.
Figure 4.22: An example sampling method for PCM
4.70
Nyquist Theorem
To reproduce the original analog signal,
one necessary condition is that the
sampling rate must be twice the highest
frequency of the analog signal.
Figure 4.23: Nyquist sampling rate for low-pass and bandpass signals
4.72
Figure 4.24: Recovery of a sine wave with different sampling rates.
4.73
Example 4.9
Telephone companies digitize voice by assuming a
maximum frequency of 4000 Hz. What is the needed sample
rate for voice transmission?
Sample rate = 2*fmax
4.74
Example 4.9
Telephone companies digitize voice by assuming a
maximum frequency of 4000 Hz. What is the needed sample
rate for voice transmission?
Sample rate = 2*fmax
Sample rate = 2 * 4KHz = 8 K samples/sec
4.75
Example 4.10
A complex low-pass signal has a bandwidth of 200 kHz.
What is the minimum sampling rate for this signal?
4.76
Example 4.10
A complex low-pass signal has a bandwidth of 200 kHz.
What is the minimum sampling rate for this signal?
Solution
Sample rate = 2 * fmax = 400 Ksamples/sec
4.77
Example 4.11
A complex bandpass signal has a bandwidth of 200 kHz.
What is the minimum sampling rate for this signal?
.
4.78
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.
4.79
PCM: Quantization
1. Analog signal is between Vmin and Vmax
PCM: Quantization
1. Analog signal is between Vmin and Vmax
2. Divide the range into L equal zones
delta = (Vmax-Vmin)/L
PCM: Quantization
1. Analog signal is between Vmin and Vmax
2. Divide the range into L equal zones
3. Normalize using the delta value.
PCM: Quantization
1. Analog signal is between Vmin and Vmax
2. Divide the range into L equal zones
3. Normalize using the delta value
4. Assign a quantized value from 0 to L-1 to
the mid-point of each zone.
Figure 4.26: Quantization and encoding of a sampled signal
4.84
SNR-db for PCM
SNR db = 6.02 nb + 1.76
nb = number of bits per sample
Example 4.12
What is the SNRdB in the example of Figure 4.26?
4.86
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) + 1.76 = 19.82 dB. Increasing the number
of levels increases the SNR.
4.87
Example 4.13
A telephone subscriber line must have an SNRdB above 40.
What is the minimum number of bits per sample?
4.88
Example 4.13
A telephone subscriber line must have an SNRdB above 40.
What is the minimum number of bits per sample?
SNR-db = 6.02 nb + 1.76 db, solve for r
4.89
Example 4.13
A telephone subscriber line must have an SNRdB above 40.
What is the minimum number of bits per sample?
SNRdb = 6.02 nb + 1.76 db, solve for r
nb = (SNRdb – 1.76 db)/6.02
nb = (40 -1.76)/6.02 = 6.4
nb = 6.4, must be at least 7 bits
4.90
Example 4.14
We want to digitize the human voice. What is the bit rate,
assuming 8 bits per sample? The human voice has
frequencies from 0 to about 4KHz.
4.91
Example 4.14
We want to digitize the human voice. What is the bit rate,
assuming 8 bits per sample? The human voice has
frequencies from 0 to about 4KHz.
Sample rate = 2 * fmax
Sample rate = 2 * 4KHz = 8Ksamples/sec
Bit rate = 8 bits/sample * 8Ksamples/sec = 64kbps
4.92
4-3 TRANSMISSION MODES
The transmission of binary data across a link can be
accomplished in either
– parallel mode
– serial mode
4.93
Figure 4.31: Data transmission modes
4.94
4.3.1 Parallel Transmission
Bit streams are sent in parallel over multiple wires.
4.95
Figure 4.32: Parallel transmission
4.96
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
4.97
Figure 4.33: Serial transmission
4.98
Serial Transmission Types
Asynchronous
Synchronous
Isochronous
Serial Transmission Types
Asynchronous – signal timing is not
important due to the start bit, stop bit,
and gap for each 8-bit frame.
inefficient due to overhead
Figure 4.34: Asynchronous transmission
4.101
Serial Transmission
Synchronous – continuous stream of bits
without overhead or gaps. Timing at both
ends must be correct.
Figure 4.35: Synchronous transmission
Direction of flow
11110111
4.103
Frame
11111011 11110110
Frame
•••
Frame
11110111
11110011