DATA COMMUNICATIONS

IT – 21012
(SEMESTER I)
 CHAPTER 4

DIGITAL TRANSMISSION

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 4.1 DIGITAL-TO-DIGITAL CONVERSION
• In this section, we explore how digital data is represented using digital signals.
• The conversion involves three techniques:
• line coding,
• block coding, and
• scrambling.
Note: Line coding is always needed;
block coding and scrambling may or may not be needed.

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 4.1.1. LINE CODING

• Line coding is the process of converting digital data to digital signals. Data may
be in the form of text, numbers, graphical images, audio, or video, are stored in
computer memory as sequences of bits. 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.

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 SIGNAL ELEMENT VS DATA ELEMENT
Data Element Signal Element
the smallest entity that can represent a the shortest unit (timewise) of a digital
piece of information signal

A Data element is the bit. A signal element carries data elements.

Data elements are being carried. Signal elements are the carriers.

A ratio r is defined the number of data elements carried by each signal element.

Define the ratio of data elements to signal elements with r = (data elements) / (signal
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elements).
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 DATA RATE VS SIGNAL RATE
Data Rate Signal Rate
the number of data elements (bits) sent in 1s the number of signal elements sent in 1s.

The unit is bits per second (bps). The unit is the baud.
The data rate is sometimes called the bit rate.
The signal rate is sometimes called the pulse
rate, the modulation rate, or the baud rate.
Goal in data communications: to increase the Goal in data communications : to decrease the
data rate signal rate

Increasing the data rate increases the speed of Decreasing the signal rate decreases the
transmission. bandwidth requirement.

The relationship between data rate and signal rate is

Save = c x N x (1/r) baud
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N is the data rate (bps); c is the case factor; S is the number of signal elements per second; and r is
the previously defined factor.
 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

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 BANDWIDTH
• The minimum bandwidth can be given as

the maximum data rate if the bandwidth of the channel is given

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 EXAMPLE 4.2

The maximum data rate of a channel 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 )

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 BASELINE WANDERING

• In decoding a digital signal, the receiver calculates a running average of the received
signal power. This average is called the baseline. The incoming signal power is
evaluated against this baseline to determine the value of the data element. A long
string of 0s or 1s can cause a drift in the baseline (baseline wandering) and make it
difficult for the receiver to decode correctly. A good line coding scheme needs to
prevent baseline wandering.

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 DC COMPONENTS

• When the voltage level in a digital signal is constant for a while, the spectrum
creates very low frequencies (results of Fourier analysis). These frequencies around
zero, called DC (direct-current) components, present problems for a system that
cannot pass low frequencies. A telephone line cannot pass frequencies below 200
Hz. For telephone systems, we need a scheme with no DC component.

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 SELF-SYNCHRONIZATION
• To correctly interpret the signals received from the sender, the receiver’s bit intervals
must correspond exactly to the sender’s bit intervals. If the receiver clock is faster or
slower, the bit intervals are not matched and the receiver might misinterpret the
signals. A self-synchronizing digital signal includes timing information in the data
being transmitted. This can be achieved if there are transitions in the signal that alert
the receiver to the beginning, middle, or end of the pulse. If the receiver’s clock is
out of synchronization, these points can reset the clock.

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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.

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 4.1.2 LINE CODING SCHEMES

• We can roughly divide line coding schemes into five broad categories, as shown in
Figure 4.4.

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 UNIPOLAR SCHEME
• In a unipolar scheme, all the signal levels are on one side of the time axis.
NRZ (Non-Return-to-Zero)
The positive voltage defines bit 1 and the zero voltage defines bit 0. It is called
NRZ because the signal does not return to zero at the middle of the bit. Compared with
its polar this scheme is very costly. The normalized power (the power needed to send
1 bit per unit line resistance) is double that for polar NRZ. For this reason, this scheme
is normally not used in data communications today.

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 POLAR SCHEMES
• In polar schemes, the voltages are on both sides of the time axis. For example, the voltage
level for 0 can be positive and the voltage level for 1 can be negative.
• Non-Return-to-Zero (NRZ)
• Return-Zero(RZ)
• Biphase (Manchester and differential Manchester)

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 Non-Return-to-Zero(NRZ)
Non-Return-to-Zero(NRZ)
Polar NRZ scheme can be implemented with two voltages(V).
Eg. –V for bit 1 and +V for bit 0.
Two levels of voltage amplitude are used. Two versions of polar NRZ: NRZ-L and
NRZ-I.
NRZ-L (NRZ-Level)
The level of the voltage determines the value of the bit. Eg. Voltage-level for 0 can be
positive and Voltage-level for 1 can be negative.
NRZ-I (NRZ-Invert)
The change or lack of change in the level of the voltage determines the value of the bit.
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If there is no change, the bit is 0; if there is a change, the bit is 1.
Note: NRZ-L and NRZ-I both have an average signal rate of N/2 Bd. 19
 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 Solution:
r = 1 (1 bit / signal)
S=c*N/r
S = ½ * (1Mbps) / 1 = .5 Mbaud = 500 Kbaud
r = log2(L)
N = 2 B log2( L) solve for B
B = ½ * (N / r ) 20

B = 1Mbps/2 = .5 MHz = 500 KHz

 RETURN- ZERO (RZ)
• The main problem with NRZ encoding occurs when the sender and receiver clocks are
not synchronized. One solution is to use the return-to-zero (RZ) scheme, which uses
three values: positive, negative, and zero. There is always a transition at the middle of
the bit: from high to zero (for 1) or from low to zero (for 0).
• The main disadvantage of RZ encoding is that it requires two signal changes to encode
a bit and therefore occupies greater bandwidth. Another problem is the complexity: RZ
uses three levels of voltage, which is more complex to create and detect. Problem
occurs when there is a sudden change of polarity in the system. This result in all 0s
interpreted as 1s and all 1s interpreted as 0s.

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 BIPHASE
Manchester Encoding
This is a combination of NRZ-L and RZ schemes (RZ – transition at the middle of
the bit). There is always a transition at the middle of the bit. Either from high to low (for 0)
or from low to high (for 1). It uses only two voltage levels. The duration of the bit is
divided into two halves. The voltage remains at one level during the first half and moves to
the other level in the second half. The transition at the middle of the bit provides
synchronization.
Differential Manchester
This is a combination of NRZ-I and RZ schemes. There is always a transition at the
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middle of the bit, but the bit-values are determined at the beginning of the bit. If the next
bit is 0, there is a transition. If the next bit is 1, there is none.
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 BIPOLAR SCHEME
• The coding scheme uses three voltage levels: positive, negative, and zero. Two
variations of bipolar encoding: AMI (Alternate Mark Inversion) and Pseudo-ternary.
• AMI
Binary 0 is represented by a neural 0 voltage. Binary 1s are represented by
alternating positive and negative voltages. It is used for some long distance
communications.
• Pseudoternary
Binary 1 is represented by a neural 0 voltage. Binary 0s are represented by
alternating positive and negative voltages.
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 MULTILEVEL SCHEMES
The goal is to send more bits per signal. We designate the different schemes by
this notation: mBnL
m = the length of the binary pattern
B = two possible data elements (0 or 1) binary data
L = number of levels
N = the length of the signal pattern
A letter is often used in place of L: B (binary) for L = 2, T (ternary) for L = 3, and Q
(quaternary) for L = 4.
In mBnL schemes, a pattern of m data elements is encoded as a pattern of n signal
elements in which 2m ≤ Ln.
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 2B1Q
• The first mBnL scheme we discuss, two binary, one quaternary (2B1Q), uses data
patterns of size 2 and encodes the 2-bit patterns as one signal element belonging to a
four-level signal. In this type of encoding m = 2, n = 1, and L = 4 (quaternary).

There are no redundant signal patterns in this scheme because 22 = 41. The 2B1Q scheme is used in28DSL
(Digital Subscriber Line) technology to provide a high-speed connection to the Internet by using subscriber
telephone lines.
 8B6T
A very interesting scheme is eight binary, six ternary (8B6T). This code is used with
100BASE-4T cable. The idea is to encode a pattern of 8 bits as a pattern of six signal
elements, where the signal has three levels (ternary). In this type of scheme, we can have
28 = 256 different data patterns and 36 = 729 different signal patterns. The mapping table
is shown in Appendix F. There are 729 - 256 = 473 redundant signal elements that
provide synchronization and error detection.

The average signal rate of the scheme is theoretically Save = ½ × N × 6/8; in practice the minimum 29

bandwidth is very close to 6N/8.

 4D-PAM5
• The last signaling scheme we discuss in this category is called four-dimensional five
level pulse amplitude modulation (4D-PAM5). The 4D means that data is sent over
four wires at the same time. It uses five voltage levels, such as -2, -1, 0, 1, and 2.
The worst signal rate for this imaginary one-dimensional version is N × 4/8, or N/2
Four wires transmit in parallel using 8B4Q over each wire. This is used for Gigabit
Ethernet.

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 MULTITRANSITION: MLT-3
• The multiline transmission, three-level (MLT-3) scheme uses three levels (+V, 0, and -
V) and three transition rules to move between the levels.
1. If the next bit is 0, there is no transition.
2. If the next bit is 1 and the current level is not 0, the next level is 0.
3. If the next bit is 1 and the current level is 0, the next level is the opposite of the last
nonzero level.

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SUMMARY OF LINE CODING SCHEMES

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 4.1.3 BLOCK CODING

Redundancy is needed to ensure synchronization and to provide some kind of

inherent error detecting. Block coding can give 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. Block coding normally involves three steps: division, substitution,
and combination. In the division step, a sequence of bits is divided into groups of m
bits. In substitution step, we substitute an m-bit group with an n-bit group. Finally, the
n-bit groups are combined to form a stream. The new stream has more bits than the
original bits.
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 4B/5B
• The four binary/five binary (4B/5B) coding scheme was designed to be used in
combination with NRZ-I.

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 EXAMPLE 4.5
We need to send data at a 1-Mbps rate. What is the minimum required bandwidth, using a
combination of 4B/5B and NRZ-I or Manchester coding?
Solution:
NRZ-I encodes one bit per signal, r = 1.
Manchester encoding encodes one bit per two signals: r = ½
C = ½ for both cases.
B=cN/r
NRZ-I : B = ½ (1.25Mbps) / 1 = 626KHz
Man : B = ½ ( 1 Mbps) / .5 = 1. MHz
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 8B/10B
• The eight binary/ten binary (8B/10B) encoding is similar to 4B/5B encoding except
that a group of 8 bits of data is now substituted by a 10-bit code. It provides greater
error detection capability than 4B/5B. The 8B/10B block coding is actually a
combination of 5B/6B and 3B/4B encoding.

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 4.1.4 SCRAMBLING
• Scrambling is used with long distance bipolar AMI line coding.
• Scrambling helps synchronize when the data stream contains a sequence of 8
consecutive zeros.

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 SCRAMBLING TECHNIQUES
• Two common scrambling techniques are B8ZS and HDB3.
• Bipolar with 8-zero substitution (B8ZS) is commonly used in North America. In this
technique, eight consecutive zero-level voltages are replaced by the sequence
000VB0VB. The V in the sequence denotes violation; this is a nonzero voltage that
breaks an AMI rule of encoding (opposite polarity from the previous). The B in the
sequence denotes bipolar, which means a nonzero level voltage in accordance with the
AMI rule. There are two cases.

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 SCRAMBLING TECHNIQUES
• High-density bipolar 3-zero (HDB3) is commonly used outside of North America. In
this technique, which is more conservative than B8ZS, four consecutive zero-level
voltages are replaced with a sequence of 000V or B00V. The reason for two different
substitutions is to maintain the even number of nonzero pulses after each substitution.
The two rules can be stated as follows:
1. If the number of nonzero pulses after the last substitution is odd, the substitution
pattern will be 000V, which makes the total number of nonzero pulses even.
2. If the number of nonzero pulses after the last substitution is even, the substitution
pattern will be B00V, which makes the total number of nonzero pulses even.
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 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.

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 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 steps:
 The analog signal is sampled.
The sampled signal is quantized.
The quantized values are encoded as streams of bits.

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 PCM - 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.
The inverse of the sampling interval is called the sampling rate or sampling
frequency and denoted by fs, where fs = 1/Ts.
To reproduce the original analog signal, one necessary condition is that the
sampling rate must be twice the highest frequency of the analog signal.

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• Example 4.9. Telephone companies digitize voice by assuming a maximum
frequency of 4000 Hz. What is the needed sample rate for voice transmission?
• Solution:
Sample rate = 2*fmax
Sample rate = 2 * 4KHz = 8 K samples/sec

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.
Sample rate = 2 * fmax = 400 Ksamples/sec
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• 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.

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 PCM - QUANTIZATION

1. Analog signal is between Vmin and Vmax

2. Divide the range into L equal zones, each of height ∆ (delta) Normalize using the
delta value

3. Assign a quantized value from 0 to L-1 to the mid-point of each zone.

4. We approximate the value of the sample amplitude to the quantized values.

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The value at the top of each sample in the graph shows the actual amplitude. In the
chart, the first row is the normalized value for each sample (actual amplitude/Δ). The
quantization process selects the quantization value from the middle of each zone. This
means that the normalized quantized values (second row) are different from the
normalized amplitudes. The difference is called the normalized error (third row). The
fourth row is the quantization code for each sample based on the quantization levels
at the left of the graph. The encoded words (fifth row) are the final products of the
conversion.

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 SNR-DB FOR PCM

SNR db = 6.02 nb + 1.76

nb = number of bits per sample

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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.
Example 4.13. A telephone subscriber line must have an SNRdB above 40. What is the minimum
number of bits per sample?
Solution:
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 57
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.
Solution:
Sample rate = 2 * f max
Sample rate = 2 * 4KHz = 8Ksamples/sec
Bit rate = 8 bits/sample * 8Ksamples/sec = 64kbps

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 4.2.2 DELTA MODULATION (DM)

• PCM is a very complex technique. Other techniques have been developed to reduce the
complexity of PCM. The simplest is delta modulation. PCM finds the value of the
signal amplitude for each sample; DM finds the change from the previous sample.
Figure 4.28 shows the process. Note that there are no code words here; bits are sent one
after another.

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 4.3 TRANSMISSION MODES

• Two ways of transmitting data over a link: Parallel Mode and Serial Mode.

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 PARALLEL TRANSMISSION
• Multiple bits are sent with each block-tick. ‘n’ bits in a group are sent
simultaneously. ‘n’ wires are used to send ‘n’ bits at one time. Each bit has its own
wire. Typically, the 8 wires are bundled in a cable with a connector at each end.
• Advantage:
Speed: Parallel transmission can increase the transfer speed by a factor of n
over serial transmission.
• Disadvantage:
Cost: Parallel transmission requires n communication lines just to transmit the
data-stream. Because this is expensive, parallel transmission is usually limited to short
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distances.
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 SERIAL TRANSMISSION
• Serial Mode
One bit is sent with each clock-tick using only a single link.
• Advantage:
Cost: Serial transmission reduces cost of transmission over parallel by a
factor of n.
• Disadvantage:
Since communication with device is parallel, following two converters are
required at interface: Parallel-to-serial converter and Serial-to-parallel converter.
Three types of serial transmission: asynchronous, synchronous and isochronous. 64
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