Digital Transmission
* Line coding is the process of converting binary data to a digital signal.
* The number of different values allowed in a signal is the signal level. The number of symbols
that represent data is the data level.
* Bit rate is a function of the pulse rate and data level.
* Line coding methods must eliminate the dc component and provide a means of synchronization
between the sender and the receiver.
* Line coding methods can be classified as unipolar, polar, or bipolar.
* NRZ, RZ, Manchester, and differential Manchester encoding are the most popular polar
encoding methods.
* AMI is a popular bipolar encoding method.
* Block coding can improve the performance of line coding through redundancy and error
correction.
* Block coding involves grouping the bits, substitution, and line coding.
* 4B/5B, 8B/10B, and 8B/6T are common block coding methods.
* Analog-to-digital conversion relies on PCM (pulse code modulation).
* PCM involves sampling, quantizing, and line coding.
* The Nyquist theorem says that the sampling rate must be at least twice the highest-frequency
component in the original signal.
* Digital transmission can be either parallel or serial in mode.
* In parallel transmission, a group of bits is sent simultaneously, with each bit on a separate line.
* In serial transmission, there is only one line and the bits are sent sequentially.
* Serial transmission can be either synchronous or asynchronous.
* In asynchronous serial transmission, each byte (group of 8 bits) is framed with a start bit and a
stop bit. There may be a variable-length gap between each byte.
* In synchronous serial transmission, bits are sent in a continuous stream without start and stop
bits and without gaps between bytes. Regrouping the bits into meaningful bytes is the
responsibility of the receiver.
Multiple Choice Quiz
(See related pages)
Results Reporter
Out of 48 questions, you answered 14 correctly, for a final grade of 29%.
14 correct (29%)
34 incorrect (71%)
0 unanswered (0%)
0 ungraded (0%)
Please answer all questions.
Your Results:
The correct answer for each question is indicated by a
1CORRECT
.
Unipolar, bipolar, and polar encoding are types of _______ encoding.
A)line
B)block
C)NRZ
D)Manchester
2INCORRECT _______ encoding has a transition at the middle of each bit.
A)RZ
B)Manchester
C)Differential Manchester
D)All the above
3INCORRECT _______ encoding has a transition at the beginning of each 0 bit.
A)RZ
B)Manchester
C)Differential Manchester
D)All the above
4INCORRECT PCM is an example of _______ conversion.
A)digital-to-digital
B)digital-to-analog
C)analog-to-analog
D)analog-to-digital
If the frequency spectrum of a signal has a bandwidth of 500 Hz with the highest
5INCORRECT frequency at 600 Hz, what should be the sampling rate, according to the Nyquist
theorem?
A)200 samples/s
B)500 samples/s
C)1000 samples/s
D)1200 samples/s
6INCORRECT The Nyquist theorem specifies the minimum sampling rate to be_______.
A)equal to the lowest frequency of a signal
B)equal to the highest frequency of a signal
C)twice the bandwidth of a signal
D)twice the highest frequency of a signal
7INCORRECT Which of the following encoding methods does not provide for synchronization?
A)NRZ-L
B)RZ
C)NRZ-I
D)Manchester
8CORRECT
Which encoding method uses alternating positive and negative values for 1s?
A)NRZ-I
B)RZ
C)Manchester
D)AMI
9INCORRECT Which quantization level results in a more faithful reproduction of the signal?
A)2
B)8
C)16
D)32
10INCORRECT Block coding can help in _______ at the receiver.
A)Synchronization
B)Error detection
C)Attenuation
D)(a) and (b)
11CORRECT
In _______ transmission, bits are transmitted simultaneously, each across its own wire.
A)Asynchronous serial
B)Synchronous serial
C)Parallel
D)(a) and (b)
12CORRECT
In _______ transmission, bits are transmitted over a single wire, one at a time.
A)asynchronous serial
B)synchronous serial
C)parallel
D)(a) and (b)
13INCORRECT In _______ transmission, a start bit and a stop bit frame a character byte.
A)asynchronous serial
B)synchronous serial
C)parallel
D)(a) and (b)
14INCORRECT In asynchronous transmission, the gap time between bytes is _______.
A)fixed
B)variable
C)a function of the data rate
D)zero
15INCORRECT
___________ conversion involves three techniques: line coding, block coding, and
scrambling.
A)Analog-to-digital
B)Digital-to-analog
C)Analog-to-analog
D)Digital-to-digital
16CORRECT
________ is the process of converting digital data to a digital signal.
A)Block coding
B)Line coding
C)Scrambling
D)None of the above
17INCORRECT _______ provides redundancy to ensure synchronization and inherent error detection.
A)Block coding
B)Line coding
C)Scrambling
D)None of the above
18CORRECT
________ is normally referred to as mB/nB coding; it replaces each m-bit group with an
n-bit group.
A)Block coding
B)Line coding
C)Scrambling
D)None of the above
19CORRECT
________ provides synchronization without increasing the number of bits.
A)Scrambling
B)Line coding
C)Block coding
D)None of the above
20INCORRECT Two common scrambling techniques are ________.
A)NRZ and RZ
B)AMI and NRZ
C)B8ZS and HDB3
D)Manchester and differential Manchester
21INCORRECT
The most common technique to change an analog signal to digital data is called
__________.
A)PAL
B)PCM
C)sampling
D)none of the above
22INCORRECT The first step in PCM is ________.
A)quantization
B)modulation
C)sampling
D)none of the above
23INCORRECT There are three sampling methods: __________.
A)quantized, sampled, and ideal
B)ideal, sampled, and flat-top
C)ideal, natural, and flat-top
D)none of the above
24INCORRECT
______ finds the value of the signal amplitude for each sample; ____ finds the change
from the previous sample.
A)DM; PCM
B)PCM; DM
C)DM; CM
D)none of the above
25INCORRECT
While there is (are) only _____ way(s) to send parallel data, there is (are) three
subclass(es) of serial transmission.
A)one; two
B)two; three
C)one; three
D)none of the above
26INCORRECT
In ______ transmission, we send 1 start bit (0) at the beginning and 1 or more stop bits
(1s) at the end of each byte.
A)synchronous
B)asynchronous
C)isochronous
D)none of the above
27INCORRECT
In _________ transmission, we send bits one after another without start or stop bits or
gaps. It is the responsibility of the receiver to group the bits.
A)synchronous
B)asynchronous
C)isochronous
D)none of the above
28INCORRECT
The ________ mode provides synchronization for the entire stream of bits must. In other
words, it guarantees that the data arrive at a fixed rate.
A)synchronous
B)asynchronous
C)isochronous
D)none of the above
29CORRECT
A _________ digital signal includes timing information in the data being transmitted.
A)self-synchronizing
B)self-modulated
C)self-transmitted
D)none of the above
30CORRECT
In decoding a digital signal, the receiver calculates a running average of the received
signal power, called the _______.
A)baseline
B)base
C)line
D)none of the above
31CORRECT
The ________ rate defines the number of data elements sent in 1s; the ______ rate is the
number of signal elements sent in 1s.
A)data; signal
B)signal; data
C)baud; bit
D)none of the above
32CORRECT
The signal rate is sometimes called the ____ rate.
A)baud
B)bit
C)signal
D)none of the above
33INCORRECT The data rate is sometimes called the ___ rate.
A)baud
B)bit
C)signal
D)none of the above
34INCORRECT
In a _____ scheme, all the signal levels are on one side of the time axis, either above or
below.
A)polar
B)bipolar
C)unipolar
D)all of the above
35CORRECT
In ______ schemes, the voltages are on the 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.
A)polar
B)bipolar
C)unipolar
D)all of the above
36INCORRECT In _____, the level of the voltage determines the value of the bit.
A)NRZ-I
B)NRZ-L
C)both (a) and (b)
D)neither (a) nor (b)
37CORRECT
In ______, the change or lack of change in the level of the voltage determines the value
of the bit.
A)NRZ-I
B)NRZ-L
C)both (a) and (b)
D)neither (a) nor (b)
38CORRECT
The idea of RZ and the idea of NRZ-L are combined into the ________ scheme.
A)Manchester
B)differential Manchester
C)both (a) and (b)
D)neither (a) nor (b)
39INCORRECT The idea of RZ and the idea of NRZ-I are combined into the ________ scheme.
A)Manchester
B)differential Manchester
C)both (a) and (b)
D)neither (a) nor (b)
40INCORRECT In _______ encoding, 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.
A)Manchester
B)differential Manchester
C)both (a) and (b)
D)neither (a) nor (b)
In ___________ there is always a transition at the middle of the bit, but the bit values are
41INCORRECT 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.
A)Manchester
B)differential Manchester
C)both (a) and (b)
D)neither (a) nor (b)
42INCORRECT
In Manchester and differential Manchester encoding, the transition at the middle of the
bit is used for __________.
A)bit transfer
B)baud transfer
C)synchronization
D)none of the above
43INCORRECT The minimum bandwidth of Manchester and differential Manchester is ____ that of NRZ.
A)the same as
B)twice
C)thrice
D)none of the above
44INCORRECT In _______encoding, we use three levels: positive, zero, and negative.
A)unipolar
B)bipolar
C)polar
D)none of the above
45INCORRECT
The _____ scheme uses data patterns of size 2 and encodes the 2-bit patterns as one
signal element belonging to a four-level signal.
A)4B5B
B)2B1Q
C)MLT-3
D)none of the above
46INCORRECT
The ______ scheme uses three levels (+V, 0, and -V) and three transition rules to move
between the levels.
A)4B5B
B)2B1Q
C)MLT-3
D)none of the above
47INCORRECT ______ substitutes eight consecutive zeros with 000VB0VB.
A)B4B8
B)HDB3
C)B8ZS
D)none of the above
48INCORRECT ______ substitutes four consecutive zeros with 000V or B00V
A)B4B8
B)HDB3
C)B8ZSf
D)none of the above
1
CHAPTER 4
Digital Transmission
Solutions to Odd-Numbered Review Questions and Exercises
Review Questions
1. The three different techniques described in this chapter are line coding, block coding,
and scrambling.
3. The data rate defines the number of data elements (bits) sent in 1s. The unit is bits
per second (bps). The signal rate is the number of signal elements sent in 1s. The
unit is the baud.
5. When the voltage level in a digital signal is constant for a while, the spectrum creates
very low frequencies, called DC components, that present problems for a system
that cannot pass low frequencies.
7. In this chapter, we introduced unipolar, polar, bipolar, multilevel, and multitransition
coding.
9. Scrambling, as discussed in this chapter, is a technique that substitutes long zerolevel
pulses with a combination of other levels without increasing the number of
bits.
11. In parallel transmission we send data several bits at a time. In serial transmission
we send data one bit at a time.
Exercises
13. We use the formula s = c × N × (1/r) for each case. We let c = 1/2.
a. r = 1 → s = (1/2) × (1 Mbps) × 1/1 = 500 kbaud
b. r = 1/2 → s = (1/2) × (1 Mbps) × 1/(1/2) = 1 Mbaud
c. r = 2 → s = (1/2) × (1 Mbps) × 1/2 = 250 Kbaud
d. r = 4/3 → s = (1/2) × (1 Mbps) × 1/(4/3) = 375 Kbaud
15. See Figure 4.1 Bandwidth is proportional to (3/8)N which is within the range in
Table 4.1 (B = 0 to N) for the NRZ-L scheme.
17. See Figure 4.2. Bandwidth is proportional to (12.5 / 8) N which is within the range
in Table 4.1 (B = N to B = 2N) for the Manchester scheme.
2
19. See Figure 4.3. B is proportional to (5.25 / 16) N which is inside range in Table 4.1
(B = 0 to N/2) for 2B/1Q.
21. The data stream can be found as
a. NRZ-I: 10011001.
b. Differential Manchester: 11000100.
c. AMI: 01110001.
23. The data rate is 100 Kbps. For each case, we first need to calculate the value f/N.
We then use Figure 4.8 in the text to find P (energy per Hz). All calculations are
approximations.
a. f /N = 0/100 = 0 → P = 0.0
b. f /N = 50/100 = 1/2 → P = 0.3
c. f /N = 100/100 = 1 → P = 0.4
d. f /N = 150/100 = 1.5 → P = 0.0
Figure 4.1 Solution to Exercise 15
Figure 4.2 Solution to Exercise 17
00000000
1111111100110011
01010101
Case a
Case b
Case c
Case d
Average Number of Changes = (0 + 0 + 8 + 4) / 4 = 3 for N = 8
B (3 / 8) N
00000000
1111111100110011
01010101
Case a
Case b
Case c
Case d
Average Number of Changes = (15 + 15+ 8 + 12) / 4 = 12.5 for N = 8
B (12.5 / 8) N
3
25. In 5B/6B, we have 25 = 32 data sequences and 26 = 64 code sequences. The number
of unused code sequences is 64 − 32 = 32. In 3B/4B, we have 23 = 8 data
sequences and 24 = 16 code sequences. The number of unused code sequences is
16 − 8 = 8.
27
a. In a low-pass signal, the minimum frequency 0. Therefore, we have
fmax = 0 + 200 = 200 KHz. → fs = 2 × 200,000 = 400,000 samples/s
b. In a bandpass signal, the maximum frequency is equal to the minimum frequency
plus the bandwidth. Therefore, we have
fmax = 100 + 200 = 300 KHz. → fs = 2 × 300,000 = 600,000 samples /s
29. The maximum data rate can be calculated as
Nmax = 2 × B × nb = 2 × 200 KHz × log24 = 800 kbps
31. We can calculate the data rate for each scheme:
Figure 4.3 Solution to Exercise 19
a. NRZ → N = 2 × B = 2 × 1 MHz = 2 Mbps
b. Manchester → N = 1 × B = 1 × 1 MHz = 1 Mbps
c. MLT-3 → N = 3 × B = 3 × 1 MHz = 3 Mbps
d. 2B1Q → N = 4 × B = 4 × 1 MHz = 4 Mbps
11 11 11 11 11 11 11 11
00 00 00 00 00 00 00 00 01 10 01 10 01 10 01 10
+3
+1
−3
−1
+3
+1
−3
−1
+3
+1
−3
−1
00 11 00 11 00 11 00 11
+3
+1
−3
−1
Case a
Case b
Case c
Case d
Average Number of Changes = (0 + 7 + 7 + 7) / 4 = 5.25 for N = 16
B (5.25 / 8) N
4
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