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 ReporterOut 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

5INCORRECTIf the frequency spectrum of a signal has a bandwidth of 500 Hz with the highest 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

21INCORRECTThe 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

25INCORRECTWhile 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

26INCORRECTIn ______ 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

27INCORRECTIn _________ 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

28INCORRECTThe ________ 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

30CORRECTIn 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

31CORRECTThe ________ 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

34INCORRECTIn 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

35CORRECTIn ______ 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)

37CORRECTIn ______, 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)

40INCORRECTIn _______ 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)

41INCORRECTIn ___________ there is always a transition at the 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.

A)Manchester

B)differential Manchester

C)

both (a) and (b)

D)neither (a) nor (b)

42INCORRECTIn 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

45INCORRECTThe _____ 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

46INCORRECTThe ______ 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 4Digital TransmissionSolutions to Odd-Numbered Review Questions and ExercisesReview Questions1. 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 bitsper second (bps). The signal rate is the number of signal elements sent in 1s. Theunit is the baud.5. When the voltage level in a digital signal is constant for a while, the spectrum createsvery low frequencies, called DC components, that present problems for a systemthat cannot pass low frequencies.7. In this chapter, we introduced unipolar, polar, bipolar, multilevel, and multitransitioncoding.9. Scrambling, as discussed in this chapter, is a technique that substitutes long zerolevelpulses with a combination of other levels without increasing the number ofbits.11. In parallel transmission we send data several bits at a time. In serial transmissionwe send data one bit at a time.Exercises13. 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 kbaudb. r = 1/2 → s = (1/2) × (1 Mbps) × 1/(1/2) = 1 Mbaudc. r = 2 → s = (1/2) × (1 Mbps) × 1/2 = 250 Kbaudd. r = 4/3 → s = (1/2) × (1 Mbps) × 1/(4/3) = 375 Kbaud15. See Figure 4.1 Bandwidth is proportional to (3/8)N which is within the range inTable 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 rangein Table 4.1 (B = N to B = 2N) for the Manchester scheme.219. 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 asa. 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 areapproximations.a. f /N = 0/100 = 0 → P = 0.0b. f /N = 50/100 = 1/2 → P = 0.3c. f /N = 100/100 = 1 → P = 0.4d. f /N = 150/100 = 1.5 → P = 0.0Figure 4.1 Solution to Exercise 15

Figure 4.2 Solution to Exercise 170 0 0 0 0 0 0 01 1 1 1 1 1 1 1 0 0 1 1 0 0 1 10 1 0 1 0 1 0 1Case aCase bCase cCase dAverage Number of Changes = (0 + 0 + 8 + 4) / 4 = 3 for N = 8B (3 / 8) N0 0 0 0 0 0 0 01 1 1 1 1 1 1 1 0 0 1 1 0 0 1 10 1 0 1 0 1 0 1Case aCase bCase cCase dAverage Number of Changes = (15 + 15+ 8 + 12) / 4 = 12.5 for N = 8B (12.5 / 8) N325. In 5B/6B, we have 25 = 32 data sequences and 26 = 64 code sequences. The numberof unused code sequences is 64 − 32 = 32. In 3B/4B, we have 23 = 8 datasequences and 24 = 16 code sequences. The number of unused code sequences is16 − 8 = 8.27a. In a low-pass signal, the minimum frequency 0. Therefore, we havefmax = 0 + 200 = 200 KHz. → fs = 2 × 200,000 = 400,000 samples/sb. In a bandpass signal, the maximum frequency is equal to the minimum frequencyplus the bandwidth. Therefore, we havefmax = 100 + 200 = 300 KHz. → fs = 2 × 300,000 = 600,000 samples /s29. The maximum data rate can be calculated asNmax = 2 × B × nb = 2 × 200 KHz × log24 = 800 kbps31. We can calculate the data rate for each scheme:Figure 4.3 Solution to Exercise 19a. NRZ → N = 2 × B = 2 × 1 MHz = 2 Mbpsb. Manchester → N = 1 × B = 1 × 1 MHz = 1 Mbpsc. MLT-3 → N = 3 × B = 3 × 1 MHz = 3 Mbpsd. 2B1Q → N = 4 × B = 4 × 1 MHz = 4 Mbps11 11 11 11 11 11 11 1100 00 00 00 00 00 00 00 01 10 01 10 01 10 01 10+3+1−3−1+3+1−3−1+3+1−3−100 11 00 11 00 11 00 11+3+1−3−1Case aCase bCase cCase dAverage Number of Changes = (0 + 7 + 7 + 7) / 4 = 5.25 for N = 16B (5.25 / 8) N4

ANIMATION FOR FIGURES

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