1. 4.1
Chapter 4
Digital Transmission

2. 4.2
Components of Data
Communication
 Data
 Analog: Continuous value data (sound, light,
temperature)
 Digital: Discrete value (text, integers,
symbols)
 Signal
 Analog: Continuously varying
electromagnetic wave
 Digital: Series of voltage pulses (square
wave)

3. 4.3
Analog Data-->Signal Options
 Analog data to analog signal
 Inexpensive, easy conversion (eg telephone)
 Used in traditional analog telephony
 Analog data to digital signal
 Requires a codec (encoder/decoder)
 Allows use of digital telephony, voice mail

4. 4.4
Digital Data-->Signal Options
 Digital data to analog signal
 Requires modem (modulator/demodulator)
 Necessary when analog transmission is used
 Digital data to digital signal
 Less expensive when large amounts of data
are involved
 More reliable because no conversion is
involved

5. 4.5
4-1 DIGITAL-TO-DIGITAL CONVERSION4-1 DIGITAL-TO-DIGITAL CONVERSION
In this section, we see how we can represent digitalIn this section, we see how we can represent digital
data by using digital signals. The conversion involvesdata by using digital signals. The conversion involves
three techniques:three techniques: line codingline coding,, block codingblock coding, and, and
scramblingscrambling. Line coding is always needed; block. Line coding is always needed; block
coding and scrambling may or may not be needed.coding and scrambling may or may not be needed.
Line Coding
Line Coding Schemes
Block Coding
Scrambling
Topics discussed in this section:Topics discussed in this section:

6. 4.6
Figure 4.1 Line coding and decoding

7. 4.7
Figure 4.2 Signal element versus data element
r = number of data elements / number of signal elements

8. 4.8
Data Rate Vs. Signal Rate
•Data rate: the number of data elements (bits) sent
in 1s (bps). It’s also called the bit rate
•Signal rate: the number of signal elements sent in 1s
(baud). It’s also called the pulse rate, the modulation
rate, or the baud rate.
We wish to:
1. increase the data rate (increase the speed of
transmission)
2. decrease the signal rate (decrease the bandwidth
requirement)
3. Worst case, best case, and average case of r
4. S = c * N / r baud

9. 4.9
Baseline wandering
Baseline: running average of the
received signal power
DC Components
Constant digital signal creates low
frequencies
Self-synchronization
Receiver Setting the clock matching the
sender’s

10. 4.10
Figure 4.3 Effect of lack of synchronization

11. 4.11
Figure 4.4 Line coding schemes

12. 4.12
Figure 4.5 Unipolar NRZ scheme

13. 4.13
Digital Encoding
of Digital Data
 Most common, easiest method is different
voltage levels for the two binary digits
 Typically, negative=1 and positive=0
 Known as NRZ-L, or nonreturn-to-zero
level, because signal never returns to zero,
and the voltage during a bit transmission is
level

14. 4.14
Differential NRZ
 Differential version is NRZI (NRZ, invert
on ones)
 Change=1, no change=0
 Advantage of differential encoding is that it
is more reliable to detect a change in
polarity than it is to accurately detect a
specific level

15. 4.15
Problems With NRZ
 Difficult to determine where one bit ends
and the next begins
 In NRZ-L, long strings of ones and zeroes
would appear as constant voltage pulses
 Timing is critical, because any drift results
in lack of synchronization and incorrect bit
values being transmitted

16. 4.16
Figure 4.6 Polar NRZ-L and NRZ-I schemes

17. 4.17
Figure 4.7 Polar RZ scheme

18. 4.18
Manchester Code
 Transition in the middle of each bit period
 Transition provides clocking and data
 Low-to-high=1 , high-to-low=0
 Used in Ethernet

19. 4.19
Differential Manchester
 Midbit transition is only for clocking
 Transition at beginning of bit period=0
 Transition absent at beginning=1
 Has added advantage of differential
encoding
 Used in token-ring

20. 4.20
Figure 4.8 Polar biphase: Manchester and differential Manchester schemes

21. 4.21
• High=0, Low=1
• No change at begin=0, Change at
begin=1
• H-to-L=0, L-to-H=1
• Change at begin=0, No change at
begin=1

22. 4.22
Bipolar schemes: AMI (Alternate Mark Inversion) and pseudoternary

23. 4.23
Multilevel Schemes
• In mBnL schemes, a pattern of m
data elements is encoded as a
pattern of n signal elements in
which 2m
≤ Ln
• m: the length of the binary pattern
• B: binary data
• n: the length of the signal pattern
• L: number of levels in the signaling
• B for l=2 binary
• T for l=3 ternary
• Q for l=4 quaternary

24. 4.24
Figure 4.10 Multilevel: 2B1Q scheme
Used in DSL

25. 4.25
Figure 4.11 Multilevel: 8B6T scheme

26. 4.26
Figure 4.13 Multitransition: MLT-3 scheme

27. 4.27
Table 4.1 Summary of line coding schemes
Polar

28. 4.28
Block Coding
• Redundancy is needed to ensure
synchronization and to provide
error detecting
• Block coding is normally referred to
as mB/nB coding
• it replaces each m-bit group with an
n-bit group
• m < n

29. 4.29
Figure 4.14 Block coding concept

30. 4.30
Figure 4.15 Using block coding 4B/5B with NRZ-I line coding scheme

31. 4.31
Table 4.2 4B/5B mapping codes

32. 4.32
Figure 4.16 Substitution in 4B/5B block coding

33. 4.33
Figure 4.17 8B/10B block encoding

34. 4.34
Scrambling
• It modifies the bipolar AMI encoding
(no DC component, but having the
problem of synchronization)
• It does not increase the number of
bits
• It provides synchronization
• It uses some specific form of bits to
replace a sequence of 0s

35. 4.35
Figure 4.19 Two cases of B8ZS scrambling technique
B8ZS substitutes eight consecutive zeros with 000VB0VB

36. 4.36
Figure 4.20 Different situations in HDB3 scrambling technique
HDB3 substitutes four consecutive zeros with 000V or B00V depending
on the number of nonzero pulses after the last substitution.

37. 4.37
4-2 ANALOG-TO-DIGITAL CONVERSION4-2 ANALOG-TO-DIGITAL CONVERSION
The tendency today is to change an analog signal toThe tendency today is to change an analog signal to
digital data.digital data.
In this section we describe two techniques,In this section we describe two techniques,
pulse code modulationpulse code modulation andand delta modulationdelta modulation..

38. 4.38
Figure 4.21 Components of PCM encoder

39. 4.39
According to the Nyquist theorem, the
sampling rate must be at least 2 times
the highest frequency contained in the
signal.
What can we get from this:
1. we can sample a signal only if the signal is
band-limited
2. the sampling rate must be at least 2 times the
highest frequency, not the bandwidth

40. 4.40
Figure 4.23 Nyquist sampling rate for low-pass and bandpass signals

41. 4.41
Figure 4.24 Recovery of a sampled sine wave for different sampling rates

42. 4.42
Figure 4.25 Sampling of a clock with only one hand

43. 4.43
An example related is the seemingly backward rotation of
the wheels of a forward-moving car in a movie.
This can be explained by under-sampling. A movie is
filmed at 24 frames per second. If a wheel is rotating
more than 12 times per second, the under-sampling
creates the impression of a backward rotation.
Example

44. 4.44
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.
Example

45. 4.45
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.
Example

46. 4.46
Figure 4.26 Quantization and encoding of a sampled signal

47. 4.47
What is the SNRdB in the example of Figure 4.26?
Solution
We have eight levels and 3 bits per sample, so
SNRdB = 6.02 x 3 + 1.76 = 19.82 dB
Increasing the number of levels increases the SNR.
Contribution of the quantization error to SNRdb
SNRdb= 6.02nb + 1.76 dB
nb: bits per sample (related to the number of level L)

48. 4.48
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
Example
Telephone companies usually assign 7 or 8 bits per
sample.

49. 4.49
PCM decoder: recovers the original signal

50. 4.50
We have a low-pass analog signal of 4 kHz. If we send the
analog signal, we need a channel with a minimum
bandwidth of 4 kHz. If we digitize the signal and send 8
bits per sample, we need a channel with a minimum
bandwidth of 8 × 4 kHz = 32 kHz.
The minimum bandwidth of the digital signal is nb
times greater than the bandwidth of the analog
signal.
Bmin= nb x Banalog

51. 4.51
DM (delta modulation) finds the change from the
previous sample
Next bit is 1, if amplitude of the analog signal is larger
Next bit is 0, if amplitude of the analog signal is smaller

52. 4.52
Figure 4.29 Delta modulation components

53. 4.53
Figure 4.30 Delta demodulation components

54. 4.54
4-3 TRANSMISSION MODES4-3 TRANSMISSION MODES
1. The transmission of binary data across a link can1. The transmission of binary data across a link can
be accomplished in either parallel or serial mode.be accomplished in either parallel or serial mode.
2. In parallel mode, multiple bits are sent with each2. In parallel mode, multiple bits are sent with each
clock tick.clock tick.
3. In serial mode, 1 bit is sent with each clock tick.3. In serial mode, 1 bit is sent with each clock tick.
4. there are three subclasses of serial transmission:4. there are three subclasses of serial transmission:
asynchronous, synchronous, and isochronous.asynchronous, synchronous, and isochronous.

55. 4.55
Figure 4.31 Data transmission and modes

56. 4.56
Figure 4.32 Parallel transmission

57. 4.57
Figure 4.33 Serial transmission

58. 4.58
Asynchronous transmission
1. We send 1 start bit (0) at the beginning and 1 or more stop bits
(1s) at the end of each byte.
2. There may be a gap between each byte.
3. Extra bits and gaps are used to alert the receiver, and allow it to
synchronize with the data stream.
4. Asynchronous here means “asynchronous at the byte level,”
but the bits are still synchronized, their durations are the same.

59. 4.59
Synchronous transmission
In synchronous 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.