Chapter 4
DigitalTransmission
4.1 Line Coding
Some Characteristics
Line Coding Schemes
Some Other Schemes
Figure 4.1 Line coding
Figure 4.2 Signal level versus data level
two
Example 1Example 1
A signal has two data levels with a pulse duration of 1ms. We calculate the pulse rate and bit rate as follows:
Pulse Rate = 1/ 10-3= 1000 pulses/sPulse Rate = 1/ 10-3= 1000 pulses/s
Bit Rate = Pulse Rate x log2 L = 1000 x log2 2 = 1000 bpsBit Rate = Pulse Rate x log2 L = 1000 x log2 2 = 1000 bps
Example 2Example 2
A signal has four data levels with a pulse duration of 1ms. We calculate the pulse rate and bit rate as follows:
Pulse Rate = = 1000 pulses/sPulse Rate = = 1000 pulses/s
Bit Rate = PulseRate x log2 L = 1000 x log2 4 = 2000 bpsBit Rate = PulseRate x log2 L = 1000 x log2 4 = 2000 bps
Figure 4.4 Lack of synchronization
Example 3Example 3
In a digital transmission, the receiver clock is 0.1 percentfaster than the sender clock. How many extra bits persecond does the receiver receive if the data rate is 1Kbps? How many if the data rate is 1 Mbps?
SolutionSolution
At 1 Kbps:
1000 bits sent 1001 bits received1 extra bps
At 1 Mbps:
1,000,000 bits sent 1,001,000 bits received1000 extra bps
Figure 4.5 Line coding schemes
Unipolar encoding uses only onevoltage level.
Note:Note:
Figure 4.6 Unipolar encoding
Polar encoding uses two voltage levels(positive and negative).Polar encoding uses two voltage levels(positive and negative).
Note:Note:
Figure 4.7 Types of polar encoding
In NRZ-L the level of the signal isdependent upon the state of the bit.In NRZ-L the level of the signal isdependent upon the state of the bit.
Note:Note:
In NRZ-I the signal is inverted if a 1 isencountered.In NRZ-I the signal is inverted if a 1 isencountered.
Note:Note:
Figure 4.8 NRZ-L and NRZ-I encoding
Figure 4.9 RZ encoding
A good encoded digital signal mustcontain a provision forsynchronization.A good encoded digital signal mustcontain a provision forsynchronization.
Note:Note:
Figure 4.10 Manchester encoding
In Manchester encoding, thetransition at the middle of the bit isused for both synchronization and bitrepresentation.In Manchester encoding, thetransition at the middle of the bit isused for both synchronization and bitrepresentation.
Note:Note:
Figure 4.11 Differential Manchester encoding
In differential Manchester encoding,the transition at the middle of the bit isused only for synchronization.The bit representation is defined by theinversion or noninversion at thebeginning of the bit.In differential Manchester encoding,the transition at the middle of the bit isused only for synchronization.The bit representation is defined by theinversion or noninversion at thebeginning of the bit.
Note:Note:
In bipolar encoding, we use threelevels: positive, zero,and negative.In bipolar encoding, we use threelevels: positive, zero,and negative.
Note:Note:
Figure 4.12 Bipolar AMI encoding
B8ZS
Bipolar With 8 Zeros Substitution
Based on bipolar-AMI
If octet of all zeros and last voltage pulsepreceding was positive encode as 000+-0-+
If octet of all zeros and last voltage pulsepreceding was negative encode as 000-+0+-
Causes two violations of AMI code
Receiver detects and interprets as octet of allzeros
HDB3
High Density Bipolar 3 Zeros
Based on bipolar-AMI
If the number of 1s since the last substitution is odd.
+ 0000- 0000
+ 000+- 000-
If the number of 1s since the last substitution is even.
+ 0000- 0000
+ -00-- +00+
B8ZS and HDB3
Figure 4.13 2B1Q
Figure 4.14 MLT-3 signal
4.2 Block Coding
Steps in Transformation
Some Common Block Codes
Figure 4.15 Block coding
Figure 4.16 Substitution in block coding
Table 4.1 4B/5B encodingTable 4.1 4B/5B encoding
Data
Code
Data
Code
0000
1111011110
1000
1001010010
0001
0100101001
1001
1001110011
0010
1010010100
1010
1011010110
0011
1010110101
1011
1011110111
0100
0101001010
1100
1101011010
0101
0101101011
1101
1101111011
0110
0111001110
1110
1110011100
0111
0111101111
1111
1110111101
Figure 4.17 Example of 8B/6T encoding
4.3 Sampling4.3 Sampling
Pulse Amplitude Modulation
Pulse Code Modulation
Sampling Rate: Nyquist Theorem
How Many Bits per Sample?
Bit Rate
Figure 4.18 PAM
Figure 4.19 Quantized PAM signal
Figure 4.20 Quantizing by using sign and magnitude
Figure 4.21 PCM
Figure 4.22 From analog signal to PCM digital code
According to the Nyquist theorem, thesampling rate must be at least 2 timesthe highest frequency.According to the Nyquist theorem, thesampling rate must be at least 2 timesthe highest frequency.
Note:Note:
Figure 4.23 Nyquist theorem
Example 4Example 4
What sampling rate is needed for a signal with abandwidth of 10,000 Hz (1000 to 11,000 Hz)?
SolutionSolution
The sampling rate must be twice the highest frequency inthe signal:
Sampling rate = 2 x (11,000) = 22,000 samples/s Sampling rate = 2 x (11,000) = 22,000 samples/s
Example 5Example 5
A signal is sampled. Each sample requires at least 12levels of precision (+0 to +5 and -0 to -5). How many bitsshould be sent for each sample?
SolutionSolution
We need 4 bits; 1 bit for the sign and 3 bits for the value.A 3-bit value can represent 23 = 8 levels (000 to 111),which is more than what we need. A 2-bit value is notenough since 22 = 4. A 4-bit value is too much because 24= 16.
Example 6Example 6
We want to digitize the human voice. What is the bit rate,assuming 8 bits per sample?
SolutionSolution
The human voice normally contains frequencies from 0to 4000 Hz.
Sampling rate = 4000 x 2 = 8000 samples/sSampling rate = 4000 x 2 = 8000 samples/s
Bit rate = sampling rate x number of bits per sample= 8000 x 8 = 64,000 bps = 64 KbpsBit rate = sampling rate x number of bits per sample= 8000 x 8 = 64,000 bps = 64 Kbps
4.4 Transmission Mode4.4 Transmission Mode
Parallel Transmission
Serial Transmission
Figure 4.24 Data transmission
Figure 4.25 Parallel transmission
Figure 4.26 Serial transmission
In asynchronous transmission, wesend 1 start bit (0) at the beginningand 1 or more stop bits (1s) at the endof each byte. There may be a gapbetween each byte.In asynchronous transmission, wesend 1 start bit (0) at the beginningand 1 or more stop bits (1s) at the endof each byte. There may be a gapbetween each byte.
Note:Note:
Asynchronous here means“asynchronous at the byte level,” butthe bits are still synchronized; theirdurations are the same.Asynchronous here means“asynchronous at the byte level,” butthe bits are still synchronized; theirdurations are the same.
Note:Note:
Figure 4.27 Asynchronous transmission
In synchronous transmission,we send bits one after another withoutstart/stop bits or gaps.It is the responsibility of the receiver togroup the bits.In synchronous transmission,we send bits one after another withoutstart/stop bits or gaps.It is the responsibility of the receiver togroup the bits.
Note:Note:
Figure 4.28 Synchronous transmission
