1432.encoding concepts

Chapter 4 Digital Transmission

4.1 Line Coding Some Characteristics Line Coding Schemes Some Other Schemes

Figure 4.2 Signal level versus data level

Figure 4.4 Lack of synchronization

Figure 4.5 Line coding schemes

Unipolar encoding uses only one voltage level. Note:

Figure 4.6 Unipolar encoding

Polar encoding uses two voltage levels (positive and negative). Note:

Figure 4.7 Types of polar encoding

In NRZ-L the level of the signal is dependent upon the state of the bit. Note:

In NRZ-I the signal is inverted if a 1 is encountered. Note:

Figure 4.8 NRZ-L and NRZ-I encoding

A good encoded digital signal must contain a provision for synchronization. Note:

Figure 4.10 Manchester encoding

In Manchester encoding, the transition at the middle of the bit is used for both synchronization and bit representation. Note:

Figure 4.11 Differential Manchester encoding

In differential Manchester encoding, the transition at the middle of the bit is used only for synchronization. The bit representation is defined by the inversion or noninversion at the beginning of the bit. Note:

In bipolar encoding, we use three levels: positive, zero, and negative. Note:

Figure 4.12 Bipolar AMI encoding

4.2 Block Coding Steps in Transformation Some Common Block Codes

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

4B5B Example In order to send information using 4B5B encoding, the data byte to be sent is first broken into two nibbles. If the byte is 0E, the first nibble is 0 and the second nibble is E. Next each nibble is remapped according to the 4B5B table. Hex 0 is remapped to the 4B5B code 11110. Hex E is remapped to the 4B5B code 11100. Other information remapping types are 5B6B and 8B10B. 11101 1111 F 11100 1110 E 11011 1101 D ... ... ... 10100 0010 2 01001 0001 1 11110 0000 0 4B5B Code (Binary)‏ Data (Hex)‏ 4B5B Encoding Table

Figure 4.16 Substitution in block coding

Table 4.1 4B/5B encoding 11010 1100 01010 0100 11011 1101 01011 0101 11100 1110 01110 0110 11101 1111 01111 0111 10111 1011 10101 0011 10100 01001 11110 Code 10110 1010 0010 10011 1001 0001 10010 1000 0000 Code Data Data

Table 4.1 4B/5B encoding (Continued)‏ 10001 K (start delimiter)‏ 01101 T (end delimiter)‏ 11001 S (Set)‏ 00111 R (Reset)‏ 11000 J (start delimiter)‏ 00100 11111 00000 Code H (Halt) I (Idle) Q (Quiet)‏ Data

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

Figure 4.17 Example of 8B/6T encoding

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

B8ZS Bipolar with 8 zero substitution4 Eight consecutive zero level voltages are replaced by the sequence 000VB0VB: V stands for voilation breaks AMI rule. Oppsite polarity from the previous. B denotes bipolar. Whch is in accordance with AMI rule.

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

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.

4.3 Sampling Pulse Amplitude Modulation Pulse Code Modulation Sampling Rate: Nyquist Theorem How Many Bits per Sample? Bit Rate

Pulse amplitude modulation has some applications, but it is not used by itself in data communication. However, it is the first step in another very popular conversion method called pulse code modulation. Note:

Figure 4.19 Quantized PAM signal

Figure 4.20 Quantizing by using sign and magnitude

Figure 4.22 From analog signal to PCM digital code

According to the Nyquist theorem, the sampling rate must be at least 2 times the highest frequency. Note:

Example 4 What sampling rate is needed for a signal with a bandwidth of 10,000 Hz (1000 to 11,000 Hz)? Solution The sampling rate must be twice the highest frequency in the signal: Sampling rate = 2 x (11,000) = 22,000 samples/s

Example 5 A signal is sampled. Each sample requires at least 12 levels of precision (+0 to +5 and -0 to -5). How many bits should be sent for each sample? Solution We need 4 bits; 1 bit for the sign and 3 bits for the value. A 3-bit value can represent 2 3 = 8 levels (000 to 111), which is more than what we need. A 2-bit value is not enough since 2 2 = 4. A 4-bit value is too much because 2 4 = 16.

Example 6 We want to digitize the human voice. What is the bit rate, assuming 8 bits per sample? Solution The human voice normally contains frequencies from 0 to 4000 Hz. Sampling rate = 4000 x 2 = 8000 samples/s Bit rate = sampling rate x number of bits per sample = 8000 x 8 = 64,000 bps = 64 Kbps

Note that we can always change a band-pass signal to a low-pass signal before sampling. In this case, the sampling rate is twice the bandwidth. Note:

4.4 Transmission Mode Parallel Transmission Serial Transmission

Figure 4.24 Data transmission

Figure 4.25 Parallel transmission

Figure 4.26 Serial transmission

In asynchronous transmission, we send 1 start bit (0) at the beginning and 1 or more stop bits (1s) at the end of each byte. There may be a gap between each byte. Note:

Asynchronous here means “asynchronous at the byte level,” but the bits are still synchronized; their durations are the same. Note:

Figure 4.27 Asynchronous transmission

In synchronous transmission, we send bits one after another without start/stop bits or gaps. It is the responsibility of the receiver to group the bits. Note:

Figure 4.28 Synchronous transmission

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