Digital Transmission Prof Choong Seon HONG Kyung Hee

Digital Transmission Prof. Choong Seon HONG Kyung Hee University 1

4 장 Digital Transmission 4. 1 Line Coding 4. 2 Block Coding 4. 3 Sampling 4. 4 Transmission Mode Kyung Hee University 2

Line Coding q Converts sequence of bits to a digital signal Kyung Hee University 3

Characteristics of Line Coding q Signal level vs. data level q Pulse rate vs. bit rate q dc components q Self-synchronization Kyung Hee University 4

Signal Level versus Data Level q Signal level – number of values allowed in a signal q Data level – number of values used to represent data Kyung Hee University Three signal levels, two data levels 5

Pulse Rate versus Bit Rate q Pulse rate – number of pulses per second q Bit rate – number of bits per second Kyung Hee University 6

Pulse Rate versus Bit Rate Example 1> A signal has two data levels with a pulse duration of 1 ms. We calculate the pulse rate and bit rate as follows: q. Pulse Rate = 1/ 10 -3= 1000 pulses/s q. Bit Rate = Pulse Rate x log 2 L = 1000 x log 2 2 = 1000 bps Kyung Hee University 7

Pulse Rate versus Bit Rate Example 2 > A signal has four data levels with a pulse duration of 1 ms. We calculate the pulse rate and bit rate as follows q. Pulse Rate = = 1000 pulses/s q. Bit Rate = Pulse. Rate x log 2 L = 1000 x log 2 4 = 2000 bps Kyung Hee University 8

DC Components q Some systems (such as transformer) will not allow passage of dc component q DC Component is just extra energy on the line, but useless Kyung Hee University 9

Self-Synchronization q Receiver’s bit intervals must correspond exactly to the sender’s bit intervals q Self-synchronizing signal includes timing information q If the receiver’s clcok is out of synchronization, these alerting points can reset the clock. Kyung Hee University 10

Self-Synchronization - Lack of synchronization Kyung Hee University 11

Self-Synchronization q Example 3 In a digital transmission, the receiver clock is 0. 1 percent faster than the sender clock. How many extra bits per second does the receiver receive if the data rate is 1 Kbps? How many if the data rate is 1 Mbps? q Solution At 1 Kbps: 1000 bits sent 1001 bits received 1 extra bps At 1 Mbps: 1, 000 bits sent 1, 000 bits received 1000 extra bps Kyung Hee University 12

Line Coding Scheme Kyung Hee University 13

Unipolar coding q Unipolar encoding uses only one voltage level. Kyung Hee University 14

Unipolar encoding disadvantages q dc component q no synchronization Kyung Hee University 15

Polar q Polar encoding uses two voltage levels (positive and negative). Kyung Hee University 16

Variations of Polar Encodings Kyung Hee University 17

Variations of Polar Encodings q In Nonreturn to Zero-level (NRZ-L) the level of the signal is dependent upon the state of the bit. q In Nonreturn to Zero-Invert (NRZ-I) the signal is inverted if a 1 is encountered. Kyung Hee University 18

NRZ-L (level) and NRZ-I (invert) encoding q Kyung Hee University 19

Return to Zero (RZ) encoding A good encoded digital signal must contain a provision for synchronization. Kyung Hee University 20

Manchester Encoding In Manchester encoding, the transition at the middle of the bit is used for both synchronization and bit representation. Kyung Hee University 21

Differential Manchester q 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. Kyung Hee University 22

Bipolar q In bipolar encoding, we use three levels: positive, zero, and negative. q Bipolar AMI encoding Kyung Hee University 23

Some Other Schemes q 2 B 1 Q (two binary, one quaternary) -1 -2 q MLT 3 (Multiline Transmission, three level (MLT-3) Kyung Hee University 24

4. 2 Block Coding q Steps in transmission Step 1: Division l divide sequence of bits into groups of m bits Step 2: Substitution l substitute an m-bit code for an n-bit group Step 3: Line Coding l create the signal Kyung Hee University 25

Block coding Kyung Hee University 26

Block coding q Substitution in block coding Kyung Hee University 27

Block coding q 4 B/5 B encoding Data Code 0000 11110 10010 0001 010011 0010 1010 10110 0011 10101 10111 0100 01010 11010 01011 11011 0110 011100 01111 11101 q no more than 1 leading 0 q no more than 2 trailing 0 s q normally encoded w/ NRZ-I (1 indicated by transition) Kyung Hee University 28

Block coding q Rest of the 5 bit codes Synchronization Error correction q Example J (start delimiter) : 11000 T (end delimiter) : 01101 S (set) : 11001 R (reset) : 00111 Kyung Hee University 29

Block coding q 8 B/6 T substitute 8 bit group w/ 6 symbol code each symbol is ternary (+1, 0, -1) 8 bit code = 28 = 256 6 bit ternary = 36 = 729 Kyung Hee University 30

4. 3 Sampling q PAM : Pulse Amplitude Modulation Kyung Hee University 31

Sampling q. 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. • Term sampling means measuring the amplitude of the signal at equal intervals. Kyung Hee University 32

Sampling q Quantized PAM Signal Kyung Hee University 33

Sampling q Quantizing by using sign and magnitude Kyung Hee University 34

PCM Kyung Hee University 35

PCM q From analog signal to PCM digital code Kyung Hee University 36

Sampling Rate and Nyquist Theorem q. According to the Nyquist theorem, the sampling rate must be at least 2 times the highest frequency. Kyung Hee University 37

Sampling Rate and Nyquist Theorem 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 Kyung Hee University 38

Sampling Rate and Nyquist Theorem 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 23 = 8 levels (000 to 111), which is more than what we need. A 2 -bit value is not enough since 22 = 4. A 4 -bit value is too much because 24 Kyung Hee = 16. University 39

Sampling Rate and Nyquist Theorem 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 =Kyung 8000 Hee x 8 = 64, 000 bps = 64 Kbps University 40

4. 4 Transmission Mode q Kyung Hee University 41

Parallel Transmission Kyung Hee University 42

Serial Transmission q Kyung Hee University 43

Asynchronous Transmission q In asynchronous transmission, we send 1 start bit (0) at the beginning and 1 or more stop bits (1 s) at the end of each byte. There may be a gap between each byte. q Asynchronous here means “asynchronous at the byte level, ” but the bits are still synchronized; their durations are the same. Kyung Hee University 44

Asynchronous Transmission Kyung Hee University 45

Synchronous Transmission q 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. Kyung Hee University 46