1 Kyung Hee University Digital Transmission

2 Kyung Hee University 4 장 Digital Transmission 4.1 Line Coding 4.2 Block Coding 4.3 Sampling 4.4 Transmission Mode

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

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

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

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

7 Kyung Hee University 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:  Pulse Rate = 1/ = 1000 pulses/s  Bit Rate = Pulse Rate x log 2 L = 1000 x log 2 2 = 1000 bps

8 Kyung Hee University 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  Pulse Rate = = 1000 pulses/s  Bit Rate = PulseRate x log 2 L = 1000 x log 2 4 = 2000 bps

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

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

11 Kyung Hee University Self-Synchronization - Lack of synchronization

12 Kyung Hee University Self-Synchronization Example 3  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?  Solution At 1 Kbps: 1000 bits sent  1001 bits received  1 extra bps At 1 Mbps: 1,000,000 bits sent  1,001,000 bits received  1000 extra bps

13 Kyung Hee University Line Coding Scheme

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

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

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

17 Kyung Hee University Variations of Polar Encodings

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

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

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

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

22 Kyung Hee University Differential Manchester  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.

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

24 Kyung Hee University Some Other Schemes  2B1Q (two binary, one quaternary)  MLT3 (Multiline Transmission, three level (MLT-3) -2

25 Kyung Hee University 4.2 Block Coding  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

26 Kyung Hee University Block coding

27 Kyung Hee University Block coding  Substitution in block coding

28 Kyung Hee University Block coding  4B/5B encoding  no more than 1 leading 0  no more than 2 trailing 0s  normally encoded w/ NRZ-I (1 indicated by transition) DataCodeDataCode

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

30 Kyung Hee University Block coding  8B/6T substitute 8 bit group w/ 6 symbol code each symbol is ternary (+1, 0, -1) 8 bit code = 2 8 = bit ternary = 3 6 = 729

31 Kyung Hee University 4.3 Sampling  PAM : Pulse Amplitude Modulation

32 Kyung Hee University Sampling  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.

33 Kyung Hee University Sampling  Quantized PAM Signal

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

35 Kyung Hee University PCM

36 PCM  From analog signal to PCM digital code

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

38 Kyung Hee University 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

39 Kyung Hee University 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 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.

40 Kyung Hee University 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 = 8000 x 8 = 64,000 bps = 64 Kbps

41 Kyung Hee University 4.4 Transmission Mode 

42 Kyung Hee University Parallel Transmission

43 Kyung Hee University Serial Transmission 

44 Kyung Hee University Asynchronous 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.  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.  Asynchronous here means “asynchronous at the byte level,” but the bits are still synchronized; their durations are the same.

45 Kyung Hee University Asynchronous Transmission

46 Kyung Hee University Synchronous 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.  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.