Chapter 4 Digital Transmission

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Presentation transcript:

Chapter 4 Digital Transmission

4 장 Digital Transmission 4.1 Line Coding 4.2 Block Coding 4.3 Sampling 4.4 Transmission Mode

Line Coding Converts sequence of bits to a digital signal

Characteristics of Line Coding Signal level vs. data level Pulse rate vs. bit rate dc components Self-synchronization

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

Pulse Rate versus Bit Rate Pulse rate – number of pulses per second Bit rate – number of bits per second

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/ 10-3= 1000 pulses/s Bit Rate = Pulse Rate x log2 L = 1000 x log2 2 = 1000 bps

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 log2 L = 1000 x log2 4 = 2000 bps

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

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 clock is out of synchronization, these alerting points can reset the clock.

Self-Synchronization - Lack of synchronization

Self-Synchronization 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

Line Coding Scheme

Unipolar coding Unipolar encoding uses only one voltage level.

Unipolar encoding disadvantages dc component no synchronization

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

Variations of Polar Encodings

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.

NRZ-L (level) and NRZ-I (invert) encoding

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

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

Differential Encoding Data represented by changes rather than levels More reliable detection of transition rather than level In complex transmission layouts it is easy to lose sense of polarity

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.

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

Some Other Schemes 2B1Q (two binary, one quaternary) MLT3 (Multiline Transmission, three level (MLT-3) -1 -2

4.2 Block Coding Steps in transmission Step 1: Division divide sequence of bits into groups of m bits Step 2: Substitution substitute an m-bit code for an n-bit group Step 3: Line Coding create the signal

Block coding

Block coding Substitution in block coding

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

Block coding Rest of the 5 bit codes Example Synchronization Error correction Example Q (quiet) : 00000 I (Idle) : 11111 H (Halt) : 00100 J (start delimiter) : 11000 K (start delimiter) : 10001 T (end delimiter) : 01101 S (set) : 11001 R (reset) : 00111

Block coding 8B/6T 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

4.3 Sampling PAM : Pulse Amplitude Modulation

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.

Sampling Quantized PAM Signal

Sampling Quantizing by using sign and magnitude

PCM

PCM From analog signal to PCM digital code -87

Sampling Rate and Nyquist Theorem According to the Nyquist theorem, the sampling rate must be at least 2 times the highest frequency. = x Hz = 2x samples/s =

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

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 = 16.

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

4.4 Transmission Mode

Parallel Transmission

Serial Transmission

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

Asynchronous Transmission

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.