Chapter 4 Digital Transmission

Digital Transmission Digital data to digital signal Line coding Analog data to digital signal PCM (Pulse code modulation)

4.1 Line Coding Some Characteristics Line Coding Schemes Some Other Schemes

Figure 4.1 Digital Transmission: Line coding Actual signal: digital pulse Digital data: Abstract Data วัตถุประสงค์ของการทำ Line coding เพื่อแทน digital data ด้วย ระดับ voltage ที่กำหนด เพื่อกำจัดหรือลดปัญหา เพื่อเพิ่มความเร็วในการส่งข้อมูล

‘1’: +5V ‘0’: 0V ‘1’: +5V or -5V ‘0’: 0V + + - two Figure 4.2 Signal level versus data level ‘1’: +5V ‘0’: 0V ‘1’: +5V or -5V two ‘0’: 0V + + -

Problems Digital data to Digital signal conversion DC component สัญญาณดิจิตอลที่มีองค์ประกอบ DC ทำให้ระดับสัญญาณเพี้ยนไป Lack of Synchronization การตรวจจับบิตข้อมูลผิดตำแหน่ง ทำให้อ่านค่าของระดับสัญญาณผิดไป ปัญหาเหล่านี้ ส่งผลให้ ตัวรับแปลงสัญญาณกลับเป็นบิตข้อมูลผิดเพี้ยนไป

Figure 4.3 DC component + + + + + + + + - - - -

DC Component Problem สัญญาณดิจิตอลที่มีองค์ประกอบ DC ทำให้ระดับสัญญาณเพี้ยนไป Stray Capacitance Coupling Voltage level is constant, spectrum creates very low frequencies These frequencies around zero called DC (direct current) component DC component present problems for system that cannot pass low frequencies or system that uses electrical coupling (via transformer)

Figure 4.4 Lack of synchronization Fast clock

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

Signal level (Element) versus Data level (Element) r means number of data elements carried by each signal element

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 duration = signal duration (sec) = 1 ms Pulse Rate = 1/Pulse duration = 1/ 10-3= 1000 pulses/s L = data level L = 2; Bit Rate = Pulse Rate x log2 L = 1000 x log2 2 = 1000 bps L = 4; Bit Rate = 1000 x log2 4 = 2000 bps จำนวน Bit (data) ต่อ 1 Signal Element (Pulse)

Figure 4.5 Line coding schemes Multi-level

Unipolar encoding uses only one voltage level (positive or negative). Figure 4.6 Unipolar encoding Note: Unipolar encoding uses only one voltage level (positive or negative). +

Polar encoding uses two voltage levels (positive and negative). Figure 4.7 Types of polar encoding Note: Polar encoding uses two voltage levels (positive and negative). Non-Return-to-Zero Return-to-Zero

In NRZ-I (NRZ-M) the signal is inverted if a 1 is encountered. Note: In NRZ-L the level of the signal is dependent upon the state of the bit. In NRZ-I (NRZ-M) the signal is inverted if a 1 is encountered.

NRZ-I: Non-Return-to-Zero-Inversion NRZ-M: Non-Return-to-Zero-Mark Figure 4.8 NRZ-L and NRZ-I encoding NRZ-L: Non-Return-to-Zero-Level ‘0’: +5V ‘1’: -5V NRZ-I: Non-Return-to-Zero-Inversion NRZ-M: Non-Return-to-Zero-Mark ‘1’: Inversion + + + + + ‘0’: Noninversion - - - -

Figure 4.9 Return-to-Zero (RZ) encoding Bipolar-Return-to-Zero (BIP) A good encoded digital signal must contain a provision for synchronization. ‘0’: -V +V ‘1’: +V -V

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

Figure 4.10 Manchester encoding +V -V

Note: 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.

For Synchronization ‘0’: Inversion ‘0’: Inversion Figure 4.11 Differential Manchester encoding For Synchronization ‘0’: Inversion ‘1’: Noninversion ‘0’: Inversion

Applications of Line Coding NRZ encoding: RS232 based protocols Manchester encoding: Ethernet networks Hard drive Differential Manchester encoding: token-ring networks NRZ-Inverted encoding: Fiber Distributed Data Interface (FDDI)

Figure 4.12 Bipolar AMI encoding AMI stands for Alternate Mark Inversion, Variation of AMI is Pseudoternary In bipolar encoding, we use three levels: positive, zero, and negative. ‘1’: Inversion ‘0’: 0 V +V

(Bipolar with Eight-Zero Substitution) Figure 5-11 B8ZS Encoding (Bipolar with Eight-Zero Substitution) 0 0 0 V B 0 V B 0 0 0 V B 0 V B 0 0 0 V B 0 V B V = Violation, B = Bipolar

Example: B8ZS Encoding 0 0 0 V B 0 V B V = Violation, B = Bipolar Figure 5-13 Example: B8ZS Encoding 0 0 0 V B 0 V B +V +V -V V = Violation, B = Bipolar

(High Density Bipolar 3-zero Encoding) Figure 5-12 Pattern 1 : 0 0 0 V HDB3 Encoding (High Density Bipolar 3-zero Encoding) Pattern 2 : B 0 0 V 0 0 0 V 0 0 0 V B 0 0 V B 0 0 V

Figure 5-14 Example: HDB3 Encoding 0 0 0 V B 0 0 V

Line Coding Multi-level coding m B n L 2m ≤ Ln. Length of binary pattern Length of signal pattern Binary data Number of Level in the signaling m B n L B (binary) for L=2 T(trenary) for L=3 Q(quaternary) for L=4 2m ≤ Ln. Line Coding Data Elements Signal Elements Multi-level coding 1) 2B1Q (2 Binary 1 Quaternary) 2) 8B/6T (8 data bits as six ternary) 3) MLT-3 (Multi-Level Transition 3

Used in ISDN 64 kbps or 128 kbps via telephone line Figure 4.13 2B1Q (2 Binary 1 Quaternary) - - Used in ISDN 64 kbps or 128 kbps via telephone line

2B1Q (two binary, one quaternary) Uses data patterns of size 2 one signal element belonging to four-level signal 2B1Q is used in DSL (Digital Subscriber Line) technology

28=256 different data patterns Figure 4.17 Example of 8B/6T encoding 36=478 different signal patterns ใช้ Signal 3 ระดับ Data 8 Bits ความยาวของ Signal = 6 478-256=222 redundant signal elements Used in 100BASET4 Ethernet (100 Mbps) via UTP in Star Topology

‘0’: Non-Inversion ‘1’: Inversion ( -V, 0, +V) First introduced by Cisco System for FDDI (Fiber Distributed Data Interface: Token Ring) Used in 100Base-TX (100 Mbit/s Ethernet)

‘0’: Non-Inversion Figure 4.14 MLT-3 signal ‘1’: Inversion ( -V, 0, +V)

4D-PAM5: Gigabit Ethernet: 1000Base-T (Four-Dimensional Five-level Pulse Amplitude Modulation) 00 01 11 10

Polar

4.2 Block Coding Steps in Transformation Some Common Block Codes

Figure 4.15 Block coding m < n Combination

Figure 4.16 Substitution in block coding 25 = 32 24 = 16

No more than one leading zero (left bit) and No more than two trailing zeros (right bits) Table 4.1 4B/5B encoding Block-coded stream does not have more than three consecutive 0s 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

Table 4.1 4B/5B encoding (Continued) Data Code 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

Pulse Code Modulation (PCM) Analog to Digital Conversion (A/D)

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

Figure 4.22 From analog signal to PCM digital code Continuous Discrete

Note: 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.

Components of PCM encoder

Three different sampling methods for PCM

According to the Nyquist theorem, Nyquist sampling rate for low-pass and bandpass signals According to the Nyquist theorem, the sampling rate must be at least 2 times the highest frequency contained in the signal.

Recovery of a sampled sine wave for different sampling rates Sampling at the Nyquist rate can create a good approximation of the original sine wave. Oversampling can also create the same approximation, but is redundant and unnecessary. Sampling below the Nyquist rate does not produce a signal that looks like the original sine wave.

Sampling of a clock with only one hand The second hand of a clock has a period of 60 s. According to the Nyquist theorem, we need to sample hand every 30 s

Figure 4.23 Nyquist theorem According to the Nyquist theorem, the sampling rate must be at least 2 times the highest frequency.

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

Examples An example of under-sampling is the seemingly backward rotation of the wheels of a forward-moving car in a movie. A movie is filmed at 24 frames per second. If a wheel is rotating more than 12 times per second, the under- sampling creates the impression of a backward rotation. Telephone companies digitize voice by assuming a maximum frequency of 4000 Hz. The sampling rate therefore is 8000 samples per second.

Example A complex low-pass signal has a bandwidth of 200 kHz. What is the minimum sampling rate for this signal? Solution The bandwidth of a low-pass signal is between 0 and f, where f is the maximum frequency in the signal. Therefore, we can sample this signal at 2 times the highest frequency (200 kHz). The sampling rate is therefore 400,000 samples per second.

Quantization and encoding of a sampled signal

Figure 4.19 Quantized PAM signal ** Signed Integer **

Figure 4.20 Quantizing by using sign and magnitude

Figure 4.21 PCM

Components of a PCM decoder

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 L >= 12 -> L = 16 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.

Example 6 We want to digitize the human voice. What is the bit rate, assuming 8 bits per sample? Solution 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

In this case, the sampling rate is twice the bandwidth. Note: 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.

The process of delta modulation

Delta modulation components

4.4 Transmission Mode Parallel Transmission Serial Transmission

Figure 4.24 Data transmission

Figure 4.25 Parallel transmission

Figure 4.26 Serial transmission Universal Asynchronous Receiver/Transmitter (UART)

Note: 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.

Figure 4.27 Asynchronous transmission

It is the responsibility of the receiver to group the bits. Note: 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.

Figure 4.28 Synchronous transmission