chapter 4 combi.pdf

ITT300 INTRODUCTION TO DATA COMMUNICATION AND NETWORKING CHAPTER 4 DATA ENCODING PART 2 ADAPTED FROM: Behrouz A. Forouzan LESSON OUTCOME In this section, we will discuss how we can represent data signal from digital signal to analog signal...

ITT300 INTRODUCTION TO DATA COMMUNICATION AND NETWORKING CHAPTER 4 DATA ENCODING PART 2 ADAPTED FROM: Behrouz A. Forouzan LESSON OUTCOME In this section, we will discuss how we can represent data signal from digital signal to analog signal and from digital signal to analog signal. Data rate vs Signal rate, carrier signal Types of Modulation - ASK, FSK and PSK (excluding for QAM) Constellation Diagram (Bit and Di-bit only) 2  DIGITAL-TO-ANALOG CONVERSION - Process of converting digital data (sequence of bits) to analog signals - At the sender, digital data are encoded into an analog signal; at the receiver, the analog signal are decoded into digital data again. 5.3 RECAPPING DATA ELEMENT AND SIGNAL ELEMENT Digital-to-analog transmission terminology Data Element vs Signal element Data element – as the smallest piece of information to be exchanged the bit. Signal element – as the smallest unit of a signal that is constant. Data Rate vs Signal Rate Data Rate = bit rate Signal Rate = baud rate 4  ASPECTS OF DIGITAL–TO-ANALOG CONVERSION Digital-to-analog transmission terminology Bit rate (N) – the number of data elements (bits) transmitted in a second Baud rate (S) – the number of signal elements transmitted in a second The relationship between bit rate and baud rate is: S = N x 1/r N = data rate/bit rate (bps) r = the number of data elements carried in one signal element Value of r in analog transmission is r = log2L where L is the type of signal element, NOT the level In the analog transmission of digital data, the baud rate is less than or equal to the bit rate. 5 Aspects of Digital–to-Analog Conversion Bandwidth Bandwidth required for analog transmission is proportional to signal rate (baud rate) and not the bit rate. The higher baud rate ~ higher the bandwidth required 6  Example 1 An analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second, find the baud rate and the bit rate. Solution S = N x 1/R R=4 S = baud rate N = bit rate The baud rate is equal to the number of signal units transmitted per second. Since 1000 signal units are transmitted per second, the baud rate is 1000 bauds. Baud rate = 1000 bauds per second (baud/s) Bit rate = S x R = 1000 x 4 = 4000 bps Example 2 The bit rate of a signal is 3000. If each signal unit carries 6 bits, what is the baud rate? Solution S = N x 1/R Baud rate = 3000 x 1 / 6 = 500 baud/s Exercise 3 Example An analog signal carries 4 bits in each signal unit. If the bit rate is 1 Mbps, find the baud rate. Solution R=4 S = baud rate S = N x 1/R N = bit rate Bit Rate = 1Mbps = 1000000 bps S = 1000000 x ¼ = 250,000bps Example 4 An analog signal has a bit rate of 8000 bps and a baud rate of 1000 baud/s. How many data elements are carried by each signal element? How many signal elements do we need? Solution S = 1000 baud/s , N = 8000 bps , R = ?, r = log2L S = N x 1/R 8 = log2L 1000 = 8000 x 1/ r L = 2^8 = 256 1000 x r = 8000 r = 8 bits 10 Types of digital – to – analog signal Modulation of Digital Data 4.11  Types of digital – to – analog signal ASK involves changing the ASK involves changing the ASK involves changing amplitude (or strength) of frequency of a carrier signal the phase of a carrier a carrier signal to to represent data. signal to represent data. represent data. Frequency change Amplitude change Amplitude and Phase Phase change Frequency and Phase constant Amplitude and constant Frequency constant 4.12  Amplitude Shift Keying (ASK) The strength (amplitude) of the carrier signal is varied to represent binary 1 or 0. Both frequency and phase remain constant while amplitude changes ASK is highly susceptible to noise interference such as electromagnetic induction (Disadvantage) Advantages; Reduction in the amount of energy required to transmit information  Amplitude Shift Keying(ASK) Amplitude Frequency Phase Figure 5.3 ASK 14 Amplitude Shift Keying (ASK) Relationship between Bit Rate, Baud rate and Bandwidth Bit rate = Baud rate in ASK So, Baud rate → S = N x 1/r Bit rate → N = S x r (where r = 1 in ASK) Bandwidth can be expressed as: B = (1+d) x S → unit is Hz where S = baud rate d is factor related to the condition of line (the value of d is between 0 and 1) if d =0, Bandwidth = Baud Rate = Bit Rate if d = 1, Bandwidth = 2 x Baud Rate = 2 x Bit Rate Relationship between baud rate and bandwidth in ASK The carrier frequency is in the middle of the spectrum (fc) Example 1 Find the minimum bandwidth for an ASK signal transmitting at 2000 per second. The transmission mode is half-duplex and assume d = 0. Solution In ASK the baud rate and bit rate are the same (assuming d = 0). The baud rate is therefore 2000. An ASK signal requires a minimum bandwidth equal to its baud rate. Therefore, the minimum bandwidth is 2000 Hz. Example 2 Given a bandwidth of 5000 Hz for an ASK signal, what are the baud rate and bit rate? (assume d=0) Solution Bandwidth = (1 + d) x S S = N x 1/r , where S is baud rate and N is bit rate In ASK the baud rate is the same as the bandwidth when d = 0, which means the baud rate is 5000. But because the baud rate and the bit rate are also the same for ASK because r = 1, the bit rate is 5000 bps. Example 3 If we have an available bandwidth of 100 kHz which spans from 200 to 300 kHz. What are the carrier frequency and the bit rate if we modulated our data by using ASK with d = 0 and if d=1? Apply B= (1+d) x S Solution The middle of the bandwidth is located at 250 kHz. This means that our carrier frequency can be at fc = 250 kHz. In ASK, bit rate = baud rate If d=0 , Bandwidth = (1+0) Baudrate= Bitrate = 100 Kbps If d=1 Bandwidth = (1 + 1) Baudrate 100 KHz = 2 Baudrate Baudrate = 50K baud/sec Because r =1, Bitrate = r x baudrate = 50 kbps Example 4 Given a bandwidth of 10,000 Hz (1000 to 11,000 Hz), draw the full-duplex ASK diagram of the system. Find the carriers and the bandwidths in each direction. Assume there is no gap between the bands in the two directions. Assuming d = 0, what is the bit rate for each direction. Solution For full-duplex ASK, the bandwidth for each direction is BW = 10000 / 2 = 5000 Hz The carrier frequencies can be chosen at the middle of each band fc1 (forward) = 1000 + 5000/2 = 3500 Hz fc1 (backward) = 11000 – 5000/2 = 8500 Hz Bandwidth = (1+d) x Baud rate , d= 0 so Baudrate = 5000 baud/sec So bit rate = baud rate = 5000 bps Solution to Example 5 Drawing for full-duplex ASK diagram of the system  Exercise 1 If we have an available bandwidth of 500 kHz and the lowest frequency is 500 KHz What are the carrier frequency and the bit rate if we modulated our data by using ASK with d = 0? Assume the transmission is half duplex. Apply Bandwidth = (1+d) x S and S = N x 1/r. Solution Available BW = 500kHz Bandwidth signal → 500kHz – 1000kHz For full-duplex ASK, the bandwidth for each direction is BW = 10000 / 2 = 5000 Hz The carrier frequencies can be chosen at the middle of each band fc1 (forward) = 500 + 500/2 = 750 kHz Bandwidth = (1+d) x Baud rate , d= 0 so Baudrate = 500 baud/sec So bit rate = baud rate = 500 bps Frequency Shift Keying The frequency of the carrier signal is varied to represent data The frequency of the modulated signal is constant for the duration of one signal element, but changes for the next signal element if the data element changes Both peak amplitude and phase remain constant for all signal elements Avoid most of the noise problems of ASK but limited to the physical capabilities of the carrier. 23 Frequency Shift Keying FSK 24  Frequency Shift Keying (FSK) Relationship between Bit Rate, Baud rate and Bandwidth Bit rate = Baud rate in FSK So, Baud rate → S = N x 1/r Bit rate → N = S x r (where r = 1 in FSK) Bandwidth can be expressed as: B = (1+d) x S + (Frequency carrier1 – Frequency carrier0) → unit is Hz where S = baud rate d is factor related to the condition of line (the value of d is between 0 and 1) if d =0, Bandwidth = Baud Rate = Bit Rate if d = 1, Bandwidth = 2 x Baud Rate = 2 x Bit Rate  Frequency Shift Keying (FSK) Relationship between Bit Rate, Baud rate and Bandwidth Bandwidth = (Frequency for carrier1 – Frequency for carrier0)+ baud rate ; assuming d=0  Example 1 Find the maximum bit rates for an FSK signal if the bandwidth of the medium is 12,000 Hz and the difference between the two carriers is 2000 Hz. Transmission is in full-duplex mode. (assuming d=0 and r = 1) Solution The transmission is full duplex, only 6000 Hz is allocated for each direction. The difference between the two carriers is 2000 Hz B = (1+d) x baud rate + (fc1 − fc0 ) 6000 = (1+0) x baud rate + 2000 Baud rate = 6000 − 2000 = 4000 But because the baud rate is the same as the bit rate, the bit rate is 4000 bps. Example 2 We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz. What should be the carrier frequency and the bit rate if we modulated our data by using FSK with d = 1? Solution The available bandwidth of 100 kHz (from 200 to 300 kHz) d=1 middle of bandwidth = 100/2 = 50khz → 2Δf to be 50 kHz B = (1+d) x baud rate + (fc1 − fc0 ) B = (1+1) x baud rate + 50 100 = 2 x baud rate + 50 baud rate = 25kbaud Bit rate = 25kbps 28 Phase shift keying (PSK) ▪ In Phase Shift Keying, the phase of the carrier signal is shifted to represent data ▪ Both peak amplitude and frequency remain constant while the phase tend to change ▪ If two different phase are used(0 and 180 degrees) the method is called 2-PSK. ▪ If four different phase are used(0,90,180,270), the method is called 4-PSK. ▪ The relationship between phase to bit value can be shown in table or by constellation diagram 5.29 Phase shift keying ▪ The common technique is 2‐PSK or Binary PSK (BPSK) – use 2 different phase ▪ The common technique is 4‐PSK or Quadrature PSK (QPSK) – use 4 different phase ▪ Not susceptible to noise degradation that affects ASK or bandwidth limitations of FSK ▪ PSK is superior to FSK because we do not need two carrier signals 5.30  CONSTELLATION DIAGRAM Define the relationship between amplitude and phase of a signal elements, particularly when using two carriers (one in-phase and one quadrate). Figure 4.16 Concept of a constellation diagram 5.31  CONSTELLATION DIAGRAM Figure 4.17 Constellation diagrams for ASK (OOK) 5.32  CONSTELLATION DIAGRAM Figure 4.18 Constellation diagrams for BPSK 5.33  CONSTELLATION DIAGRAM Figure 4.19 Constellation diagrams for QPSK 5.34  ITT300 INTRODUCTION TO DATA COMMUNICATION AND NETWORKING CHAPTER 4 DATA ENCODING PART 1 ADAPTED FROM: Behrouz A. Forouzan JUST TO RECAP Both analog and digital information can be encoded as either analog or digital signal. Modulation is the process of encoding source onto a carrier signal with frequency. 4.2  DIGITAL-TO-DIGITAL CONVERSION In this section, we will discuss how we can represent digital data by using digital signals. Simplest encoding scheme – assign one Signal Element: For digital; voltage level to binary one and another to A voltage pulse of binary zero. constant amplitude Digital signal: Discrete, discontinuous voltage pulses Each pulse is a signal element Binary data encoded into a signal element 4.3  DIGITAL-TO-DIGITAL CONVERSION Involves 3 techniques: Line coding Block coding Scrambling Line coding is always needed; block coding and scrambling may or may not be needed. 4.4  LINE CODING Line coding is the process of converting digital data to digital signals. At the sender, digital data are encoded into a digital signal. At the receiver, the digital data are recreated by decoding the digital signal. 4.5 LINE CODING AND DECODING  SIGNAL ELEMENT VS DATA ELEMENT In data communications, our goal is to send data element. A data element is the smallest entity that can represent a piece of information: this is the bit. Data elements are being carried In digital data communications, a signal element carries data elements. A signal element is the shortest unit of a digital signal(timewise). signal elements are the carriers. 4.6  SIGNAL ELEMENT VS DATA ELEMENT In data communications, our goal is to send data element. The smallest entity is One bit of data 1 bit element is carried by One data element is 2 signal. carried by one signal So, r = ½ (or r=0.5) element. The symbol for data element is r r=1/1, r=1 Two data element is 3 signal element carried by one carries 4 data signal element. element (4 bits) r= 2/1 r=4/3 4.7  DATA RATE vs SIGNAL RATE Data Rate = Bit Rate No. of data element sent in 1s; unit in bps Signal Rate a.k.a pulse rate, modulation rate or baud rate No. of signal element send in 1s, unit is the baud In data communication Increase data rate increase the speed of transmission Decrease signal rate decreases bandwidth requirement 4.8 DATA RATE vs SIGNAL RATE (cont…) Formula to show the relationship between data rate and signal rate as: S = c × N × 1/r baud N = data rate (bps) c = case factor (varies for each case) S = number of signal elements r = previously define factor **r = the number of bits per signal elements 4.9  Example 1 A signal is carrying data in which one data element is encoded as one signal element ( r = 1). If the bit rate is 100 kbps, what is the average value of the baud rate if c is between 0 and 1? *r=1 mean 1 signal element is carry one data element Solution S = c × N × 1/r = 0.5 × 100000 × 1/1 = 50000 baud = 50 kbaud 4.10  Example 2 A signal is carrying data in which one data element is encoded as one signal element ( r = 1). If the bit rate is 100 kbps, what is the lowest value of the baud rate if c is between 0 and 1? Solution S = c × N × 1/r = 0 × 100000 × 1/1 = 0 baud = 0 kbaud 4.11  Example 3 A signal is carrying data in which one data element is encoded as one signal element ( r = 1). If the bit rate is 100 kbps, what is the highest value of the baud rate if c is between 0 and 1? Solution S = c × N × 1/r = 1 × 100000 × 1/1 = 100 000 baud = 100 kbaud 4.12  Example 4 A signal is carrying data in which one data element is encoded as one signal element ( r = 1). If the bit rate is 250 kbps, what is the value of the baud rate if c is 1? Solution S = c × N × 1/r = 1 × 250000 × 1/1 = 25000 baud = 25 kbaud 4.13  DC COMPONENT Some line coding leave a residual (balance) direct‐current (dc) component This component is an undesirable components with reasons : Extra energy residing on the line and useless Signal is distorted / create errors in the output when pass through a system that does not allow passage of a dc component 4.14 SIGNALS WITH AND WITHOUT DC COMPONENT To mitigate these issues, several line coding techniques aim to minimize or eliminate the DC component. Techniques like Manchester encoding, differential Manchester encoding, and 4B/5B encoding are used to ensure a more balanced line signal that is less prone to these problems. 4.15  SELF-SYNCHRONIZATION To correctly interpret the signals received from the sender, the receiver’s bit interval (time required to send one single bit) must correspond exactly to the sender’s bit intervals. If the receiver clock is faster or slower, the bit intervals are not matched, and the receiver might misinterpret the signals. Biphase encoding technique have synchronization. 4.16 EFFECT OF LACK OF SYNCHRONIZATION 4.17 LINE CODING SCHEMES Unipolar encoding signal does not switch polarity. Polar encoding uses two levels of voltages: one positive and one negative. Bipolar encoding uses three voltage levels: positive, zero, and negative. 4.18  POLAR Types : Non-Return-to-Zero (NRZ) NRZ-L and NRZ-I Return-to-Zero (RZ) Biphase Manchester & Differential Manchester 4.19 UNIPOLAR NRZ A +ve voltage represents a binary one, and a zero voltage represents a binary zero. This Encoding Algorithm was easy to implement and required low bandwidth, but it failed to present error correction and loss of synchronization was likely to occur. 4.20  POLAR NRZ SCHEME NRZ‐L (Non-return-to-Zero-level) The value of the signal is always either +ve or –ve Bit 0 = +ve voltage level , bit 1 = –ve voltage level There is no voltage zero in this NRZ-L 0 = high level 1 = low level 4.21 POLAR - NRZ-L SCHEMES In NRZ-L the level of the voltage determines the value of the bit. It is called NRZ because the signal does not return to zero at the middle of the bit. Disadvantages : DC component, lack of synchronization *r = 1, 1 data signal is used to carry 1 data element 4.22  POLAR BIPHASE: MANCHESTER SCHEME In Manchester encoding, the duration of the bit is divided into two halves. Use an inversion at the middle of each bit interval for both synchronization and bit representation No DC component because each bit has a positive and negative voltage The drawback is the signal rate. The signal rate for Manchester and differential Manchester is double that for NRZ. 4.23  POLAR BIPHASE: MANCHESTER SCHEME In Manchester encoding, bit 1 = negative to positive, bit 0 = positive to negative bit 0 = + to - *r = 1/2, 2 data signal is used to carry 1 data element bit 1 = - to + 4.24