SIGNAL ENCODIN TECHNIQUES ReviewRecap Lecture 20 Overview n

SIGNAL ENCODIN TECHNIQUES Review/Recap Lecture 20

Overview n n n n Differential Encoding Modem Functions Digital Vs Analog (Advantages of Digital Over Analog) Binary Amplitude Shift Keying (Limitations) NRZ-L QPSK Vs. OQPSK QAM 2

Differential Encoding Q: - What is differential encoding? 3

Way of looking at techniques Data Medium Digital Analog Digital NRZ Manchester Differential Manchester ASK FSK PSK modems Analog Pulse Coded Modulation (digitized voice) AM/FM radio Television

Encoding Techniques n n Encoding schemes deal with how to transport bits over the physical media… We must deal and manage many issues to well represent and interpret the bits: n n Timing of bits (start and end, duration, Signal levels) Clocking ( Synchronizing transmitter and receive, External clock, Sync mechanism based on signal) Error detection, Signal interference and Noise immunity Cost and complexity 5

Encoding Techniques n Encoding techniques depend on the type of data to transmit and the medium being used: n n Digital data on digital signal (our focus in this brief description. . . ) Analog data on digital signal Digital data on analog signal Analog data on analog signal 6

Digital Data, Digital Signal n Digital signal: n n n Discrete, discontinuous voltage pulses Each pulse is a signal element Binary data encoded into signal elements 7

A Simple Encoding Scheme n n n 0 is Vo (some voltage) 1 is V 1 Example : let us encode 10011101 Clock Vo V 1 8

Problem? n n n Let us now try to encode two bytes that come one 2 seconds after the other: 10011101 and 00001101 What is the problem with this encoding scheme? What is the solution? 9

Bipolar Encoding (RZ Signal) +0. 85 V n 1 is 0 is n Example : let us encode 01100010 n 0 0 Hight. V + No. V –» 1 bit Low. V + No. V –» 0 bit -0. 85 V Clock 10

Problem? n n The problem with this encoding scheme is that we have to return to 0 each time the transmission is done which generates some complexity and cost in implementing and managing this technique We should look for a better technique… We have 2 alternatives: Encoding of 01100010 Clock 11

Alternative 1: Non-Return to Zero n n Two different voltages for 0 and 1 bits Voltage constant during bit interval n n no transition i. e. no return to zero voltage e. g. Absence of voltage for zero, constant positive voltage for one More often, negative voltage for one value and positive for the other This is NRZ also known as Manchester Encoding. . . 12

Alternative 1: NRZ-Manchester Encoding +0. 85 V n 1 is 0 is n Example : let us encode 01100010 n 0 -0. 85 V 0 Hight. V + Low. V –» 1 bit Low. V + High. V –» 0 bit Clock 13

Alternative 2: Differential Encoding n n n Data represented by changes trends rather than levels of voltage More reliable detection of transition rather than level In complex transmission layouts it is easy to lose sense of polarity 14

Differential Manchester Encoding n n n 0 is 1 is 0 (Presence of transition) 0 (Abscence of transition) A transition in the middle of the bit is required anyway… Example : let us encode 10011101 Clock 15

Unipolar NRZ 1 0 1 1 1 0 0 Polar NRZ-Inverted (Differential Encoding) Bipolar Encoding Manchester Encoding Differential Manchester Encoding 16

Differential Encoding n n n 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 17

Pro and Cons If you have a DC component (i. e. NRZ), you must preserve it and therefore your transmission channel must have a DC (galvanic) path. Try that with a satellite! That's a pretty long cable out into space. There are work-arounds; DC restorers and such for NRZ signals, but there must still be limits placed on the amount of zeros that can be transmitted at-a-time. Coding schemes that have a clock embedded into the signal (i. e. manchester) have no DC component but require a higher bandwidth channel because both the signal and the clock take up bandwidth. These signals can be passed through transformers, AC coupled amplifiers, and other non-galvanic paths. This is important if you need isolation (i. e. for safety) or if the transmitter and receiver are at locations where different power sources exist (common mode rejection). AC coupled amplifiers are almost always cheaper than DC coupled amplifiers. 18

Pros and Cons Embedded clocks provide very good sync between Tx and Rx, because you can extract the exact clock you need for the signal. NRZ does not have an embedded clock so it is difficult to regenerate a clock at the Rx end. Miller coding also contains an embedded clock, but uses only half the bandwidth as manchester coding. In general the differential types offer better noise immunity than the non-differential types. The drawback is that you need differential Tx and Rx amplifiers, which are generally more expensive. 19

Differential Encoding Q: - What is differential encoding? Ans: - In differential encoding, the signal is decoded by comparing the polarity of adjacent signal elements rather than determining the absolute value of a signal element. 20

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Modem Functionality Q: - What function does a modem perform? 22

Digital to Analog n n n Many carrier facilities are analog Many transmission media are also analog (microwave, radio) We can carry digital values over analog signals We must ‘encode’ the digital data over the digital signal This is what is done with modems digital Modem Analog Modem = Modulation/Demodulation

Modem Hardware for Modulation and Demodulation n n A mechanism that accepts a sequence of data bits and applies modulation to a carrier wave according to the bits is called a modulator A mechanism that accepts a modulated carrier wave and recreates the sequence of data bits that was used to modulate the carrier is called a demodulator Transmission of data requires a modulator at one end of the transmission medium and a demodulator at the other Most communication systems are full duplex n which means each location needs both a modulator to send data and a demodulator to receive data

Modem Hardware for Modulation and Demodulation n n Users would like to keep cost low and make the pair of devices easy to install and operate Manufacturers combine modulation and demodulation mechanisms into a single device n n n called a modem (modulator and demodulator) Figure 10. 9 illustrates how a pair of modems use a 4 -wire connection to communicate Modems are designed to provide communication over long distances

Modem Hardware for Modulation and Demodulation

Optical and Radio Frequency Modems n Modems are also used with other media n n including Radio Frequency (RF) transmission and optical fibers A pair of RF modems can be used to send data via radio A pair of optical modems can be used to send data across a pair of optical fibers Modems can use entirely different media, but the principle remains the same: n n at the sending end, a modem modulates a carrier at the receiving end, data is extracted from the modulated carrier

Dialup Modems n A dialup modem uses an audio tone n n A dialup modem uses data to modulate an audible carrier n n n as with conventional modems, the carrier is modulated at the sending end and demodulated at the receiving end which is transmitted to the phone system The chief difference between dialup and conventional modems arises from the lower bandwidth of audible dialup modems Interior of a modern telephone system used today is digital n n n The phone system digitizes the incoming audio Transports a digital form internally Converts the digitized version back to analog audio for delivery The receiving modem demodulates the analog carrier Extracts the original digital data

Dialup Modems n Figure illustrates the ironic use of analog and digital signals by dialup modems n n n a dialup modem is usually embedded in a computer Term internal modem to denote an embedded device Term external modem to denote a separate physical device

QAM Applied to Dialup n n QAM is also used with dialup modems as a way to maximize the rate at which data can be sent Figure 10. 11 shows the bandwidth available on a dialup connection Most telephone connections transfer frequencies between 300 and 3000 Hz A given connection may not handle the extremes well n n n Thus, to guarantee better reproduction and lower noise, dialup modems use frequencies between 600 and 3000 Hz It means the available bandwidth is 2400 Hz A QAM scheme can increase the data rate dramatically

QAM Applied to Dialup

V. 32 and V. 32 bis Dialup Modems n n Consider the V. 32 and V. 32 bis standards Figure on next page illustrates the QAM constellation for n a V. 32 modem that uses 32 combinations of ASK and PSK n n a V. 32 bis modem uses 128 combinations of ASK and PSK n n n to achieve a data rate of 9600 bps in each direction to achieve a data rate of 14, 400 bps in each direction Figure 10. 13 illustrates the constellation Sophisticated signal analysis is needed to detect the minor change that occurs from a point in the constellation to a neighboring point

QAM for V. 32 Dialup Modems n n Consider the V. 32 and V. 32 bis standards Figure on next page illustrates the QAM constellation for n a V. 32 modem that uses 32 combinations of ASK and PSK n to achieve a data rate of 9600 bps in each direction n a V. 32 bis modem uses 128 combinations of ASK and PSK n to achieve a data rate of 14, 400 bps in each direction

QAM for V. 32 bis Dialup Modems n n Figure illustrates the constellation Sophisticated signal analysis is needed to detect the minor change that occurs from a point in the constellation to a neighboring point

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Modems n n Encode digital values over analog circuits To encode data, modems use combinations of n n n Amplitude Shift Keying (ASK) Frequency Shift Keying (FSK) Phase Shift Keying (PSK) Typically, a carrier frequency is provided and various SHIFT KEYING is applied For half duplex we could have only one carrier For full duplex, we need two carriers, one for each direction

Modems (Standards) n n There have been many modem standards over time Early modems used a form of frequency shifting – Ex 300 bps full duplex modem n Originating modem n n n Sine wave at 1070 Hz for a 0 bit Sine wave at 1270 Hz for a 1 bit Answering modem n n Sine wave at 2025 Hz for a 0 bit Sine wave at 2225 Hz for a 1 bit

Modems n n Later modems used combinations of shift keying by combining PSK and ASK Some common 9600 bps modems used n n n 12 phase shifts at 1 amplitude 4 phase shift at a second amplitude Combination of 16 different states Called Quadrature Amplitude Modulation (QAM) Baud rate was 2400 cycles per second Each of the 16 states represented 4 bits

Modem Constellation

Modems Standard n n n n V. 21 V. 22 bis V. 23 V. 32 V. 34 V. 90 V. 92 Baud rate Bit rate Modulation Technique 300 FSK 600/1200 PSK 600 1200/2400 QAM 1200 FSK 2400 4800/9600 QAM/TCM 2400 28, 800 2400 56, 000

Modems n n To improve performance, compression and error correction standards developed Two compression standards in in vogue n n n V. 42 bis MNP 5 Two error correction standards n n V. 42 MNP 4

Modem Functionality Q: - What function does a modem perform? Ans: - A modem converts digital information into an analog signal, and conversely. 42

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Digital Vs. Analog Q: - Indicate three major advantages of digital transmission over analog transmission. 44

Analog vs. Digital Transmission n Transmissions can also be either analog or digital. n n n Analog transmissions, like analog data, vary continuously. Examples of analog data being sent using analog transmissions are broadcast TV and radio. Digital transmissions are made of square waves with a clear beginning and ending. Computer networks send digital data using digital transmissions. Data can be converted between analog and digital formats. n n When digital data is sent as an analog transmission modem (modulator/demodulator) is used. When analog data is sent as a digital transmission, a codec (coder/decoder) is used. 45

Data Type vs. Transmission Type Analog Transmission Digital Transmission Analog Data Radio, Broadcast PCM & Video TV standards using codecs Digital Data Modem-based LAN Cable Communications Standards 46

Advantages of Digital Transmission n Digital transmission: n n n produces fewer errors than analog transmission. Because the transmitted data is binary (1 s and 0 s), it is easier to detect and correct errors. permits higher transmission rates. Optical fiber, for example, is designed for digital transmission. is more efficient. It’s possible to send more data through a given circuit using digital rather than analog transmission. is more secure since it is easier to encrypt. Integrating voice, video and data on the same circuit is also far simpler with digital transmission since signals made up of digital data are easier to combine. 47

Digital Over Analog n n n Digital signals do not get corrupted by noise etc. You are sending a series of numbers that represent the signal of interest (i. e. audio, video etc. ) Digital signals typically use less bandwidth. This is just another way to say you can cram more information (audio, video) into the same space. Digital can be encrypted so that only the intended receiver can decode it (like pay per view video, secure telephone etc. ) 48

Analog Signals Digital data, analog signal: Some transmission media, such as optical fiber and satellite, will only propagate analog signals. Analog data, analog signal: Analog data are easily converted to an analog signal. 49

Digital Signals Data Digital data, digital signal: In general, the equipment for encoding digital data into a digital signal is less complex and less expensive than digitalto analog Equipment. Analog data, digital signal: Conversion of analog data to digital form permits the use of modern digital transmission and switching equipment for analog data. 50

Analog and Digital Transmission 51

Q: - Indicate three major advantages of digital transmission over analog transmission. n Ans: - Cost, capacity utilization, and security and privacy are three major advantages enjoyed by digital transmission over analog transmission. 52

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Amplitude Shift Keying n Q: - How are binary values represented in amplitude shift keying, and what is the limitation of this approach? 54

Types of Analog to Digital Conversion Types of digital-to-analog conversion

Amplitude Shift Keying n encode 0/1 by different carrier amplitudes n n usually have one amplitude zero susceptible to sudden gain changes inefficient used for: n n up to 1200 bps on voice grade lines very high speeds over optical fiber

Amplitude Shift Keying n n Values represented by different amplitudes of carrier Usually, one amplitude is zero n n n i. e. presence and absence of carrier is used Susceptible to sudden gain changes Inefficient 57

Example of ASK n Bit Values 00 01 10 11 Amplitude A 1 A 2 A 3 A 4 58

Example of ASK Amplitude Shifting Keying (four amplitudes), two bits per baud 59

Amplitude Shift Keying (ASK) n n n ASK is implemented by changing the amplitude of a carrier signal to reflect amplitude levels in the digital signal. For example: a digital “ 1” could not affect the signal, whereas a digital “ 0” would, by making it zero. The line encoding will determine the values of the analog waveform to reflect the digital data being carried.

Bandwidth of ASK n n n The bandwidth B of ASK is proportional to the signal rate S. B = (1+d)S “d” is due to modulation and filtering, lies between 0 and 1.

Binary ASK Binary amplitude shift keying

Binary ASK Implementation of binary ASK

ASK Q: - How are binary values represented in amplitude shift keying, and what is the limitation of this approach? n Ans: - With amplitude-shift keying, binary values are represented by two different amplitudes of carrier frequencies. This approach is susceptible to sudden gain changes and is rather inefficient. 64

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NRZ-L Q: - What is NRZ-L? What is a major disadvantage of this data encoding approach? 66

Line Encoding Line coding schemes

Line Encoding n n Error detection - errors occur during transmission due to line impairments. Some codes are constructed such that when an error occurs it can be detected. For example: a particular signal transition is not part of the code. When it occurs, the receiver will know that a symbol error has occurred.

Line Encoding n n Noise and interference - there are line encoding techniques that make the transmitted signal “immune” to noise and interference. This means that the signal cannot be corrupted, it is stronger than error detection.

Line Encoding n Complexity - the more robust and resilient the code, the more complex it is to implement and the price is often paid in baud rate or required bandwidth.

Unipolar n n n All signal levels are on one side of the time axis - either above or below NRZ - Non Return to Zero scheme is an example of this code. The signal level does not return to zero during a symbol transmission. Scheme is prone to baseline wandering and DC components. It has no synchronization or any error detection. It is simple but costly in power consumption.

Unipolar NRZ scheme

Polar - NRZ n n n The voltages are on both sides of the time axis. Polar NRZ scheme can be implemented with two voltages. E. g. +V for 1 and -V for 0. There are two versions: n n NZR - Level (NRZ-L) - positive voltage for one symbol and negative for the other NRZ - Inversion (NRZ-I) - the change or lack of change in polarity determines the value of a symbol. E. g. a “ 1” symbol inverts the polarity a “ 0” does not.

NRZ-L and NRZ-I Polar NRZ-L and NRZ-I schemes

NRZ-L and NRZ-I n n In NRZ-L the level of the voltage determines the value of the bit. In NRZ-I the inversion or the lack of inversion determines the value of the bit. NRZ-L and NRZ-I both have an average signal rate of N/2 Bd. NRZ-L and NRZ-I both have a DC component problem and baseline wandering, it is worse for NRZ-L. Both have no self synchronization &no error detection. Both are relatively simple to implement. 75

Nonreturn to Zero-Level (NRZ-L) n n easiest way to transmit digital signals is to use two different voltages for 0 and 1 bits voltage constant during bit interval n no transition (no return to zero voltage) absence of voltage for 0, constant positive voltage for 1 more often, a negative voltage represents one value and a positive voltage represents the other(NRZ-L)

NRZ-L Q: - What is NRZ-L? What is a major disadvantage of this data encoding approach? Ans: - Non return-to-zero-level (NRZ-L) is a data encoding scheme in which a negative voltage is used to represent binary one and a positive voltage is used to represent binary zero. A disadvantage of NRZ transmission is that it is difficult to determine where one bit ends and the next bit begins. n 77

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QPSK and OQPSK Q: - What is the difference between QPSK and offset QPSK? 79

Quadrature PSK n more efficient use if each signal element represents more than one bit n n uses phase shifts separated by multiples of /2 (90 o) each element represents two bits split input data stream in two and modulate onto carrier and phase shifted carrier can use 8 phase angles and more than one amplitude n 9600 bps modem uses 12 angles, four of which have two amplitudes

QPSK and OQPSK Modulators

QPSK and OQPSK n n In QPSK signaling, the bit transitions of the even and odd bit streams occur at the same time instants. but in OQPSK signaling, the even and odd bit Streams, m. I(t) and m. Q(t), are offset in their relative alignment by one bit period (half-symbol period)

QPSK and OQPSK Q: - What is the difference between QPSK and offset QPSK? Ans: - The difference is that offset QPSK introduces a delay of one bit time in the Q stream 83

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QAM Q: - What is QAM? 85

PSK

PSK Constellation

4 -PSK

4 -PSK Characteristics

8 -PSK Characteristics

PSK Bandwidth The minimum BW for PSK transmission is the same as that required for ASK transmission Max baud rates of ASK and PSK are the same for a given BW, but the bit rates could be 2 or more times greater

QAM n n n Quadrature Amplitude Modulation Combined ASK and PSK If there are x variations in phase and y variations in amplitude, it will give us x times y possible variations.

4 -QAM and 8 -QAM Constellations 4 possible variations 8 possible variations

Time domain for 8 -QAM Signal

16 -QAM Constellation – Different configurations 16 out of 36/32 possible variations are utilized – to ensure readability Greater ratio of phase shift to amplitude handles noise best 1 st figure – ITU-T recommendation 2 nd figure – OSI recommendation

Quadrature Amplitude Modulation n n QAM used on asymmetric digital subscriber line (ADSL) and some wireless combination of ASK and PSK logical extension of QPSK send two different signals simultaneously on same carrier frequency n n use two copies of carrier, one shifted 90° each carrier is ASK modulated two independent signals over same medium demodulate and combine for original binary output

QAM Q: - What is QAM? . Ans: - QAM takes advantage of the fact that it is possible to send two different signals simultaneously on the same carrier frequency, by using two copies of the carrier frequency, one shifted by 90° with respect to the other. For QAM, each carrier is ASK modulated. n 97

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Summary n n n n Differential Encoding ? ? Functions Performed by Modem Digital Advantages Over Analog) Limitations of Binary Amplitude Shift Keying Disadvantages of NRZ-L QPSK and OQPSK Difference What is QAM

Complimentary Session for (Signal Encoding & Antennas) 100

Complimentary Session(Multiple FSK (MFSK)) Where Question 101

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Q: - A sine wave is to be used for two different signaling schemes: (a) PSK; (b) QPSK. The duration of a signal element is 10 -5 s. If the received signal is o : ' the following form and if the measured noise power at the receiver is 2. 5 X 10 -8 , , vatts, determine the Eb/N 0 (in d. B) for each case. Solution: - 103

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Q Q: - What SNR ratio is required to achieve a bandwidth efficiency , of 1. 0 for ASK, FSK, PSK, and QPSK? Assume that the required bit error rate is 10 -6 105

Q/A 106

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NRZ-L Q: - An NRZ-L signal is passed through a filter with r = 0. 5 and then mqdulated onto a carrier. The data rate is 2400 bps. Evaluate the bandwidth for ASK and FSK. For FSK assume that the two frequencies used are 50 k. Hz and 55 k. Hz. Sol 108

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Bandwidth Q: - Assume that a telephone line channel is equalized to allow bandpass data transmission over a frequency range of 600 to 3000 Hz. The available bandwidth is 2400 Hz. For r = 1 , evaluate the required bandwidth for 2400 bps QPSK and 4800 -bps, eight level multilevel signaling. Is the bandwidth adequate? 110

Bandwidth Q: - Assume that a telephone line channel is equalized to allow bandpass data transmission over a frequency range of 600 to 3000 Hz. The available bandwidth is 2400 Hz. For r = 1 , evaluate the required bandwidth for 2400 bps QPSK and 4800 -bps, eight level multilevel signaling. Is the bandwidth adequate? 111

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Q Q: - A microwave transmitter has an output of 0. 1 W at 2 GHz. Assume that this transmitter is used in a microwave communication system where the transmitting and receiving antennas are parabolas, each 1. 2 m in diameter. a. What is the gain of each antenna in decibels? b. Taking into account antenna gain, what is the effective radiated power of the transmitted signal? c. If the receiving antenna is located 24 km from the transmitting antenna over a free space path, find the available signal power out of the receiving antenna in d. Bm units 113

Q: - A microwave transmitter has an output of 0. 1 W at 2 GHz. Assume that this transmitter is used in a microwave communication system where the transmitting and receiving antennas are parabolas, each 1. 2 m in diameter. a. What is the gain of each antenna in decibels? b. Taking into account antenna gain, what is the effective radiated power of the transmitted signal? c. If the receiving antenna is located 24 km from the transmitting antenna over a free space path, find the available signal power out of the receiving antenna in d. Bm units 114

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Q Q: - Determine the height of an antenna for a TV station that must be able to reach customers up to 80 km away. Ans: - 116