Encoding and Modulating Data Signal from analog to digital and from digital to analog

Encoding and Modulating Data Signal from analog to digital and from digital to analog

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  ENCODING AND MODULATING ENCODINGAND MODULATING -Digital-to-digital conversion -Digital-to-digital conversion (encoding digital data into digital signal) (encoding digital data into digital signal) -Analog-to-digital conversion -Analog-to-digital conversion (digitizing analog signal) (digitizing analog signal) -Digital-to-analog conversion -Digital-to-analog conversion (modulating a digital signal) (modulating a digital signal) -Analog-to-analog conversion -Analog-to-analog conversion (modulating analog signal) (modulating analog signal) Chapter 4
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  Encoding Encoding Coding is theprocess of embedding clocks into a given data stream and producing a signal that can be transmitted over a selected medium. • Transmitter is responsible for "encoding" i.e. inserting clocks into data according to a selected coding scheme • Receiver is responsible for "decoding" i.e. separating clocks and data from the incoming embedded stream. • Systems that use coding are synchronous systems . • We must encode data into signals to send them from one place to another. • A signal needs to be manipulated in such a way so that it contains identifiable changes that are recognizable to the sender and receiver. • First the information must be translated into agreed upon patterns of 0s and 1s. • E.g. Text is translated into either one of two character codes: ASCII or EBCDIC (Appendix A). • Then, since there are two types of signals digital and analog, there are 4 possible encoding techniques that can be used on the data: • Digital-to-digital, Digital-to-Analog, Analog-to-analog, Analog-to-digital.
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  Digital-to-Digital Encoding Digital-to-Digital Encoding •The binary signals created by your computer (DTE) are translated into a sequence of voltage pulses that can be sent through the transmission medium. • Binary signals have two basic parameters: amplitude and duration. • As the number of bits sent per unit of time increases, the bit duration decreases. • The three most common methods of encoding used are: unipolar , polar , and bipolar .
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  Unipolar  The simplestand most primitive type of encoding is Unipolar encoding.  Typically, one voltage level stands for binary 0 and another voltage level for binary 1.  Polarity refers to whether you have a positive or a negative pulse.  Unipolar encoding uses only one polarity, only one of the two binary states is encoded, usually the 1.  Two problems with unipolar encoding: DC component and synchronization .
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  UNIPOLAR ENCODING Unipolar encodinguses only one voltage level.
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  •DC Component •Average amplitudeof a unipolar encoded signal is nonzero. •This creates a direct current (DC component) -- shifts the zero level -- that cannot travel through some media (e.g. microwave). •Synchronization •The change in voltage for each bit is what allows a digital encoding system to indicate changes in bit type. •Long strings of zeros and ones do not produce any transitions which may create problems in error detection and recovery.
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  Lack of synchronization
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  Polar encoding  Polarencoding uses two levels (positive and negative) of amplitude.  Polar encoding eliminates some of the DC residual problem, because the average voltage level on the line is reduced.  The power to transmit this signal is one half that of unipolar signal.  Several types: NRZ, RZ, and biphase.
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  POLAR ENCODING Polar encodinguses two voltage levels Polar encoding uses two voltage levels (positive and negative). (positive and negative).
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  NRZ • Non-return toZero (NRZ) -- signal is always positive or negative. • Two main types of NRZ: NRZ-L and NRZ-I NRZ-L • NRZ-L: signal never returns to zero voltage, and the value during a bit time is a level voltage. • Good for short and well- shielded transmission paths. In NRZ-L the level of the signal is In NRZ-L the level of the signal is dependent upon the state of the bit. dependent upon the state of the bit.
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  NRZ-I – NRZ-I :invert on ones – The transition between a positive and negative voltage represents a 1 bit. – Provides more synchronization than NRZ-L because there is a transition for each 1 bit. In NRZ-I the signal is inverted if a 1 is In NRZ-I the signal is inverted if a 1 is encountered. encountered.
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  NRZ-L and NRZ-Iencoding
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  RZ encoding •Tries tosolve the problem of losing synchronization due to long strings of consecutive 1s or 0s. •Signal change during each bit promotes synchronization. •Positive voltage=1 •Negative voltage=0 •Signal returns to zero halfway through the bit interval.
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  Biphase Biphase  Signal changesat the middle of the bit interval, does not return to zero, goes to opposite pole.  Good solution to synchronization problem  Two types of biphase encoding used in networks: Manchester and Differential Manchester
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  Manchester (or diphaseor biphase encoding) This code is self-clocking. Provides a transition for every bit in the middle of the bit cell. This transition is used only to provide clocking. +ve to -ve transition for a "0" bit -ve to +ve transition for a "1" bit Residual DC component is eliminated by having both polarities for every bit. This scheme is used in Ethernet and IEEE 802.3 compliant LANs
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  Manchester Encoding In Manchesterencoding, the In Manchester encoding, the transition at the middle of the bit is transition at the middle of the bit is used for both synchronization and bit used for both synchronization and bit representation. representation.
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  Differential Manchester Coding Code is self-clocking  Transition for every bit in the middle of the bit cell  Transition at the beginning of the bit cell if the next bit is " 0 "  NO Transition at the beginning of the bit cell if the next bit is " 1 "  Used in Token Ring or IEEE 802.5-compliant LANs.
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  Differential Manchester encoding Indifferential Manchester encoding, the In differential Manchester encoding, the transition at the middle of the bit is used transition at the middle of the bit is used only for synchronization. only for synchronization. The bit representation is defined by the The bit representation is defined by the inversion or noninversion at the beginning inversion or noninversion at the beginning of the bit. of the bit.
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  Analog-to-Digital Encoding Analog-to-Digital Encoding Thechallenge is to transform potentially infinite values in an analog message to digital without losing data (e.g. voice on CD) Pulse Amplitude Modulation (PAM) • Samples the analog message and generates pulses based on the sampling. • Sampling-- measures the amplitude of the signal at equal intervals.
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  Analog-to-Digital Encoding Analog-to-Digital Encoding PulseCode Modulation (PCM) •Uses 3-4 processes to create a digital signal: PAM (sampling), quantization (discrete amplitudes +/- value), binary encoding, and digital-to-digital encoding. •PCM is the sampling method used to digitize voice in T-line transmission in North American telecommunications system (codecs). •PCM sampling rate - twice the highest frequency of the original signal - to ensure the accurate reproduction of an original analog signal using PAM •Fig. 5-21 Nyquist Theorem
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  Digital-to-Analog Encoding Digital-to-Analog Encoding Usedin transmitting data from one computer to another across a public access phone line Bit Rate and Baud Rate • Bit rate is the number of bits transmitted per second. • Bit rate is always >/= to the baud rate • Baud rate is the number of signal units per second required to send those bits. Carrier signal • A high-frequency signal that acts as a basis for the information signal - by sender • Digital information is encoded by modulating the signal's: amplitude, frequency, or phase . Bit rate=Baud rate X No. of bits per signal element
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  Example 1 Example 1 Ananalog 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 Solution Baud rate = 1000 bauds per second (baud/s) Baud rate = 1000 bauds per second (baud/s) Bit rate = 1000 x 4 = 4000 bps Bit rate = 1000 x 4 = 4000 bps
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  Example 2 Example 2 Thebit rate of a signal is 3000. If each signal unit carries 6 bits, what is the baud rate? Solution Solution Baud rate = 3000 / 6 = 500 baud/s Baud rate = 3000 / 6 = 500 baud/s
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  Amplitude Shift Keying(ASK) •To represent binary signals, the amplitude is varied - 1 or 0. •Keying means turning a transmitter on and off. •Highly susceptible to noise interference. •Noise -- random electrical signals (voltages) that tend to generate errors in transmission; introduced into a line by heat from circuit components, or natural disturbances .
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  Relationship between baud rateand bandwidth in ASK
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  Example 3 Example 3 Findthe minimum bandwidth for an ASK signal transmitting at 2000 bps. The transmission mode is half- duplex. Solution Solution In ASK the baud rate and bit rate are the same. 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.
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  Example 4 Example 4 Givena bandwidth of 5000 Hz for an ASK signal, what are the baud rate and bit rate? Solution Solution In ASK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But because the baud rate and the bit rate are also the same for ASK, the bit rate is 5000 bps.
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  Frequency Shift Keying(FSK) • Frequency is varied to represent binary 1 or 0. • Noise interference not a problem because it's looking for frequency changes and doesn't care about voltage spikes.
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  Phase Shift Keying(PSK)  Phase is varied to represent binary 1 or 0.  Limited by the ability of the equipment to detect small differences in phase. This limits its potential bit rate.
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  Quadrature Amplitude Modulation (QAM) Means combining ASK and PSK in such a way that we have a maximum contrast between each bit, dibit (one-pair), quadbit (two-pair), and so on.  Theoretically, any measurable number of changes in amplitude can be combined with any measurable number of changes in phase.  Uses more phase shifts than amplitude shifts to reduce noise susceptibility.
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  Bit and baud
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  Time domain foran 8-QAM signal
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  Analog-to-Analog Encoding Analog-to-Analog Encoding Threeways: Amplitude Modulation (AM), Frequency Modulation (FM), and Phase Modulation (PM). Amplitude Modulation (AM) • The carrier's signal is modulated so that amplitude varies with the changing amplitude of the signal. • Fig. 5.42 • The bandwidth of an AM signal is equal to twice the bandwidth of the modulating signal and covers a range centered around the carrier frequency. • AM radio stations need a minimum bandwidth of 10 Khz. Frequency Modulation (FM) • The frequency of the carrier signal is modulated to follow the changing voltage level (amplitude) of the modulating signal. • The bandwidth of an FM signal is equal to 10 times the bandwidth of the modulating signal. • An FM station needs a bandwidth of 200 KHz (0.2 MHz).
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  Phase Modulation (PM) •Thephase of the carrier signal is modulated to follow the changing voltage level (amplitude) of the modulating signal. •Used as an alternative to frequency modulation.
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  Figure 5-40 WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998 Analog to Analog Encoding
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  Types of analog-to-analogmodulation
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  Amplitude Modulation
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  AM Bandwidth
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  AM Band Allocation
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  Example Example We have anaudio signal with a bandwidth of 4 KHz. What is the bandwidth needed if we modulate the signal using AM? Ignore FCC regulations. Solution Solution An AM signal requires twice the bandwidth of the original signal: BW = 2 x 4 KHz = 8 KHz
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  Figure 5-45 WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998 Frequency Modulation
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  Figure 5-46 WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998 FM Bandwidth
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  The bandwidth ofa stereo audio signal is usually 15 KHz. Therefore, an FM station needs at least a bandwidth of 150 KHz. The FCC requires the minimum bandwidth to be at least 200 KHz (0.2 MHz). Note: Note:
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  FM Band Allocation
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  Example Example We have anaudio signal with a bandwidth of 4 MHz. What is the bandwidth needed if we modulate the signal using FM? Ignore FCC regulations. Solution Solution An FM signal requires 10 times the bandwidth of the original signal: BW = 10 x 4 MHz = 40 MHz