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Completing Lecture 3 and 4 Chapter 3 and 4 Handout #3

Completing Lecture 3 and 4 Chapter 3 and 4 Handout #3. Dr. Clincy Professor of CS. DIGITAL-TO-DIGITAL CONVERSION. Can represent digital data by using digital signals. The conversion involves three techniques: line coding – converting bit sequences to signals

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Completing Lecture 3 and 4 Chapter 3 and 4 Handout #3

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  1. Completing Lecture 3 and 4Chapter 3 and 4Handout #3 Dr. Clincy Professor of CS Lecture

  2. DIGITAL-TO-DIGITAL CONVERSION Can represent digital data by using digital signals. The conversion involves three techniques: line coding – converting bit sequences to signals block coding – adding redundancy for error detection scrambling– deals with the long zero-level pulse issue Line coding is always needed; Block coding and scrambling may or may not be needed. Lecture

  3. Line coding and decoding At Tx - Digital data represented as codes is converted to a digital signal via an encoder At Rx – Digital signal is converted back to digital codes via a decoder Lecture

  4. Signal element versus data element Data element - smallest entity representing info Signal element – shortest unit of a digital signal (carriers) r – is the ratio of # of data elements carried per signal element Example of adding extra signal elements for synchronization Example of increasing data rate Lecture

  5. Data Rate Versus Signal Rate Data rate (or bit rate) - # of data elements (or bits) transmitted in 1 second – bits-per-second is the unit Signal rate (pulse rate or baud rate) - # of signal elements transmitted in 1 second – baud is the unit OBJECTIVE ALWAYS: increase data rate while decreasing signal rate – more “bang” for the “buck” Is it intuitive that if you had a data pattern of all 0s or 1s, it would effect the signal rate ? Therefore to relate data-rate with signal-rate, the pattern matters. Worst Case Scenario – we need the maximum signaling rate (alternating 1/0s) Best Case Scenario – we need the minimum signaling rate (all 1/0s) Focus on average case S = c x N x 1/r N – data rate (bps) c – case factor S - # of signal elements r – ratio of data to signal Lecture

  6. Example 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? Solution We assume that the average value of c is 1/2 . The baud rate is then Lecture

  7. Bandwidth Now we understand what baud rate is And we understand what bit rate (or data rate) is Baud rate - # of carriers on the transport Data rate - # of passengers (or bits) in the carriers With this, we clearly see that baud rate effects bandwidth usage Signaling changes relate to frequency changes – therefore the bandwidth is proportionate with the baud rate: Bmin = c x N x 1/r or Nmax = 1/c x B x r minimum bandwidth maximum data rate (given the bandwidth) N – data rate C – case factor This formula is consistent with Nyquist formula r – data to signal ratio Lecture

  8. Example The maximum data rate of a channel (see Chapter 3) is Nmax = 2 × B × log2 L (defined by the Nyquist formula). Does this agree with the previous formula for Nmax? Solution A signal with L levels actually can carry log2L bits per level. If each level corresponds to one signal element and we assume the average case (c = 1/2), then we have Lecture

  9. Decoding Issue 1 Keep in mind the Rx decodes the digital signal – how is it done ? • Rx determines a “moving average” of the signal’s power or voltage levels • This average is called the baseline • Then the Rx compares incoming signal power to this average (or baseline) • If higher than the baseline, could be a 1 • If lower than the baseline, could be a 0 • In using such a technique, is it intuitive that long runs of 0s or 1s could skew the average (baseline) ?? – this is called baseline wandering (effects Rx’s ability to decode correctly) Lecture

  10. Decoding Issue 2 Effect of lack of synchronization For the Rx, to correctly read the signal, both the Tx and Rx “bit intervals” must be EXACT Example of Rx timing off – therefore decoding the wrong data from the signal To fix this, the Tx could insert timing info into the data that synchs the Rx to the start, middle and end of a pulse – these points could reset an out-of-synch Rx Lecture

  11. Example 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, the receiver receives 1001 bps instead of 1000 bps. At 1 Mbps, the receiver receives 1,001,000 bps instead of 1,000,000 bps. NOTE: Keep in mind that a FASTER clock means SHORTER intervals Lecture

  12. Chapter 3 and 4Handout #3 Dr. Clincy Professor of CS Lecture

  13. Line coding scheme categories Lecture

  14. Unipolar NRZ scheme Data Signal Voltages on one side of the axis Positive voltage signifies 1 Almost zero voltage signifies 0 Power needed to send 1 bit unit of resistance Lecture

  15. Polar NRZ-L and NRZ-I schemes (non-return-to-zero) change no change Voltages on both sides of the axis NRZ-L (level) version – voltage level determines the bit value NRZ-I (invert) version – voltage change or no-change determines the bit value (no change = 0, change = 1) Lecture

  16. Polar RZ scheme Uses 3 values: positive, negative and zero Signal changes Not between bits BUT during the bit H-to-L in middle for 1 L-to-H in middle for 0 Positioning occurs at the beginning of the period Lecture

  17. Polar biphase: Manchester and Differential Manchester Schemes Manchester: H-to-L=0, L-to-H=1 Differential Manchester: H-to-L or L-to-H at begin=0, No change at begin=1 Lecture

  18. Bipolar schemes: AMI and pseudoternary Bipolar encoding uses 3 voltage levels: positive, negative and zero. One data element is at ZERO, while the others alternates between negative and positive Alternate Mark Inversion (AMI) scheme – neutral zero voltage is 0 and alternating positive and negative voltage represents 1 Pseudoternary scheme – vice versa from the AMI scheme Lecture

  19. Multilevel Schemes These schemes attempt to increase the number of bits per baud Given m data elements, could produce 2m data patterns Given L levels, could produce Ln combinations of signal patterns (where n is the length of the signal patterns) If 2m = Ln, each data pattern is encoded into one signal pattern (1-to-1) If 2m < Ln, data patterns use a subset of signal patterns – could use the extra signal patterns for fixing baseline wandering and error detection Classify these codes as mBnL where: m – length of the binary pattern B – means Binary data n – length of the signal pattern L - # signaling levels (letters in place of L: B=2, T=3 and Q=4) Lecture

  20. Multilevel: 2B1Q scheme 2B1Q Data patterns of size 2 bits Encodes 2-bit patterns in one signal element 4 levels of signals If previous level was positive and the next level becomes +3, represents 01 If previous level was positive and the next level becomes -3, represents 11 Lecture

  21. Multilevel: 8B6T scheme Data patterns of size 8 bits Encodes 8-bit patterns in six signal elements Using 3 levels of signal Lecture

  22. Multilevel: 4D-PAM5 scheme 4-dimensional five-level pulse amplitude modulation scheme Instead of transmitting in serial form – parts of the code are in sent in parallel over 4 wires (versus 1 wire) In this particular case, it would take ¼ less time to transmit Lecture

  23. Multitransition: MLT-3 scheme Multi-line transmission, three-level scheme Uses three levels and three transition rules to jump between levels: - if the next bit is 0, there is no transition - if the next bit is 1 and the current level is not 0, the next level is 0 - if the next bit is 1 and the current level is 0, the next level is the opposite of the last nonzero level Lecture

  24. Block coding concept Block coding provides redundancy for synchronization and error detection Block coding changes a block of m bits into a block of n bits (where n>m) Block coding is also called mB/nB encoding Lecture

  25. Using block coding 4B/5B with NRZ-I line coding scheme Fixes that PROBLEM of long stream of 0s Use 4B/5B to change the long stream of 0s prior to using NRZ-I For example, for 4B/5B encoding, 4-bit groups or replaced with 5-bit groups and those 5-bit groups are re-combined – NOTE: the 5-bit code could be completely different from the original 4-bit code Lecture

  26. 4B/5B mapping codes Because the 5-bit code has 25 = 32 codes, the extra codes can be used for control sequences and error detection For example, for 4B/5B encoding, 4-bit groups or replaced with 5-bit groups and those 5-bit groups are re-combined – NOTE: the 5-bit code could be completely different from the original 4-bit code Lecture

  27. 8B/10B block encoding If there are more consecutive 0s over 1s (or vice versa), controller detects and complements either the 0s or 1s – uses 768 redundant bit groups for this Eight binary, ten binary encoding scheme 8-bit codes replaced with 10-bit codes Provide greater error detection 5 most significant bits are fed to 5B/6B encoder 3 least significant bits are fed to 3B/4B encoder Done to simplify mapping table Lecture

  28. Scrambling The biphase encoding schemes are suited for long-distance communication due to bandwidth requirement. However, bipolar AMI encoding is good because of the narrow bandwidth requirement – however, long streams of 0s could throw off the synchronization In dealing with synchronization issue, we could substitute long zero-level pulses with a combination of other levels to provide synchronization This is called scrambling Lecture

  29. AMI used with scrambling Unlike block coding, scrambling is done at the SAME time encoding is done System inserts the require pulses based on “scrambling rules” Two techniques: (1) bipolar with 8-zero substitution (B8ZS), (2) high-density bipolar 3-zero (HDB3) Lecture

  30. Two cases of B8ZS scrambling technique Takes 8 consecutive zeros and replace with 000VB0VB where V denotes violation (a non-zero voltage not in accordance with the AMI rule) and B denotes bipolar (a non-zero voltage in accordance with AMI rule) Recall the Bipolar AMI scheme on page 110 [Alternate Mark Inversion (AMI) scheme – neutral zero voltage is 0 and alternating positive and negative voltage represents 1] Lecture

  31. Different situations in HDB3 scrambling technique For HDB3, 4 consecutive zeros are replaced with 000V or B00V With the two choices, an even number of non-zero pulses can be maintained Rule 1: if the # of non-zero pulses is odd after the last substitution, use pattern 000V – which will make the total number even Because # of non-zero pulses here is even, used B00V. Now we have only 1 non-zero pulse (odd), so use 000V Since there are no non-zero pulses after the 2nd substitution, the 3rd substitution is B00V because this is an even case Rule 2: if the # of non-zero pulses is even after the last substitution, use pattern B00V – which will make the total number even Lecture

  32. Chapter 5Handout #4 Dr. Clincy Professor of CS Lecture

  33. Digital-to-analog conversion Based on the digital data, the Modulator changes characteristics of the “controllable” analog signal (bandpass analog signal) on the transmitter side to represent the digital data Demodulator interprets the analog signal in re-creating the digital data on the receiver side Terminology: “modulating digital data into an analog signal” The analog signal we can control ? Sine Wave, Carrier Signal, Periodic Signal Lecture

  34. Types of digital-to-analog conversion Change amplitude to represent a bit Change frequency to represent a bit Change phase to represent a bit Combination of changing both amplitude and phase to represent a set of bits Lecture

  35. Recall • For digital transmission, bit rate (data rate) and signal rate (baud rate) relationship was • S = N x 1/r where r = # of data elements per signal element and N is the data rate in bps (and S is the signaling or baud rate) • For analog, r = log2L where L is the type of signal (versus level) Lecture

  36. Example An analog signal carries 4 bits per signal element. If 1000 signal elements are sent per second, find the bit rate. Solution In this case, r = 4, S = 1000, and N is unknown. We can find the value of N from Lecture

  37. Example An analog signal has a bit rate of 8000 bps and a baud rate of 1000 baud. How many data elements are carried by each signal element? How many signal elements do we need? Solution In this example, S = 1000, N = 8000, and r and L are unknown. We find first the value of r and then the value of L. Lecture

  38. Binary amplitude shift keying changing the original amplitude Explain this not changing the original amplitude Bandwidth (B) is proportional to the signal rate (S) and depending on the modulation and filtering process, the required bandwidth can range between S to 2S (where middle bandwidth is fc). The value of d relates to the modulation and filtering process B = (1 + d) x S Lecture

  39. Example 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 = 1? Solution The middle of the bandwidth is located at 250 kHz. This means that our carrier frequency can be at fc = 250 kHz. We can use the formula for bandwidth to find the bit rate (with d = 1 and r = 1). Lecture S = N * 1/r

  40. Binary frequency shift keying changing the original frequency Use two different carrier frequencies, f1 and f2, for 0 and 1 Explain this not changing the original frequency Lecture

  41. Example 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 midpoint of the band is at 250 kHz. We choose 2Δf to be 50 kHz; this means The difference (delta) between the two frequencies Lecture

  42. Binary phase shift keying changing the original phase Explain this not changing the original phase Lecture

  43. QPSK and its implementation QPSK – Quadrature Phase Shift Keying Use 2 bits in each signal element – decreases baud rate and bandwidth Uses 4 possible phases (versus 2) 2 composite signals are created Because the 2 signals are using the same bandwidth – each signal has ½ bandwidth Lecture

  44. Example Find the bandwidth for a signal transmitting at 12 Mbps for QPSK. The value of d = 0. Solution For QPSK, 2 bits is carried by one signal element. This means that r = 2. So the signal rate (baud rate) is S = N × (1/r) = 6 Mbaud. With a value of d = 0, we have B = S = 6 MHz. B = (1 + d) x S Lecture

  45. Concept of a constellation diagram Helps define the amplitude and phase of a signal element the amplitude of the 2nd carrier Peak Amplitude Phase This is the amplitude given using one carrier Only use 1 carrier and phase is static and 2 amplitude levels Only use 1 carrier and 1 amplitude and 2 phases (0o and 180o) Uses 2 carriers and 1 amplitude and 4 phases (45o, 135o, -45o, -135o) Lecture

  46. Constellation diagrams for some QAMs QAM – Quadrature Amplitude Modulation For QPSK, we only changed the phase For QAM, we change both the phase and amplitude Has a 0 amplitude and a positive amplitude (with 2 carriers) Has a negative amplitude and a positive amplitude (with 2 carriers) Has 2 positive amplitudes (with 2 carriers) Has 4 negative levels and 4 positive levels (with 2 carriers) Lecture

  47. ANALOG TO ANALOG Analog-to-analog conversion is the representation of analog information by an analog signal. One may ask why we need to modulate an analog signal; it is already analog. Modulation is needed if the medium is bandpass in nature or if only a bandpass channel is available to us. Bandpass – signal being shifted to a particular range Lowpass – signal that IS NOT shifted to a particular range Lecture

  48. Types of analog-to-analog modulation Lecture

  49. Amplitude modulation Vary the amplitude of the carrier signal to mimic the changing voltage levels (amplitude) of the modulating signal result The total bandwidth required for AM can be determined from the bandwidth of the audio signal: BAM = 2B. Lecture

  50. Frequency modulation Vary the frequency of the carrier signal to mimic the changes in voltage level (amplitude) of the modulating signal result The total bandwidth required for FM can be determined from the bandwidth of the audio signal: BFM = 2(1 + β)B. Would be given Lecture

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