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Transmission and Impairments: Baseband and Broadband

Learn about the concepts of baseband and broadband transmission, channel capacity, impairments such as attenuation, distortion, and noise, and the Shannon equation for determining channel capacity. Also, explore Nyquist equation, bandwidth, throughput, latency, and the bandwidth-delay product.

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Transmission and Impairments: Baseband and Broadband

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  1. Chapter 2 and 3Handout #2 4 of 10 Dr. Clincy Professor of CS Review Exam 3 Will create D2L Final Project folders after today’s lecture Lecture 2

  2. Baseband Transmission • In sending the digital signal over channel without changing the digital signal to an analog signal • Use low-pass channel – meaning the bandwidth can be as low as zero • Typical: 2 computers directly connected In baseband transmission, the required bandwidth is proportional to the bit rate; if we need to send bits faster, we need more bandwidth (the frequency will need to increase) Lecture 2

  3. Broadband Transmission • Broadband transmission or modulation means changing the digital signal to an analog signal for transmission • Modulation allows us to use a bandpass channel – a channel where the bandwidth doesn’t start at zero • Bandpass channels are more available than low-pass channels Lecture 2

  4. Modulation of a digital signal for transmission on a Bandpass channel (Broadband transmission) Lecture 2

  5. Channel Capacity • As we know, impairments limits the actual data rate realized • The actual rate realized at which data can be transmitted over a given path, under given conditions is called Channel Capacity • Four concepts • Data rate – the rate, in bps, the data can be communicated • Bandwidth – constrained by the Tx and transport medium – expressed in cycles per second or Hertz • Noise – average level of noise over the communication path • Error rate – the rate in which erroneous bits are received Lecture 2

  6. Impairments Lecture 2

  7. Attenuation Loss of energy – the signal can lose energy as it travels and try to overcome the resistance of the medium Decibel (dB) is a unit of measure that measures a signal’s lost or gain of strength – can be expressed in power or voltage dB = 10 log10 [P2/P1] = 20 log10 [V2/V1] Samples of the power or voltage taken at times 1 and 2. Lecture 2

  8. Distortion Distortion is when the signal changes its form. The each signal that makes up a composite signal could have different propagation speeds across the SAME medium – because of this, the different signals could have different delays (arriving at the receiver) – this causes a distortion. Lecture 2

  9. Noise Thermal Noise - the uncontrollable or random motion of electrons in the transport medium which creates an extra signal (not sent by the transmitter) Induced Noise – undesired devices acting as a transmitting antenna and those signals being picked up Cross Talk Noise – effect of one wire crossing another wire Impulse Noise – spikes in energy (ie lightning) Lecture 2

  10. Signal to Noise Ratio SNR = avg-signal-power/avg-noise-power High SNR – good (less corruption) Low SNR – bad (more noise than good power) SNR is described in Decibels (dB) SNRdB = 10 log10 SNR Lecture 2

  11. Shannon Equation • Shannon’s equation is used to determine the actual capacity of a channel given noise exist • C = B log2 (1 + SNR) • B = Bandwidth • C= Channel Capacity • SNR = Signal-to-noise ratio Actual ratio Lecture 2

  12. Chapter 3 and 4Handout #3 Dr. Clincy Professor of CS 5 of 10 Lecture 2

  13. Nyquist Equation • Given no noise, determine the maximum bit rate • BitRate = 2 x B x log2 L • B is the bandwidth of the channel • L is the # signal levels used • BitRate unit is bps (bits per second) Having 2 levels is reliable because a Rx can interpret 2 levels – suppose you had 64 levels – less reliable or more complex to interpret Lecture 2

  14. Bandwidth Bandwidth is a measure of performance Bandwidth in hertz – range of frequencies Bandwidth in bps – bps a channel can handle (D/A case here (ie. Modem)) Lecture 2

  15. Throughput • Throughput is a measure of performance – how fast data can flow through a network • Bandwidth could be what the channel could handle however, Throughput would be the amount that actually flowed through • Bandwidth – potential • Throughput – actual Lecture 2

  16. Latency Latency is a performance measure – how long it takes an message to completely arrive to the receiver Latency consist of propagation time (time for a bit to travel from Tx to the Rx) Propagation time = distance/propagation-speed transmission time (time for a message to be sent) Transmission time = message-size/bandwidth queuing time (time each intermediate node holds the message) processing time (time each node spends processing the message) Note: if message is small, more bandwidth exists and therefore, the latency is more of propagation time versus transmission time Lecture 2

  17. The bandwidth-delay product defines the number of bits that can fill the link. • This important when dealing with “full duplex” and being concerned about sending data to the Rx and receiving acknowledgments back from the Rx at the same time – before sending the next set of data Lecture 2

  18. Filling the link with bits for case 1 In other words, there can be no more than 5 bits at any time on the link. Lecture 2

  19. Filling the link with bits in case 2 5 bps 25 bps In other words, there can be no more than 25 bits at any time on the link. Lecture 2

  20. Chapter 4: Digital Transmission Physical Layer Issues

  21. Data Vs Signal • Fully explain the difference between signaland data before getting into the details Today’s Lecture (Digital Transmission) DATASIGNAL D D A Next Lecture (Analog Transmission) DATA SIGNAL A D A Lecture 2

  22. 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

  23. 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

  24. CS3501 Exam 3 Results • Average Score = 54 (Average Grade = 75) • Score SD = 18 Grading Scaled Used: • 100-83 A-grade (2 students) • 82-64 B-grade (6 students) • 63-45 C-grade (7 students) • 44-26 D-grade (9 students) • 25-7 F-grade (0 students) In getting your grade logged, be sure and pass back the exam after we go over them

  25. Chapter 4Handout #3 6 of 10 Dr. Clincy Professor of CS Your next test will cover lectures 21 – 26 You should use a calculator You can view the handouts via your laptop or you can print them - you shouldn’t use the browser Lecture 2

  26. 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

  27. 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

  28. 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

  29. 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

  30. 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

  31. 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

  32. 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

  33. 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)

  34. 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

  35. 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

  36. Line coding scheme categories

  37. 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

  38. 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)

  39. 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

  40. 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

  41. 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

  42. 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)

  43. 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

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

  45. 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

  46. 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

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