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ECEN5533 Modern Commo Theory Lesson #11 1 October 2013 Dr. George Scheets

ECEN5533 Modern Commo Theory Lesson #11 1 October 2013 Dr. George Scheets. Read 2.5 - 2.8 Exam #1 (Covers Chapter 1, 5, and a bit of 2) Remote DL: No later 3 October Design # 1 3 October September (Local) No later than 10 October (Remote DL) -1 per working day late fee

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ECEN5533 Modern Commo Theory Lesson #11 1 October 2013 Dr. George Scheets

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  1. ECEN5533 Modern Commo TheoryLesson #11 1 October 2013Dr. George Scheets • Read 2.5 - 2.8 • Exam #1 (Covers Chapter 1, 5, and a bit of 2) • Remote DL: No later 3 October • Design #1 • 3 October September (Local) • No later than 10 October (Remote DL) • -1 per working day late fee • Quiz 1 Final Results • Hi = 19.4, Low = 7.3, Ave = 14.35, σ = 3.64

  2. ECEN5533 Modern Commo TheoryLesson #12 3 October 2013Dr. George Scheets • Problems: 2.5, 2.8, 2.12, 2.14 • Design #1 • 3 October September (Local) • No later than 10 October (Remote DL) • -1 per working day late fee • Quiz #2 • 24 October (Local) • No later than 31 October (Remote DL)

  3. ECEN5533 Modern Commo TheoryLesson #13 8 October 2013Dr. George Scheets • Read: 3.1 • Problems: 2.15 - 2.19 • Design #1 • No later than 10 October (Remote DL) • -1 per working day late fee • Corrected Tests (Variable due dates) • Quiz #2 • 24 October (Local) • No later than 31 October (Remote DL)

  4. Modern Communication TheoryECEN5533 Lesson #14 10 October 2013 • Read: 3.2 • Problems: 3.1, 3.2, 3.5, 3.6 • Design #1 due 10 October (Remote DL) • Corrected Tests (Variable due dates) • Quiz #2 • 24 October (Local) • No later than 31 October (Remote DL)

  5. ECEN5533 Modern Commo TheoryDr. George Scheets Lesson #15 15 October 2013 • Read Section 4.1 - 4.3 • Problems: 3.13, 3.14, 4.1, 4.2 • Corrected Exam #1 due • 1-2 weeks after you get them back (DL) • Quiz #2, 30 October (Live) • No later than 6 November (DL)

  6. ECEN5533 Modern Commo TheoryDr. George Scheets Lesson #16 17 October 2013 • Read Section 4.4 - 4.5 • Problems: 4.4, 4.6, 4.8, 4.9 • Reworked Exam #1 due today (remote) • Reworked Design #1 due various dates • Late reworks accepted @ -1 per working day • Quiz #2 • 24 October (Local) • < 31 October (Remote)

  7. ECEN5533 Modern Commo TheoryDr. George Scheets Lesson #17 22 October 2013 • Read Section 4.9 - 4.10, Skim Design #2 • Problems: 4.14, 4.16, 4.18, 4.19 • Reworked Design #1 due various dates • Late reworks accepted @ -1 per working day • Quiz #2 (Focuses on Chapters 2 – 4, 5 Digital) • 24 October (Local) • < 31 October (Remote)

  8. Symbol (Bit) Detection Low Pass Filter Source Decoder estimate of analog input discrete time signal estimate bit stream ^ x(t) receiver side Decide 1 or 0 received signal r(t) = x(t) + n(t)

  9. Single Sample Detector Switch closes every T seconds • Best to sample in middle of bit interval VHI Comparator ^ Hold x(t) x(t) + n(t)  VLOW

  10. Single Sample Detector: SNR = 1 Threshold is placed midway between nominal Logic 1 and 0 values. 4.5 0 4.5 0 20 40 60 80 100 0 k 99 Detected sequence = 0011010111 at the receiver, but there were some near misses.

  11. Single Sample Detector fR(r) Area = P[Logic 1] Area = P[Logic 0] r volts E[Logic 0] E[Logic 1] pdf spread is a function of Noise PowerAverage Logic 0/1 value is a function of Signal Power

  12. Fall 2002 Tcom Systems Final • 'Average' based on 1 test chosen at random126.00 out of 150 • 'Average' based on 10 tests chosen randomly109.44 out of 150 • The more points in an average, the better. • Actual Midterm Average106.85 out of 150

  13. 4.5 0 4.5 0 20 40 60 80 100 0 k 99 Matched Filter Detector: SNR = 1 Orange Bars are average voltage over that symbol interval. Averages are less likely to be wrong. SSD P(BE) = 0.1587, 10 S.I. Samples P(BE) = 0.000783

  14. Multiple Sample Detector fR(r) r volts E[Logic 0] E[Logic 1] PDF of single sample points.PDF of averages.Voltage estimates based on averages less likely to be way off.

  15. Binary Matched Filter Detector Integrate over T Decide 1 or 0 r(t) + n(t) s1(t) - so(t) ^ x(t) Decide 1 or 0 Box has: Sampler: Samples once per bit, at end of bit interval Comparator: Compares sample voltage to a threshold Hold: Holds voltage for one symbol interval.

  16. M-Ary Signaling One of M possible symbols is transmitted every T seconds. M is usually a power of 2 Log2M bits/symbol M = 256 symbols?Each symbol can represent 8 bits

  17. M-Ary Signaling Bandwidth required Function of symbols/second Function of symbol shape The more rapidly changing is the symbol, the more bandwidth it requires. An M-Ary signal with the same symbol rate and similar symbol shape as a Binary signal uses essentially the same bandwidth. More bits can be shoved down available bandwidth

  18. Digital Example: Binary Signaling Serial Bit Stream (a.k.a. Random Binary Square Wave) One of two possible pulses is transmitted every T seconds. This is a 1 watt signal. Symbol voltages are 2 volts apart. volts +1 0 time -1 T If T = .000001 seconds, then this 1 MBaud waveform moves 1 Mbps. (Baud = symbols per second)

  19. Example:M-Ary Signal One of M possible symbols is transmitted every T seconds.EX) 4-Ary signaling. This is also a 1 watt signal. Note each symbol can represent 2 bits. Some symbol voltages are 0.89 volts apart. volts +1.34 If T = .000001 seconds, then this 1 MBaud signal moves 2 Mbps. +.45 time -.45 Same BW as previous binary signal. -1.34 T

  20. M-Ary Signaling • No Free LunchFor equal-power signals • As M increases, signals get closer together... • ... and receiver detection errors increase

  21. PDF for Noisy Binary Signal fR'(r') area under each bell shaped curve = 1/2 -1 +1 r' volts Threshold is where PDF's cross (0 volts) Mean values are 1 volt from the threshold.

  22. PDF for Noisy 4-ary Signal area under each bell shaped curve = 1/4 fR'(r') Gray Code is best. 00 01 11 10 -1.34 -0.45 +0.45 +1.34 r' volts Threshold is where PDF's cross (-0.89, 0, +0.89 volts) Six tails are on wrong side of thresholds. Area of tails = P(Symbol Error)

  23. Where is this used? • M-Ary signaling used in narrow bandwidth environments... • ... preferably with a high SNR • Example: Dial-up phone modems • ... sometimes with a not so high SNR • Digital Cell Phones

  24. Intersymbol Interference Given a chunk of bandwidth There is a maximum tolerable symbol rate Nyquist showed... In theory, so long as the bandwidth is at least half the symbol rate, ISI can be eliminated. Requires Ideal (unrealizable) Filters. Rs symbols/second can be moved in Rs/2 Hertz

  25. Inter-Symbol Interference In Practice, bandwidth generally > half the symbol rate to make ISI tolerable. Closer to Rs symbols/second in Rs Hertz Example) V.34 Modems (33.6 Kbps)3429 symbols/second in about 3500 Hertz Example) 14.4 Kbps ModemsBandwidth about 3.5 KHzSymbol Rate of 2400 symbols/secondUses 64 QAM: 6 bits per symbol

  26. ISI due to Brick-Wall Filtering 4.5 non-causal smearing z k 0 z2 k Equalizer can undo some of this. 4.5 0 20 40 60 80 100 120 140 0 k 127

  27. Equalization Seeks to reverse effects of channel filtering H(f) Ideally Hequalizer(f) = 1/H(f) Result will be flat spectrum Not always practical if parts of |H(f)| have small magnitude Adaptive Filters frequently used

  28. System with Multipath|H(f)| = .9 - .4e -jω0.13

  29. Required Equalizer Filter|Heq(f)| = 1/|H(f)|

  30. |Heq(f)| = 1 / (.9 - .4e -jω0.13 ) |Heq3(f)| = 1.111 + 0.4444e-jω0.13 + 0.1975e -jω0.26+ ... Transfer function of a 3 tap FIR Equalizing filter.

  31. Tapped Delay Line Equalizera.k.a. FIR Filter and Moving Average Filter Output Input Σ 1.111 0.1975 0.4444 |H(f)*Heq3(f)| Delay 0.13 sec Delay 0.26 sec Ideally |H(f)Heq(f)| = 1 Was 0.5 < |H(f)| < 1.3 Now 0.9 < |H(f)Heq3(f)| < 1.1

  32. Radio Waves

  33. Phasor Representations Phasor comparison of AM, FM, and PM Geometrical Representation of Signals & Noise Useful for M-Ary symbol packing If symbol rate & shape are unchanged, bandwidth required is a function of the number of dimensions (axes)

  34. Coherent Matched Filter Detection • On-Off Binary Amplitude Shift Keying • Best P(BE) = Q( [E/No]0.5 ), where E is the average Energy per Bit • Anti-Podal Binary Phase Shift Keying • Best P(BE) = Q( [2E/No]0.5 ) • Binary Frequency Shift Keying with Optimum Frequency Offset of 0.75R Hertz • Best P(BE) = Q( [1.22E/No]0.5 ) • Null-to-Null Bandwidth Required? • ASK & PSK require same amount (2R Hz) • FSK (2.75R Hz for optimum performance)

  35. Syncing to Anti-podal BPSK • Message m(t) = + & - 1volt peak pulses • x(t) = m(t)cos(2πfct) • Square x(t) to get x(t)2 = m(t)2cos2(2πfct) • Band pass filter to get double frequency term • cos(2π2fct) • Run thru hard limiter to get 2fc Hz square wave • Run thru divide by 2 counter to get fc Hz square wave • Filter out harmonics to get fc Hz cosine

  36. Fiber Optics... • Fiber Optic Link Analysis for ON-OFF noncoherent ASK • Thermal Noise on the Fiber is NOT a problem • Noiseless Detection No pulse transmitted: No far side photons Pulse transmitted: Number of far side photons has discrete Poisson Distribution Mistake occurs when pulse transmitted and number of far side photons = 0. • Real World Detectors are plagued by thermal noise

  37. Fiber Optics... • Conditional densities have different variances • We will assume equal variances & Gaussian densities

  38. Fiber Optic Networks • E[Logic 0] = 0 volts • E[Logic 1] = [2*Pelectric]0.5 • Noise Power = 12*k*T°*R • P(Bit Error ≈ Q( [SNR/2]0.5 ) • This is an approximation.Actual P(BE) equation is more complex.

  39. Channel Capacity (C) • Bandwidth & SNR impact realizable bit rate • Bandwidth required • function of symbol shape and symbol rate • Ability to reliably detect symbols • function of # of symbols ("M" in M-Ary) & SNR • Maximum bit rate that can be reliably shoved down a connection • Error free commo theoretically possible if actual bit rate R < C. Not possible if R > C.

  40. Channel Capacity (C) • Bandwidth, Bit Rate, SNR, and BER related • Channel Capacity defines relationship C = Maximum reliable bit rate C = W*Log2(1 + SNR) bps Bandwidth impacts the maximum Baud rate

  41. Channel Capacity (C) • Bandwidth, Bit Rate, SNR, and BER related • Channel Capacity defines relationship C = Maximum reliable bit rate C = W*Log2(1 + SNR) bps Bandwidth impacts the maximum Baud rate SNR impacts the maximum number of different symbols (the "M" in M-ary) that can reliably be detected.

  42. Channel Capacity (C) • Ex) 6 MHz TV RF Channel (42 dB SNR)C = 6,000,000 *Log2(1 + 15,849) = 83.71 Mbps • Ex) 64 KHz Fiber Bandwidth & Tbps bit rate1 Tbps = 64,000* Log2(1 + SNR) My calculator can't generate a high enough SNR... Bogus Claim! • Ex) Tbps long distance over Power LinesLow Bandwidth, Low SNR... Bogus Claim!

  43. Trading off SNR for Capacity C • Channel Capacity Defines the Limit • C = W*Log2(1 + SNR) bps • Suppose at/near C limit & no extra BW available and... • Current SNR = 10 (C = W3.459) ? • Need to bump SNR up to 120 to double bps (C = W6.919) • Current SNR = 120? • Need to bump SNR up to 14,640 to double bps (C = W13.84) • Current SNR = 14,640? • Need to bump SNR up to 214.4M to double bps (C = W27.68) • To increase C by factor of 8 • Can't increase W? • Increase SNR by factor of 21,435,888

  44. Trading off W for Capacity C • Channel Capacity Defines the Limit • C = W*Log2(1 + SNR) bps • C = W*Log2(1 + Psignal/Pnoise) • C = W*Log2(1 + Psignal/(NoW)) • Suppose at/near C limit & you're power limited... • Current SNR = 10; C = W*Log2(1 + 10) = W3.459 • Doubling W yields C = 2W*Log2(1 + 5) = W5.170 • Doubling W again yields C = 4W*Log2(1 + 2.5) = W7.229 • As W → ∞, C → W14.43To increase C by factor of 8 requires C = W3.459*8 = W27.68Can't do this only by increasing W!

  45. Trading off W & Signal Power for Capacity C • Channel Capacity Defines the Limit • C = W*Log2(1 + Psignal/(NoW)) • Suppose at/near C limit & Current SNR = 10C = W*Log2(1 + 10) = W3.45 • Increasing both W & Signal Power can yield more reasonable solutions • Increasing W by a factor of 8 yields C = 8W*Log2(1 + 1.25) = W9.359 • Bumping Psignal by 8.008 yieldsC = 8W*Log2(1 + 10.01) = W27.69Result: Channel Capacity increases by factor of 8

  46. Operating well below the Channel Capacity limit? • Channel Capacity Defines the Limit • C = W*Log2(1 + SNR) bps • Doubling the signal power will generally allow the bit rate to be doubled, or nearly so. • Provided Sufficient BW Available Eb/No = EIRP - R + Other Terms (dB)(Digital Link Equation)

  47. Eb/No versus C/W bps per Hz • Want 30 bps/Hz?Need Eb/No > 75.53 dB • Want 3 bps/Hz?Need Eb/No > 3.68 dB • Want 1 bps/Hz?Need Eb/No > 0 dB • Note as C/W → 0Required Eb/No > -1.6 dBa.k.a. Shannon Limit

  48. Entropy • Average number of bits/symbol required to represent a set of symbols • Based on symbol probability • Lower bound regarding the amount of data compression possible • Equation presented does NOT account for spatial or time redundancies • Can be represented as conditional probabilites

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