1. ECEN5533 Modern Commo Theory Lesson #11 23 September 2014 Dr. George Scheets n Read 2.5 - 2.8 n Quiz #1 rework due 1 week after return (DL) n Exam #1: DL no later 25 September u 23 < Initial scores < 80 n Design Problem #1 u Due 25 September (Live) u Not Later than 2 October (DL)

2. ECEN5533 Modern Commo Theory Lesson #1225 September 2014 Dr. George Scheets n Problems: 2.5, 2.8, 2.12, 2.14 n Exam #1: DL no later 25 September u Corrections due 2 October (Live) n Design Problem #1 u Due 25 September (Live) u Not Later than 2 October (DL)

3. ECEN5533 Modern Commo Theory Lesson #1330 September 2014 Dr. George Scheets  Read: 3.1  Problems: 2.15 - 2.19 n Exam #1 Corrections due u 2 October (Live) u 1 Week after return (DL) n Design Problem #1 u Due Not Later than 2 October (DL)

4. ECEN5533 Modern Commo Theory Lesson #142 October 2014 Dr. George Scheets  Read: 3.2  Problems: 3.1, 2, 5, & 6  Exam #1 Corrections due  Today (Live)  1 Week after return (DL)  Design Problem #1  Today (DL)  Rework due 9 October (Live)  Quiz #2  16 October (Live)  Not later than 23 October (DL)

5. ECEN5533 Modern Commo Theory Dr. George Scheets Lesson #15 7 October 2014 n Read Section 4.1 - 4.3 n Problems: 3.13, 3.14, 4.1, 4.2 n Design Problem #1 u Rework due 9 October (Live) n Quiz #2 u 16 October (Live)

6. ECEN5533 Modern Commo Theory Dr. George Scheets Lesson #16 9 October 2014 n Read Section 4.4 - 4.5 n Problems: 4.4, 6, 8, 9 n Design Problem #1 u Rework due 9 October n Quiz #2 u 16 October

7. ECEN5533 Modern Commo Theory Dr. George Scheets Lesson #17 14 October 2014 n Problems: Old Quiz #2 n Quiz #2 next time u Chapters 2 – 4 Up to & excluding Fiber Optic Systems

8. MegaMoron Communications n Matthew Gaalswyk & Michael Butler u promoted to Senior Engineer I n $489.6M n 2.2 GHz center frequency u Antenna placed in middle of city u Hi Gain 2.315 over 295 degrees u Low Gain 1.711 over 65 degrees n EIRP = 84.00 dBW n Receiver u LNA: F= 1.73 dB, G = 60 dB u IC: G = 26.56 dB Lo High

9. Quantization n Best Quantizer SNR occurs when round-off voltages occur with equal probability n If input is Uniformly Distributed... u Uniform Quantizer offers best SNR Quantize n If input is NOT Uniformly Distributed... u Best SNR Quantize occurs when F Step sizes are small in domain of likely voltages F Step sizes are large in domain of unlikely voltages

10. Quantizers… n Quantizer SNR Input voltage mapped to representative value Causes round-off error Uniform quantizer error tends to be Uniformly Distributed Uniformly Distributed u So long as input PDF symmetrical about 0 Wired Voice Telephone System uses Companding

11. Source Coders... n Pulse Code Modulation Each sample is independently quantized and coded n Differential PCM The difference between successive samples is transmitted n Source Decoder Yields distorted replica of transmitter analog input voltage

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

13. Single Sample Detector Comparator Switch closes every T seconds  Hold x(t) + n(t) x(t) ^ n Best to sample in middle of bit interval V HI V LOW

14. Single Sample Detector: SNR = 1 020406080100 0 4.5 990k Threshold is placed midway between nominal Logic 1 and 0 values. Detected sequence = 0011010111 at the receiver, but there were some near misses.

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

16. Fall 2002 Tcom Systems Final n 'Average' based on 1 test chosen at random 126.00 out of 150 n 'Average' based on 10 tests chosen randomly 109.44 out of 150 u The more points in an average, the better. n Actual Midterm Average 106.85 out of 150

17. Matched Filter Detector: SNR = 1 020406080100 0 4.5 990k 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

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

19. Binary Matched Filter Detector r(t) + n(t) s 1 (t) - s o (t) Integrate over T Decide 1 or 0 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.

20. Binary Matched Filter Detector r(t) + n(t) Analog Or Digital Filter h(t) = x(T-t) Decide 1 or 0 x(t) ^ Alternatively, could just have a filter matched to the pulse. The output over time of this filter and the previous differ, except at the sample time T, they're the same.

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

22. System with Multipath n h(t) = 0.9δ(t) – 0.4δ(t - 0.13) n |H(f)| =.9 -.4e -jω0.13

23. Required Equalizer Filter |H eq (f)| = 1/|H(f)|

24. H eq (f) = 1 / (.9 -.4e -jω0.13 ) H eq3 (f) = 1.111 + 0.4938e -jω0.13 + 0.2194e -jω0.26 +... Impulse Response of a 3 tap FIR Equalizing filter. h(t) = 1.111δ(t) + 0.4938δ(t – 0.13) + -.2194δ(t – 0.26)

25. Tapped Delay Line Equalizer a.k.a. FIR Filter and Moving Average Filter 1.1110.21940.4938 Delay 0.13 sec Delay 0.26 sec Σ Input Output Ideally |H(f)H eq (f)| = 1 Was 0.5 < |H(f)| < 1.3 Now 0.9 < |H(f)H eq3 (f)| < 1.1 |H(f)*H eq3 (f)|

26. Time Domain (3 Tap Equalizer) System Input System Output Multipath Equalizer Output

27. Tapped Delay Line Equalizer 8 Taps 1.1110.0038060.4938 Delay 0.13 sec Delay 0.91 sec Σ Input Output

28. Time Domain (8 Tap Equalizer) System Input System Output Equalizer Output

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

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

31. ISI due to Brick-Wall Filtering 020406080100120140 0 4.5 z k z2 k 1270k smearing non-causal Equalizer can undo some of this.

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

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

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

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

36. M-Ary Signaling n No Free Lunch For equal-power signals u As M increases, signals get closer together... u... and receiver detection errors increase

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

38. PDF for Noisy 4-ary PAM Signal f R' (r') r' volts +0.45 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) +1.34-0.45-1.34 area under each bell shaped curve = 1/4 00 01 11 10 Gray Code is best.

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

40. Rough Rule of Thumb n Want antenna aperture > 1/4 wavelength u To get relatively efficient EM radiation n Wavelength λ = (speed of EM wave) (frequency of EM wave) n Baseband pulses → Huge Antenna

41. Radio Waves

42. Syncing to Anti-podal BPSK n Message m(t) = + & - 1volt peak pulses u x(t) = m(t)cos(2πf c t) n Square x(t) to get x(t) 2 = m(t) 2 cos 2 (2πf c t) n Band pass filter to get double frequency term u cos(2π2f c t) n Run thru hard limiter to get 2f c Hz square wave n Run thru divide by 2 counter to get f c Hz square wave n Filter out harmonics to get f c Hz cosine

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

44. Coherent Matched Filter Detection n On-Off Binary Amplitude Shift Keying u Best P(BE) = Q( [E/N o ] 0.5 ), where E is the average Energy per Bit n Anti-Podal Binary Phase Shift Keying u Best P(BE) = Q( [2E/N o ] 0.5 ) n Binary Frequency Shift Keying with Optimum Frequency Offset of 0.75R Hertz u Best P(BE) = Q( [1.22E/N o ] 0.5 ) n Null-to-Null Bandwidth Required? u ASK & PSK require same amount (2R Hz) u FSK (2.75R Hz for optimum performance)

45. Fiber Optics... n Fiber Optic Link Analysis for ON-OFF noncoherent ASK n Thermal Noise on the Fiber is NOT a problem n 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. n Real World Detectors are plagued by thermal noise

46. Fiber Optics... n Conditional densities have different variances n We will assume equal variances & Gaussian densities

47. Fiber Optic Networks n E[Logic 0] = 0 volts n E[Logic 1] = [2*P electric ] 0.5 n Noise Power = 12*k*T°*R n P(Bit Error ≈ Q( [SNR/2] 0.5 ) u Single Sample Detector approximation. u A Matched Filter Detector would be better. u Actual P(BE) equation is more complex.

48. Channel Capacity (C) n Bandwidth & SNR impact realizable bit rate u Bandwidth required F function of symbol shape and symbol rate u Ability to reliably detect symbols F function of # of symbols ("M" in M-Ary) & SNR n Maximum bit rate that can be reliably shoved down a connection u Error free commo theoretically possible if actual bit rate R C.

49. Channel Capacity (C) n Bandwidth, Bit Rate, SNR, and BER related n Channel Capacity defines relationship C = Maximum reliable bit rate C = W*Log 2 (1 + SNR) bps Bandwidth impacts the maximum Baud rate

50. Channel Capacity (C) n Bandwidth, Bit Rate, SNR, and BER related n Channel Capacity defines relationship C = Maximum reliable bit rate C = W*Log 2 (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.

51. Channel Capacity (C) n Ex) 6 MHz TV RF Channel (42 dB SNR) C = 6,000,000 *Log 2 (1 + 15,849) = 83.71 Mbps n Ex) 64 KHz Fiber Bandwidth & Tbps bit rate 1 Tbps = 64,000* Log 2 (1 + SNR) My calculator can't generate a high enough SNR... Bogus Claim! n Ex) Tbps long distance over Power Lines Low Bandwidth, Low SNR... Bogus Claim!