ECEN5533 Modern Commo Theory Lesson #18. 22 February 2016 Dr ECEN5533 Modern Commo Theory Lesson #18 22 February 2016 Dr. George Scheets Read 3.1 Problems 2.15 - 2.17 Design Problem #1 due 26 February Quiz #2 on 7 March Exam #1 final results Hi = 97, Low = 77, ave = 85.75, σ = 8.30 A > 88, B > 75

ECEN5533 Modern Commo Theory Lesson #19. 24 February 2016 Dr ECEN5533 Modern Commo Theory Lesson #19 24 February 2016 Dr. George Scheets Read 3.2 Problems 2.18, 2.19, 3.1 Design Problem #1 due 26 February Quiz #2 on 7 March

ECEN5533 Modern Commo Theory Lesson #20. 26 February 2016 Dr ECEN5533 Modern Commo Theory Lesson #20 26 February 2016 Dr. George Scheets Problems 3.2, 5, & 6 Design Problem #1 due 26 February Quiz #2 on 7 March

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.

Binary Matched Filter Detector Analog Or Digital Filter h(t) = x(T-t) Decide 1 or 0 r(t) + n(t) ^ 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.

Symbol Detection Matched Filter P(BE) < Single Sample Detector P(BE) Single Sample Detector P(BE) independent of Bit Rate If system remains wideband (square pulse in & out) Sampling done at baseband Matched Filter P(BE) gets worse as Bit Rate ↑ Eventually → Single Sample Detector P(BE) Multiply r(t) by s1(t) - s0(t) prior to integration RF system? s1(t) – s0(t) is a sinusoid Integrator sees product of two sinusoids Get a difference frequency term = baseband pulses Get a double frequency term, filtered out by integrator

Intersymbol Interference Given a chunk of bandwidth There is a maximum tolerable symbol rate Nyquist showed Rs symbols/second can be moved in Rs/2 Hertz Using sinc pulses instead of square pulses Real World Closer to Rs symbols/second in Rs Hertz EX) V.34 Modems (33.6 Kbps) 3429 symbols/second in about 3500 Hertz

ISI due to Brick-Wall Filtering 4.5 z k z2 k 4.5 20 40 60 80 100 120 140 k

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

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? Each symbol can represent 8 bits

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

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

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

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

Digital Example: Binary Signaling This is a _ watt signal. Sampled voltage values are 2 volts apart. 1 time +1 volts -1 T If T = .000001 seconds, then this 1 MBaud waveform moves 1 Mbps. (Baud = symbols per second)

Example: 4-Ary Signal 1 This is also a _ watt signal. Some sample voltages are 0.89 volts apart. In the presence of noise, will have more errors here. 1 volts +1.34 If T = .000001 seconds, then this 1 MBaud signal moves 2 Mbps. +.45 -.45 time -1.34 T

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

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

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)

Example:M-Ary Signal One of M possible symbols is transmitted every T seconds. EX) 4-Ary signaling. Note each symbol can represent 2 bits. volts +1.34 If T = .000001 seconds, then this 1 MBaud signal moves 2 Mbps. +.45 -.45 time -1.34 T

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)

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

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

Required Equalizer Filter |Heq(f)| = 1/|H(f)| Heq(f) = 1 / (. 9 - Required Equalizer Filter |Heq(f)| = 1/|H(f)| Heq(f) = 1 / (.9 - .4e -jω0.13 )

Heq(f) = 1 / (. 9 -. 4e -jω0. 13 ) Heq3(f) = 1. 111 + 0. 4938e-jω0. 13 Heq(f) = 1 / (.9 - .4e -jω0.13 ) Heq3(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)

Tapped Delay Line Equalizer a.k.a. FIR Filter or Moving Average Filter Input Output 1.111 |H(f)*Heq3(f)| Σ 0.4938 Delay 0.13 sec 0.2194 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

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

Tapped Delay Line Equalizer 8 Taps Input Output 1.111 Σ 0.4938 Delay 0.13 sec 0.003806 Delay 0.91 sec

Time Domain (8 Tap Equalizer) System Input Multipath Output Equalizer Output

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

Radio Waves

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)