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:

ECEN5533 Modern Commo TheoryLesson #11 23 September 2014Dr. George Scheets

ECEN5533 Modern Commo TheoryLesson #11 23 September 2014Dr. George Scheets Read 2.5 - 2.8Read 2.5 - 2.8 Quiz #1 rework due 1 week after return (DL)Quiz #1 rework due 1 week after return (DL) Exam #1: DL no later 25 SeptemberExam #1: DL no later 25 September

23 23 << Initial scores Initial scores << 80 80 Design Problem #1Design Problem #1

Due 25 September (Live)Due 25 September (Live) Not Later than 2 October (DL)Not Later than 2 October (DL)



Problems: 2.5, 2.8, 2.12, 2.14Problems: 2.5, 2.8, 2.12, 2.14 Exam #1: DL no later 25 SeptemberExam #1: DL no later 25 September

Corrections due 2 October (Live)Corrections due 2 October (Live) Design Problem #1Design Problem #1

Due 25 September (Live)Due 25 September (Live) Not Later than 2 October (DL)Not Later than 2 October (DL)



Read: 3.1Read: 3.1 Problems: 2.15 - 2.19Problems: 2.15 - 2.19 Exam #1 Exam #1 Corrections due Corrections due

2 October (Live)2 October (Live) 1 Week after return (DL)1 Week after return (DL)

Design Problem #1Design Problem #1 Due Not Later than 2 October (DL)Due Not Later than 2 October (DL)

ECEN5533 Modern Commo TheoryLesson #14 2 October 2014Dr. George Scheets

ECEN5533 Modern Commo TheoryLesson #14 2 October 2014Dr. George Scheets

Read: 3.2Read: 3.2 Problems: 3.1, 2, 5, & 6Problems: 3.1, 2, 5, & 6 Exam #1 Corrections due Exam #1 Corrections due

Today (Live)Today (Live) 1 Week after return (DL)1 Week after return (DL)

Design Problem #1Design Problem #1 Today (DL)Today (DL) Rework due 9 October (Live)Rework due 9 October (Live)

Quiz #2Quiz #2 16 October (Live)16 October (Live) Not later than 23 October (DL)Not later than 23 October (DL)

ECEN5533 Modern Commo TheoryDr. George ScheetsLesson #15 7 October 2014


Read Section 4.1 - 4.3Read Section 4.1 - 4.3 Problems: 3.13, 3.14, 4.1, 4.2Problems: 3.13, 3.14, 4.1, 4.2 Design Problem #1Design Problem #1

Rework due 9 October (Live)Rework due 9 October (Live) Quiz #2Quiz #2

16 October (Live)16 October (Live)



Read Section 4.4 - 4.5Read Section 4.4 - 4.5 Problems: 4.4, 6, 8, 9Problems: 4.4, 6, 8, 9 Design Problem #1Design Problem #1

Rework due 9 OctoberRework due 9 October Quiz #2Quiz #2

16 October16 October



Problems: Old Quiz #2Problems: Old Quiz #2 Quiz #2 next timeQuiz #2 next time

Chapters 2 – 4Chapters 2 – 4Up to & excluding Fiber Optic SystemsUp to & excluding Fiber Optic Systems

MegaMoron CommunicationsMegaMoron Communications Matthew Gaalswyk & Michael ButlerMatthew Gaalswyk & Michael Butler

promoted to Senior Engineer Ipromoted to Senior Engineer I $489.6M$489.6M 2.2 GHz center frequency2.2 GHz center frequency

Antenna placed in middle of cityAntenna placed in middle of city Hi Gain 2.315 over 295 degreesHi Gain 2.315 over 295 degrees Low Gain 1.711 over 65 degreesLow Gain 1.711 over 65 degrees

EIRP = 84.00 dBWEIRP = 84.00 dBW Receiver Receiver

LNA: F= 1.73 dB, G = 60 dBLNA: F= 1.73 dB, G = 60 dB IC: G = 26.56 dBIC: G = 26.56 dB

Lo

High

QuantizationQuantization

Best Quantizer SNR occurs when round-off Best Quantizer SNR occurs when round-off voltages occur with equal probability voltages occur with equal probability

If input is Uniformly Distributed...If input is Uniformly Distributed... Uniform Quantizer offers best SNRUniform Quantizer offers best SNRQuantizeQuantize

If input is NOT Uniformly Distributed...If input is NOT Uniformly Distributed... Best SNRBest SNRQuantizeQuantize occurs when occurs when

Step sizes are small in domain of likely voltagesStep sizes are small in domain of likely voltages Step sizes are large in domain of unlikely voltagesStep sizes are large in domain of unlikely voltages

Quantizers…Quantizers…

Quantizer SNRQuantizer SNRInput voltage mapped to representativeInput voltage mapped to representative value valueCauses round-off errorCauses round-off errorUniform quantizer error tends to beUniform quantizer error tends to be

Uniformly DistributedUniformly Distributed So long as input PDF symmetrical about 0So long as input PDF symmetrical about 0

Wired Voice Telephone System Wired Voice Telephone System uses Compandinguses Companding

Source Coders...Source Coders... Pulse Code ModulationPulse Code Modulation

Each sample is Each sample is independently independently quantized quantized and codedand coded

Differential PCMDifferential PCM The The differencedifference between successive samples between successive samples is transmittedis transmitted

Source DecoderSource Decoder Yields distorted replica of transmitter Yields distorted replica of transmitter analog input voltageanalog input voltage

Low PassFilter

estimateof analog

input

discrete time signal

estimate

receiver side

SourceDecoder

bitstream

Decide1 or 0

received signalr(t) = x(t) + n(t)

x(t)^

Symbol (Bit) DetectionSymbol (Bit) Detection

Single Sample DetectorSingle Sample Detector

Comparator

Switch closesevery T seconds

Holdx(t) + n(t)

x(t)^

Best to sample in middle of bit intervalBest to sample in middle of bit interval

VHI

VLOW

Single Sample Detector: SNR = 1Single Sample Detector: SNR = 1

0 20 40 60 80 100

0

4.5

4.5

990 k

Threshold is placed midway between nominal Logic 1 and 0 values.

Detected sequence = 0011010111 at the receiver,but there were some near misses.

Single Sample DetectorSingle Sample Detector

fR(r)

r voltsE[Logic 1]E[Logic 0]

pdf spread is a function of Noise PowerAverage Logic 0/1 value is a function of Signal Power

Area = P[Logic 1]Area = P[Logic 0]

Fall 2002 Tcom Systems FinalFall 2002 Tcom Systems Final

'Average' based on 1 test chosen at random'Average' based on 1 test chosen at random126.00 out of 150126.00 out of 150

'Average' based on 10 tests chosen randomly'Average' based on 10 tests chosen randomly109.44 out of 150109.44 out of 150 The more points in an average, the better.The more points in an average, the better.

Actual Midterm AverageActual Midterm Average106.85 out of 150106.85 out of 150

Matched Filter Detector: SNR = 1Matched Filter Detector: SNR = 1

0 20 40 60 80 100

0

4.5

4.5

990 k

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

Multiple Sample DetectorMultiple Sample Detector

fR(r)

r voltsE[Logic 1]E[Logic 0]

PDF of single sample points. PDF of averages.Voltage estimates based on averages less likely to be way off.

Binary Matched Filter DetectorBinary Matched Filter Detector

r(t) + n(t)

s1(t) - so(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 thresholdHold: Holds voltage for one symbol interval.

Binary Matched Filter DetectorBinary Matched Filter Detector

r(t) + n(t)

AnalogOr

DigitalFilter

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.

EqualizationEqualization Seeks to reverse effects of channel Seeks to reverse effects of channel

filtering H(f)filtering H(f) Ideally HIdeally Hequalizerequalizer(f) = 1/H(f)(f) = 1/H(f)

Result will be flat spectrumResult will be flat spectrum Not always practical if parts of |H(f)| have Not always practical if parts of |H(f)| have

small magnitudesmall magnitude Adaptive Filters frequently usedAdaptive Filters frequently used

System with MultipathSystem with Multipath

0 20 40 60 80 1000.4

0.6

0.8

1

1.2

1.4

Hf i

i

h(t) = 0.9h(t) = 0.9δδ(t) – 0.4(t) – 0.4δδ(t - 0.13)(t - 0.13) |H(f)| = .9|H(f)| = .9 - - .4e .4e -j-jωω0.130.13

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

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

0 20 40 60 80 1000.5

1

1.5

2

1

Hf i

i

Heq(f) = 1 / (.9 - .4e -jω0.13 ) Heq3(f) = 1.111 + 0.4938e-jω0.13 + 0.2194e -jω0.26 + ...

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 Equalizera.k.a. FIR Filter and Moving Average Filter

Tapped Delay Line Equalizera.k.a. FIR Filter and Moving Average Filter

1.111

0.2194

0.4938Delay

0.13 sec

Delay0.26 sec

ΣInput Output

Ideally |H(f)Heq(f)| = 1Was 0.5 < |H(f)| < 1.3Now 0.9 < |H(f)Heq3(f)| < 1.1

|H(f)*Heq3(f)|

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

System OutputMultipath

Equalizer Output

Tapped Delay Line Equalizer8 Taps

Tapped Delay Line Equalizer8 Taps

1.111

0.003806

0.4938Delay

0.13 sec

Delay0.91 sec

ΣInput Output

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

System Output

Equalizer Output

Intersymbol InterferenceIntersymbol Interference Given a chunk of bandwidthGiven a chunk of bandwidth

There is a maximum tolerable symbol rateThere is a maximum tolerable symbol rate Nyquist showed...Nyquist showed...

In theory, so long as the bandwidth is at least In theory, so long as the bandwidth is at least half the symbol rate, ISI can be eliminated.half the symbol rate, ISI can be eliminated. Requires Ideal (unrealizable) Filters.Requires Ideal (unrealizable) Filters. Requires Sync pulses Requires Sync pulses Rs symbols/second can be moved in Rs/2 HertzRs symbols/second can be moved in Rs/2 Hertz

Inter-Symbol InterferenceInter-Symbol Interference

In Practice, bandwidth generally > half In Practice, bandwidth generally > half the symbol rate to make ISI tolerable.the symbol rate to make ISI tolerable. Closer to Rs symbols/second in Rs HertzCloser to Rs symbols/second in Rs Hertz Example) V.34 Modems (33.6 Kbps)Example) V.34 Modems (33.6 Kbps)

3429 symbols/second in about 3500 Hertz3429 symbols/second in about 3500 Hertz Example) 14.4 Kbps ModemsExample) 14.4 Kbps Modems

Bandwidth about 3.5 KHzBandwidth about 3.5 KHzSymbol Rate of 2400 symbols/secondSymbol Rate of 2400 symbols/secondUses 64 QAM: 6 bits per symbolUses 64 QAM: 6 bits per symbol

ISI due to Brick-Wall FilteringISI due to Brick-Wall Filtering

0 20 40 60 80 100 120 140

0

4.5

4.5

zk

z2k

1270 k

smearingnon-causal

Equalizer can undo some of this.

M-Ary SignalingM-Ary Signaling

One of M possible symbols is transmitted everyOne of M possible symbols is transmitted everyT seconds.T seconds.

M is usually a power of 2M is usually a power of 2

LogLog22M bits/symbolM bits/symbol M = 256 symbols?M = 256 symbols?

Each symbol can represent 8 bitsEach symbol can represent 8 bits

M-Ary SignalingM-Ary Signaling Bandwidth requiredBandwidth required

Function of symbols/secondFunction of symbols/second Function of symbol shapeFunction of symbol shape

The more rapidly changing is the symbol, The more rapidly changing is the symbol, the more bandwidth it requires.the more bandwidth it requires.

An M-Ary signal with the same symbol rate and An M-Ary signal with the same symbol rate and similar symbol shape as a Binary signal uses similar symbol shape as a Binary signal uses essentially the same bandwidth.essentially the same bandwidth.

More bits can be shoved down available More bits can be shoved down available bandwidthbandwidth

Digital Example: Binary SignalingDigital Example: Binary Signaling Serial Bit Stream Serial Bit Stream

(a.k.a. Random Binary Square Wave)(a.k.a. Random Binary Square Wave) One of two possible pulses is transmitted every T seconds.One of two possible pulses is transmitted every T seconds. This is a 1 watt signal. Symbol voltages are 2 volts apart.This is a 1 watt signal. Symbol voltages are 2 volts apart.

time

+1

volts

0

-1

T

If T = .000001 seconds, then this1 MBaud waveform moves 1 Mbps.(Baud = symbols per second)

Example:M-Ary SignalExample:M-Ary Signal One of M possible symbols is transmitted everyOne of M possible symbols is transmitted every

T seconds.T seconds.EX) 4-Ary PAM signaling. This is also a 1 watt signal. EX) 4-Ary PAM signaling. This is also a 1 watt signal. Note each symbol can represent 2 bits. Note each symbol can represent 2 bits. Some symbol voltages are 0.89 volts apart.Some symbol voltages are 0.89 volts apart.

time

+1.34volts

+.45

T

If T = .000001 seconds, thenthis 1 MBaud signal moves 2 Mbps.

-.45

-1.34Same BW asprevious binary signal.

M-Ary SignalingM-Ary Signaling

No Free LunchNo Free LunchFor equal-power signalsFor equal-power signals As M increases, signals get closer As M increases, signals get closer

together...together... ... and receiver detection errors increase... and receiver detection errors increase

PDF for Noisy Binary SignalPDF for Noisy Binary Signal

fR'(r')

r' volts+1-1

Threshold is where PDF's cross (0 volts)Mean values are 1 volt from the threshold.

area undereach bell

shaped curve= 1/2

PDF for Noisy 4-ary PAM SignalPDF for Noisy 4-ary PAM Signal

fR'(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 undereach bell

shaped curve= 1/4

00 01 11 10

GrayCodeis best.

Where is this used?Where is this used?

M-Ary signaling used in narrow M-Ary signaling used in narrow bandwidth environments...bandwidth environments...

... preferably with a high SNR... preferably with a high SNR Example: Dial-up phone modemsExample: Dial-up phone modems

... sometimes with a not so high SNR... sometimes with a not so high SNR Digital Cell PhonesDigital Cell Phones

Rough Rule of ThumbRough Rule of Thumb

Want antenna aperture Want antenna aperture >> 1/4 wavelength 1/4 wavelength To get relatively efficient EM radiationTo get relatively efficient EM radiation

Wavelength Wavelength λλ = = (speed of EM wave(speed of EM wave) ) (frequency of EM wave)(frequency of EM wave)

Baseband pulses → Huge AntennaBaseband pulses → Huge Antenna

Radio WavesRadio Waves

Syncing to Anti-podal BPSKSyncing to Anti-podal BPSK

Message m(t) = + & - 1volt peak pulsesMessage m(t) = + & - 1volt peak pulses x(t) = m(t)cos(2x(t) = m(t)cos(2ππffcct)t)

Square x(t) to get x(t)Square x(t) to get x(t)22 = m(t) = m(t)22coscos22(2(2ππffcct)t)

Band pass filter to get double frequency termBand pass filter to get double frequency term cos(2cos(2ππ2f2fcct)t)

Run thru hard limiter to get 2fRun thru hard limiter to get 2fcc Hz square wave Hz square wave

Run thru divide by 2 counter to get fRun thru divide by 2 counter to get fcc Hz square wave Hz square wave

Filter out harmonics to get fFilter out harmonics to get fcc Hz cosine Hz cosine

Phasor RepresentationsPhasor Representations

Phasor comparison of AM, FM, and PMPhasor comparison of AM, FM, and PM Geometrical Representation of Signals &Geometrical Representation of Signals &

Noise Noise Useful for M-Ary symbol packingUseful for M-Ary symbol packing If symbol rate & shape are unchanged, If symbol rate & shape are unchanged,

bandwidth required is a function of the bandwidth required is a function of the number of dimensions (axes)number of dimensions (axes)

Coherent Matched Filter DetectionCoherent Matched Filter Detection On-Off Binary Amplitude Shift KeyingOn-Off Binary Amplitude Shift Keying

Best P(BE) = Q( [E/Best P(BE) = Q( [E/NNoo]]0.5 0.5 ), ),

where E is the average Energy per Bitwhere E is the average Energy per Bit Anti-Podal Binary Phase Shift KeyingAnti-Podal Binary Phase Shift Keying

Best P(BE) = Q( [2E/Best P(BE) = Q( [2E/NNoo]]0.5 0.5 ))

Binary Frequency Shift Keying with Optimum Frequency Offset Binary Frequency Shift Keying with Optimum Frequency Offset of 0.75R Hertzof 0.75R Hertz Best P(BE) = Q( [1.22E/Best P(BE) = Q( [1.22E/NNoo]]0.5 0.5 ))

Null-to-Null Bandwidth Required?Null-to-Null Bandwidth Required? ASK & PSK require same amount (2R Hz)ASK & PSK require same amount (2R Hz) FSK (2.75R Hz for optimum performance)FSK (2.75R Hz for optimum performance)

Fiber Optics...Fiber Optics... Fiber Optic Link Analysis for ON-OFF noncoherent Fiber Optic Link Analysis for ON-OFF noncoherent

ASKASK Thermal Noise on the Fiber is NOT a problem Thermal Noise on the Fiber is NOT a problem Noiseless DetectionNoiseless Detection

No pulse transmitted: No far side photonsNo pulse transmitted: No far side photonsPulse transmitted: Number of far side photons has Pulse transmitted: Number of far side photons has

discrete Poisson Distributiondiscrete Poisson DistributionMistake occurs when pulse transmitted and Mistake occurs when pulse transmitted and

number of far side photons = 0.number of far side photons = 0. Real World Detectors are plagued by thermal noiseReal World Detectors are plagued by thermal noise

Fiber Optics...Fiber Optics...

Conditional densities have different Conditional densities have different variancesvariances

We will assume equal variances & We will assume equal variances & Gaussian densitiesGaussian densities

Fiber Optic NetworksFiber Optic Networks

E[Logic 0] = 0 voltsE[Logic 0] = 0 volts E[Logic 1] = [2*PE[Logic 1] = [2*Pelectricelectric]]0.50.5

Noise Power = 12*k*T°*RNoise Power = 12*k*T°*R P(Bit Error ≈ Q( [SNR/2]P(Bit Error ≈ Q( [SNR/2]0.50.5 ) )

Single Sample Detector approximation.Single Sample Detector approximation. A Matched Filter Detector would be better.A Matched Filter Detector would be better. Actual P(BE) equation is more complex.Actual P(BE) equation is more complex.

Channel Capacity (C)Channel Capacity (C) Bandwidth & SNR impact realizable bit rateBandwidth & SNR impact realizable bit rate

Bandwidth requiredBandwidth required function of symbol shape and symbol ratefunction of symbol shape and symbol rate

Ability to reliably detect symbolsAbility to reliably detect symbols function of # of symbols ("M" in M-Ary) & SNR function of # of symbols ("M" in M-Ary) & SNR

Maximum bit rate that can be Maximum bit rate that can be reliablyreliably shoved shoved down a connection down a connection Error free commo theoretically possible if actual bit Error free commo theoretically possible if actual bit

rate R rate R << C. Not possible if R > C. C. Not possible if R > C.

Channel Capacity (C)Channel Capacity (C) Bandwidth, Bit Rate, SNR, and BER relatedBandwidth, Bit Rate, SNR, and BER related Channel Capacity defines relationshipChannel Capacity defines relationship

C = Maximum reliable bit rate C = Maximum reliable bit rate C = W*Log C = W*Log22(1 + SNR) bps(1 + SNR) bps

Bandwidth impacts the maximum Baud rate

Channel Capacity (C)Channel Capacity (C) Bandwidth, Bit Rate, SNR, and BER relatedBandwidth, Bit Rate, SNR, and BER related Channel Capacity defines relationshipChannel Capacity defines relationship

C = Maximum reliable bit rate C = Maximum reliable bit rate C = W*Log C = W*Log22(1 + SNR) bps(1 + SNR) bps

Bandwidth impacts the maximum Baud rate

SNR impacts the maximum number ofdifferent symbols (the "M" in M-ary)

that can reliably be detected.

Channel Capacity (C)Channel Capacity (C) Ex) 6 MHz TV RF Channel (42 dB SNR)Ex) 6 MHz TV RF Channel (42 dB SNR)

C = 6,000,000 *LogC = 6,000,000 *Log22(1 + 15,849) = 83.71 Mbps (1 + 15,849) = 83.71 Mbps

Ex) 64 KHz Fiber Bandwidth & Tbps bit rateEx) 64 KHz Fiber Bandwidth & Tbps bit rate1 Tbps = 64,000* Log1 Tbps = 64,000* Log22(1 + SNR) (1 + SNR)

My calculator can't generate a high enoughMy calculator can't generate a high enoughSNR... Bogus Claim!SNR... Bogus Claim!

Ex) Tbps long distance over Power LinesEx) Tbps long distance over Power LinesLow Bandwidth, Low SNR... Bogus Claim! Low Bandwidth, Low SNR... Bogus Claim!

Documents

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: