ECEN4523 Commo TheoryECEN4523 Commo TheoryLecture #26 19 October 2015Lecture #26 19 October 2015Dr. George ScheetsDr. George Scheetswww.okstate.edu/elec-engr/scheets/ecen4533www.okstate.edu/elec-engr/scheets/ecen4533

ECEN4523 Commo TheoryECEN4523 Commo TheoryLecture #26 19 October 2015Lecture #26 19 October 2015Dr. George ScheetsDr. George Scheetswww.okstate.edu/elec-engr/scheets/ecen4533www.okstate.edu/elec-engr/scheets/ecen4533

Read 6.2 (skim quantization material)Read 6.2 (skim quantization material) Problems: 6.1-5, 9 & 6.2-9Problems: 6.1-5, 9 & 6.2-9 Quiz #6, 23 October (Live) Quiz #6, 23 October (Live)

Remote DL students Remote DL students << 30 October 30 October Exam #2, 30 October (Live) Exam #2, 30 October (Live)

Remote DL students Remote DL students << 6 November 6 November Design Problem, due 6 November (Live) Design Problem, due 6 November (Live)

Remote DL students Remote DL students << 13 November 13 November

ECEN4523 Commo TheoryECEN4523 Commo TheoryLecture #27 21 October 2015Lecture #27 21 October 2015Dr. George ScheetsDr. George Scheetswww.okstate.edu/elec-engr/scheets/ecen4533www.okstate.edu/elec-engr/scheets/ecen4533

ECEN4523 Commo TheoryECEN4523 Commo TheoryLecture #27 21 October 2015Lecture #27 21 October 2015Dr. George ScheetsDr. George Scheetswww.okstate.edu/elec-engr/scheets/ecen4533www.okstate.edu/elec-engr/scheets/ecen4533

Read 7.1 & 7.2 Read 7.1 & 7.2 Problems: 6.2-2, 7.2-3 & 4Problems: 6.2-2, 7.2-3 & 4 Quiz #6, 23 October (Live) Quiz #6, 23 October (Live)

Remote DL students Remote DL students << 30 October 30 October Exam #2, 30 October (Live) Exam #2, 30 October (Live)

Remote DL students Remote DL students << 6 November 6 November Design Problem, due 6 November (Live) Design Problem, due 6 November (Live)

Remote DL students Remote DL students << 13 November 13 November

Design ProblemDesign Problem

Simulate above Analog Commo SystemSimulate above Analog Commo System Design Post Filter Design Post Filter

Minimize Average e(t)Minimize Average e(t)22 a.k.a. MSE a.k.a. MSE

M Tap Finite Impulse Response Post Filter

M Tap Finite Impulse Response Post Filter

W0

W(M-1)

W1Delay? * ts

Delay? * ts

ΣInput Output

H(f) desired?Find h(t).

Wn = h(nts).

Figure of MeritMSE*M 0.63

Lower is Better.

GradingGrading Real World RFP: Real World RFP:

1 team gets full credit1 team gets full credit Everyone else gets a zeroEveryone else gets a zero

Partial credit Partial credit Awarded on Quizzes & TestsAwarded on Quizzes & Tests NOT AWARDED ON DESIGN PROJECTS!NOT AWARDED ON DESIGN PROJECTS!

Real world designs don't get partial creditReal world designs don't get partial credit Either Work or They Don'tEither Work or They Don't

Double check your work!!! Double check your work!!!

Digitizing an Analog SignalDigitizing an Analog Signal

Sample Sample Continuous Time → Discrete TimeContinuous Time → Discrete Time fs > 2(Signal BW) with Ideal Samplerfs > 2(Signal BW) with Ideal Sampler Practically speaking, fs ≈ 2.2(Signal BW)Practically speaking, fs ≈ 2.2(Signal BW)

Quantize Quantize Continuous Voltage → Discrete VoltageContinuous Voltage → Discrete Voltage Introduces Round-Off ErrorIntroduces Round-Off Error We'll focus on PCMWe'll focus on PCM

Pulse Code ModulationPulse Code Modulation

What most off-the-shelf A/D Converters doWhat most off-the-shelf A/D Converters do Rounds voltage to one of L possible valuesRounds voltage to one of L possible values L usually a power of 2L usually a power of 2 Each Voltage assigned equal length code wordEach Voltage assigned equal length code word

Binary 1's and 0'sBinary 1's and 0's N = logN = log22L bits per wordL bits per word

EX) L = 256 voltages? N = 8 bitsEX) L = 256 voltages? N = 8 bits

Example) Coding aMicrophone Output

Example) Coding aMicrophone Output

time (sec)

m(t) volts (air pressure)

Energy from about 500 - 3,500 Hz.

A/D ConvertorA/D Convertor

time (sec)

m(t) volts (air pressure)

Step #1)Sample the waveform at rate > 2*Max Frequency.Telephone voice is sampled at 8,000 samples/second.

1/8000 second

A/D ConvertorA/D Convertor Legacy Wired Telephone System uses PCMLegacy Wired Telephone System uses PCM

Pulse Code ModulationPulse Code Modulation One of N possible equal length Code One of N possible equal length Code Words is Words is assigned to each Voltageassigned to each VoltageN Typically a Power of 2N Typically a Power of 2LogLog22N bits per code wordN bits per code word Wired Phone System: N = 256 & 8 bits/wordWired Phone System: N = 256 & 8 bits/word Compact Disk: N = 65,536 & 16 bits/wordCompact Disk: N = 65,536 & 16 bits/word

A/D Convertor. 1 bit/sample.A/D Convertor. 1 bit/sample.

time (sec)

Example) N = 2. Assign 0 or 1 to voltage.

0 < Voltage < +5v, Assign Logic 1-5v < Voltage < 0, Assign Logic 0

3.62 v, output a 1

t1

Bit Stream Out = 1111110000111...

A/D Convertor. 1 bit/sample.A/D Convertor. 1 bit/sample.Example) N = 2. Assign 0 or 1 to voltage.

Far side gets... 1111110000111 (13 samples)Need to output 13 voltages.What does a 1 represent? A 0?

Receive a 1? Output +2.5 v (mid-range)Receive a 0? Output -2.5 v (mid-range)

Hold the voltage until next sample

0 < Voltage < +5v, Assign Logic 1-5v < Voltage < 0, Assign Logic 0

A/D Convertor. 1 bit/sample.A/D Convertor. 1 bit/sample.

Input to the transmitter.Output at the receiver.

Considerable Round-Off error exists.

+2.5 v

-2.5 v

time (sec)

Example) N = 4. Assign 00, 01, 10 or 11.

2.5 < Voltage < 5 , Assign 110 < Voltage < 2.5, Assign 10-2.5 < Voltage < 0, Assign 00-5 < Voltage < -2.5, Assign 01

3.62 v, Assign 11

t1

Bit Stream Out =11111011111100 000000101011...

+2.5 v

-2.5 v

A/D Convertor. 2 bits/sampleA/D Convertor. 2 bits/sample

A/D Convertor. 2 bits/sample.A/D Convertor. 2 bits/sample.

Input to the transmitter.Output at the receiver.

Receive 11? Output 3.75vReceive 10? Output 1.25vReceive 00? Output -1.25vReceive 01? Output -3.75vReduced Round-Off error exists.

+3.75 v

+1.25 v

-1.25 v

-3.75 v

A/D Convertor. 1 bit/sample.A/D Convertor. 1 bit/sample.

Input to the transmitter.Output at the receiver.

Considerable Round-Off error exists.

+2.5 v

-2.5 v

Circuit Switched Voice (POTS)Circuit Switched Voice (POTS) Bandwidth ≈ 3,500 HertzBandwidth ≈ 3,500 Hertz A/D ConverterA/D Converter

samples voice 8,000 times/secondsamples voice 8,000 times/second rounds off voice to one of 256 voltage levelsrounds off voice to one of 256 voltage levels transmits 8 bits per sample to far sidetransmits 8 bits per sample to far side

D/A ConverterD/A Converter receives 8 bit code wordreceives 8 bit code word outputs one of 256 voltage levels for 1/8000th secondoutputs one of 256 voltage levels for 1/8000th second

64,000 bps64,000 bps

Compact DiskCompact Disk Bandwidth ≈ 20,000 HertzBandwidth ≈ 20,000 Hertz A/D ConverterA/D Converter

samples voice 44,100 times/secondsamples voice 44,100 times/second rounds off voice to one of 65,536 voltage levelsrounds off voice to one of 65,536 voltage levels transmits 16 bits per sample to far sidetransmits 16 bits per sample to far side

D/A ConverterD/A Converter receives 16 bit code wordreceives 16 bit code word outputs one of 65,536 voltage levels for 1/44100th outputs one of 65,536 voltage levels for 1/44100th

secondsecond 705,600 bps705,600 bps

Sampling & Quantizing ExamplesSampling & Quantizing Examples fs = 16 KHzfs = 16 KHz

4096 quantiles 4096 quantiles 256 quantiles (approximate phone quality)256 quantiles (approximate phone quality) 32 quantiles 32 quantiles 4 quantiles (generally 2 levels used!)4 quantiles (generally 2 levels used!)

4096 quantiles4096 quantiles fs = 16 KHzfs = 16 KHz fs = 8 KHz fs = 8 KHz fs = 2 KHzfs = 2 KHz fs = 1 KHzfs = 1 KHz

1/8th Second of Voice1/8th Second of Voice

1/8th Second of Voice1/8th Second of Voice

1/8th Second of Voice1/8th Second of Voice

Sampling & Quantizing ExamplesSampling & Quantizing Examples fs = 16 KHzfs = 16 KHz

4096 quantiles 4096 quantiles 256 quantiles (approximate phone quality)256 quantiles (approximate phone quality) 32 quantiles 32 quantiles 4 quantiles (generally 2 levels used!)4 quantiles (generally 2 levels used!)

4096 quantiles4096 quantiles fs = 16 KHz fs = 16 KHz fs = 8 KHz (some interference)fs = 8 KHz (some interference) fs = 2 KHzfs = 2 KHz fs = 1 KHzfs = 1 KHz

Fourier TransformsFourier Transforms Are 1 to 1 mappingsAre 1 to 1 mappings

x(t) has one and only one X(f) volts/Hzx(t) has one and only one X(f) volts/Hz RRXX((ττ) has one and only one S) has one and only one SXX(f) watts/Hz(f) watts/Hz

Many different x(t) can map to same RMany different x(t) can map to same RXX((ττ)) Random Bit StreamsRandom Bit Streams SinusoidsSinusoids Random Noise, etc. Random Noise, etc.

Mapping of x(t) to SMapping of x(t) to SXX(f) is many to one(f) is many to one Why SWhy SXX(f) is useful for commo analysis (f) is useful for commo analysis

Basic Block DiagramBasic Block Diagram

source: Figure 1.2, Lathi & Ding, MODERN DIGITAL AND ANALOG COMMUNICATION SYSTEMS

This is all in the telecom Physical Layer.

Basic Block Diagram (Digital)Basic Block Diagram (Digital)

source: Figure 1.2, Lathi & Ding, MODERN DIGITAL AND ANALOG COMMUNICATION SYSTEMS

This is all in the telecom Physical Layer.

Input TransducerInput Transducer Analog Signal In?Analog Signal In?

SamplerSampler PCM…PCM…

QuantizeQuantize Assign binary code word (1's and 0's)Assign binary code word (1's and 0's)

……or Application Specific Coderor Application Specific Coder Voice or VideoVoice or Video Generate binary outputGenerate binary output

Digital Signal in?Digital Signal in? May want to clean up and retimeMay want to clean up and retime

Basic Block Diagram (Digital)Basic Block Diagram (Digital)

source: Figure 1.2, Lathi & Ding, MODERN DIGITAL AND ANALOG COMMUNICATION SYSTEMS

This is all in the telecom Physical Layer.

Transmitter (Digital)Transmitter (Digital) Binary Bit Stream InBinary Bit Stream In

May Not Want to Transmit BinaryMay Not Want to Transmit Binary Can reduce RF BW by going M-AryCan reduce RF BW by going M-Ary Transmitting M possible symbolsTransmitting M possible symbols

Multiple bits mapped to each symbolMultiple bits mapped to each symbol EX) 90 Mbps → 10 bits/symbol (M = 2EX) 90 Mbps → 10 bits/symbol (M = 21010 =1,024) =1,024)

Baud rate = 9 M symbols/secondBaud rate = 9 M symbols/second

24 bit color 224 = 16.78 M colors24 bit color 224 = 16.78 M colors

256 Colors256 Colors

16 Colors16 Colors

Manchester Pulse X(f)Manchester Pulse X(f)

Fourier Transform of 1/2 Pulse

Fourier Transform of Manchester Pulse

Ethernet Uses Manchester CodingEthernet Uses Manchester Coding

time

+1

volts

0

-1

T

0 0

LogicOne

LogicZero

All symbolshave a transition

in the middle.

Ethernet Uses Manchester CodingEthernet Uses Manchester Coding

time

+1

volts

0

-1

T

High Pass Filters Emphasize Change

High Pass Filter OutputHigh Pass Filter Output

time

+1

0

-1

Rectify (Absolute Value)Rectify (Absolute Value)

time

+1

0

-1T

Result always has pulses T seconds apart.Useful for receiver synchronization.

Serial Bit Stream: NRZ CodingSerial Bit Stream: NRZ Coding

time

+1

volts

0

-1

T

0 0

LogicOne

LogicZero

Called ‘Non Return to Zero’because voltage never dwells

on zero volts.

T