ecen5533 modern commo theory lesson 11 1 october 2013 dr george scheets n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
ECEN5533 Modern Commo Theory Lesson #11 1 October 2013 Dr. George Scheets PowerPoint Presentation
Download Presentation
ECEN5533 Modern Commo Theory Lesson #11 1 October 2013 Dr. George Scheets

Loading in 2 Seconds...

play fullscreen
1 / 49

ECEN5533 Modern Commo Theory Lesson #11 1 October 2013 Dr. George Scheets - PowerPoint PPT Presentation


  • 103 Views
  • Uploaded on

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

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'ECEN5533 Modern Commo Theory Lesson #11 1 October 2013 Dr. George Scheets' - ivi


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
ecen5533 modern commo theory lesson 11 1 october 2013 dr george scheets
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
ecen5533 modern commo theory lesson 12 3 october 2013 dr george scheets
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)
ecen5533 modern commo theory lesson 13 8 october 2013 dr george scheets
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)
modern communication theory ecen5533 lesson 14 10 october 2013
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)
ecen5533 modern commo theory dr george scheets lesson 15 15 october 2013
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)
ecen5533 modern commo theory dr george scheets lesson 16 17 october 2013
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)
ecen5533 modern commo theory dr george scheets lesson 17 22 october 2013
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)
symbol bit detection
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)

single sample detector
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

single sample detector snr 1
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.

single sample detector1
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

fall 2002 tcom systems final
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
matched filter detector snr 1

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

multiple sample detector
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.

binary matched filter detector
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.

m ary signaling
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

m ary signaling1
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
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)

example m ary signal
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

m ary signaling2
M-Ary Signaling
  • No Free LunchFor equal-power signals
    • As M increases, signals get closer together...
    • ... and receiver detection errors increase
pdf for noisy binary signal
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.

pdf for noisy 4 ary signal
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)

where is this used
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
intersymbol interference
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

inter symbol interference
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

isi due to brick wall filtering
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

equalization
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

h eq f 1 9 4e j 0 13 h eq3 f 1 111 0 4444e j 0 13 0 1975 e j 0 26
|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.

tapped delay line equalizer a k a fir filter and moving average filter
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

phasor representations
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)

coherent matched filter detection
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)
syncing to anti podal bpsk
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
fiber optics
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
fiber optics1
Fiber Optics...
  • Conditional densities have different variances
  • We will assume equal variances & Gaussian densities
fiber optic networks
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.
channel capacity c
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.
channel capacity c1
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

channel capacity c2
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.

channel capacity c3
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!
trading off snr for capacity c
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
trading off w for capacity c
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!
trading off w signal power for capacity c
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
operating well below the channel capacity limit
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)
eb no versus c w bps per hz
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
entropy
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