Loading in 2 Seconds...

ECEN5533 Modern Commo Theory Lesson #11 1 October 2013 Dr. George Scheets

Loading in 2 Seconds...

- By
**ivi** - Follow User

- 104 Views
- Uploaded on

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 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 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 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 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 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 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 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

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

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

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 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

- '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

0

4.5

0

20

40

60

80

100

0

k

99

Matched Filter Detector: SNR = 1Orange 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

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

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

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 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.

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

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 Signaling

- No Free LunchFor equal-power signals
- As M increases, signals get closer together...
- ... and receiver detection errors increase

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

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?

- 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

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

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

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

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

|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 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 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

- 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

- 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 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 Optics...

- Conditional densities have different variances
- We will assume equal variances & Gaussian densities

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)

- 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 (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 (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 (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

- 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

- 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

- 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?

- 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

- 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

- 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

Download Presentation

Connecting to Server..