EE521 Analog and Digital Communications

April 12, 2005 Week 13 1

EE521 Analog and Digital CommunicationsJames K. Beard, Ph. D.

[email protected]

Tuesday, March 29, 2005

http://astro.temple.edu/~jkbeard/

Week 13 2April 12, 2005

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Week 13 3April 12, 2005

Essentials Text: Bernard Sklar, Digital Communications,

Second Edition SystemView Office

E&A 349 Tuesday afternoons 3:30 PM to 4:30 PM & before class MWF 10:30 AM to 11:30 AM

Term Projects Due April 19 (Next Week) Final Exam Scheduled

Tuesday, May 10, 6:00 PM to 8:00 PM Here in this classroom

Week 13 4April 12, 2005

Today’s Topics

Term Project Quiz

Main Quiz Backup Quiz

Term Project Discussion (as time permits)

Week 13 5April 12, 2005

Quiz Question 1 Parts I and II

Criteria for a signal to be a power/energy signal Finite energy == energy signal Finite power == power signal

Equations

2*

2

*

1lim

T

x TT

x

P x t x t dtT

E x t x t dt

Week 13 6April 12, 2005

Question 1 Part III: Power Spectrum of an Energy Signal Energy spectrum is simply magnitude squared of Fourier

transform of energy signal Fourier transform of energy signal not defined Use the Fourier transform of the autocorrelation function

for the power spectrum

2*

2

1lim

exp 2

T

x TT

x x

R x t x t dtT

G f R j f d

Week 13 7April 12, 2005

Question 2 Part I

This is a quadrature demodulator The output x0(t)

Frequency shifted to baseband Negative frequency image

Shifted to -2.f0

Attenuated by LPF Has a bandwidth of B/2

LPF filter Bandpass flat to B/2 Stopband start by 2.f0 - B/2 Leak-through from xB(t) may be considered too

Week 13 8April 12, 2005

Question 2 Part II (1 of 2)

Sample after the LPFNyquist sample rate is 2.B/2 or BWaveform preservation to bandwidth W>B

may be considered Sample at I.F.

Sample rate equations

0 04 12 ,

2 1 2s

f ff B k

k B

Week 13 9April 12, 2005


The L.O. at the sample times

Decimation opportunitiesOutput sample rate should be about BDecimation by 2 may be possible

0

2 1

0

1 2 1

4

exp 2 1

ss

ik i k

s

kT

f f

j f T i j j

Week 13 10April 12, 2005

Question 3

Antipodal pulses Amplitude is 1 volt, duration is T seconds Bandwidth is 1/T, noise PSD is N0

BER is

What is Eb/N0?

0

2 TBER Q

N

Week 13 11April 12, 2005

Question 4 Part I (1 of 2)

Mean number of bit errors per unit time is

BER for four errors per hour at 1 MB/s

ER BER R

96 1

sec1/ 15 min 60

min1.1 10

10 sec

ERBER

R

Week 13 12April 12, 2005

Question 4 Part I (2 of 2)

Using the base equation

The approximation given provides us with

0

2 bEBER Q

N

0 0

25.98, 17.76 12.5 dBb bE E

N N

Week 13 13April 12, 2005

Question 4 Part II

The SNR equation

Bandwidth of 1.2 MHz, bit rate 1 MBPS0

bERSNR

BW N

117.76 14.67 11.66 dB

1.2SNR

Week 13 14April 12, 2005

Question 5 Part I

A linear block code has a generator matrix

Code words are found by left-multiplying by all 16 combinations of bits

1 1 1 1 0 0 0

1 0 1 0 1 0 0

0 1 1 0 0 1 0

1 1 0 0 0 0 1

G

Week 13 15April 12, 2005

Code VectorsVector Message Code 0000 0 0 0 0 0 0 0 V1 0001 1 1 0 0 0 0 1 V2 0010 0 1 1 0 0 1 0

0011 1 0 1 0 0 1 1 V4 0100 1 0 1 0 1 0 0

0101 0 1 1 0 1 0 1 0110 1 1 0 0 1 1 0 0111 0 0 0 0 1 1 1

V8 1000 1 1 1 1 0 0 0 1001 0 0 1 1 0 0 1 1010 1 0 0 1 0 1 0 1011 0 1 0 1 0 1 1 1100 0 1 0 1 1 0 0 1101 1 0 0 1 1 0 1 1110 0 0 1 1 1 1 0 1111 1 1 1 1 1 1 1

Week 13 16April 12, 2005


The code is systemic because the last four columns are an identity matrix

The parity array portion of the generator matrix is the first three columns

1 1 1

1 0 1

0 1 1

1 1 0

P

Week 13 17April 12, 2005


The parity check matrix augments the matrix P along the other index with an identity matrix

3

1 0 0 1 1 0 1

| 0 1 0 1 0 1 1

0 0 1 1 1 1 0

TH I P

Week 13 18April 12, 2005

Question 5 Part III

The syndrome is S=r.HT

The received data vector r is {1,1,0,1,1,0,1}

The syndrome is {0,1,0} Received data has a bit error Corrected data vector is {1,0,0,1,1,0,1} Decoded message is {1,1,0,1}

Week 13 19April 12, 2005

Question 5 Part IV

The minimum Hamming distance between codes Hamming distance between codewords == Hamming

weight of their sum Hamming weight of a codeword == Hamming

distance from the all-zeros codeword Closed on subtraction means all differences are equal

to one of the codewords Thus, the smallest Hamming weight is the

minimum Hamming distance For our problem this is 3

Week 13 20April 12, 2005

Question 5 Part V

Error-detecting and correcting capability

For a dmin of 3Correct 1Detect 2

1mind corrected detected

Week 13 21April 12, 2005

Question 6 ( 1 of 3)

Calculate the probability of message error for a (24,12) linear block code using 12-bit data sequences. Assume that the code corrects up to two bit errors per block and that the base BER is 10-3.

Week 13 22April 12, 2005

Question 6 (2 of 3)

Probability of a message error is the probability of 3 or more bit errors out of 24

Binomial distribution

Incomplete beta function

| , 1

, 1

NN ss

B es a

p

NP n a p N p p

s

I a N a

11

0

, 1x

bax

a bI a b t t dt

a b

Week 13 23April 12, 2005

Question 6 (3 of 3)

Probability of 3 errors is 1.98192E-06 Probability of 3 or more errors is

1.99238E-06 Differences is about 0.5% Improvement in BER is a factor of 505

Week 13 24April 12, 2005

Backup Quiz Question 1 Part I

See main quiz Question 2 Equation for the output

LPF Bandpass flat to B/2+Δf Stopband start by 2.f0 - B/2-Δf

Leak-through from xB(t) may be considered too

0 exp 2x t s t j f t

Week 13 25April 12, 2005

Backup Quiz Question I Part IISystemView

0

0

10e+3

10e+3

20e+3

20e+3

-40

-60

-80

-100

-120

Mag in d

B

Frequency in Hz (dF = 762.9e-3 Hz)

Cx FFT: 20 Log | FFT| Output R + jOutput I

UNSAMPLED OUTPUT

Week 13 26April 12, 2005

From Inspection of Output

Use cursor readout in SystemView Spectrum is down 30 dB at about 3700 Hz Passband to 3700 Hz Stopband starts by 16.5 kHz to attenuate image

at 20 kHz Operations

LPF to specifications Sample at two times 3900 Hz

Week 13 27April 12, 2005

System

Week 13 28April 12, 2005

Critical Block Parameters

LPF Parameters: Operator: Linear Sys Butterworth Lowpass IIR 4 Poles Fc = 6e+3 Hz Quant Bits = None Init Cndtn = 0 DSP Mode Disabled Max Rate = 100e+3 Hz

Sampler Parameters: Operator: Sampler Interpolating Rate = 7.8e+3 Hz Aperture = 0 sec Aperture Jitter = 0 sec Max Rate = 7.8e+3 Hz

Week 13 29April 12, 2005

Spectrum of Sampled OutputSystemView

-2e+3

-2e+3

0

0

2e+3

2e+3

-30

-50

-70

-90

Mag in

dB


Cx FFT : 20 Log |FFT | Out R + jOut I Mixed Radix

Week 13 30April 12, 2005

Backup Quiz Question I Part III

Sample the 10,000 Hz signal. Use the lowest sample rate that preserves the signal.

See main quiz Question 2 Part II

0

0

1 10,000 10.8513

2 7,400 2

13,333 , 14

2 1 40,000 , 0s

fk

B

Hz kff

k Hz k

Week 13 31April 12, 2005

Input Signal at 10,000 HzSystemView

5e+3

5e+3

7.5e+3

7.5e+3

10e+3

10e+3

12.5e+3

12.5e+3

15e+3

15e+3

-30

-50

-70

-90

Mag in

dB


Cx FFT: 20 Log | FFT | Input R + jInput I

Week 13 32April 12, 2005

Sampled at 13,333 HzSystemView

-5e+3

-5e+3

-2.5e+3

-2.5e+3

0

0

2.5e+3

2.5e+3

5e+3

5e+3

-30

-50

-70

-90

-110

-130

-150

Mag in

dB


Cx FFT : 20 Log |FFT | Samp R + jSamp I Mixed Radix

WHOOPS

Week 13 33April 12, 2005

Sampled at 40,000 HzSystemView

-16e+3

-16e+3

-8e+3

-8e+3

0

0

8e+3

8e+3

16e+3

16e+3

-40

-60

-80

-100

-120

Mag in

dB


Cx FFT : 20 Log |FFT | Samp R + jSamp I Mixed Radix

Week 13 34April 12, 2005

Parameters

Sample at 40,000 Hz Frequency-shift down to baseband

10,000 Hz frequency shiftComplex LO (digital quadrature demodulator)

Downsample to 8,000 Hz complex

Week 13 35April 12, 2005

Output Before DownsamplingSystemView

-16e+3

-16e+3

-8e+3

-8e+3

0

0

8e+3

8e+3

16e+3

16e+3

-40

-60

-80

-100

-120

Mag in

dB


Cx FFT : 20 Log |FFT | Mixed R + jMixed I

Week 13 36April 12, 2005

Output Downsampled to 8 kHz Complex Without Filtering

SystemView

-2e+3

-2e+3

0

0

2e+3

2e+3

-30

-40

-50

-60

-70

-80

-90

Mag in

dB


Cx FFT: 20 Log | FFT | Out R + jOut I Mixed Radix

Week 13 37April 12, 2005

Final System Block Diagram

Week 13 38April 12, 2005

Backup Quiz 2 Question 2

See main Quiz 2 Question 5 The received data vector r is

{1,0,1,0,1,1,1} The syndrome is {1,0,1} Received data has a bit error Corrected data vector is {1,0,1,0,0,1,1} Decoded message is {0,0,1,1} Other answers same as those of main quiz

EE521 Analog and Digital Communications

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