1. April 12, 2005Week 13 1 EE521 Analog and Digital Communications James K. Beard, Ph. D. jkbeard@temple.edu Tuesday, March 29, 2005 http://astro.temple.edu/~jkbeard/

2. Week 132 April 12, 2005 Attendance

3. Week 133 April 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

4. Week 134 April 12, 2005 Today’s Topics Term Project Quiz  Main Quiz  Backup Quiz Term Project Discussion (as time permits)

5. Week 135 April 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

6. Week 136 April 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

7. Week 137 April 12, 2005 Question 2 Part I This is a quadrature demodulator The output x 0 (t)  Frequency shifted to baseband  Negative frequency image Shifted to -2. f 0 Attenuated by LPF  Has a bandwidth of B/2 LPF filter  Bandpass flat to B/2  Stopband start by 2. f 0 - B/2  Leak-through from x B (t) may be considered too

8. Week 138 April 12, 2005 Question 2 Part II (1 of 2) Sample after the LPF  Nyquist sample rate is 2. B/2 or B  Waveform preservation to bandwidth W>B may be considered Sample at I.F.  Sample rate equations

9. Week 139 April 12, 2005 Question 2 Part II (2 of 2) The L.O. at the sample times Decimation opportunities  Output sample rate should be about B  Decimation by 2 may be possible

10. Week 1310 April 12, 2005 Question 3 Antipodal pulses Amplitude is 1 volt, duration is T seconds Bandwidth is 1/T, noise PSD is N 0 BER is What is E b /N 0 ?

11. Week 1311 April 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

12. Week 1312 April 12, 2005 Question 4 Part I (2 of 2) Using the base equation The approximation given provides us with

13. Week 1313 April 12, 2005 Question 4 Part II The SNR equation Bandwidth of 1.2 MHz, bit rate 1 MBPS

14. Week 1314 April 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

15. Week 1315 April 12, 2005 Code Vectors

16. Week 1316 April 12, 2005 Question 5 Part II (1 of 2) 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

17. Week 1317 April 12, 2005 Question 5 Part II (2 of 2) The parity check matrix augments the matrix P along the other index with an identity matrix

18. Week 1318 April 12, 2005 Question 5 Part III The syndrome is S=r. H T 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}

19. Week 1319 April 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

20. Week 1320 April 12, 2005 Question 5 Part V Error-detecting and correcting capability For a d min of 3  Correct 1  Detect 2

21. Week 1321 April 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.

22. Week 1322 April 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

23. Week 1323 April 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

24. Week 1324 April 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. f 0 - B/2-Δf  Leak-through from x B (t) may be considered too

25. Week 1325 April 12, 2005 Backup Quiz Question I Part II UNSAMPLED OUTPUT

26. Week 1326 April 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

27. Week 1327 April 12, 2005 System

28. Week 1328 April 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

29. Week 1329 April 12, 2005 Spectrum of Sampled Output

30. Week 1330 April 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

31. Week 1331 April 12, 2005 Input Signal at 10,000 Hz

32. Week 1332 April 12, 2005 Sampled at 13,333 Hz WHOOPS

33. Week 1333 April 12, 2005 Sampled at 40,000 Hz

34. Week 1334 April 12, 2005 Parameters Sample at 40,000 Hz Frequency-shift down to baseband  10,000 Hz frequency shift  Complex LO (digital quadrature demodulator) Downsample to 8,000 Hz complex

35. Week 1335 April 12, 2005 Output Before Downsampling

36. Week 1336 April 12, 2005 Output Downsampled to 8 kHz Complex Without Filtering

37. Week 1337 April 12, 2005 Final System Block Diagram

38. Week 1338 April 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