1 / 27

DATA COMMUNICATION 2-dimensional transmission

DATA COMMUNICATION 2-dimensional transmission. A.J. Han Vinck May 1, 2003. we describe orthogonal signaling 2-dimensional transmission model. Content. „orthogonal“ binary signaling. 2 signals S 1 (t) S 2 (t) in time T Example: Property : orthogonal energy E. T. T.

joy
Download Presentation

DATA COMMUNICATION 2-dimensional transmission

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. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. DATA COMMUNICATION2-dimensional transmission A.J. Han Vinck May 1, 2003

  2. we describe orthogonal signaling 2-dimensional transmission model Content

  3. „orthogonal“ binary signaling 2 signals S1 (t) S2 (t) in time T Example: Property: orthogonal energy E T T

  4. Quadrature Amplitude Modulation: QAM 1 0 0 S(t) 1 1 0

  5. QAM receiver 1/0 +/- r(t) 1/0 +/- r(t) = S(t) + n(t) Note: sin(x)sin(x) = ½ (1 – cos (2x) ) sin(x)cos(x) = ½ sin (2x)

  6. about the noise

  7. about the noise Conclusion: n1 and n2 are Gaussian Random Variables zero mean uncorrelated (and thus statistically independent (f(x,y) =f(x)f(y) ) with variance 2.

  8. Geometric presentation (1)

  9. Geometric presentation (2) 11 10 00 01 ML receiver:find maximum p(r|s) min p(n) decision regions

  10. performance From Chapter 1: P(error) =

  11. extension 4-QAM  2 bits 16-QAM  4 bits/s Channel 2 Channel 1

  12. Geometric presentation (2) 1 equal density transmitted 2 noise vector n received The noise vector n has length |n| = ( 12+22) ½ n has a spherically symmetric distribution!

  13. Geometric presentation (1) Prob (error) = Prob(length noise vector > d/2) d/2 r r‘

  14. Error probability for coded transmission The error probabiltiy is similar to the 1-dimensional situation: We have to determine the minimum d2Euclidean between any two codewords Example: C d2Euclidean = C‘

  15. Error probability The two-code word error probability is then given by:

  16. modulation schemes On-off FSK 8-PSK  3 bits/s 1 bit/symbol 1 bit/symbol 4-QAM  2 bits 16-QAM  4 bits/s

  17. transmitted symbol energy energy: per information bit must be the same FSK

  18. performance d/2 From Chapter 1: P(error) = FSK

  19. Coding with same symbol speed In k symbol transmissions, we transmit k information bits. We use a rate ½ code In k symbol transmissions, we transmit k bits ML receiver:

  20. Famous Ungerböck coding In k symbol transmissions transmit We can transmit 2k information bits and k redundant digits In k symbol transmissions transmit 2k digits Hence, we can use a code with rate 2/3 with the same energy per info bit!

  21. modulator info ci 23 encoder Signal mapper ci{000,001,010,...111}

  22. example transmit 00 00 00 10 10 01 Parity even Parity odd 11 11 11 or 00 01 00 01 10 11 10 11 Decoder: 1) first detect whether the parity is odd or even 2) do ML decoding given the parity from 1) Homework: estimate the coding gain

  23. Example: Frequency Shift Keying-FSK Transmit: s(1):= s(0):= Note: FSK

  24. Modulator/demodulator m modulator S(t) m r(t) Select largest demodulator m

  25. Ex: Binary Phase Shift Keying-BPSK Transmit: s(1):= s(0):= m m‘ > or < 0?  

  26. On-off BFSK BPSK Modulation formats

  27. PERFORMANCE 10-1 10-2 10-3 10-4 10-5 10-6 10-7 Error rate On-off BPSKQPSK 5 10 15 Eb/N0 dB

More Related