Multidimensional signals in baseband digital transmission
Upcoming SlideShare
Loading in...5
×
 

Multidimensional signals in baseband digital transmission

on

  • 1,212 views

 

Statistics

Views

Total Views
1,212
Views on SlideShare
1,212
Embed Views
0

Actions

Likes
1
Downloads
18
Comments
0

0 Embeds 0

No embeds

Accessibility

Categories

Upload Details

Uploaded via as Microsoft PowerPoint

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

Multidimensional signals in baseband digital transmission Multidimensional signals in baseband digital transmission Presentation Transcript

  • Baseband Digital Transmission Multidimensional Signals
  • Multidimensional Vs. Multiamplitude Multiamplitude signal is One-dimensional 00 01 10 11 -3d -d d 3d  With One basis signal: g(t) Multidimensional signal  Can be defined using multidimensional orthogonal signals
  • Multidimensional Orthogonal Signals Many ways to construct In this text, Construction of a set of M=2k waveforms si(t), i=1,2,…,M-1  Mutual Orthogonality  Equal Energy  Using Mathematics T ∫ s (t ) s (t )dt = ℑδ , i, k = 0,1,..., M − 1 0 i k ik 1, i = k  where Kronecker delta is δ ik =  0, i ≠ k
  • Example of M=2 2 Multiamplitude Vs. Multidimensional s0 s1 s2 s3 3d s3 A d s2 -d s1 T -3d s0 T All signal has identical energy  T A2T ℑ= ∫ si2 (t )dt = , i, k = 0,1,..., M − 1 0 M See figure 5.27, page 202, for signal constellation
  • Receiver for AWGN Channel Received signal from AWGN channel  r (t ) = si (t ) + n(t ), m = 0,1,..., M − 1, 0 ≤ t ≤ T White Gaussian process With power spectrum N0/2 Receiver decides which of M signal waveform was transmitted by observing r(t)  Optimum receiver minimize Probability of Error
  • Optimum Receiver for AWGN Channel Needs M Correlators (or Matched Filters) s 0 (t) Samples at t=T r(t) t ∫ ( ) dτ 0 r0 s M-1 (t) Detector Output Decision t ∫ ( ) dτ 0 r M-1 T ri = ∫ r (t ) si (t )dt , i = 0,1,..., M − 1 0
  • Signal Correlators Let s0(t) is transmitted T T  r0 = ∫ r (t ) s0 (t )dt = ∫ {s0 (t ) + n(t )}s0 (t )dt 0 0 T T = ∫ s (t )dt + ∫ s0 (t )n(t )dt = ℑ + n0 2 0 0 0 T T  ri = ∫ r (t ) s0 (t )dt = ∫ {s0 (t ) + n(t )}si (t )dt 0 0 T T = ∫ s0 (t ) si (t )dt + ∫ si (t )n(t )dt = ni , i ≠ 0 0 0 Orthogonal Noise component (Gaussian Process)  Zero mean and Variance N TℑN 2 ∫ σ = E (n ) = 2 s (t )dt = 2 i 0 2 i 0 0 2
  • pdf of correlator output Let s0(t) is transmitted P( r0 | s0 (t )) s 0 (t) Samples at t=T 1 − ( r0 −ℑ )2 / 2σ 2 = e 2πσs 0 (t)+n(t) t ∫ ( ) dτ 0 r0 ℑ s M-1 (t) t ∫ ( ) dτ 0 r M-1 0 P(ri | s0 (t ))  Mean of correlator output 1 2 / 2σ 2 E[r0] = ℑ, E[ri] = 0 = e − ri 2πσ 
  • Optimum Detector Let s0(t) is transmitted  The probability of correct decision  Is probability that r0 > ri  Pc = P(r0 > r1 , r0 > r2 ,..., r0 > rM −1 )  The average probability of symbol error  PM = 1 − Pc = 1 − P(r0 > r1 , r0 > r2 ,..., r0 > rM −1 ) 1 ∞ ∫ 2 M −1 − ( y − 2 ℑ / N 0 ) / 2 = {1 − [1 − Q( y )] e dy} 2π −∞  For M=2 case (Binary Orthogonal signal) ℑb  P2 = Q ( ), ℑb = ℑ for binary N0
  • More on Probability error Average probability of symbol error  Same even if si(t) is changed  Numerically evaluation of integral Converting probability of symbol error to probability of a binary digit error  2k −1 Pb = k PM 2 −1 Symbol error probability Bit error probability  Figure 5.29 page 206  As M64, we need small SNR to get a given probability of error
  • Biorthogonal Signals Biorthogonal  For M=2k multidimensional signal  M/2 signals are orthogonal  M/2 signals are negative of above signals  M signals having M/2 dimensions Example of M = 22 s 0 (t) s 1 (t) s 3 (t)= -s 0 (t) s 4 (t)= -s 1 (t) T/2 T/2 TA A -A -A T/2 T/2 T Symbol interval T=kT b
  • Biorthogonal Signals Signal Constellation  s0 (t ) = ( ℑ , 0, 0,..., 0) sM / 2 = (− ℑ, 0, 0,..., 0) s1 (t ) = (0, ℑ , 0,..., 0) sM / 2 +1 = (0, − ℑ , 0,..., 0) L sM / 2 −1 (t ) = (0, 0, 0,..., ℑ ) sM −1 = (0, 0, 0,..., − ℑ ) Examples  M=2 : Antipodal signal M=4 (0, ℑ ) (− ℑ , 0) ( ℑ , 0) (− ℑ , 0) ( ℑ , 0) (0, − ℑ )
  • Receiver for AWGN Channel Received signal from AWGN channel  r (t ) = si (t ) + n(t ), m = 0,1,..., M − 1, 0 ≤ t ≤ T White Gaussian process With power spectrum N0/2 Receiver decides which of M signal waveform was transmitted by observing r(t)  Optimum receiver minimize Probability of Error
  • Optimum Receiver for AWGN Channel Needs M/2 correlator (or matched filter) s 0 (t) Samples at t=T r(t) t ∫ ( ) dτ 0 r0 s M/2-1 (t) Detector Output Decision t ∫ ( ) dτ 0 r M/2-1 T M ri = ∫ r (t ) si (t )dt , i = 0,1,..., −1 0 2
  • Signal Correlators Let s0(t) is transmitted (i = 0,1, … ,M/2 – 1) T T  r0 = ∫ r (t ) s0 (t )dt = ∫ {s0 (t ) + n(t )}s0 (t )dt 0 0 T T = ∫ s (t )dt + ∫ s0 (t )n(t )dt = ℑ + n0 2 0 0 0 T T  ri = ∫ r (t ) s0 (t )dt = ∫ {s0 (t ) + n(t )}si (t )dt 0 0 T T = ∫ s0 (t ) si (t )dt + ∫ si (t )n(t )dt = ni , i ≠ 0 0 0 Orthogonal Noise component (Gaussian Process)  Zero mean and Variance N TℑN 2 ∫ σ = E (n ) = 2 s (t )dt = 2 i 0 2 i 0 0 2
  • pdf of correlator output  Let s0(t) is transmitted (for M/2 –1 correlator) P(r0 | s0 (t )) s 0 (t) Samples at t=T 1 − ( r0 −ℑ )2 / 2σ 2 = e 2πσs 0 (t)+n(t) t ∫ ( ) dτ 0 r0 ℑ s M/2-1 (t) t ∫ ( ) dτ 0 r M/2-1 0  Mean of correlator output P(ri | s0 (t ))  E[r0] = ℑ, E[ri] = 0 1 2 / 2σ 2 = e − ri 2πσ
  • The detector Select correlator output whose magnitude |ri| is largest M  rj = max{ ri }, i = 0,1,..., −1 i 2 But, we don’t distinguish si(t) and –si(t)  By observing |ri |  We need sign information  si(t) if ri > 0  - si(t) if ri < 0
  • Probability of Error Probability of correct decision ∞ 1 r0 / ℑN 0 / 2 1  Pc = ∫ [ ∫ e − x2 / 2 M −1 dx ] e − ( r0 −ℑ )2 / 2σ 2 dr0 0 2π − r0 / ℑN 0 / 2 2πσ Probability of symbol error  PM = 1 − Pc See figure 5.33 page 213  Note M=2 and M=4 case  Symbol error probability and Bit-error probability
  • Homework Illustrative problem  5.10, 5.11 Problems  5.11, 5.13