Upcoming SlideShare
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Standard text messaging rates apply

# Multidimensional signals in baseband digital transmission

878

Published on

Published in: Education
1 Like
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

Views
Total Views
878
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
21
0
Likes
1
Embeds 0
No embeds

No notes for slide

### Transcript

• 1. Baseband Digital Transmission Multidimensional Signals
• 2. 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
• 3. 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
• 4. 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
• 5. 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
• 6. 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
• 7. 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
• 8. 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 
• 9. 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
• 10. 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
• 11. 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
• 12. 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 )
• 13. 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
• 14. 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
• 15. 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
• 16. 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
• 17. 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
• 18. 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
• 19. Homework Illustrative problem  5.10, 5.11 Problems  5.11, 5.13