Communication Systems Engineering Episode 1 Part 10 ppsx

... 10 01 Send T = 11 010 1 011 11 010 1 011 Receive T’ = 11 010 1 011 (no errors) No way of knowing how many errors occurred or which bits are In error 10 01 010 00 10 01 00 011 01 10 01 010 01 10 01 ... c 1 = x 1 + x 2 + x 3 c 2 = x 2 + x 3 + x 4 c 3 = x 1 + x 2 + x 4 Example r = 3, G = 10 01 M = 11 010 1 => M2 r = 11 010 100 0 1...

Ngày tải lên: 07/08/2014, 12:21

... M M 1 1 1 ∑ P i Log( ) = ∑ P i Log( ) + ∑ P i Log( ) ≤ 0 , equality when P i = i =1 MP i i =1 P i i =1 M M M M 1 That is , ∑ P i Log ( ) ≤ ∑ P i Log(M) = Log(M) i =1 P i i =1 Eytan ... experiment – X = x 1 with probability p – X = x 2 with probability (1- p) – H(X) = pLog 2 (1/ p) + (1- p)Log 2 (1/ (1- p)) = H b (p) – H(X) is maximized with p =1/ 2, H b...

Ngày tải lên: 07/08/2014, 12:21

... Lecture 4-Feb L1 6-Feb L2 11 -Feb L3 13 -Feb L4 18 -Feb 20-Feb L5 25-Feb L6 27-Feb L7 4-Mar L8 6-Mar L9 11 -Mar L10 13 -Mar L 11 18 -Mar L12 20-Mar L13 25-Mar 27-Mar 1- Apr L14 Eytan Modiano ... 8-Apr L16 10 -Apr L17 15 -Apr L18 17 -Apr L19 22-Apr 24-Apr L20 29-Apr L 21 1- May L22 6-May L23 8-May L24 13 -May L25 15 -May L26 5 /19 - 5/23 Topic Reading Packet...

Ngày tải lên: 07/08/2014, 12:21

... the Fourier transform of the original separated by 1/ Ts intervals -1/ Ts -w w 1/ Ts 2/Ts Eytan Modiano Slide 10 Proof, continued • If 1/ Ts > 2W then the replicas of X(f) will not overlap ... Slide 6 jt jf Rectangle pulse t  1 ||< 1/ 2  t / / Π() =  12 | t |= 12   0 otherwise 12 Π ∞ 2 π / F[(t)] = ∫ −∞ Π(t)e − jft dt = ∫ 12 e − j 2 π ft dt /...

Ngày tải lên: 07/08/2014, 12:21

... N=4, – D = 0 .11 75, H(x) = 1. 911 – Recall: uniform quantizer, D= 0 .11 88, H(x) = 1. 904 (slight improvement) 0.9 816 1. 51 0.4528 -0.9 816 -0.4528 Eytan Modiano Slide 12 - 1. 51 ∆∆ ∆ EX Example ... first (given region (a i -1 , a i )) – Minimize distortion Eytan Modiano Slide 9 ∈∈ ∈ ×× Pulse code modulation voice Quantizer µ -law Uniform Q Sampler encoder 011...

Ngày tải lên: 07/08/2014, 12:21

... ≤ ⇒≤≤ ⇒ ⇒ =       <+ − − == ∑∑ log( ) log( ) log( ) log( ) 11 2 21 11 1 11 n i =       <+ =< +       =+ ≤< + == ∑∑ log( ) log( ) , , log( ) ( ) . , () () 11 1 1 11 1 11 pp Now npnp p HX Hence HX n HX ii ii i k i i i k Eytan ... 0 .1 a 5 0 .1 0.3 0.25 0.25 0.2 0.3 0.25 0.45 + + + 0.55 0.45 + 1. 0 1 0 0 1 0 1 0 1 Letter Codeword a 1...

Ngày tải lên: 07/08/2014, 12:21

... coding achieves 1 bit difference between adjacent levels • Example M= 8 (can be generalized) A 1 000 A 2 0 01 A 3 011 A 4 010 A 5 11 0 A 6 11 1 A 7 10 1 Eytan Modiano A 8 10 0 Slide ... Two-dimensional signals • S i = (S i1 , S i2 ) • Set of signal points is called a constellation S 1 = (1, 1) S 2 =( -1, -1) S 3 = (1, -1) S 4 =( -1, 1) • 2-D constel...

Ngày tải lên: 07/08/2014, 12:21

... points 10 -1 P e 10 -5 12 14 antipodal orthogonal 3dB E b /N 0 (dB) Eytan Modiano Slide 25 ∈∈ ∈ Signal Detection • After matched filtering we receive r = S m + n – S m ∈ {S 1 , ... continued g(t) 0 => S 1 = g(t) 1 => S 2 = -g(t) A Eytan Modiano S(t) Y(t) T 2T 3T “S 1 (t)” T “S 2 (t)” T T “Y 1 (t)” “Y 2 (t)” 2T 2T Slide 14 Detection...

Ngày tải lên: 07/08/2014, 12:21

... received (E b /N 0 ) d = 10 dB • PAM modulation P b = Q( E N2 0 b / ) • Repeater: Received (E b /N 0 ) u/d = 1/ 2 (E b /N 0 ) u = 10 dB - 3dB = 7dB – => Pb = 5x10 -4 from table 7.55 or ... attenuation P R1 = P T /L, P T2 = P R1 A, P R2 = P T2 /L, … P N1 = P N , P N2 = P N1 A/L + P N , … Let A = L => P RK = P T /L, P NK = KP N P RK /P NK = P T /LKP N = 1/ K...

Ngày tải lên: 07/08/2014, 12:21

... 0000 010 1 10 10 11 11 Syndrome 10 00 11 01 0 010 011 1 10 010 0 00 01 111 0 10 11 01 110 0 10 01 011 0 0 011 11 Suppose 011 1 is received, S = 10 , co-set leader = 10 00 Decode: C = 011 1 + 10 00 = 11 11 G HH T =       =       =             10 10 010 1 10 10 010 1 10 01 10 01 Eytan ... code, n = 6, k = 3, R = 1/ 2...

Ngày tải lên: 07/08/2014, 12:21