Communication Systems Engineering Episode 1 Part 2 ppsx

... a 1 , a 2 , b 1 , and b 2 , whose output aneroid capsule [ ENG ] A thin, disk-shaped box or capsule, usually metallic, partially evacuatedis 0 only if either a 1 and a 2 or b 1 and b 2 are 1. Abbreviated ... vere bl} 10 , the length of the longer side is 2 Ϫ(2N Ϫ 1) /4 artificial voice [ ENG ACOUS ] 1. Small loud- meters, while the length of the shorter side is sp...

Ngày tải lên: 21/07/2014, 15:20

... assembly 11 //SYS 21 / //INTEGRAS/B&H/PRS/FINALS_0 7-0 5-0 3/0750654376-CH0 01. 3D – 14 – [1 18 /18 ] 8.5 .20 03 9 : 12 PM Fig. 1. 15 General classification of joining processes. 14 A strategic view //SYS 21 / //INTEGRAS/B&H/PRS/FINALS_0 7-0 5-0 3/0750654376-CH0 01. 3D ... //SYS 21 / //INTEGRAS/B&H/PRS/FINALS_0 7-0 5-0 3/0750654376-CH0 01. 3D – 12 – [1 18 /18 ]...

Ngày tải lên: 21/07/2014, 16:21

... www.english-test.net Test 22 TOEIC Vocabulary / Word by Meaning / Test # 22 Q1 adj. prepared; set; eager; willing; quick; fast (a) ample (b) comfortable (c) ready (d) few Q2 adj. present; opposite ... www.english-test.net Test 42 TOEIC Vocabulary / Word by Meaning / Test # 42 Q1 n. tip; bonus; recompense (a) blue sky laws (b) sharp (c) gratuity (d) improvement Q2 v. to carry; to...

Ngày tải lên: 21/07/2014, 22:20

... 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

... 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 (1/ 2) = 1 Not surprising ... Rules for entropy 1. Chain rule: H(X 1 , , X n ) = H(X 1 ) + H(X 2 |X 1 ) + H(X 3 |X 2 ,X 1 ) + …+ H(X n |X n -1 …X 1 ) 2. H(X,Y) = H(X) + H(Y|X) = H(Y) + H(X|...

Ngày tải lên: 07/08/2014, 12:21

... 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 / e − π − e π π f ... F 1 [ X δ ( f )T s Π( )] 2W ∞ t () = ∑ xnT s )Sinc( − n)xt ( n =−∞ T s Eytan Modiano Slide 11 Properties of the Fourier transform ã Linearity x1(t) <=>...

Ngày tải lên: 07/08/2014, 12:21

... E.g., N =4, D = 0 .11 75, H(x) = 1. 911 – Recall: uniform quantizer, D= 0 .11 88, H(x) = 1. 9 04 (slight improvement) 0.9 816 1. 51 0 .45 28 -0.9 816 -0 .45 28 Eytan Modiano Slide 12 - 1. 51 EX ... 3 fx Uniform quantizer example ã N =4, X~N(0 ,1) x () = 2 πσ 1 e − x 2 / 2 σ 2 , σ 2 = 1 ã From table 6.2, =0.9957, D=0 .11 88, H(Q)= 1. 9 04 Notice th...

Ngày tải lên: 07/08/2014, 12:21

... 0. 25 a 3 0. 25 a 4 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 11 a 2 10 a 3 01 a 4 0 01 a 5 000 n bits symbol HX p p Shannon Fanocodes ... holds!) n 1 n 2 n 3 n 4 n 5 Eytan Modiano Slide 11 Huffman code example A = {a 1 ,a 2 ,a 3 , a 4 , a 5 } and p = {0.3, 0. 25, 0. 25,...

Ngày tải lên: 07/08/2014, 12:21

... coding achieves 1 bit difference between adjacent levels ã 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 ... PAM Eg x y ,4 − QAM ⇒ A m , A m ∈ { + / − 1 } x y 16 − QAM ⇒ A m , A m ∈ { + / − 1, + / − 3 } Eytan Modiano Slide 16 -3 -1 1 3 -3 -1 1 3 Bandpass signals...

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 ã After matched filtering we receive r = S m + n S m {S 1 , S M } ... Notes on Q(x) Q(0) = 1/ 2 – Q(-x) = 1- Q(x) – Q( ∞ ) = 0, Q(- ∞ ) =1 – If X is N(m, σ 2 ) Then P(X>x) = Q((x-m)/ ) ã Example: Pe = P[r<0|S1 was sent) f | (| s1) ~ N( E b...

Ngày tải lên: 07/08/2014, 12:21

... 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 (P R1 /P N1 ) Received ... (E b /N 0 ) d = 10 dB ã PAM modulation P b = Q( E N2 0 b / ) ã Repeater: Received (E b /N 0 ) u/d = 1/ 2 (E b /N 0 ) u = 10 dB - 3dB = 7dB – => Pb = 5x10 -4 from...

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 ... = 6, k = 3, R = 1/ 2 000 →→ →→ 000000 10 0 →→ →→ 10...

Ngày tải lên: 07/08/2014, 12:21

... 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...

Ngày tải lên: 07/08/2014, 12:21

Ngày tải lên: 07/08/2014, 12:21

Ngày tải lên: 07/08/2014, 12:21