Communication Systems Engineering Episode 1 Part 4 docx

... kalиənиdər} ture of 1 gram of water from 14 .5Њ to 15 .5ЊCat calibrating tank [ ENG ] A tank having known ca- a constant pressure of 1 standard atmosphere; pacity used to check the volumetric accuracy of equal ... escape of air from the ər} working chamber of a pneumatic caisson. board-foot [ ENG ] Unit of volume in measuring { blo ¯ au ˙ t} lumber; equals 14 4 cubic...

Ngày tải lên: 21/07/2014, 15:20

... and displayed on the ANSYS Graphics” window. Ch02-H6875.tex 24 /11 /2006 17 : 2 page 49 2 .4 Solution stage 49 A Figure 2 . 14 Type of analysis definition. Select the type of analysis that is appropriate ... beam on the ANSYS Graphics window as shown in Figure 3.7. Ch02-H6875.tex 24 /11 /2006 17 : 2 page 46 46 Chapter 2 Overview of ANSYS structure and visual capabilities...

Ngày tải lên: 06/08/2014, 11:21

... 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

... 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

... seconds 0 2 4 6 8 10 12 14 16 18 20 0 0 .2 0 .4 0.6 0.8 ALOHA SCHEMES TDMA (10 USERS) Perfect Scheduling (M/M/1) Reservation with 20 % overhead packet length 24 00 bits transmission ... for mixed voice and data Slot 1 2 3 4 5 6 frame 1 frame 2 frame 3 frame 4 frame 5 15 3 20 2 15 7 3 9 7 9 7 9 18 7 3 15 9 6 18 idle idle idle idle 2...

Ngày tải lên: 07/08/2014, 12:21

Ngày tải lên: 07/08/2014, 12:21

... occur? C) Find the generator matrix for the (7 ,3) block code based on the above generator. (note that a (7 ,3) code has 3 information bits and 4 check bits). D) Using the generator matrix from ... MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Aeronautics and Astronautics 16 .36 : Comm. Sys. Engineering Date Issued: April 17 Problem Set No. 8 Date Due: April ... code? Probl...

Ngày tải lên: 07/08/2014, 12:21