Communication Systems Engineering Episode 2 Part 5 ppsx

... 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

... 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

... 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

... Back 7 ARQ SN 0 3 4 5 t 1 6 RN 0 1 2 3 5 Window (0,6) (1, 7) (5 ,11 ) (2, 8) (3,9) Node A Node B 2 0 5 Packets delivered 0 1 2 3 4 5 ã Note that packet RN -1 must be accepted at B before ... developed for X .25 networks ã They all use bit oriented framing with flag = 011 111 10 ã They all use a 16 -bit CRC for error detection ã They all use Go Back N ARQ wi...

Ngày tải lên: 07/08/2014, 12:21

... that we can tolerate Circuits P B 20 1% 15 8% 7 30% Eytan Modiano Slide 22 Delay formulas ã M/G/1 ã M/M/1 ã M/D/1 DX=+ à à / DX=+ à à / ( )2 DX X =+ à 2 21(/) Delay components: Service (transmission) ... arrivals in T It can be shown that, E[n] = T E[n ] = T + ( T) = E[(n-E[n]) ] = E[n ]-E[n] = T 22 22 22 λ λλ σλ Eytan Modiano Slide 19 Example (fast food restaurant) ã Custo...

Ngày tải lên: 07/08/2014, 12:21

... group transmits (1 ,2, 3, 4) (1 ,2, 3) 4 success collision (2, 3) collision idle collision (2, 3) success success Notice that after the idle slot, collision between (2, 3) was sure to happen ... the first success is 2, so the expected number of slots to transmit 2 packets is 3 slots Throughput over the 3 slots = 2/ 3 – In practice above algorithm cannot real...

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

... 1 2 3 4 N = {1 ,2, 3,4} A = {(1 ,2) , (2, 1),(1,4), (4 ,2) , (4,3),(3 ,2) } ã Directed walk: (4 ,2, 1,4,3 ,2) ã Directed path: (4 ,2, 1) ã Directed cycle: (4 ,2, 1,4) ã Data networks are best represented ... n2, ,nk) in which each adjacent node pair is an arc. ã A path is a walk with no repeated nodes. 1 2 4 3 1 2 4 3 Walk (1 ,2, 3,4 ,2) Path (1 ,2, 3,4) Eytan Modian...

Ngày tải lên: 07/08/2014, 12:21

... CONGESTION CONTROL BER EFFICIENCY 0 0.1 0 .2 0.3 0.4 0.5 0 .6 0.7 0.8 0.9 1 1.00E-07 1.00E- 06 1.00E-05 1.00E-04 1.00E-03 1,544 64 KBPS 16 KBPS 2. 4 KBPS KBPS ã TCP assumes dropped packets ... addresses) – Allocate a block of contiguous addresses E.g., 1 92. 4. 16. 1 - 1 92. 4. 32. 155 Bundles 16 class C addresses The first 20 bits of the address field are the...

Ngày tải lên: 07/08/2014, 12:21

Ngày tải lên: 07/08/2014, 12:21

Ngày tải lên: 07/08/2014, 12:21

Ngày tải lên: 07/08/2014, 12:21

Ngày tải lên: 07/08/2014, 12:21

Ngày tải lên: 07/08/2014, 12:21

Ngày tải lên: 07/08/2014, 12:21