Communication Systems Engineering Episode 2 Part 4 pps
Communication Systems Engineering Episode 1 Part 1 pps
... 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
Communication Systems Engineering Episode 1 Part 4 docx
... 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
Communication Systems Engineering Episode 1 Part 7 pps
... 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
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...
Ngày tải lên: 07/08/2014, 12:21
Communication Systems Engineering Episode 2 Part 1 pptx
... 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
Communication Systems Engineering Episode 2 Part 2 pptx
... 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
Communication Systems Engineering Episode 2 Part 3 pps
... 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
Communication Systems Engineering Episode 2 Part 4 pps
... 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
Communication Systems Engineering Episode 2 Part 5 ppsx
... 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
Communication Systems Engineering Episode 2 Part 6 pps
... 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
