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

... 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 3 3 6 Eytan Modiano Slide 17 ... 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...

Ngày tải lên: 07/08/2014, 12:21

... and 1fails (I.e., d31=infinity) – Node 3 will update D3 = d 32 + D2 = 3 – In the next step node 2 will update: D2 = d 23+ D3 = 4 – It will take nearly 100 iterations before node 2 converges on the ... Slide 26 Graphs • A graph G = (N,A) is a finite nonempty set of nodes and a set of node pairs A called arcs (or links or edges) 1 2 3 1 2 3 4 N = {1 ,2, 3} N = {1...

Ngày tải lên: 07/08/2014, 12:21

... 4.2BSD - 19 83 First widely available release – 4.3BSD Tahoe - 1988 Slow start and congestion avoidance – 4.3BSD Reno - 1990 Header compression – 4.4BSD - 19 93 Multicast support, RFC 1 32 3 ... 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 same and are esse...

Ngày tải lên: 07/08/2014, 12:21

... function. { impəls rispa ¨ ns } has the standard value of 9.80665 m/s 2 or approx- imately 32 . 1 739 8 ft/s 2 equal to 33 86 .38 864 034 1 impulse sealing [ ENG ] Heat-sealing of plastic materials by applying ... absolute temperature equal ter above 100ЊF (38 ЊC) and noting the time it to 1 /27 3. 16 of the absolute temperature of the takes to cool from 100 to 95ЊF (38 to 35 ЊC) or...

Ngày tải lên: 21/07/2014, 15:20

... Given) ISO BS UNS DIN CuNi30MnlFe CN 107 C71500 CuNi30Fe 2. 08 82 Copper Minimum Rem. Rem. Rem. Rem. Maximum Nickel Minimum 29 .0 30 .0 29 .0 30 .0 Maximum 32 . 0 32 . 0 33 .0 32 . 0 Iron Minimum 0.4 0.4 0.4 ... C38500), and Tin Brasses (C 420 00, C4 430 0, C44500, C46400) in Different Chemical Environments Environment/alloy 11000 122 00 22 000 23 000 26 000 28 000 36 000 38 500...

Ngày tải lên: 05/08/2014, 09:20

... L19 22 -Apr 24 -Apr L20 29 -Apr L21 1-May L 22 6-May L 23 8-May L24 13- May L25 15-May L26 5/19 - 5 / 23 Topic Reading Packet communications, DLC, error checking using CRC Tanenbaum 3 ARQ ... Slide 25 Course Syllabus Date 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 L11 18-Mar L...

Ngày tải lên: 07/08/2014, 12:21

... ||< 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 = = Sin() = Sinc f − jf π f 2 π () ... − j 2 π f o t xt() cos( 2 π f o t) = () 2 π () 2 ( x tHence,( ) cos( 2 π f o t) ⇔ Xf − f o ) + X( f + f o ) 2 • Example: x(t)= sinc(t), F[sinc(t)] = Π (f)...

Ngày tải lên: 07/08/2014, 12:21

... for M-PAM M 2 − 1 M 2 − 1 E av = 3 E g => E bav = 3Log 2 () E g M E = 3Log 2 () E bav M g M 2 − 1  Log M M E bav − 2 2 0 6 1 () ( N)  Pe P e = 2Q   , P eb ... ∫ (( τ )) 2 d τ ∫ hT − ττ T  ∫ τ s 2 ( ) d s SNR =  0 T  ≤ 0 T 0 = 2 ∫ (( τ )) 2 d τ = 2 E s N 0 ∫ hT − t)dt N 0 ∫ hT − t ) dt N 0 0 N 0 2 ( 2...

Ngày tải lên: 07/08/2014, 12:21

... check is a modulo 2 sum of some of the data bits Example: 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 = 1001 M = 110101 => M2 r = 110101000 ... of 1 ,2, or 3 errors (d > 3) 2) All bursts of errors of r or fewer bits 3) Random large numbers of errors with prob. 1 -2 -r • Standard DLC's use a CRC with r=16 with...

Ngày tải lên: 07/08/2014, 12:21