Communication Systems Engineering Episode 2 Part 8 docx
Engineering Tribology Episode 2 Part 8 docx
... Phase Lubrication of Oleic Acid and TCP, Wear , Vol. 21 4, 19 98, pp. 20 7 -21 1. TEAM LRN 4 02 ENGINEERING TRIBOLOGY 0 0.1 0 .2 0.3 à 0 100 20 0 300 Temperature [C] FIGURE 8. 54 Experimental friction characteristic ... Mech. Engrs. Publ., London, 1963, pp. 70 -80 . 37 A. Dyson, Scuffing, A Review, Tribology International, Vol. 8, 1975, Part 1: pp. 77 -87 , Part 2: pp. 117...
Ngày tải lên: 05/08/2014, 09:19
Handbook of Corrosion Engineering Episode 2 Part 8 docx
... 9/1/99 6:15 Page 8 42 83 3 Corrosion Inhibitors 10.1 Introduction 83 3 10 .2 Classification of Inhibitors 83 4 10 .2. 1 Passivating (anodic) 83 6 10 .2. 2 Cathodic 83 7 10 .2. 3 Organic 83 7 10 .2. 4 Precipitation ... Solution Corrosion current, Corrosion rate, TCA, ppm R p , ⍀иcm 2 mAиcm 2 mmиy Ϫ1 Efficiency, % 0 14 1.55 18. 0 0 25 0 35 0. 62 7 .2 60 1000 143 0.1 52...
Ngày tải lên: 05/08/2014, 09:20
Communication Systems Engineering Episode 1 Part 3 docx
... 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
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 8 pdf
... 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
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
