Communication Systems Engineering Episode 1 Part 5 ppt

Dictionary of Engineering Episode 1 Part 5 ppt

Dictionary of Engineering Episode 1 Part 5 ppt

... and the total momentum column of mercury 1 centimeter high, having a of the system is zero. Also known as center density of 13 .59 51 grams per cubic centimeter, of momentum coordinate system. ... ska ¯ l} ədиe ¯ } centigram [ MECH ] Unit of mass equal to 0. 01 center of inertia See center of mass. { senиtər əv gram or 10 5 kilogram. Abbreviated cg. inərиshə } { sen...

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Communication Systems Engineering Episode 1 Part 1 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...

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Communication Systems Engineering Episode 1 Part 2 ppsx

Communication Systems Engineering Episode 1 Part 2 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|...

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Communication Systems Engineering Episode 1 Part 3 docx

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) <=>...

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Communication Systems Engineering Episode 1 Part 4 docx

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...

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Communication Systems Engineering Episode 1 Part 5 ppt

Communication Systems Engineering Episode 1 Part 5 ppt

... 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,...

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Communication Systems Engineering Episode 1 Part 6 potx

Communication Systems Engineering Episode 1 Part 6 potx

... 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...

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Communication Systems Engineering Episode 1 Part 7 pps

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...

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Communication Systems Engineering Episode 1 Part 8 pdf

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...

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Communication Systems Engineering Episode 1 Part 9 pptx

Communication Systems Engineering Episode 1 Part 9 pptx

... 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

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Communication Systems Engineering Episode 1 Part 10 ppsx

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...

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Communication Systems Engineering Episode 2 Part 2 pptx

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...

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Communication Systems Engineering Episode 2 Part 5 ppsx

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...

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