Communication Systems Engineering Episode 3 Part 4 pdf

Communication Systems Engineering Episode 3 Part 4 pdf

Communication Systems Engineering Episode 3 Part 4 pdf

... occur? C) Find the generator matrix for the (7 ,3) block code based on the above generator. (note that a (7 ,3) code has 3 information bits and 4 check bits). D) Using the generator matrix from ... MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Aeronautics and Astronautics 16 .36 : Comm. Sys. Engineering Date Issued: April 17 Problem Set No. 8 Date Due: April ... code? Probl...
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Communication Systems Engineering Episode 1 Part 4 docx

Communication Systems Engineering Episode 1 Part 4 docx

... can be used to represent 4 ∆ quantization levels – Soon we will learn that you only need H(Q) bits ∆ − ∆ R 1 R 2 R 3 R 4 q 1 = -3 /2 q 4 = 3 /2 q 2 =∆/2 q 3 =∆/2 Eytan Modiano Slide ... A 2 A 2 [] = 2 A ∫ − A xdx = 3 A 2 / 3 A 2 / 3 SQNR = ∆ 2 / 12 = ( 2 AN) 2 / 12 = N 2 ,(∆ = 2 A / N) / Eytan Modiano Slide 5 Lecture 4: Quantization Eyt...
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Communication Systems Engineering Episode 1 Part 8 pdf

Communication Systems Engineering Episode 1 Part 8 pdf

... 3dB = 7dB – => Pb = 5x10 -4 from table 7.55 or 7.58 • Regenerator: P b (up) = P b (down) = 3x10 -6 – (from table with (E b /N 0 ) d = 10dB) – Hence P b (up/down) ~ 2 P b (up) ~ 6x3x10 -6 ... – Hence a doubling of the diameter D increases gain by a factor of 4 Eytan Modiano Slide 4 Repeaters Tx P T2 + A P T3 + A + Rx P N P N P N • A repeater simply amplifies the sig...
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Communication Systems Engineering Episode 2 Part 4 pps

Communication Systems Engineering Episode 2 Part 4 pps

... 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 ... D Tp / (D Tp + τ + 2 τ (e-1)) • Let β = τ / D Tp => Efficiency ≈ 1/(1 +4. 4 β ) = λ < 1/(1 +4. 4 β ) Eytan Modiano Slide 7 Token ring issues • Fairness: Can a...
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Communication Systems Engineering Episode 1 Part 3 docx

Communication Systems Engineering Episode 1 Part 3 docx

... t F[(t)] δ F[(t)] = ∫ −∞ δ (t)e − jft dt = e 0 = 1 δ () t δ () ⇔ 1 0 Eytan Modiano Slide 4 1 Lecture 3: The Sampling Theorem Eytan Modiano AA Dept. Eytan Modiano Slide 1 The Fourier Transform ... f )e jft df • Notation: X(f) = F[x(t)] X(t) = F-1 [X(f)] x(t) � X(f) Eytan Modiano Slide 3 Notes about Sampling Theorem • When sampling at rate 2W the reconstruction filt...
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