3 j g proakis m saleki communication systems engineering(2nd edition)

... Recording 6.8.1 6.8.2 6.9 The JPEG Image-Coding Standard 6.10 Further Reading Problems 31 6 Digital Audio in Telephone Transmission Systems, 31 7 Digital Audio Recording, 31 9 32 3 32 7 32 7 DIGITAL TRANSMISSION ... Historical Review Elements of an Electrical Communication System 1.2.1 1.2.2 Digital Communication System, Early Work in Digital Communications, 10 1 .3 12 1.4 Mathema...

Ngày tải lên: 05/05/2015, 00:05

Ngày tải lên: 27/06/2014, 05:20

... Leon W Digital & analog communication systems / Leon W Couch, II.—8th ed p cm ISBN-13: 978-0-13-291538-0 (alk paper) ISBN-10: 0-13-291538-3 (alk paper) Telecommunication systems Digital communications ... for Analog Systems Comparison with Baseband Systems, 531 AM Systems with Product Detection, 532 AM Systems with Envelope Detection, 533 DSB-SC Systems, 535 SSB Sys...

Ngày tải lên: 05/04/2014, 12:54

Ngày tải lên: 14/05/2014, 11:44

... 0.4 0.2 0.2 0 -0 .2 -0 .2 -0 .4 -0 .4 -0 .6 -0 .6 -0 .8 -0 .8 -1 0.2 0.4 0.6 0.8 1.2 1.4 1.6 1.8 0.2 0.4 0.6 0.8 -1 0.2 0.4 0.6 0.8 1.4 1.6 1.8 0.8 0.6 0.4 0.2 -0 .2 -0 .4 -0 .6 -0 .8 -1 Problem 3. 4 y(t) = ... 1500)] 10 + [δ(f − fc − 30 00) + δ(f − fc + 30 00) +δ(f + fc − 30 00) + δ(f + fc + 30 00)] The next figure depicts the spectrum of u(t)...

Ngày tải lên: 12/08/2014, 16:21

... ≤ ⇒ i=1 n αi βi ≤ αi i=1 i=1 n βi i=1 Equality holds if αi = kβi , for i = 1, , n ∗ ∗ 2) The second equation is trivial since |xi yi | = |xi ||yi | To see this write xi and yi in polar jθxi ... it is observed both signals have the same envelope but there is a phase reversal at t = for the second signal Am2 (t) cos(2πf0 t) (right plot) This discontinuity is shown clearly in the next figure

Ngày tải lên: 05/03/2014, 19:20

... SOLUTIONS MANUAL Communication Systems Engineering Second Edition John G Proakis Masoud Salehi Prepared by Evangelos Zervas Upper Saddle ... ≤ ⇒ i=1 n αi βi ≤ αi i=1 i=1 n βi i=1 Equality holds if αi = kβi , for i = 1, , n ∗ ∗ 2) The second equation is trivial since |xi yi | = |xi ||yi | To see this write xi and yi in polar jθxi ... it is observed both signals have the same en...

Ngày tải lên: 28/03/2014, 17:20

... ≤ ⇒ i=1 n αi βi ≤ αi i=1 i=1 n βi i=1 Equality holds if αi = kβi , for i = 1, , n ∗ ∗ 2) The second equation is trivial since |xi yi | = |xi ||yi | To see this write xi and yi in polar jθxi ... it is observed both signals have the same envelope but there is a phase reversal at t = for the second signal Am2 (t) cos(2πf0 t) (right plot) This discontinuity is shown clearly in the next figure

Ngày tải lên: 27/06/2014, 08:20

... ≤ ⇒ i=1 n αi βi ≤ αi i=1 i=1 n βi i=1 Equality holds if αi = kβi , for i = 1, , n ∗ ∗ 2) The second equation is trivial since |xi yi | = |xi ||yi | To see this write xi and yi in polar jθxi ... it is observed both signals have the same envelope but there is a phase reversal at t = for the second signal Am2 (t) cos(2πf0 t) (right plot) This discontinuity is shown clearly in the next figure

Ngày tải lên: 27/06/2014, 17:20

... SOLUTIONS MANUAL Communication Systems Engineering Second Edition John G Proakis Masoud Salehi Prepared by Evangelos Zervas Upper Saddle ... publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book The author and publisher shall not be liable in any event for ... ≤ ⇒ i=1 n αi βi ≤ αi i=1 i=1 n β...

Ngày tải lên: 27/06/2014, 17:20

... ≤ ⇒ i=1 n αi βi ≤ αi i=1 i=1 n βi i=1 Equality holds if αi = kβi , for i = 1, , n ∗ ∗ 2) The second equation is trivial since |xi yi | = |xi ||yi | To see this write xi and yi in polar jθxi ... it is observed both signals have the same envelope but there is a phase reversal at t = for the second signal Am2 (t) cos(2πf0 t) (right plot) This discontinuity is shown clearly in the next figure

Ngày tải lên: 27/06/2014, 23:20

... − 4f cos(2πf0 (1 + 2n)t)dt + f0 4f0 cos(2πf0 (1 − 2n)t)dt − 4f 1 1 4f 4f sin(2πf0 (1 − 2n)t)| 10 sin(2πf0 (1 + 2n)t)| 10 + 2π (1 + 2n) 2π (1 − 2n) 4f0 4f0 n 1 ( 1) + π (1 + 2n) (1 − 2n) = = 9) ... zm,I ) i =1 m =1 n n i =1 ≤ = 1 2 2 (zi,R + zi,I ) (zm,R + zm,I ) i =1 m =1 n (zi,R i =1 Thus n zi i =1 n 1 + zi,I ) 2 (zm,R + zm,I ) n m =1 n ≤ (zi,R + zi,I ) i =1 n n zi...

Ngày tải lên: 12/08/2014, 16:21

... = 4 4 = 34 28 72 p(Y = 0) = 4 = 24 1 p(X = 1) = 4 p(X = 2) = p(X = 3) = 4 p(X = 4) = 4 4 3 4 33 28 p(Y = 1) = 1 = 33 28 p(Y = 2) = 2 = 3 4 28 p(Y = 3) = 4 = 28 4 p(Y = 4) = = = 24 = 24 = 24 = ... = 14 KHz fk3 = 18 KHz fk4 = 22 KHz fk5 = 26 KHz fk6 = 30 KHz fk7 = 34 KHz fk8 = 38 KHz fk9 = 42 KHz fk10 = 46 KHz fl1 = 290 KHz fl2 = 330 KHz fl3 = 370 KHz fl4 = 41 0 KHz...

Ngày tải lên: 12/08/2014, 16:21

... x,t,a,q,pi,p,b1,b2,b3,b4,b5 PARAMETER (p=.2316419d+00, b1=.3198 153 0d+00, 84 ∞ √ 2r + + b2 =-. 356 563782d+00, b3=1.781477937d+00, b4 =-1 .821 255 978d+00, b5=1.330274429d+00) Cpi=4.*atan(1.) C-INPUT PRINT*, ’Enter -x-’ READ*, ... 1 .59 × 10 1 .58 7 × 10−1 −2 1 .5 6.68 × 10 6.6 85 × 10−2 2.28 × 10−2 2.276 × 10−2 −3 2 .5 6.21 × 10 6.214 × 10−3 −3 1. 35 × 10 1. 351 × 10−3 3 .5 2.33 ×...

Ngày tải lên: 12/08/2014, 16:21

... 10 · 1.256 = −3.97 18 ˆ √ x x5 = −ˆ12 = − 10 · 0.9424 = −2. 980 1 ˆ √ x x6 = −ˆ11 = − 10 · 0.65 68 = −2.0770 ˆ √ x x7 = −ˆ10 = − 10 · 0. 388 1 = −1.2273 ˆ √ x x8 = −ˆ9 = − 10 · 0.1 284 = −0.4060 ˆ The ... the rate-distortion bound, we obtain R= σ2 = 3.2 186 log2 D 5) The distortion of the 16-level optimal quantizer is D16 = σ · 0.01154 whereas that of the 8- level optimal quantizer is D8 =...

Ngày tải lên: 12/08/2014, 16:21