By Kostas Stamatiou Proakis Salehi
Electronic Communications is a vintage e-book within the zone that's designed for use as a senior or graduate point textual content. The textual content is versatile and will simply be utilized in a one semester direction or there's sufficient intensity to hide semesters. Its complete nature makes it an outstanding publication for college kids to maintain for reference of their specialist careers. This all-inclusive advisor offers a very good creation to the research and layout of electronic verbal exchange structures. comprises professional insurance of recent issues: Turbocodes, Turboequalization, Antenna Arrays, electronic mobile platforms, and Iterative Detection. handy, sequential association starts with a glance on the heritage and class of channel versions and builds from there.
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Extra info for Digital Communications Fifth Edition - Instructor Solution Manual
The triangular-shaped spectrum S(f ) = 1 − √ W , |f | ≤ W may be obtained by convolving the rectangular-shaped spectrum S1 (f ) = 1/ W , |f | ≤ W/2. Then : ∞ ∞ fk (t)fj (t)dt = y(t)dt = Y (f )|f =0 −∞ −∞ where Y (f ) is the Fourier transform of y(t). Since : y(t) = fk (t)fj (t) ←→ Y (f ) = Fk (f ) ∗ Fj (f ). 52 Beq = 1 G ∞ 0 |H(f )|2 df For the filter shown in Fig. P2-12 we have G = 1 and ∞ Beq = 0 |H(f )|2 df = B PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation.
5/T. Clearly the spectrum for the rectangular pulse has a narrower main lobe. However, it has higher sidelobes. 14 1. Bn = In + In−1 . Hence : In In−1 Bn 1 1 2 1 −1 0 −1 −1 1 0 −1 −2 The signal space representation is given in the following figure, with P (Bn = 2) = P (Bn = −2) = 1/4, P (Bn = 0) = 1/2. Bn o I o ✲ -2 0 2 2. w and SBB (f ) = = ∞ m=−∞ RBB (m) exp(−j2πf mT ) 2 [1 + cos 2πf T ] = 4 cos 2 πf T = 2 + exp(j2πf T ) + exp(−j2πf T ) PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved.
05 0 -5 -4 -3 -2 -1 0 fT 1 2 3 4 5 3. 44 1 2 = 2 ⇒ f3db = T (πf3db T ) PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Digital Communications Fifth Edition - Instructor Solution Manual by Kostas Stamatiou Proakis Salehi