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Unformatted text preview: Solutions Manual to Accompany Digital Communications
Second Edition Bernard Sklar Prentice Hall PTR
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“t '6 113 1.20 (d) The normalized spectrum, me/ 6,1106), appears as: 0.?5 6 6 6 6 9.105 92105 94105 45105 .810 110‘5 LUZ10610410 Loo10 1.0810 1.110 f Applying numerical methods with Mathcad ®, the twosided 99%
bandwidth can be found by numerical integration as:
J106+ f1 1H Ge(f)df
EGAN where fl is found to be equal to 103 kHz. Thus, the two—sided 99%
bandwidth is equal to 206 kHz. Since this bandwidth corresponds to the given spectrum (with
signaling rate : 10,000 symbols/s), normalizing it relative to one symbol per second, yields the twoSided 99% bandwidth as
(206x103/104) = 20.6 Hz for one symbol per second, or in general
the 99% bandwidth in terms of the signaling rate, R, is 20.6xR Hz. 2 0.99 1.20 (e)
35dB Bandwidth: 35—dB attenuation :> 10'35 = 3.16 X 10“ 114 115 ...
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 Spring '08
 Nagaraj,SV

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