通信原理第4-5ç« ä&f

通信原理第4-5ç« ä&f

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通信原理第 4-5 章作业部分答案 4-2 解: 1 200 f Hz = 2 320 f Hz = 500 200 Hz ± 500 320 Hz ± 所以通过截止频率为 400 Hz 低通,留下的频率有 180 ,200 ,300 ,320 H zH z 4-5 解: 8 q S N = 4-10 解: 二进制: 2 log 64 8 48 B R symbol kHz kBd = 四进制: 4 log 64 8 24 B R symbol kHz kBd = 2/ p x Q Δ= 22 2 12288 /12 pp q xx N == Δ 12288 64 3 = 4-14 解: 1) ,段落码取 101 360 256 360 512 << 256 104 = 21 (2 2 ) 16 96 + ×= 段内码取 0110 输出为 11010110 对应为 352 ,量化误差: 360 352 8 = 2) ,所以对应均匀量化码为 86 352 2 2 2 =++ 5 00101100000 5-8 解: 2 1 3600 1200 1 ) cos ( 1200) 1200 2400 3600 2400 2400 0 fH z Gf f H z f H z z π (= 1 1200 6 1 ( ) ( ) cos ( 1200) 1200 2400 6 2400 0 2400 TR f Hf Hf f f f < <
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2 21 {( ) } () () 1 0 nc R En t S f H f d f W == 2 5-9 解: 2 1 2 2 1 1 4 1 14 14 (1) ( ) ( ) ( ) ( ) 1 () c o s ( ) 2 (1 ) 3 1 [ c o s ( ) ] 2 ()c o s( ) 1 0 2 1 11 0 10 cT R TR T nc R nc T T nc T Gf H fH fH f Hf Hf T f T f T MP A M M SW NS f H f d fS f T f T Sf T f T d f W S N π +∞ −∞ = < d f = ×
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This note was uploaded on 02/28/2012 for the course CS 466 taught by Professor 蒋铃鸽 during the Fall '11 term at Shanghai Jiao Tong University.

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通信原理第4-5ç« ä&f

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