第八章+答案

第八章+答案

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第八章 8-2 证: 0 00 0 0 0 0 0 () 2 [ cos( ) ( )] cos 2 cos( )cos( ) 21 [cos( ) cos(2 )] 2 1 cos( ) sin( )cos( ) im t T St St ht Atn t t T A td T A tt T AA t TT ωθ ω ωτ θ ωτ τ d θω τ θ =∗ =+ + + + + 上式第二项近似为零 t T = 0 St 0 cos( ) AT 所以输出包络峰值为 A 8-6 解: (1) 2 10 0 1 cos cos sin( ) sin( ) ()() T At t d t E ρω ωω = −+ 0 ρ = n T π −= m T += 当第二项近似为零,则 m T 0 0 sin( 2 ) (2 ) mT πω = 0 0 d d = ,取最小值。 ) 2 t g ωπ ( ) 4.493 Tr a d min 0.217 = − 2( ) 2 f T Δ=− = . 7 (2) 0 = 2 11 2224 e EA Pe r f c e r f c NN == T 0.217 =− 2 1 (1 ) 1 1.217 ( 222 4 e P erfc erfc ) T 能量是原来的 1.217 倍,误码率减小性能提高。 8-15 采用 8 PSK 调制传输 数据,最小理论带宽是多少? 4800 / bs
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4800 / R bs = 2 1600 log s R R baud M == 2 3200 s B RH Z = = 8-20 某数字微波通信系统,载频为 ,要求在 2 GHz 20 MHz 带宽内传输 34 / M 数字信号。 若采用升余弦基带信号及四相调制,确定: (1) 滚降系数
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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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第八章+答案

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