通信原理前三章ä

通信原理前三章ä

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通信原理前三章作业答案: 1 2 解: ( 1 )( 1 , 1 1 , 0 1 1 1 1 0 0 21 1 2 1 3 331 03 4 5 yx PP P P P =+ =×+ ×= ) x ( 0 0 , 0 0 , 1 0 0 0 0 1 1 11 9 2 3 2 5 P P P =+=× + × ) x 1 3 解:( 1 2 18000( ) 60( ) 300( ) log 2( / ) B bit R s Baud bit symbol == 2 2 18000( ) 60( ) 100( ) log 8( / ) B bit R s Baud bit symbol 1 5 解: 2 1 2222 2 2 1 log ( ) 1111 1 0.3log ( ) 0.2log ( ) 0.2log ( ) 0.1log ( ) 0.15log ( ) 0.05log ( ) 0.3 0.2 0.2 0.1 0.15 0.05 2.41( / ) n i i i Hp p bit symbol = = =++++ + = 1 1 7 解:每个象素的信源熵: 10 2 1 1 log (10) 3.32( / ) 10 H bit symbol 每秒钟传输的象素数: 525 625 30 9843750(symbol) n =×× = 每秒钟传输信息量: 3.32( / )*9843750(symbol)=32.7M(bit) I H n bit symbol =∗=
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2 3 证明 : () [ ] [][ ] [][ ] 00 cos sin cos sin 0 Ext EA t B t EAE t EBE t ωω ω =+ ⎡⎤ ⎣⎦ = 12 1 2 01 02 22 2 2 2 (, ) [ () () ] [( cos sin )( cos sin )] [ cos cos sin sin cos sin cos sin ] [cos cos sin sin ] [cos ( )] () x x Rtt Extxt t A t E A t tB t tA B t B t t Ett tt Et t R ωωωω σω τ = + + + =− = 所以写 x(t) 是广义平稳过程。 2 6 证明:( 1 10 0 [ ( )] [ ( )cos ] [() ] c o s ES t t t = = ] 不等于 0 ,是非平稳的; ] 等于 0 ,讨论 x R 1 20 [()c o s ( )( )c o s ( ) ] t xt t 2 ()c o s ( )c o s ( ) x R i t 有关,所以非平稳的 2 0 0 [ ( )] [ ( )cos( )] [ ( )] [cos( )] 1 cos( ) 2 0 x t Ext E t mt π d θ ωθ θθ = t 无关 2 () 1 2 2 1 2 2 1 2 0 1 0 2 2 0 ] o s ( o s ( ) ] [()( ) ][ c o s ( o s ( ) ] 1 () c o s ( ) 2 1 o s () 2 St x x R tt EStSt t xt t Ext xt E t t Rt
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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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通信原理前三章ä

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