ch6&ch7答案

ch6&ch7ç­”æ&iexcl...

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6-4 解: 1 0 0 () ( ) ( ) T i St S ST t d τ ττ =− + 0 tT << 0 0 22 2 sin sin ( ) cos sin 24 2 cos 2 t nn Sa a T t d TT an t a n t t Tn T t t T π ππ + 2 Tt T 2 0 2 (2 )cos 2 T t T t 图可参见 6 11 2 2 0 1 2 ST aT = 0 4 T n −= 0 2 0 1 2 3 2 n T ω = 1 2 dn d T ωπ = 1 2 d d nT = 所以 n 越大 1/T 精度要求越高。 6-5 解: 1 0 2 ( ) 2 m T At ht ST t T T ≤< = ≤≤ 2 3 2 0 2 1) 2 T jf t mm t j f t Hf hf e d t A ee −− = 2 2 2 0 2 0 2 (3 2 ) 2 3 (4 3 ) 2 3 ) 2 2 0 T t T At T T AT t Tt T ATt t T else −≤ = < < < 图略。 4
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0 22 00 2 0 4 2 0 () ( ) 2(1 cos( )) sin 2 2 nm jf T NN Sf A f Hf N A ef f fT NA f π == =− = t 5 2 max EA d T 7-2 解: 1 y > ,显然判决为 1 H 1 y 2 1 2 2 1 0 0 1 exp( ) 2 2 1 1 2 y H fyH H σ πσ = > < 0 2 1 ln 2 H y H > < 2 2 2 ln 2 y ≤− 1 y > ,判为 1 H 2 ln 1 2 y ≤≤ ,判为 0 H 2 2 >
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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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ch6&amp;amp;ch7&amp;ccedil;&amp;shy;”&amp;aelig;&amp;iexcl...

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