第五章作业答&aeli

第五章作业答&aeli

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5-13 】解: n n n x 1 sin ] [ 的频谱为: 1 1 0 1 ) ( j e X 设零值插入后的信号为 ] [ 1 n x ,则 3 0 3 ] 3 [ ] [ ] [ ) 3 ( 1 k n k n n x n x n x 应有, ) ( ) ( 3 1 j j e X e X 1 5 3 1 时, ) ( 1 j e X 的结果如题 5.13 图( a )所示, 频谱在 3 2 k 处,有宽度为 3 2 1 ,幅度为 1 的矩形脉冲。 ] [ 1 n x 经低通滤波后,将保留 ) ( 1 j e X k 2 处的矩形脉冲,滤除其它地方的脉冲, ) ( ) ( ) ( 1 j j j e H e X e W 最后对 ] [ n w 进行抽取, ] 5 [ ] [ n w n y ] [ n y 的频谱如题 5.12 图( b )所示, 可得 , n n n y 5 3 5 sin ] [ 1 2 5 3 1 时,与( 1 )的情况类似, ) ( 1 j e X 的频谱在位于 3 2 k ,处有宽度为 3 2 1 ,幅度为 1 的矩形脉冲,
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这时脉冲宽度的范围为 3 2 ~ 5 2 ] [ 1 n x 经低通滤波后,将保留 ) ( 1 j e X k 2 处的矩形脉冲,滤除其它地方的脉冲, 但脉冲宽度被限制在 5 的范围内。 最后对 ] [ n w 进行抽取, ] [ n y 的频谱 ) ( j e Y 将被限制在 的范围内, 即对所有的 均有, 5 1 ) ( j e Y 这时 , 5 ] [ ] [ n n y 5-14 】解: ] [ n x 的最高频率为 14 3 M
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