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2021年春季学期-信号与系统-第十四次作业参考答案-第一小题参考答案

系统 2021 信号 作业 第一 春季 参考答案 学期
2023-09-11 14:15:20 时间

本文是 2021年春季学期-信号与系统-第十四次作业参考答案 中各小题的参考答案。

 

§01 第一小题

1、用闭式表达以下有限长序列的DFT： ( 1 )      x [ n ] = δ [ n ] \left( 1 \right)\,\,\,\,x\left[ n \right] = \delta \left[ n \right] (1)x[n]=δ[n] ( 2 )      x [ n ] = δ [ n − n 0 ] ,      ( 0 < n 0 < N ) \left( 2 \right)\,\,\,\,x\left[ n \right] = \delta \left[ {n - n_0 } \right],\,\,\,\,\left( {0 < n_0 < N} \right) (2)x[n]=δ[n−n0​],(0<n0​<N) ( 3 )      x [ n ] = a n R N [ n ] \left( 3 \right)\,\,\,\,x\left[ n \right] = a^n R_N \left[ n \right] (3)x[n]=anRN​[n] ( 4 )      x [ n ] = e j ω 0 n R N [ n ] \left( 4 \right)\,\,\,\,x\left[ n \right] = e^{j\omega _0 n} R_N \left[ n \right] (4)x[n]=ejω0​nRN​[n]

▓ 求解

（1）解答：
x ( n ) = δ ( n )            x\left( n \right) = \delta \left( n \right)\;\;\;\;\; x(n)=δ(n)

X ( k ) = ∑ n = 0 N − 1 x [ n ] e − j 2 π N n k = 1 X\left( k \right) = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]e^{ - j{{2\pi } \over N}nk} } = 1 X(k)=n=0∑N−1​x[n]e−jN2π​nk=1

（2）解答：
x ( n ) = δ ( n − n 0 ) ,        ( 0 < n 0 < N )            x\left( n \right) = \delta \left( {n - n_0 } \right),\,\,\,\,\,\,\left( {0 < n_0 < N} \right)\;\;\;\;\; x(n)=δ(n−n0​),(0<n0​<N)

X ( k ) = ∑ n = 0 N − 1 x [ n ] e − j 2 π N n k   = e − j 2 π N n 0 k X\left( k \right) = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]e^{ - j{{2\pi } \over N}nk} } \, = e^{ - j{{2\pi } \over N}n_0 k} X(k)=n=0∑N−1​x[n]e−jN2π​nk=e−jN2π​n0​k

（3）解答：
x ( n ) = a n R N ( n )            x\left( n \right) = a^n R_N \left( n \right)\;\;\;\;\; x(n)=anRN​(n)

X ( k ) = ∑ n = 0 N − 1 a n R N [ n ] W n k = ∑ n = 0 N − 1 a W k = 1 − ( a W k ) N 1 − a W k   = 1 − a N 1 − a e − j 2 π N k X\left( k \right) = \sum\limits_{n = 0}^{N - 1} {a^n R_N \left[ n \right]W^{nk} } = \sum\limits_{n = 0}^{N - 1} {aW^k } = {{1 - \left( {aW^k } \right)^N } \over {1 - aW^k }}\, = {{1 - a^N } \over {1 - ae^{ - j{{2\pi } \over N}k} }} X(k)=n=0∑N−1​anRN​[n]Wnk=n=0∑N−1​aWk=1−aWk1−(aWk)N​=1−ae−jN2π​k1−aN​

（4）解答：
x ( n ) = e j ω 0 n R N ( n )            x\left( n \right) = e^{j\omega _0 n} R_N \left( n \right)\;\;\;\;\; x(n)=ejω0​nRN​(n)

X ( k ) = a = e j ω 0 1 − e j ω 0 N 1 − e j ( ω 0 − 2 π N k ) X\left( k \right)\mathop = \limits^{a = e^{j\omega _0 } } {{1 - e^{j\omega _0 N} } \over {1 - e^{j\left( {\omega _0 - {{2\pi } \over N}k} \right)} }} X(k)=a=ejω0​1−ej(ω0​−N2π​k)1−ejω0​N​

 

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