Words You Always Have to Look Up Popular in Wordplay See All More Words with Remarkable Origins 8 Words for Lesser-Known Musical Instruments Birds Say the Darndest Things 10 Words from Taylor Swift Songs (Merriam's Version) 10 Scrabble Words Without Any Vowels ...
More Commonly Misspelled Words Words You Always Have to Look Up Popular in Wordplay See All More Words with Remarkable Origins 8 Words for Lesser-Known Musical Instruments Birds Say the Darndest Things 10 Words from Taylor Swift Songs (Merriam's Version) ...
inexpiable Thesaurus in·ex·pi·a·ble (ĭn-ĕk′spē-ə-bəl) adj. 1.Impossible to expiate or atone for:inexpiable crimes. 2.ObsoleteImplacable. in·ex′pi·a·blyadv. American Heritage® Dictionary of the English Language, Fifth Edition. Copyright © 2016 by Houghton Mifflin...
‘)。*exp (1i*2*pi*fc*t); 翻译结果3复制译文编辑译文朗读译文返回顶部 s_tilde=(uoft(delay+(1:length(t))).')。* exp(1i*2*pi*fc*t) ; 翻译结果4复制译文编辑译文朗读译文返回顶部 *s_tilde=(uoft(delay+(1:length(t))).'). exp(1i*2*PI**T); 翻译结果5复制译文编辑译文朗读译文...
所以-(a0+a1+...+aN) 等于式子 x^N-1 的x^(N-1)的系数,所以为0. 分析总结。 对于expi2pirnn从n0到nn1求和为何当r不等于mn时其求和为零结果一 题目 关于傅立叶变换推导对于exp(i*2*pi*r*n/N)从n=0到n=N-1求和,为何当r不等于mN时,其求和为零?m是任意整数 答案 以下是一个证明方法:...
求翻译:s_tilde=(uoft(delay+(1:length(t))).').*exp(1i*2*pi*fc*t);是什么意思?待解决 悬赏分:1 - 离问题结束还有 s_tilde=(uoft(delay+(1:length(t))).').*exp(1i*2*pi*fc*t);问题补充:匿名 2013-05-23 12:26:38 s_tilde=(uoft(delay+(1:length(t))).')。* exp(1i*2...
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注意在复数范围内,1的方根可不止一个!而1^[theta/(2*pi)]可看作1的2pi/theta次方根.若2pi/theta是有理数,设2pi/theta=m/n,则1^[theta/(2*pi)]有t=lcm(m,n)个结果:{exp{i*2kpi/(2pi/theta)}|0<=k<=t-1}=exp(i*k*theta)|0<=k<=t-1};若2pi/theta不是有理数,则...
对于exp(i*2*pi*r*n/N)从n=0到n=N-1求和,为何当r不等于mN时,其求和为零?m是任意整数 扫码下载作业帮搜索答疑一搜即得 答案解析 查看更多优质解析 解答一 举报 以下是一个证明方法:把n=0对应的数记为a0,n=1对应的记为a1,等等.那么这N个数为方程x^N=1的所有解.也就是说x^N - 1 = (x-a0)...
注意在复数范围内,1的方根可不止一个!而1^[theta/(2*pi)]可看作1的2pi/theta次方根.若2pi/theta是有理数,设2pi/theta=m/n,则1^[theta/(2*pi)]有t=lcm(m,n)个结果:{exp{i*2kpi/(2pi/theta)}|0<=k<=t-1}=exp(i*k*theta)|0<=k<=t-1};若2pi/theta不是有理数,则1^[theta/(...