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请将正弦序列x(n)=e-2nsin(0.6n+π/3)表示成为复指数序列 我要具体的答案!不要说一些概念
人气:168 ℃ 时间:2020-07-12 14:51:43
解答
x(n)=e^(-2n)sin(0.6n+π/3)
因为:sin(0.6n+π/3)=(-i/2)(e^(-i)(0.6n+π/3)+e^(i)(0.6n+π/3))
所以:
x(n)=(-i/2)e^(-2n)(e^(-i)(0.6n+π/3)+e^(i)(0.6n+π/3))
=(-i/2)[e^(-2n-i(0.6n+π/3))+e^(-2n+i(0.6n+π/3)]
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