x ncos0.5πn加0.2sin0.2πnn0⋯9求出x n的离散傅立叶变换并画出其幅度谱 x(n)=cos(0.5πn)+0.2sin(0.2πn),n=0,⋯9，求出x(n)的离散傅立叶... 大小：395B | 2020-07-18 05:00:05
