解: x,()=[1+cos(2r×100r)]cos(2π×600i) =c0s(2π×600t) +2c0s(2元×701)+2c0s(2rx500) (1)抽样频率应为 f≥2×700=1400Hz (2)抽样时间间隔应为 T1=1 71400 0.00072Sec=0.72ms 2024/10/21 31(1)抽样频率应为 2 700 1400 s f Hz = 解: (2)抽样时间间隔应为 1 1 0.00072 0.72 1400 s T Sec ms f = = = x t t t a ( ) = + 1 cos 2 100 cos 2 600 ( ) ( ) ( ) ( ) ( ) cos 2 600 1 1 cos 2 700 cos 2 500 2 2 t t t = + + 2024/10/21 31