例2试求正弦量f()=-10si(的振幅H初相p与频率 解:将正弦量表达式化为基本飛式: 5兀 f(t)=10sin(100xt--+丌)=10sin(1007t+-) 6 6 5兀兀 =10cos(100nt+ )=10cos(100t+ 6 3 所以Fn=10,g=T/3rad, O=100πrad/s,f=a/2兀=50Hz例2 试求正弦量 的振幅Fm 、初相与频率 f 。 ) 6 ( ) 10sin(100  f t = −  t − 解：将正弦量表达式化为基本形式： ) 6 5 ) 10sin(100 6 ( ) 10sin(100     f t =  t − + = t + ) 3 ) 10cos(100 6 2 5 10cos(100     =  t + − = t + 所以 Fm = 10，  = /3 rad,  = 100 rad/s, f = /2 = 50Hz