I(x)=1(1+cos2rox)=1(+Mcos2zox) 所以Ms1 0≤MS1 Ix)=1+Mcos2元ax ( ) (1 cos 2 ) (1 cos 2 ) 0 0 0 x I M x I I I x I a = +   = +   因为Ia≤ I0 所以M≤1 0≤M≤1 I(x) =1+ M cos2 x