例计算下列两正弦量的相位差。解 (1)i1()=10c0s(1007x+37/4)p=3m/4-(-/2)=5z/4>0 2(t)=10c0s(100t-x/2)→q=27-57/4=37/4 2)i1(1)=10c0s(1007t+30)i2(t)=10c0s(100-105) ()=10si(100-15)g=30-(-1065)=135° (3)u1()=10c0s(1007t+30) 01≠2 u2(t)=10c0s(2007t+45)不能比较相位差 (4)i()=5c0s(1007t-30)2(t)=3cos(100m-150) i4()=-3c0s(100+30)q=-30-(-150°)=12 0 两个正弦量进行相位比较时应满足同频率、同函数、同符 号,且在主值范围比较。 以理交通大学例 计算下列两正弦量的相位差。 ( ) 10sin(100 15 ) (2) ( ) 10cos(100 30 ) 0 2 0 1 = − = + i t t i t t   ( ) 10cos(100 2) (1) ( ) 10cos(100 3 4) 2 1     = − = + i t t i t t ( ) 10cos(200 45 ) (3) ( ) 10cos(100 30 ) 0 2 0 1 = + = + u t t u t t   ( ) 3cos(100 30 ) (4) ( ) 5cos(100 30 ) 0 2 0 1 = − + = − i t t i t t   解 j = 3 4−(− 2) = 5 4  0 j = 2 − 5 4 = 3 4 0 0 0 j = 30 − (−105 ) = 135 0 0 0 j = −30 − (−150 ) = 120 ( ) 10cos(100 105 ) 0 i 2 t = t − 不能比较相位差 w1 w2 ( ) 3cos(100 150 ) 0 i 2 t = t − 两个正弦量进行相位比较时应满足同频率、同函数、同符 号，且在主值范围比较