作业帮sin (ωt) / cos (ωt) = sin (ωt) / sin(ωt + π/2) = e ^ (-j.π/2) = -j 对么?sin (ωt) / cos (ωt) = sin (ωt) / sin(ωt + π/2) = e -j.π/2 = cos(-π/2) + j.sin (-π/2) = -j ,tan (ωt) = -j 那么 tan (ωt) = -j 但是sin (ωt) = (e j.ωt - e -j.
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/30 01:38:56
sin (ωt) / cos (ωt) = sin (ωt) / sin(ωt + π/2) = e ^ (-j.π/2) = -j 对么?sin (ωt) / cos (ωt) = sin (ωt) / sin(ωt + π/2) = e -j.π/2 = cos(-π/2) + j.sin (-π/2) = -j ,tan (ωt) = -j 那么 tan (ωt) = -j 但是sin (ωt) = (e j.ωt - e -j.
sin (ωt) / cos (ωt) = sin (ωt) / sin(ωt + π/2) = e ^ (-j.π/2) = -j 对么?
sin (ωt) / cos (ωt) = sin (ωt) / sin(ωt + π/2) = e -j.π/2 = cos(-π/2) + j.sin (-π/2) = -j ,
tan (ωt) = -j 那么 tan (ωt) = -j
但是
sin (ωt) = (e j.ωt - e -j.ωt ) / 2j
cos(ωt) = (e j.ωt + e -j.ωt ) / 2
sin (ωt) / cos (ωt) = (e j.ωt - e -j.ωt )/ (e j.ωt + e -j.ωt ) .(1 / j) = (e j.ωt - e -j.ωt )/ (e j.ωt + e -j.ωt ) .(-j) ≠ -j
看这里 wiki 里面就用的这个等式 Electrical_impedance
This tells us that the ratio of AC voltage amplitude to AC current amplitude across a capacitor is ,and that the AC voltage leads the AC current across a capacitor by -90 degrees (or the AC current leads the AC voltage across a capacitor by 90 degrees)
sin (ωt) / cos (ωt) = sin (ωt) / sin(ωt + π/2) = e ^ (-j.π/2) = -j 对么?sin (ωt) / cos (ωt) = sin (ωt) / sin(ωt + π/2) = e -j.π/2 = cos(-π/2) + j.sin (-π/2) = -j ,tan (ωt) = -j 那么 tan (ωt) = -j 但是sin (ωt) = (e j.ωt - e -j.
你是不是不明白为什么tan (ωt) = -j 这个等式在数学上是不成立的,但我想你这里是要表示电路里的电流,电压的超前和滞后问题,所以在sin (ωt) / cos (ωt) = sin (ωt) / sin(ωt + π/2) = e ^ (-j.π/2) = -j 这个等式中的sin (ωt) / sin(ωt + π/2) = e ^ (-j.π/2) 这一步实际上是sin (ωt) / sin(ωt + π/2) =Im[e^(jwt)/e^(jwt+π/2)]而实际电路中,我们一般只取复频域中有实际意义的那部分,所以才会有上述等式,不知道我说的对不对,o(∩_∩)o
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sin (ωt) / cos (ωt) = sin (ωt) / sin(ωt + π/2) = e -j.π/2 ,这一步没有依据，故
sin (ωt) / cos (ωt) = -j ,是不对的，