N nabliat6 New Member Feb 11, 2010 #1 **broken link removed** oiler formula says cosx+isinx but here they multiply by minus the imaginary part of the phasor representation the coefficient of sine is minus the imaginary part why??
**broken link removed** oiler formula says cosx+isinx but here they multiply by minus the imaginary part of the phasor representation the coefficient of sine is minus the imaginary part why??
M MrAl Well-Known Member Most Helpful Member Feb 11, 2010 #2 Hi there, Note that: -3*j amps converts to: 3*sin(wt) It does *NOT* convert to: -3*sin(wt) Note: (-3*j)*j*sin(wt)=3*sin(wt) ie: (-3*j)*j=3 Last edited: Feb 11, 2010
Hi there, Note that: -3*j amps converts to: 3*sin(wt) It does *NOT* convert to: -3*sin(wt) Note: (-3*j)*j*sin(wt)=3*sin(wt) ie: (-3*j)*j=3
H Hayato Member Feb 11, 2010 #4 It would be much easier if you post the whole problem. From where did you get the sine and cosine arguments? Observe that sine comes with j and cosine comes alone. So, you could consider, sinx = -3R/[R²+(1/2C -2L)²], and cosx = (2L-1/2C)/[R²+(1/2C -2L)²]
It would be much easier if you post the whole problem. From where did you get the sine and cosine arguments? Observe that sine comes with j and cosine comes alone. So, you could consider, sinx = -3R/[R²+(1/2C -2L)²], and cosx = (2L-1/2C)/[R²+(1/2C -2L)²]