|MadSci Network: Engineering|
Hi Danny, That's a good question. The roots alone don't tell the whole story. Here's one way to look at it. The sin term can be rewritten using a sum/difference trigonometric identity: sin(wt + P) = sin(wt)cos(P) + cos(wt)sin(P) Then your original time domain equation can be written as: Ge^(-at)sin(wt + P) = Ge^(-at)[sin(wt)cos(P) + cos(wt)sin(P)] Taking the Laplace transform of the right side yields: cos(P)w/[(s+a)^2 + w^2] + sin(P)(s+a)/[(s+a)^2 + w^2] The tangent of angle P is tan(P) = sin(P)/cos(P) P can be written as: P = arctan[sin(P)/cos(P)] So what you need to do is rewrite your "s" domain equation in the form above, factor out the terms for cos(P) and sin(P) and take the arctangent of the ratio. I hope that helps. madhu.com - madhu.com/blog
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