### Re: My question concerns Laplace Transforms in electronic circuit analysis.

Date: Fri Nov 24 13:11:56 2006
Posted By: Madhu Siddalingaiah, Physicist, author, consultant
Area of science: Engineering
ID: 1163122276.Eg
Message:
```
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.

```

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