       Re: Inverse Laplace output format?

• To: mathgroup at smc.vnet.net
• Subject: [mg28785] Re: [mg28769] Inverse Laplace output format?
• From: BobHanlon at aol.com
• Date: Sat, 12 May 2001 20:18:18 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```f = (11s^2 + 172s + 700)/((s + 2)(s^2 + 12s + 100));

ans = FullSimplify[ExpToTrig[InverseLaplaceTransform[f, s, t]]]

(6*Cos[8*t] + 5*E^(4*t) +
8*Sin[8*t])/E^(6*t)

You want to find b such that
(6 Cos[8t] + 8 Sin[8t] == 10 Cos[8t - b])

((Cos[x-b]//TrigExpand) /. x -> 8t)

Cos[b]*Cos[8*t] + Sin[b]*Sin[8*t]

Off[Solve::ifun];

b /. Flatten[Solve[{10*Cos[b] == 6, 10*Sin[b] == 8}, b]]

ArcCos[3/5]

(%/Degree)//N

53.13010235415598

Bob Hanlon

In a message dated 2001/5/12 1:50:08 AM, chilibean2in at netzero.net writes:

>I have the following inverse Laplace -
>
>f = (11s^2 + 172s + 700)/((s + 2)(s^2 + 12s + 100))
>InverseLaplaceTransform[f, s, t]
>
>but the answer is in exponential form.
>
>I understand the ExpToTrig command but I want to get the
>answer in "cos theta in degree" form -
>
>5Exp[-2t] + 10Exp[-6t]cos(8t - 53.13)
>
>How do I do that because when I use the ExpToTrig I get
>
>(3 - 4 \[ImaginaryI]) Cosh[(6 + 8 \[ImaginaryI]) t] - (3 - 4 \[ImaginaryI])
>Sinh[(6 + 8 \[ImaginaryI]) t]
>
>and don't know what to do to get it to my "cos theta in degree" form.
>

```

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