RE: Inverse Laplace output format?
- To: mathgroup at smc.vnet.net
- Subject: [mg28795] RE: [mg28769] Inverse Laplace output format?
- From: "David Park" <djmp at earthlink.net>
- Date: Sun, 13 May 2001 03:28:44 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Jose, I love these "surgery on expressions" problems! Here is one path to the solution. In[102]:= f = (11*s^2 + 172*s + 700)/((s + 2)*(s^2 + 12*s + 100)) InverseLaplaceTransform[f, s, t] ComplexExpand[%] Collect[%, E^(-6*t)] % /. a_*Cos[x_] + b_*Sin[x_] :> Sqrt[a^2 + b^2]*Cos[x - ArcTan[b/a]] % /. ArcTan[x_] :> N[ArcTan[x]/Pi]*180*HoldForm[Degree] Out[102]= (700 + 172*s + 11*s^2)/((2 + s)*(100 + 12*s + s^2)) Out[103]= (3 + 4*I)*E^((-6 - 8*I)*t) + (3 - 4*I)*E^((-6 + 8*I)*t) + 5/E^(2*t) Out[104]= 5/E^(2*t) + (6*Cos[8*t])/E^(6*t) + (8*Sin[8*t])/E^(6*t) Out[105]= 5/E^(2*t) + (6*Cos[8*t] + 8*Sin[8*t])/E^(6*t) Out[106]= 5/E^(2*t) + (10*Cos[8*t - ArcTan[4/3]])/E^(6*t) Out[107]= 5/E^(2*t) + (10*Cos[8*t - 53.13010235415597*HoldForm[Degree]])/E^(6*t) The last expression looks much better in StandardForm. ArcTan[4/3] is the exact expression for 53.13 Degree and you should usually skip the last step. David Park djmp at earthlink.net http://home.earthlink.net/~djmp/ > From: Jose Lopez Jr. [mailto:chilibean2in at netzero.net] To: mathgroup at smc.vnet.net > > 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. >