Dear Mitch:

We used 'analytical' essentially meaning that there was a closed formula for it, particularly in the numerical evaluation lesson to distinguish between the two.

The parameter s, in the integral definition of the Laplace transform is complex, and this is the variable in the Laplace transform. That is where complex numbers come from.

Thank you Phil. This is a very nice real LIVE example.

Thanks Juan,

But there is a ComplexPlot3D. For example:

ComplexPlot3D[(z^2 + 1)/(z^2 - 1), {z, -2 - 2 I, 2 + 2 I}]

Carl

I have signed up to take the quizzes for certification in the Laplace Transform DSG. I have the downloaded the textbook, the daily lessons, and the exercises (with solutions). But where are the quizzes? (I do not see any links to quizzes in the download section for the study group.)

Thanks!

Dear Graham:

Both seem fine to me. For problem 6, I will suggest comparing your answer with the answers provided considering that s>0 in this case. For example, you can try the command FullSimplify[yourAnswer==answerCompared, s>0] or you can cheat a bit to very using InverseLaplaceTransform on each of the posible aswers provided. For 10, make sure to set the WorkingPrecision to 10 as an option in your call to LaplaceTransform.

Dear Carl:

I believe that in signal analysis the Fourier transform is the way to go. Note that in the Laplace transform s is complex and in fact if h(t) is the Heaviside function and s=c+i y, then the Laplace transform of f(t) is equal to the Fourier transform of exp(-c t) h(t)f(t) depending on y. This is actually used to get the inversion formula for the Laplace transform using the Fourier transform. For more on this I can refer you to the documentation page of the FourierTransform function in the Wolfram language, where I included a subsection on signals and systems in the Applications section.

Dear James:

WorkingPrecision->n causes all internal computations to be done to at most n-digit precision. This does not imply that the answer will be correct to that number of digits. The final results you get may have much lower precision.

Instead of comparing WorkingPrecision->10 with WorkingPrecision->20, we should compare each to infinite precision. This will require to have an analytic representation for the transform. Consider exercise 1 in the same lesson:

exact = LaplaceTransform[
   Cos[Sqrt[t^2 - 4]]/Sqrt[t^2 - 4] UnitStep[t - 2], t, s] /. 
  s -> 17/10

approximate = 
 LaplaceTransform[Cos[Sqrt[t^2 - 4]]/Sqrt[t^2 - 4] UnitStep[t - 2], t,
   1.7]

approximateWP10 = 
 LaplaceTransform[Cos[Sqrt[t^2 - 4]]/Sqrt[t^2 - 4] UnitStep[t - 2], t,
   1.7, WorkingPrecision -> 10]

approximateWP20 = 
 LaplaceTransform[Cos[Sqrt[t^2 - 4]]/Sqrt[t^2 - 4] UnitStep[t - 2], t,
   1.7, WorkingPrecision -> 20]

Now:

exact - approximate=4.50937*10^-10

exact - approximateWP10=-1.14043*10^-12

exact - approximateWP20=-3.46945*10^-18