I have a calculator with CAS ( https://en.wikipedia.org/wiki/Computer_algebra_system ) and in particular I have a Casio Algebra FX 2.0 Plus ( https://en.wikipedia.org/wiki/List_of_computer_algebra_systems ).

I need to compute Laplace / Zeta / Fourier Transform.

My calculator is able to calculate laplace transforms, in fact if I go in the CAS section and insert:

"$∫ (sinX * e-SX, X)$"

it gives me as a result:

" $-e^(-s·x)·(COS (x) + s · SIN (x)) / (s ^ 2 + 1) $"

then evaluating: $LIM (- e ^(-sx)·(COS (x) + s · SIN (x)) / (s ^ 2 + 1), x, ∞) - LIM (- e^(-s·x)·(COS (x) + s · SIN (x)) / (s ^ 2 + 1), x, 0 ) = 1 / (s ^ 2 + 1)$.

And indeed

$ ℒ_x [sin (x) * step (x)] (s) = 1 / (s ^ 2 + 1)$.

I tried to do it even with more complex expressions, and I always get the correct result.

Unfortunately, I can not find a way to calculate the Z-Transform or the Fourier Transform, using the CAS of my calculator .

At this point I wondered if you knew:

OR 1) A method to calculate also Z-transform and Fourier-Transform by using the CAS of my calculator.

OR 2) Calculated the laplace transform of a function (since, as proven, my calculator is able to do this), then easily deduce from it the Z-Transform and the Fourier-Transform.

OR 3) Since my calculator also accepts C / C ++ programs, do you know any program in this language for the calculation of Fourier Transform / Zeta Transform? Thanks everyone in advance for the help.


Laplace Transform: https://en.wikipedia.org/wiki/Laplace_transform

Z-Transfrom: https://en.wikipedia.org/wiki/Z-transform

Fourier Trnasform: https://en.wikipedia.org/wiki/Fourier_transform


I do not have this calculator, but according to the manual, you can compute generalized integrals (it supports definite integration and provides the infinity symbol). Thus, you can compute the Laplace transform in a single and simpler command using the definition. If the Fourier transform is defined (or equivalently if the imaginary axis is included in the ROC for the Laplace transform), you can use the substitution $ s = j \omega$ to find the Fourier Transform.

Now, Z-transform applies to discrete signals, so there is no direct connection. The only connection I know between the Laplace transform and the Z-transform is that the second appears when you ideally sample a function, take its Laplace transform and set $z = e^{sT}$, where T is the sampling period (see Starred Transform). However, this will not assist you in finding the Z-transform efficiently. Nevertheless, you should be able to use the "sum" function in your calculator to attempt to find the Z-transform directly from its definition.

https://support.casio.com/pdf/004/algebra_plus_Ch07.pdf https://en.wikipedia.org/wiki/Starred_transform

  • $\begingroup$ HI Sotiris, Thank you for your interest. As regard laplace transform is concerned, everything is okay: my calculator can calculate it through the definition of Laplace transform. As regard the transfrom-Z: I find the only possible way to go for the starred transfrom: from G (S) to G * (S) to G (Z). It will be able to be the only valid solution to apply with my calculator and it works. $\endgroup$ Jan 16 '19 at 19:45
  • $\begingroup$ As regard the Fourier transform: Are you sure that it is enough to evaluate the Laplace transform in s = jw? For example, if my starting function is: f (t) = 2 e^ (- t) * step (t) => ℒ_t [2 e^ (- t) θ (t)] (s) = 2 / (1 + s) => ℱ_t [2 e ^ (- t) θ (t)] (ω) = ℒ_t [2 e^ (- t) θ (t)] (s) when s = jw => 2 / (1 + jw) which is corrected. But if my starting function were: f (x) = sin (x) => ℒ_x [sin (x) θ (x)] (s) = 1 / (s ^ 2 + 1) and when s = jw => 1 / ((jw) ^ 2 + 1) ≠ ℱ_x [sin (x)] (ω) = i sqrt (π / 2) δ (ω - 1) - i sqrt (π / 2) δ (ω + 1). How should I do in these cases? Many thanks in advance. $\endgroup$ Jan 16 '19 at 19:45
  • $\begingroup$ Hi Salvatore! It's my pleasure. :) Now, for your first comment, I find it strange that you can go through the starred transform using your calculator but if you managed to do it its great. However, if your computer supports infinite sums you should be able to use the Z-transformation directly. $\endgroup$
    – Sotiris
    Jan 18 '19 at 21:48
  • $\begingroup$ As far as your second question is concerned, notice that I told you that you can use $s=j \omega$ if the Fourier transform is defined (equivalently if the imaginary axis is included in the Lapalce transform's ROC). Note that this is not the case for $sin(x)$ since the definition integral diverges. That is why, in order to talk about its "Fourier Transfrom" you have to extend the context of the original definition to include generalized functions. But for these, the substitution $s = j \omega$ no longer works. $\endgroup$
    – Sotiris
    Jan 18 '19 at 21:59
  • $\begingroup$ @Hi Sotiris! Unfortunately my calculator does not support infinite sums, that is why I have to go for the starred transform. My calculator does not automatically support the starred trasfrom, but it's me to insert it into input, but it is a good way to get Z-Transform from L-Transform. $\endgroup$ Jan 20 '19 at 15:04

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