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The matrix formulation of the (discrete) Fourier transform for a signal 5 terms long, can be illustrated as follows:

Signal or time domain

$\left( \begin{array}{ccccc} 1 & 1 & 1 & 1 & 1 \end{array} \right)$

Dicrete Fourier Cosine Transform

$\left( \begin{array}{ccccc} 1 & 1 & 1 & 1 & 1 \\ 1 & \frac{1}{4} \left(-1+\sqrt{5}\right) & \frac{1}{4} \left(-1-\sqrt{5}\right) & \frac{1}{4} \left(-1-\sqrt{5}\right) & \frac{1}{4} \left(-1+\sqrt{5}\right) \\ 1 & \frac{1}{4} \left(-1-\sqrt{5}\right) & \frac{1}{4} \left(-1+\sqrt{5}\right) & \frac{1}{4} \left(-1+\sqrt{5}\right) & \frac{1}{4} \left(-1-\sqrt{5}\right) \\ 1 & \frac{1}{4} \left(-1-\sqrt{5}\right) & \frac{1}{4} \left(-1+\sqrt{5}\right) & \frac{1}{4} \left(-1+\sqrt{5}\right) & \frac{1}{4} \left(-1-\sqrt{5}\right) \\ 1 & \frac{1}{4} \left(-1+\sqrt{5}\right) & \frac{1}{4} \left(-1-\sqrt{5}\right) & \frac{1}{4} \left(-1-\sqrt{5}\right) & \frac{1}{4} \left(-1+\sqrt{5}\right) \end{array} \right)$

Spectrum or frequency domain

$\left( \begin{array}{c} 5 \\ 0 \\ 0 \\ 0 \\ 0 \end{array} \right)$

Where the signal or time domain is multiplied element wise with each row in the Fourier transform matrix. Taking the row sums of the resulting matrix gives the spectrum or frequency domain.

Is there a corresponding matrix formulation for the Laplace transform? And if so what does it look like for a 5 times 5 matrix?

Hopefully this is not a waste of space on the math SE site with yet another question on the Laplace transform.

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What do you call the Laplace transform of a signal of finite length? – Did Dec 19 '12 at 16:08
I don't know. Is there no such thing? I mean does the signal to be Laplace transformed have to be Infinite? I solved problems at university using the Laplace transform lookup list in Schaums outline Mathematical Handbook. Later I learned about the discrete Fourier transform and its matrix formulation. Since the Fourier transform is, if I am correct, the Laplace transform with a complex variable - I thought the Laplace transform must have a matrix formulation also. – Mats Granvik Dec 19 '12 at 16:15
A crucial difference, as you know if you pondered its structure a bit, is that DFT of vectors of size $n$ is based on the $n$th roots of unity. The Laplace domain being the real line (instead of the unit circle) has no nontrivial such roots. But, say, you are asking a question about discrete Laplace transform without a definition of the notion? – Did Dec 19 '12 at 16:20
So the Z transform that is said to be the discrete Laplace transform is simply this: $X(z) = \mathcal{Z}\{x[n]\} = \sum_{n=0}^{\infty} x[n] z^{-n}$ ? Which is more like a power series? I have never calculated any thing with the Z-transform, and I therefore have no feel/feeling for have it behaves. Edit seconds later: I am still lost. I need a program or spelled out calculation example of what the Z or Laplace transform is. – Mats Granvik Dec 19 '12 at 16:26
The Z-transform is the discrete Laplace transform--and as such it acts on infinite sequences, not vectors of finite length. – Did Dec 19 '12 at 16:30

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