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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?2012-12-19
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    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.2012-12-19
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    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?2012-12-19
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    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.2012-12-19
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    The Z-transform **is** the discrete Laplace transform--and as such it acts on infinite sequences, not vectors of finite length.2012-12-19
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    Loosely, the Laplace transform extends the Fourier transform to a wider range of signals. Analogously, the z transform extends the discrete-time Fourier transform (DTFT) to a wider range of signals. If you deal with signals of fixed period (or finite length, which may be taken to be the period), the convergence issues are somewhat simpler - behaviour as $t \to \infty$ is a non-issue. The impetus to extend the equivalent transforms to a wider set is not the same. The equivalences then are Fourier series and the discrete Fourier transform (DFT) (why its not called series is beyond me).2012-12-19
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    (cont., sorry for the ramble): At the end of the day, however, you need to ask what operations are convenient for you (convolution, computational speed, etc.) and the choose accordingly.2012-12-19

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