Assume $$f=\sum_{k=0}^{N-1} c_k\cdot E^{k}$$ where the vector $E^k$ is $$E^k = (e^{2 \pi i k\cdot M}(0),e^{2 \pi i k\cdot M}(1),\cdots,e^{2 \pi i k\cdot M}(N-1))$$ (M is a constant and e represents the exponential function).
If we compute the inner product $f\cdot E^k$ we obtain $N\cdot c_k$.
So we change the basis in which we represent $f$ (from values at different points to "a combination of the vectors $E^k, k=\overline{0,N-1}$").
This is what the discrete Fourier transform does, correct?
Is there a similar explanation for the Laplace transform?
Considering that $$L(f'(t)) = sL(f(t))$$ and $$L(\int_0^tf(t)\cdot dt) = \frac1sL(f(t))$$ the Laplace transform has something to do with the coefficients of f expressed in the new basis ($E^k, k=\overline{0,N-1}$), no?
First yes, the discrete Fourier transform can be formulated in that way, as transformation between the coefficient sequence of a polynomial and the value tuple of an equidistant sample point set on the unit circle.
For the Z-transform, as the discrete Laplace transform, things are different, since it transforms a sequence into a formal power series which is, as a data structure, still the same sequence.