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Matt L.
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There is indeed a transform called discrete Laplace transform and it is of course closely related to the $\mathcal{Z}$-transform. The (unilateral) discrete Laplace transform of a sequence $f_n$ is defined by (cf. link)

$$\mathcal{L}_T\{f_n\}=\sum_{n=0}^{\infty}f_ne^{-snT}$$

with some $T>0$. The discrete Laplace transform can be interpreted as the Laplace transform of a sampled function $f(t)\cdot\sum_n\delta(t-nT)$ with $f_n=f(nT)$.

In practice it is not convenient to have the factor $e^{sT}$ appear all the timeeverywhere. If one substitutes $z=e^{sT}$, the discrete Laplace transform is called (unilateral) $\mathcal{Z}$-transform:

$$\mathcal{Z}\{f_n\}=\sum_{n=0}^{n=\infty}f_nz^{-n}$$$$\mathcal{Z}\{f_n\}=\sum_{n=0}^{\infty}f_nz^{-n}$$

The same can obviously be done for the bilateral versions of the transforms, where the integrals and the sums start at $-\infty$.

There is indeed a transform called discrete Laplace transform and it is of course closely related to the $\mathcal{Z}$-transform. The (unilateral) discrete Laplace transform of a sequence $f_n$ is defined by (cf. link)

$$\mathcal{L}_T\{f_n\}=\sum_{n=0}^{\infty}f_ne^{-snT}$$

with some $T>0$. The discrete Laplace transform can be interpreted as the Laplace transform of a sampled function $f(t)\cdot\sum_n\delta(t-nT)$ with $f_n=f(nT)$.

In practice it is not convenient to have the factor $e^{sT}$ appear all the time. If one substitutes $z=e^{sT}$, the discrete Laplace transform is called (unilateral) $\mathcal{Z}$-transform:

$$\mathcal{Z}\{f_n\}=\sum_{n=0}^{n=\infty}f_nz^{-n}$$

The same can obviously be done for the bilateral versions of the transforms, where the integrals and the sums start at $-\infty$.

There is indeed a transform called discrete Laplace transform and it is of course closely related to the $\mathcal{Z}$-transform. The (unilateral) discrete Laplace transform of a sequence $f_n$ is defined by (cf. link)

$$\mathcal{L}_T\{f_n\}=\sum_{n=0}^{\infty}f_ne^{-snT}$$

with some $T>0$. The discrete Laplace transform can be interpreted as the Laplace transform of a sampled function $f(t)\cdot\sum_n\delta(t-nT)$ with $f_n=f(nT)$.

In practice it is not convenient to have the factor $e^{sT}$ appear everywhere. If one substitutes $z=e^{sT}$, the discrete Laplace transform is called (unilateral) $\mathcal{Z}$-transform:

$$\mathcal{Z}\{f_n\}=\sum_{n=0}^{\infty}f_nz^{-n}$$

The same can obviously be done for the bilateral versions of the transforms, where the integrals and the sums start at $-\infty$.

Source Link
Matt L.
  • 92.5k
  • 10
  • 81
  • 184

There is indeed a transform called discrete Laplace transform and it is of course closely related to the $\mathcal{Z}$-transform. The (unilateral) discrete Laplace transform of a sequence $f_n$ is defined by (cf. link)

$$\mathcal{L}_T\{f_n\}=\sum_{n=0}^{\infty}f_ne^{-snT}$$

with some $T>0$. The discrete Laplace transform can be interpreted as the Laplace transform of a sampled function $f(t)\cdot\sum_n\delta(t-nT)$ with $f_n=f(nT)$.

In practice it is not convenient to have the factor $e^{sT}$ appear all the time. If one substitutes $z=e^{sT}$, the discrete Laplace transform is called (unilateral) $\mathcal{Z}$-transform:

$$\mathcal{Z}\{f_n\}=\sum_{n=0}^{n=\infty}f_nz^{-n}$$

The same can obviously be done for the bilateral versions of the transforms, where the integrals and the sums start at $-\infty$.