Difference between revisions of "Unilateral Laplace transform of delta derivative"
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==Theorem== | ==Theorem== | ||
− | If $\mathbb{T}$ is a time scale, then | + | If $\mathbb{T}$ is a time scale and $f \colon \mathbb{T} \rightarrow \mathbb{C}$, then |
$$\mathscr{L}_{\mathbb{T}}\{f^{\Delta}\}(z;s) = -f(s) + z\mathscr{L}\{f\}(z),$$ | $$\mathscr{L}_{\mathbb{T}}\{f^{\Delta}\}(z;s) = -f(s) + z\mathscr{L}\{f\}(z),$$ | ||
where $\mathscr{L}_{\mathbb{T}}$ denotes the [[unilateral Laplace transform]] and $f^{\Delta}$ denotes the [[delta derivative]] of $f$. | where $\mathscr{L}_{\mathbb{T}}$ denotes the [[unilateral Laplace transform]] and $f^{\Delta}$ denotes the [[delta derivative]] of $f$. |
Revision as of 15:11, 21 January 2023
Theorem
If $\mathbb{T}$ is a time scale and $f \colon \mathbb{T} \rightarrow \mathbb{C}$, then $$\mathscr{L}_{\mathbb{T}}\{f^{\Delta}\}(z;s) = -f(s) + z\mathscr{L}\{f\}(z),$$ where $\mathscr{L}_{\mathbb{T}}$ denotes the unilateral Laplace transform and $f^{\Delta}$ denotes the delta derivative of $f$.