Difference between revisions of "Unilateral convolution"
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<strong>Theorem:</strong> Suppose that $\hat{f}$ has partial $\Delta$-derivatives of all orders. Then | <strong>Theorem:</strong> Suppose that $\hat{f}$ has partial $\Delta$-derivatives of all orders. Then | ||
$$\dfrac{\partial^k \hat{f}}{\Delta^k t} (t,t)=f^{\Delta^k}(t_0).$$ | $$\dfrac{\partial^k \hat{f}}{\Delta^k t} (t,t)=f^{\Delta^k}(t_0).$$ | ||
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<div class="mw-collapsible-content"> | <div class="mw-collapsible-content"> | ||
<strong>Proof:</strong> █ | <strong>Proof:</strong> █ | ||
Revision as of 13:47, 20 January 2023
For $t \in \mathbb{T}$, the convolution on a time scale is defined by the formula $$(f*g)(t,s)=\displaystyle\int_{s}^t \hat{f}(t,\sigma(\xi))g(\xi)\Delta \xi,$$ where $\hat{f}$ denotes the solution of the shifting problem. The classic definition of the convolution using a shift in the integrand is not appropriate for time scales, since a time scale is not closed under addition and subtraction, but this definition does reduce to the classical definition in the cases of $\mathbb{T}=\mathbb{R}$ and $\mathbb{T}=\mathbb{Z}$.
Properties
Covolution theorem for unilateral Laplace transform
Unilateral convolution is associative
Delta derivative of unilateral convolution
Shift of unilateral convolution
Theorem: Suppose that $\hat{f}$ has partial $\Delta$-derivatives of all orders. Then $$\dfrac{\partial^k \hat{f}}{\Delta^k t} (t,t)=f^{\Delta^k}(t_0).$$
Proof: █
Theorem: Define $u_a(t)= \left\{\begin{array}{ll} 0 &; t < a \\ 1 &; t \geq a \end{array} \right..$ Then $$\mathscr{L}_{\mathbb{T}}\{u_s \hat{f}(\cdot,s) \}(z) = e_{\ominus z}(s,t_0)\mathscr{L}_{\mathbb{T}}\{f\}(z).$$
Proof: █
