Difference between revisions of "Unilateral convolution"

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Let $f,g \colon \mathbb{R} \rightarrow \mathbb{R}$ be [[Lebesgue integrable]] on $\mathbb{R}$. The classical (i.e. [[time scale]] $\mathbb{T}=\mathbb{R}$) convolution of $f$ and $g$ is the function $f*g \colon \mathbb{R} \rightarrow \mathbb{R}$ given by
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For $t \in \mathbb{T}$, the convolution on a [[time scale]] is defined by the formula
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$$(f*g)_{\mathbb{R}}(t)=\displaystyle\int_{\mathbb{R}} f(\tau)g(t-\tau) \mathrm{d}\tau.$$
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$$(f*g)_{\mathbb{T}}(t,s)=\displaystyle\int_{s}^t \hat{f}(t,\sigma(\xi))g(\xi)\Delta \xi,$$
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The reason the convolution is of interest is because of the so-called convolution theorem for the classical Laplace transform:
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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.
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$$\mathscr{L}_{\mathbb{R}}\{f*g\}(z)=\mathscr{L}_{\mathbb{R}}\{f\}(z)\mathscr{L}_{\mathbb{R}}\{g\}(z).$$
 
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Let $\mathbb{T}$ be any time scale and $f,g \colon \mathbb{T} \rightarrow \mathbb{C}$ be [[delta integrable]] on $\mathbb{T}$. We cannot simply use the definition of the convolution for time scales because an arbitrary time scale is not closed under addition and subtraction. The integrand $f(s)g(t-s) \mathrm{d}s$ will be written in the time scale convolution as $\hat{f}(t,\sigma(s))g(s) \Delta s$. Thus we define for $t \in \mathbb{T}$,
 
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$$(f*g)_{\mathbb{T}}(t,s)=\displaystyle\int_{s}^t \hat{f}(t,\sigma(\xi))g(\xi)\Delta \xi.$$
 
  
 
=Properties=
 
=Properties=
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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[[Covolution theorem for unilateral Laplace transform]]
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<strong>Theorem:</strong> The following formula holds:
 
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$$\widehat{(f*g)}(t,s)=\displaystyle\int_s^t \hat{f}(t,\sigma(\xi))\hat{g}(\xi,s) \Delta \xi,$$
 
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where $\widehat{(f*g)}$ denotes the solution of the [[shifting problem]] and $(f*g)$ denotes the [[convolution]].
 
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<div class="mw-collapsible-content">
 
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<strong>Proof:</strong> █
 
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</div>
 
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</div>
 
  
 
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">

Revision as of 13:36, 20 January 2023

For $t \in \mathbb{T}$, the convolution on a time scale is defined by the formula $$(f*g)_{\mathbb{T}}(t,s)=\displaystyle\int_{s}^t \hat{f}(t,\sigma(\xi))g(\xi)\Delta \xi,$$ 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.

Properties

Covolution theorem for unilateral Laplace transform

Theorem: The convolution is associative: $$(f*g)*h = f*(g*h).$$

Proof: █

Theorem: If $f$ is $\Delta$-differentiable, then $$(f*g)^{\Delta}=f^{\Delta}*g+f(t_0)g.$$ If $g$ is $\Delta$-differentiable, then $$(f*g)^{\Delta}=f*g^{\Delta}+fg(t_0).$$

Proof: █

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: (Convolution theorem) The following formula holds: $$\mathscr{L}_{\mathbb{T}}\{f*g\}(z)=\mathscr{L}\{f\}(z) \mathscr{L}\{g\}(z),$$ where $\mathscr{L}$ denotes the Laplace transform and $f*g$ denotes the convolution.

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: █

See also

Shifting problem

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