Difference between revisions of "Unilateral convolution"

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For $t \in \mathbb{T}$, the convolution on a [[time scale]] is defined by the formula
 
For $t \in \mathbb{T}$, the convolution on a [[time scale]] is defined by the formula
$$(f*g)_{\mathbb{T}}(t,s)=\displaystyle\int_{s}^t \hat{f}(t,\sigma(\xi))g(\xi)\Delta \xi,$$
+
$$(f*g)(t,s)=\displaystyle\int_{s}^t \hat{f}(t,\sigma(\xi))g(\xi)\Delta \xi,$$
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.
+
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=
 
=Properties=
[[Covolution theorem for unilateral Laplace transform]]
+
[[Covolution theorem for unilateral Laplace transform]]<br />
[[Convolution is associative]]
+
[[Unilateral convolution is associative]]<br />
 +
[[Delta derivative of unilateral convolution]]<br />
 +
[[Shift of unilateral convolution]]<br />
  
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
+
=See also=
<strong>Theorem:</strong> If $f$ is [[delta derivative|$\Delta$-differentiable]], then
+
[[Shifting problem]]
$$(f*g)^{\Delta}=f^{\Delta}*g+f(t_0)g.$$
 
If $g$ is [[delta derivative|$\Delta$-differentiable]], then
 
$$(f*g)^{\Delta}=f*g^{\Delta}+fg(t_0).$$
 
<div class="mw-collapsible-content">
 
<strong>Proof:</strong>  █
 
</div>
 
</div>
 
 
 
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<strong>Theorem:</strong> Suppose that $\hat{f}$ has partial $\Delta$-derivatives of all orders. Then
 
$$\dfrac{\partial^k \hat{f}}{\Delta^k t} (t,t)=f^{\Delta^k}(t_0).$$
 
<div class="mw-collapsible-content">
 
<strong>Proof:</strong>  █
 
</div>
 
</div>
 
 
 
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<strong>Theorem:</strong> (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]].
 
<div class="mw-collapsible-content">
 
<strong>Proof:</strong>  █
 
</div>
 
</div>
 
  
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
+
=References=
<strong>Theorem:</strong> Define $u_a(t)= \left\{\begin{array}{ll} 0 &; t < a \\
 
1 &; t \geq a \end{array} \right..$ Then
 
$$\mathscr{L}_{\mathbb{T}}\{u_s \hat{f}(\cdot,s) \}(z) = e_{\ominus z}(s,t_0)\mathscr{L}_{\mathbb{T}}\{f\}(z).$$
 
<div class="mw-collapsible-content">
 
<strong>Proof:</strong>  █
 
</div>
 
</div>
 
  
=See also=
+
[[Category:Definition]]
[[Shifting problem]]
 

Latest revision as of 15:21, 21 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

See also

Shifting problem

References