Difference between revisions of "Unilateral convolution"

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(Properties)
(Properties)
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<strong>Theorem:</strong> (Convolution theorem) The following formula holds:
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<strong>Theorem:</strong> (Convolution theorem)  
$$\mathscr{L}_{\mathbb{T}}\{f*g\}(z)=\mathscr{L}\{f\}(z) \mathscr{L}\{g\}(z),$$
 
where $\mathscr{L}$ denotes the [[Laplace transform]] and $f*g$ denotes the [[convolution]].
 
 
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<strong>Proof:</strong>  █  
 
<strong>Proof:</strong>  █  

Revision as of 13:43, 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.

Properties

Covolution theorem for unilateral Laplace transform
Unilateral convolution is associative
Delta derivative 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: (Convolution theorem)

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