|
|
| (18 intermediate revisions by the same user not shown) |
| Line 1: |
Line 1: |
| − | 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
| + | For $t \in \mathbb{T}$, the convolution on a [[time scale]] is defined by the formula |
| − | $$(f*g)_{\mathbb{R}}(t)=\displaystyle\int_{\mathbb{R}} f(\tau)g(t-\tau) \mathrm{d}\tau.$$ | + | $$(f*g)(t,s)=\displaystyle\int_{s}^t \hat{f}(t,\sigma(\xi))g(\xi)\Delta \xi,$$ |
| − | The reason the convolution is of interest is because of the so-called convolution theorem for the classical Laplace transform:
| + | 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 $\mathbb{T}=\mathbb{Z}$. |
| − | $$\mathscr{L}_{\mathbb{R}}\{f*g\}(z)=\mathscr{L}_{\mathbb{R}}\{f\}(z)\mathscr{L}_{\mathbb{R}}\{g\}(z).$$
| |
| − | | |
| − | 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 $t \in \mathbb{T}$,
| |
| − | $$(f*g)_{\mathbb{T}}(t;t_0)=\displaystyle\int_{t_0}^t \hat{f}(t,\sigma(s))g(s)\Delta s.$$ | |
| | | | |
| | =Properties= | | =Properties= |
| − | <div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
| + | [[Covolution theorem for unilateral Laplace transform]]<br /> |
| − | <strong>Theorem:</strong> The following formula holds:
| + | [[Unilateral convolution is associative]]<br /> |
| − | $$\widehat{(f*g)}(t,s)=\displaystyle\int_s^t \hat{f}(t,\sigma(\xi))\hat{g}(\xi,s) \Delta \xi,$$
| + | [[Delta derivative of unilateral convolution]]<br /> |
| − | where $\widehat{(f*g)}$ denotes the solution of the [[shifting problem]] and $(f*g)$ denotes the [[convolution]].
| + | [[Shift of unilateral convolution]]<br /> |
| − | <div class="mw-collapsible-content">
| |
| − | <strong>Proof:</strong> █
| |
| − | </div>
| |
| − | </div>
| |
| − | | |
| − | <div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
| |
| − | <strong>Theorem:</strong> The convolution is associative:
| |
| − | $$(f*g)*h = f*(g*h).$$
| |
| − | <div class="mw-collapsible-content">
| |
| − | <strong>Proof:</strong> █
| |
| − | </div>
| |
| − | </div> | |
| − | | |
| − | <div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
| |
| − | <strong>Theorem:</strong> If $f$ is [[delta derivative|$\Delta$-differentiable]], then
| |
| − | $$(f*g)^{\Delta}=f^{\Delta}*g+f(t_0)g.$$
| |
| − | 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>Corollary:</strong> The following formula holds:
| |
| − | $$\displaystyle\int_{t_0}^t \hat{f}(t,\sigma(\xi))\Delta \xi=\displaystyle\int_{t_0}^t f(\xi) \Delta \xi,$$
| |
| − | where $\hat{f}$ denotes the solution of the [[shifting problem]].
| |
| − | <div class="mw-collapsible-content">
| |
| − | <strong>Proof:</strong> █
| |
| − | </div>
| |
| − | </div> | |
| | | | |
| − | <div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
| + | =See also= |
| − | <strong>Theorem:</strong> Suppose that $\hat{f}$ has partial $\Delta$-derivatives of all orders. Then
| + | [[Shifting problem]] |
| − | $$\dfrac{\partial^k \hat{f}}{\Delta^k t} (t,t)=f^{\Delta^k}(t_0).$$
| |
| − | <div class="mw-collapsible-content">
| |
| − | <strong>Proof:</strong> █
| |
| − | </div>
| |
| − | </div>
| |
| | | | |
| − | <div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
| + | =References= |
| − | <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">
| + | [[Category:Definition]] |
| − | <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>
| |
