Difference between revisions of "Unilateral convolution"
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Revision as of 18:35, 8 February 2016
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 $$(f*g)_{\mathbb{R}}(t)=\displaystyle\int_{\mathbb{R}} f(\tau)g(t-\tau) \mathrm{d}\tau.$$ The reason the convolution is of interest is because of the so-called convolution theorem for the classical Laplace transform: $$\mathscr{L}_{\mathbb{R}}\{f*g\}(z)=\mathscr{L}_{\mathbb{R}}\{f\}(z)\mathscr{L}_{\mathbb{R}}\{g\}(z).$$
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
Theorem: The following formula holds: $$\widehat{(f*g)}(t,s)=\displaystyle\int_s^t \hat{f}(t,\sigma(\xi))\hat{g}(\xi,s) \Delta \xi,$$ where $\widehat{(f*g)}$ denotes the solution of the shifting problem and $(f*g)$ denotes the convolution.
Proof: █
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: █
Corollary: 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.
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: █
