Difference between revisions of "Covolution theorem for unilateral Laplace transform"
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(Created page with "==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 so...") |
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The following formula holds: | The following formula holds: | ||
$$\widehat{(f*g)}(t,s)=\displaystyle\int_s^t \hat{f}(t,\sigma(\xi))\hat{g}(\xi,s) \Delta \xi,$$ | $$\widehat{(f*g)}(t,s)=\displaystyle\int_s^t \hat{f}(t,\sigma(\xi))\hat{g}(\xi,s) \Delta \xi,$$ | ||
− | where $\widehat{(f*g)}$ denotes the solution of the [[shifting problem]] and $(f*g)$ denotes the [[convolution]]. | + | where $\widehat{(f*g)}$ denotes the solution of the [[shifting problem]] and $(f*g)$ denotes the [[unilateral convolution]]. |
==Proof== | ==Proof== |
Revision as of 13:42, 20 January 2023
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 unilateral convolution.