Why is the region of the integral in convolution truncated to t when the functions are zero in negative t?

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See https://en.wikipedia.org/wiki/Convolution#Definition where the equation is mentioned:



$(f * g )(t) = int_0^t f(tau) g(t - tau), dtau text for f, g : [0, infty) to mathbbR$



But the basic equation is:



$(f * g )(t) , stackrelmathrmdef= int_-infty^infty f(tau) g(t - tau) , dtau \
= int_-infty^infty f(t-tau) g(tau), dtau.
$



So what happened to the integral from $t$ to $infty$?



Is it just a hack to make the value more local and symmetrica? Is there an actual equivalence? Is it just a typo?




integration convolution




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    See https://en.wikipedia.org/wiki/Convolution#Definition where the equation is mentioned:



    $(f * g )(t) = int_0^t f(tau) g(t - tau), dtau text for f, g : [0, infty) to mathbbR$



    But the basic equation is:



    $(f * g )(t) , stackrelmathrmdef= int_-infty^infty f(tau) g(t - tau) , dtau \
    = int_-infty^infty f(t-tau) g(tau), dtau.
    $



    So what happened to the integral from $t$ to $infty$?



    Is it just a hack to make the value more local and symmetrica? Is there an actual equivalence? Is it just a typo?




    integration convolution




    share|cite|improve this question





















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      See https://en.wikipedia.org/wiki/Convolution#Definition where the equation is mentioned:



      $(f * g )(t) = int_0^t f(tau) g(t - tau), dtau text for f, g : [0, infty) to mathbbR$



      But the basic equation is:



      $(f * g )(t) , stackrelmathrmdef= int_-infty^infty f(tau) g(t - tau) , dtau \
      = int_-infty^infty f(t-tau) g(tau), dtau.
      $



      So what happened to the integral from $t$ to $infty$?



      Is it just a hack to make the value more local and symmetrica? Is there an actual equivalence? Is it just a typo?




      integration convolution




      share|cite|improve this question











      See https://en.wikipedia.org/wiki/Convolution#Definition where the equation is mentioned:



      $(f * g )(t) = int_0^t f(tau) g(t - tau), dtau text for f, g : [0, infty) to mathbbR$



      But the basic equation is:



      $(f * g )(t) , stackrelmathrmdef= int_-infty^infty f(tau) g(t - tau) , dtau \
      = int_-infty^infty f(t-tau) g(tau), dtau.
      $



      So what happened to the integral from $t$ to $infty$?



      Is it just a hack to make the value more local and symmetrica? Is there an actual equivalence? Is it just a typo?





      integration convolution





      share|cite|improve this question










      share|cite|improve this question




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      asked Aug 2 at 22:45









      ubershmekel

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          When I hit the submit button I figured it out. Because $f$ and $g$ are zero for all negative $t$ values, any non-local value will be zero as well because $g(t - tau)$ will be zero for all $tau > t$ which causes $f(tau) g(t - tau)$ to be zero.






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            When I hit the submit button I figured it out. Because $f$ and $g$ are zero for all negative $t$ values, any non-local value will be zero as well because $g(t - tau)$ will be zero for all $tau > t$ which causes $f(tau) g(t - tau)$ to be zero.






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              up vote
              0
              down vote













              When I hit the submit button I figured it out. Because $f$ and $g$ are zero for all negative $t$ values, any non-local value will be zero as well because $g(t - tau)$ will be zero for all $tau > t$ which causes $f(tau) g(t - tau)$ to be zero.






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                up vote
                0
                down vote










                up vote
                0
                down vote









                When I hit the submit button I figured it out. Because $f$ and $g$ are zero for all negative $t$ values, any non-local value will be zero as well because $g(t - tau)$ will be zero for all $tau > t$ which causes $f(tau) g(t - tau)$ to be zero.






                share|cite|improve this answer















                When I hit the submit button I figured it out. Because $f$ and $g$ are zero for all negative $t$ values, any non-local value will be zero as well because $g(t - tau)$ will be zero for all $tau > t$ which causes $f(tau) g(t - tau)$ to be zero.







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                share|cite|improve this answer



                share|cite|improve this answer








                edited Aug 3 at 16:33


























                answered Aug 2 at 22:46









                ubershmekel

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