Convolution of exponential and cosine












-1












$begingroup$


How to calculate this convolution $h(t)= e^{-at}astcos(wt)$



I have tried to put $ e^{-at}$ as my $f(t-r)$ and $cos(wt)$ as $g(t)$ , and I have found $h(t)=e^{-at} ast L(cos(wt))$.



But I don't know if I am right.






convolution







share|cite|improve this question











$endgroup$












  • $begingroup$
    Just take Laplace transforms and use that.
    $endgroup$
    – Sean Roberson
    Jan 6 at 7:37










  • $begingroup$
    Thanks,but is is it right?
    $endgroup$
    – Rmili Jad
    Jan 6 at 7:39










  • $begingroup$
    Not quite. The Laplace transform of the convolution is easy, and inverting it should be no problem. Alternatively, attempt the integral via parts.
    $endgroup$
    – Sean Roberson
    Jan 6 at 7:40










  • $begingroup$
    Thanks.If i want to use the Laplace transform then inverting in it , we will have the multiplication of 2 Dirac distributions, but i don't know what is the multiplication of 2 Dirac distributions.
    $endgroup$
    – Rmili Jad
    Jan 6 at 7:59
















-1












$begingroup$


How to calculate this convolution $h(t)= e^{-at}astcos(wt)$



I have tried to put $ e^{-at}$ as my $f(t-r)$ and $cos(wt)$ as $g(t)$ , and I have found $h(t)=e^{-at} ast L(cos(wt))$.



But I don't know if I am right.






convolution







share|cite|improve this question











$endgroup$












  • $begingroup$
    Just take Laplace transforms and use that.
    $endgroup$
    – Sean Roberson
    Jan 6 at 7:37










  • $begingroup$
    Thanks,but is is it right?
    $endgroup$
    – Rmili Jad
    Jan 6 at 7:39










  • $begingroup$
    Not quite. The Laplace transform of the convolution is easy, and inverting it should be no problem. Alternatively, attempt the integral via parts.
    $endgroup$
    – Sean Roberson
    Jan 6 at 7:40










  • $begingroup$
    Thanks.If i want to use the Laplace transform then inverting in it , we will have the multiplication of 2 Dirac distributions, but i don't know what is the multiplication of 2 Dirac distributions.
    $endgroup$
    – Rmili Jad
    Jan 6 at 7:59














-1












-1








-1





$begingroup$


How to calculate this convolution $h(t)= e^{-at}astcos(wt)$



I have tried to put $ e^{-at}$ as my $f(t-r)$ and $cos(wt)$ as $g(t)$ , and I have found $h(t)=e^{-at} ast L(cos(wt))$.



But I don't know if I am right.






convolution







share|cite|improve this question











$endgroup$




How to calculate this convolution $h(t)= e^{-at}astcos(wt)$



I have tried to put $ e^{-at}$ as my $f(t-r)$ and $cos(wt)$ as $g(t)$ , and I have found $h(t)=e^{-at} ast L(cos(wt))$.



But I don't know if I am right.






convolution




convolution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 7:41









John Wayland Bales

13.9k21238




13.9k21238










asked Jan 6 at 7:04









Rmili JadRmili Jad

1




1












  • $begingroup$
    Just take Laplace transforms and use that.
    $endgroup$
    – Sean Roberson
    Jan 6 at 7:37










  • $begingroup$
    Thanks,but is is it right?
    $endgroup$
    – Rmili Jad
    Jan 6 at 7:39










  • $begingroup$
    Not quite. The Laplace transform of the convolution is easy, and inverting it should be no problem. Alternatively, attempt the integral via parts.
    $endgroup$
    – Sean Roberson
    Jan 6 at 7:40










  • $begingroup$
    Thanks.If i want to use the Laplace transform then inverting in it , we will have the multiplication of 2 Dirac distributions, but i don't know what is the multiplication of 2 Dirac distributions.
    $endgroup$
    – Rmili Jad
    Jan 6 at 7:59


















  • $begingroup$
    Just take Laplace transforms and use that.
    $endgroup$
    – Sean Roberson
    Jan 6 at 7:37










  • $begingroup$
    Thanks,but is is it right?
    $endgroup$
    – Rmili Jad
    Jan 6 at 7:39










  • $begingroup$
    Not quite. The Laplace transform of the convolution is easy, and inverting it should be no problem. Alternatively, attempt the integral via parts.
    $endgroup$
    – Sean Roberson
    Jan 6 at 7:40










  • $begingroup$
    Thanks.If i want to use the Laplace transform then inverting in it , we will have the multiplication of 2 Dirac distributions, but i don't know what is the multiplication of 2 Dirac distributions.
    $endgroup$
    – Rmili Jad
    Jan 6 at 7:59
















$begingroup$
Just take Laplace transforms and use that.
$endgroup$
– Sean Roberson
Jan 6 at 7:37




$begingroup$
Just take Laplace transforms and use that.
$endgroup$
– Sean Roberson
Jan 6 at 7:37












$begingroup$
Thanks,but is is it right?
$endgroup$
– Rmili Jad
Jan 6 at 7:39




$begingroup$
Thanks,but is is it right?
$endgroup$
– Rmili Jad
Jan 6 at 7:39












$begingroup$
Not quite. The Laplace transform of the convolution is easy, and inverting it should be no problem. Alternatively, attempt the integral via parts.
$endgroup$
– Sean Roberson
Jan 6 at 7:40




$begingroup$
Not quite. The Laplace transform of the convolution is easy, and inverting it should be no problem. Alternatively, attempt the integral via parts.
$endgroup$
– Sean Roberson
Jan 6 at 7:40












$begingroup$
Thanks.If i want to use the Laplace transform then inverting in it , we will have the multiplication of 2 Dirac distributions, but i don't know what is the multiplication of 2 Dirac distributions.
$endgroup$
– Rmili Jad
Jan 6 at 7:59




$begingroup$
Thanks.If i want to use the Laplace transform then inverting in it , we will have the multiplication of 2 Dirac distributions, but i don't know what is the multiplication of 2 Dirac distributions.
$endgroup$
– Rmili Jad
Jan 6 at 7:59










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