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
edited Jan 6 at 7:41
John Wayland Bales
13.9k21238
asked Jan 6 at 7:04
Rmili JadRmili Jad
1
$endgroup$
add a comment |
-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
edited Jan 6 at 7:41
John Wayland Bales
13.9k21238
asked Jan 6 at 7:04
Rmili JadRmili Jad
1
$endgroup$
$begingroup$
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
add a comment |
-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
edited Jan 6 at 7:41
John Wayland Bales
13.9k21238
asked Jan 6 at 7:04
Rmili JadRmili Jad
1
$endgroup$
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
edited Jan 6 at 7:41
John Wayland Bales
13.9k21238
asked Jan 6 at 7:04
Rmili JadRmili Jad
1
edited Jan 6 at 7:41
John Wayland Bales
13.9k21238
asked Jan 6 at 7:04
Rmili JadRmili Jad
1
edited Jan 6 at 7:41
John Wayland Bales
13.9k21238
edited Jan 6 at 7:41
John Wayland Bales
13.9k21238
edited Jan 6 at 7:41
John Wayland Bales
13.9k21238
13.9k21238
asked Jan 6 at 7:04
Rmili JadRmili Jad
1
asked Jan 6 at 7:04
Rmili JadRmili Jad
1
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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