Convolutions of two functionsDefine uniform B-spline basis functions via continuous convolutionHow Heaviside...
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Convolutions of two functions
Define uniform B-spline basis functions via continuous convolutionHow Heaviside step function changes limits of integrationHow to obtain the convolution directly (not graphical) of the two functions $e^{-t}u(t)$ and $e^{-2t}u(t)$?Convolution of Two Shifted FunctionsConvolution integral involving two Heaviside functionsConvolution Integral - Shift propertyQuestion on the uniqueness of the densitiesprobability of two numbers picked from two distributions giving a fixed number multiplicationDifficult convolutions in probability problemsExample of non-zero functions with identically zero convolution
$begingroup$
I am having trouble understanding how you take the convolution of two functions.
For example, if $f_1(x) = x$, with x ranging from [0,X] how do I solve for $f_2(x)$ when
$$f_2(x) = int_{0}^{infty} f_1(x')f_1(x-x') dx'.$$
Do I need to use an indicator function?
Thanks
integration probability-distributions convolution
asked Mar 13 at 16:15
Calvin XuCalvin Xu
11
$endgroup$
add a comment |
$begingroup$
I am having trouble understanding how you take the convolution of two functions.
For example, if $f_1(x) = x$, with x ranging from [0,X] how do I solve for $f_2(x)$ when
$$f_2(x) = int_{0}^{infty} f_1(x')f_1(x-x') dx'.$$
Do I need to use an indicator function?
Thanks
integration probability-distributions convolution
asked Mar 13 at 16:15
Calvin XuCalvin Xu
11
$endgroup$
1
$begingroup$
This integral does not converge for your choice of $f_1$.
$endgroup$
– Mars Plastic
Mar 13 at 16:22
$begingroup$
Also: I suppose $*$ denotes multiplication and not convolution?
$endgroup$
– Mars Plastic
Mar 13 at 17:01
$begingroup$
yes you are correct
$endgroup$
– Calvin Xu
Mar 14 at 17:03
add a comment |
$begingroup$
I am having trouble understanding how you take the convolution of two functions.
For example, if $f_1(x) = x$, with x ranging from [0,X] how do I solve for $f_2(x)$ when
$$f_2(x) = int_{0}^{infty} f_1(x')f_1(x-x') dx'.$$
Do I need to use an indicator function?
Thanks
integration probability-distributions convolution
asked Mar 13 at 16:15
Calvin XuCalvin Xu
11
$endgroup$
I am having trouble understanding how you take the convolution of two functions.
For example, if $f_1(x) = x$, with x ranging from [0,X] how do I solve for $f_2(x)$ when
$$f_2(x) = int_{0}^{infty} f_1(x')f_1(x-x') dx'.$$
Do I need to use an indicator function?
Thanks
integration probability-distributions convolution
integration probability-distributions convolution
asked Mar 13 at 16:15
Calvin XuCalvin Xu
11
asked Mar 13 at 16:15
Calvin XuCalvin Xu
11
edited Mar 14 at 17:02
Calvin Xu
asked Mar 13 at 16:15
Calvin XuCalvin Xu
11
asked Mar 13 at 16:15
Calvin XuCalvin Xu
11
asked Mar 13 at 16:15
Calvin XuCalvin Xu
11
11
1
$begingroup$
This integral does not converge for your choice of $f_1$.
$endgroup$
– Mars Plastic
Mar 13 at 16:22
$begingroup$
Also: I suppose $*$ denotes multiplication and not convolution?
$endgroup$
– Mars Plastic
Mar 13 at 17:01
$begingroup$
yes you are correct
$endgroup$
– Calvin Xu
Mar 14 at 17:03
add a comment |
1
$begingroup$
This integral does not converge for your choice of $f_1$.
$endgroup$
– Mars Plastic
Mar 13 at 16:22
$begingroup$
Also: I suppose $*$ denotes multiplication and not convolution?
$endgroup$
– Mars Plastic
Mar 13 at 17:01
$begingroup$
yes you are correct
$endgroup$
– Calvin Xu
Mar 14 at 17:03
1
1
$begingroup$
This integral does not converge for your choice of $f_1$.
$endgroup$
– Mars Plastic
Mar 13 at 16:22
$begingroup$
This integral does not converge for your choice of $f_1$.
$endgroup$
– Mars Plastic
Mar 13 at 16:22
$begingroup$
Also: I suppose $*$ denotes multiplication and not convolution?
$endgroup$
– Mars Plastic
Mar 13 at 17:01
$begingroup$
Also: I suppose $*$ denotes multiplication and not convolution?
$endgroup$
– Mars Plastic
Mar 13 at 17:01
$begingroup$
yes you are correct
$endgroup$
– Calvin Xu
Mar 14 at 17:03
$begingroup$
yes you are correct
$endgroup$
– Calvin Xu
Mar 14 at 17:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
in your example you sustitute $f_1(x')=x'$ and $f_1(x-x')=x-x'$ in an analagous way and solve the integral having the $x'$ as your integral's variable.
If the integral converges, you will get a result in terms of x and it will be the convolution of $f_1$ with itself. if the integral dows not converge, you can't make the convolution.
answered Mar 13 at 17:07
Carlos Adrian Perez EstradaCarlos Adrian Perez Estrada
414
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
in your example you sustitute $f_1(x')=x'$ and $f_1(x-x')=x-x'$ in an analagous way and solve the integral having the $x'$ as your integral's variable.
If the integral converges, you will get a result in terms of x and it will be the convolution of $f_1$ with itself. if the integral dows not converge, you can't make the convolution.
answered Mar 13 at 17:07
Carlos Adrian Perez EstradaCarlos Adrian Perez Estrada
414
$endgroup$
add a comment |
$begingroup$
in your example you sustitute $f_1(x')=x'$ and $f_1(x-x')=x-x'$ in an analagous way and solve the integral having the $x'$ as your integral's variable.
If the integral converges, you will get a result in terms of x and it will be the convolution of $f_1$ with itself. if the integral dows not converge, you can't make the convolution.
answered Mar 13 at 17:07
Carlos Adrian Perez EstradaCarlos Adrian Perez Estrada
414
$endgroup$
add a comment |
$begingroup$
in your example you sustitute $f_1(x')=x'$ and $f_1(x-x')=x-x'$ in an analagous way and solve the integral having the $x'$ as your integral's variable.
If the integral converges, you will get a result in terms of x and it will be the convolution of $f_1$ with itself. if the integral dows not converge, you can't make the convolution.
answered Mar 13 at 17:07
Carlos Adrian Perez EstradaCarlos Adrian Perez Estrada
414
$endgroup$
in your example you sustitute $f_1(x')=x'$ and $f_1(x-x')=x-x'$ in an analagous way and solve the integral having the $x'$ as your integral's variable.
If the integral converges, you will get a result in terms of x and it will be the convolution of $f_1$ with itself. if the integral dows not converge, you can't make the convolution.
answered Mar 13 at 17:07
Carlos Adrian Perez EstradaCarlos Adrian Perez Estrada
414
answered Mar 13 at 17:07
Carlos Adrian Perez EstradaCarlos Adrian Perez Estrada
414
answered Mar 13 at 17:07
Carlos Adrian Perez EstradaCarlos Adrian Perez Estrada
414
answered Mar 13 at 17:07
Carlos Adrian Perez EstradaCarlos Adrian Perez Estrada
414
414
add a comment |
add a comment |
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$begingroup$
This integral does not converge for your choice of $f_1$.
$endgroup$
– Mars Plastic
Mar 13 at 16:22
$begingroup$
Also: I suppose $*$ denotes multiplication and not convolution?
$endgroup$
– Mars Plastic
Mar 13 at 17:01
$begingroup$
yes you are correct
$endgroup$
– Calvin Xu
Mar 14 at 17:03