Two dimensional Fourier Transform and Convolution of a periodic function
I want to calculate the Fourier transform of the following function
$$
frac{1}{T}e^{iomega_n t}e^{-iomega_m t^prime}Theta(t-t^prime),
$$
where $omega_n=frac{2pi}{T}n$ ($nin mathbb{Z}$) and $Theta$ is the Heaviside step function
$$
Theta(t-t^prime)=
begin{cases}
1, ,, t-t^primegeq 0\
0, ,, t-t^prime< 0, .
end{cases}
$$
Since it depends on two variables the transformation reads as follows
$$
frac{1}{T^2}int_0^Tdt, dt^prime , e^{i(omega_n-omega_p) t}e^{i(omega_l-omega_m) t^prime}Theta(t-t^prime),
$$
the Fourier transform only depends on the frequency differences, so I define $Delta_{np}equivomega_n-omega_p$ and $Delta_{lm}equivomega_l-omega_m$.
The convolution of a periodic function is defined as
$$
(f*g)(t)=frac{1}{sqrt{T}}int_0^Tdt^prime f(t^prime)g(t-t^prime).
$$
Hence, I interpret the two dimensional Fourier transform as a one dimensional Fourier transform of a convolution where $f(t^prime)=e^{i(omega_l-omega_m) t^prime}$ and $g(t-t^prime)=Theta(t-t^prime)$. With the substitutions $q=t-t^prime$ ($dq=dt$) and $t=t^prime+q$, the integrals can be written as the product of Fourier transforms
$$
frac{1}{T^2}int_0^Tdq, Theta(q)e^{iDelta_{np} q}int_0^Tdt^prime, e^{i(Delta_{lm}+Delta_{np}) t^prime}.
$$
With this substitution, the step function can be omitted because we integrate from $0$ to the fundamental period $T$. Now, if one of the frequency differences are non zero, then the integral vanishes and if both are zero then the integral is rendered $1$. Hence,
$$
frac{1}{T^2}int_0^Tdq, Theta(q)e^{iDelta_{np} q}int_0^Tdt^prime, e^{i(Delta_{lm}+Delta_{np}) t^prime}=delta_{n,p}delta_{m,l}.
$$
However, if I go back to my original integral and insert $Delta_{np}=Delta_{lm}=0$, I get
$$
frac{1}{T^2}int_0^Tdtint_0^T dt^prime , Theta(t-t^prime)=frac{1}{T^2}int_0^Tdtint_0^t dt^prime=frac{1}{2}.
$$
Now which result is the correct one? I have been calculating all day now to figure a way out of this ambiguity. How do I calculate this Fourier Transform of the convolution??
Thank you in advance
calculus complex-analysis fourier-transform convolution
asked Nov 26 at 18:49
mr. curious
909
|
show 1 more comment
I want to calculate the Fourier transform of the following function
$$
frac{1}{T}e^{iomega_n t}e^{-iomega_m t^prime}Theta(t-t^prime),
$$
where $omega_n=frac{2pi}{T}n$ ($nin mathbb{Z}$) and $Theta$ is the Heaviside step function
$$
Theta(t-t^prime)=
begin{cases}
1, ,, t-t^primegeq 0\
0, ,, t-t^prime< 0, .
end{cases}
$$
Since it depends on two variables the transformation reads as follows
$$
frac{1}{T^2}int_0^Tdt, dt^prime , e^{i(omega_n-omega_p) t}e^{i(omega_l-omega_m) t^prime}Theta(t-t^prime),
$$
the Fourier transform only depends on the frequency differences, so I define $Delta_{np}equivomega_n-omega_p$ and $Delta_{lm}equivomega_l-omega_m$.
The convolution of a periodic function is defined as
$$
(f*g)(t)=frac{1}{sqrt{T}}int_0^Tdt^prime f(t^prime)g(t-t^prime).
$$
Hence, I interpret the two dimensional Fourier transform as a one dimensional Fourier transform of a convolution where $f(t^prime)=e^{i(omega_l-omega_m) t^prime}$ and $g(t-t^prime)=Theta(t-t^prime)$. With the substitutions $q=t-t^prime$ ($dq=dt$) and $t=t^prime+q$, the integrals can be written as the product of Fourier transforms
$$
frac{1}{T^2}int_0^Tdq, Theta(q)e^{iDelta_{np} q}int_0^Tdt^prime, e^{i(Delta_{lm}+Delta_{np}) t^prime}.
$$
With this substitution, the step function can be omitted because we integrate from $0$ to the fundamental period $T$. Now, if one of the frequency differences are non zero, then the integral vanishes and if both are zero then the integral is rendered $1$. Hence,
$$
frac{1}{T^2}int_0^Tdq, Theta(q)e^{iDelta_{np} q}int_0^Tdt^prime, e^{i(Delta_{lm}+Delta_{np}) t^prime}=delta_{n,p}delta_{m,l}.
$$
However, if I go back to my original integral and insert $Delta_{np}=Delta_{lm}=0$, I get
$$
frac{1}{T^2}int_0^Tdtint_0^T dt^prime , Theta(t-t^prime)=frac{1}{T^2}int_0^Tdtint_0^t dt^prime=frac{1}{2}.
$$
Now which result is the correct one? I have been calculating all day now to figure a way out of this ambiguity. How do I calculate this Fourier Transform of the convolution??
Thank you in advance
calculus complex-analysis fourier-transform convolution
asked Nov 26 at 18:49
mr. curious
909
Ok, thank you for the quick reply. (I guess $h$ is the step function and $delta$ the $delta$ distribution.) How can I use the FT of the step function, when it is multiplied by another. Afterall, I am transforming a product of functions.
– mr. curious
Nov 26 at 18:58
The Fourier transform of $H(x_1,x_2) = h(frac{x_1-x_2}{sqrt{2}})$ is $widehat{H}(xi_1,xi_2)=delta(frac{xi_1+xi_2}{sqrt{2}})widehat{h}(frac{xi_1-xi_2}{sqrt{2}})$
– reuns
Nov 26 at 19:01
Ok, I see. One could consider the frequency differences simply as the Fourier exponentials. But, the function under consideration is a product of two periodic functions and the Heaviside step function. So my integral is calculate over one period, instead of $-infty$ to $infty$....Does that matter?
– mr. curious
Nov 26 at 19:16
I don't understand where you see a convolution. Assume $G(x_1,x_2) = h_1(x_1)h_2(x_2)$, find its Fourier transform, use a change of variable to obtain the FT of $H(x_1,x_2) = h_1(frac{x_1-x_2}{sqrt{2}})h_2(frac{x_1+x_2}{sqrt{2}})$, extend to the case $h_1,h_2$ are distributions
– reuns
Nov 26 at 19:21
The convolution (for periodic functions) is seen in the third and fourth equation $(e^{iDelta_{l,m}t})*(Theta(t))=1/sqrt{T}int_0^Te^{iDelta_{l,m}t^prime}Theta(t-t^prime)$.
– mr. curious
Nov 26 at 19:42
|
show 1 more comment
I want to calculate the Fourier transform of the following function
$$
frac{1}{T}e^{iomega_n t}e^{-iomega_m t^prime}Theta(t-t^prime),
$$
where $omega_n=frac{2pi}{T}n$ ($nin mathbb{Z}$) and $Theta$ is the Heaviside step function
$$
Theta(t-t^prime)=
begin{cases}
1, ,, t-t^primegeq 0\
0, ,, t-t^prime< 0, .
end{cases}
$$
Since it depends on two variables the transformation reads as follows
$$
frac{1}{T^2}int_0^Tdt, dt^prime , e^{i(omega_n-omega_p) t}e^{i(omega_l-omega_m) t^prime}Theta(t-t^prime),
$$
the Fourier transform only depends on the frequency differences, so I define $Delta_{np}equivomega_n-omega_p$ and $Delta_{lm}equivomega_l-omega_m$.
The convolution of a periodic function is defined as
$$
(f*g)(t)=frac{1}{sqrt{T}}int_0^Tdt^prime f(t^prime)g(t-t^prime).
$$
Hence, I interpret the two dimensional Fourier transform as a one dimensional Fourier transform of a convolution where $f(t^prime)=e^{i(omega_l-omega_m) t^prime}$ and $g(t-t^prime)=Theta(t-t^prime)$. With the substitutions $q=t-t^prime$ ($dq=dt$) and $t=t^prime+q$, the integrals can be written as the product of Fourier transforms
$$
frac{1}{T^2}int_0^Tdq, Theta(q)e^{iDelta_{np} q}int_0^Tdt^prime, e^{i(Delta_{lm}+Delta_{np}) t^prime}.
$$
With this substitution, the step function can be omitted because we integrate from $0$ to the fundamental period $T$. Now, if one of the frequency differences are non zero, then the integral vanishes and if both are zero then the integral is rendered $1$. Hence,
$$
frac{1}{T^2}int_0^Tdq, Theta(q)e^{iDelta_{np} q}int_0^Tdt^prime, e^{i(Delta_{lm}+Delta_{np}) t^prime}=delta_{n,p}delta_{m,l}.
$$
However, if I go back to my original integral and insert $Delta_{np}=Delta_{lm}=0$, I get
$$
frac{1}{T^2}int_0^Tdtint_0^T dt^prime , Theta(t-t^prime)=frac{1}{T^2}int_0^Tdtint_0^t dt^prime=frac{1}{2}.
$$
Now which result is the correct one? I have been calculating all day now to figure a way out of this ambiguity. How do I calculate this Fourier Transform of the convolution??
Thank you in advance
calculus complex-analysis fourier-transform convolution
asked Nov 26 at 18:49
mr. curious
909
I want to calculate the Fourier transform of the following function
$$
frac{1}{T}e^{iomega_n t}e^{-iomega_m t^prime}Theta(t-t^prime),
$$
where $omega_n=frac{2pi}{T}n$ ($nin mathbb{Z}$) and $Theta$ is the Heaviside step function
$$
Theta(t-t^prime)=
begin{cases}
1, ,, t-t^primegeq 0\
0, ,, t-t^prime< 0, .
end{cases}
$$
Since it depends on two variables the transformation reads as follows
$$
frac{1}{T^2}int_0^Tdt, dt^prime , e^{i(omega_n-omega_p) t}e^{i(omega_l-omega_m) t^prime}Theta(t-t^prime),
$$
the Fourier transform only depends on the frequency differences, so I define $Delta_{np}equivomega_n-omega_p$ and $Delta_{lm}equivomega_l-omega_m$.
The convolution of a periodic function is defined as
$$
(f*g)(t)=frac{1}{sqrt{T}}int_0^Tdt^prime f(t^prime)g(t-t^prime).
$$
Hence, I interpret the two dimensional Fourier transform as a one dimensional Fourier transform of a convolution where $f(t^prime)=e^{i(omega_l-omega_m) t^prime}$ and $g(t-t^prime)=Theta(t-t^prime)$. With the substitutions $q=t-t^prime$ ($dq=dt$) and $t=t^prime+q$, the integrals can be written as the product of Fourier transforms
$$
frac{1}{T^2}int_0^Tdq, Theta(q)e^{iDelta_{np} q}int_0^Tdt^prime, e^{i(Delta_{lm}+Delta_{np}) t^prime}.
$$
With this substitution, the step function can be omitted because we integrate from $0$ to the fundamental period $T$. Now, if one of the frequency differences are non zero, then the integral vanishes and if both are zero then the integral is rendered $1$. Hence,
$$
frac{1}{T^2}int_0^Tdq, Theta(q)e^{iDelta_{np} q}int_0^Tdt^prime, e^{i(Delta_{lm}+Delta_{np}) t^prime}=delta_{n,p}delta_{m,l}.
$$
However, if I go back to my original integral and insert $Delta_{np}=Delta_{lm}=0$, I get
$$
frac{1}{T^2}int_0^Tdtint_0^T dt^prime , Theta(t-t^prime)=frac{1}{T^2}int_0^Tdtint_0^t dt^prime=frac{1}{2}.
$$
Now which result is the correct one? I have been calculating all day now to figure a way out of this ambiguity. How do I calculate this Fourier Transform of the convolution??
Thank you in advance
calculus complex-analysis fourier-transform convolution
calculus complex-analysis fourier-transform convolution
asked Nov 26 at 18:49
mr. curious
909
asked Nov 26 at 18:49
mr. curious
909
asked Nov 26 at 18:49
mr. curious
909
asked Nov 26 at 18:49
mr. curious
909
asked Nov 26 at 18:49
mr. curious
909
909
Ok, thank you for the quick reply. (I guess $h$ is the step function and $delta$ the $delta$ distribution.) How can I use the FT of the step function, when it is multiplied by another. Afterall, I am transforming a product of functions.
– mr. curious
Nov 26 at 18:58
The Fourier transform of $H(x_1,x_2) = h(frac{x_1-x_2}{sqrt{2}})$ is $widehat{H}(xi_1,xi_2)=delta(frac{xi_1+xi_2}{sqrt{2}})widehat{h}(frac{xi_1-xi_2}{sqrt{2}})$
– reuns
Nov 26 at 19:01
Ok, I see. One could consider the frequency differences simply as the Fourier exponentials. But, the function under consideration is a product of two periodic functions and the Heaviside step function. So my integral is calculate over one period, instead of $-infty$ to $infty$....Does that matter?
– mr. curious
Nov 26 at 19:16
I don't understand where you see a convolution. Assume $G(x_1,x_2) = h_1(x_1)h_2(x_2)$, find its Fourier transform, use a change of variable to obtain the FT of $H(x_1,x_2) = h_1(frac{x_1-x_2}{sqrt{2}})h_2(frac{x_1+x_2}{sqrt{2}})$, extend to the case $h_1,h_2$ are distributions
– reuns
Nov 26 at 19:21
The convolution (for periodic functions) is seen in the third and fourth equation $(e^{iDelta_{l,m}t})*(Theta(t))=1/sqrt{T}int_0^Te^{iDelta_{l,m}t^prime}Theta(t-t^prime)$.
– mr. curious
Nov 26 at 19:42
|
show 1 more comment
Ok, thank you for the quick reply. (I guess $h$ is the step function and $delta$ the $delta$ distribution.) How can I use the FT of the step function, when it is multiplied by another. Afterall, I am transforming a product of functions.
– mr. curious
Nov 26 at 18:58
The Fourier transform of $H(x_1,x_2) = h(frac{x_1-x_2}{sqrt{2}})$ is $widehat{H}(xi_1,xi_2)=delta(frac{xi_1+xi_2}{sqrt{2}})widehat{h}(frac{xi_1-xi_2}{sqrt{2}})$
– reuns
Nov 26 at 19:01
Ok, I see. One could consider the frequency differences simply as the Fourier exponentials. But, the function under consideration is a product of two periodic functions and the Heaviside step function. So my integral is calculate over one period, instead of $-infty$ to $infty$....Does that matter?
– mr. curious
Nov 26 at 19:16
I don't understand where you see a convolution. Assume $G(x_1,x_2) = h_1(x_1)h_2(x_2)$, find its Fourier transform, use a change of variable to obtain the FT of $H(x_1,x_2) = h_1(frac{x_1-x_2}{sqrt{2}})h_2(frac{x_1+x_2}{sqrt{2}})$, extend to the case $h_1,h_2$ are distributions
– reuns
Nov 26 at 19:21
The convolution (for periodic functions) is seen in the third and fourth equation $(e^{iDelta_{l,m}t})*(Theta(t))=1/sqrt{T}int_0^Te^{iDelta_{l,m}t^prime}Theta(t-t^prime)$.
– mr. curious
Nov 26 at 19:42
Ok, thank you for the quick reply. (I guess $h$ is the step function and $delta$ the $delta$ distribution.) How can I use the FT of the step function, when it is multiplied by another. Afterall, I am transforming a product of functions.
– mr. curious
Nov 26 at 18:58
Ok, thank you for the quick reply. (I guess $h$ is the step function and $delta$ the $delta$ distribution.) How can I use the FT of the step function, when it is multiplied by another. Afterall, I am transforming a product of functions.
– mr. curious
Nov 26 at 18:58
The Fourier transform of $H(x_1,x_2) = h(frac{x_1-x_2}{sqrt{2}})$ is $widehat{H}(xi_1,xi_2)=delta(frac{xi_1+xi_2}{sqrt{2}})widehat{h}(frac{xi_1-xi_2}{sqrt{2}})$
– reuns
Nov 26 at 19:01
The Fourier transform of $H(x_1,x_2) = h(frac{x_1-x_2}{sqrt{2}})$ is $widehat{H}(xi_1,xi_2)=delta(frac{xi_1+xi_2}{sqrt{2}})widehat{h}(frac{xi_1-xi_2}{sqrt{2}})$
– reuns
Nov 26 at 19:01
Ok, I see. One could consider the frequency differences simply as the Fourier exponentials. But, the function under consideration is a product of two periodic functions and the Heaviside step function. So my integral is calculate over one period, instead of $-infty$ to $infty$....Does that matter?
– mr. curious
Nov 26 at 19:16
Ok, I see. One could consider the frequency differences simply as the Fourier exponentials. But, the function under consideration is a product of two periodic functions and the Heaviside step function. So my integral is calculate over one period, instead of $-infty$ to $infty$....Does that matter?
– mr. curious
Nov 26 at 19:16
I don't understand where you see a convolution. Assume $G(x_1,x_2) = h_1(x_1)h_2(x_2)$, find its Fourier transform, use a change of variable to obtain the FT of $H(x_1,x_2) = h_1(frac{x_1-x_2}{sqrt{2}})h_2(frac{x_1+x_2}{sqrt{2}})$, extend to the case $h_1,h_2$ are distributions
– reuns
Nov 26 at 19:21
I don't understand where you see a convolution. Assume $G(x_1,x_2) = h_1(x_1)h_2(x_2)$, find its Fourier transform, use a change of variable to obtain the FT of $H(x_1,x_2) = h_1(frac{x_1-x_2}{sqrt{2}})h_2(frac{x_1+x_2}{sqrt{2}})$, extend to the case $h_1,h_2$ are distributions
– reuns
Nov 26 at 19:21
The convolution (for periodic functions) is seen in the third and fourth equation $(e^{iDelta_{l,m}t})*(Theta(t))=1/sqrt{T}int_0^Te^{iDelta_{l,m}t^prime}Theta(t-t^prime)$.
– mr. curious
Nov 26 at 19:42
The convolution (for periodic functions) is seen in the third and fourth equation $(e^{iDelta_{l,m}t})*(Theta(t))=1/sqrt{T}int_0^Te^{iDelta_{l,m}t^prime}Theta(t-t^prime)$.
– mr. curious
Nov 26 at 19:42
|
show 1 more comment
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014753%2ftwo-dimensional-fourier-transform-and-convolution-of-a-periodic-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014753%2ftwo-dimensional-fourier-transform-and-convolution-of-a-periodic-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Ok, thank you for the quick reply. (I guess $h$ is the step function and $delta$ the $delta$ distribution.) How can I use the FT of the step function, when it is multiplied by another. Afterall, I am transforming a product of functions.
– mr. curious
Nov 26 at 18:58
The Fourier transform of $H(x_1,x_2) = h(frac{x_1-x_2}{sqrt{2}})$ is $widehat{H}(xi_1,xi_2)=delta(frac{xi_1+xi_2}{sqrt{2}})widehat{h}(frac{xi_1-xi_2}{sqrt{2}})$
– reuns
Nov 26 at 19:01
Ok, I see. One could consider the frequency differences simply as the Fourier exponentials. But, the function under consideration is a product of two periodic functions and the Heaviside step function. So my integral is calculate over one period, instead of $-infty$ to $infty$....Does that matter?
– mr. curious
Nov 26 at 19:16
I don't understand where you see a convolution. Assume $G(x_1,x_2) = h_1(x_1)h_2(x_2)$, find its Fourier transform, use a change of variable to obtain the FT of $H(x_1,x_2) = h_1(frac{x_1-x_2}{sqrt{2}})h_2(frac{x_1+x_2}{sqrt{2}})$, extend to the case $h_1,h_2$ are distributions
– reuns
Nov 26 at 19:21
The convolution (for periodic functions) is seen in the third and fourth equation $(e^{iDelta_{l,m}t})*(Theta(t))=1/sqrt{T}int_0^Te^{iDelta_{l,m}t^prime}Theta(t-t^prime)$.
– mr. curious
Nov 26 at 19:42