Mixing
Exponential Fourier transform and derivative
$begingroup$
The Fourier transform of right handed exponential function $e^{-alpha t}$ is given by
$$mathcal{F}left[ e^{-alpha t}right] = frac{1}{alpha + j omega}.$$
Its derivative is
$$frac{d}{dt} e^{-alpha t} = -alpha e^{-alpha t}.$$
So the Fourier transform of derivative is
$$mathcal{F}left[ frac{d}{dt}e^{-alpha t}right] = -frac{alpha}{alpha + j omega}.$$
But if I use the Fourier transform derivative formula
$$j omega F(omega) = frac{j omega}{alpha + j omega}.$$
These results are different. Where is my mistake?
fourier-transform
edited Jan 19 at 21:21
Glorfindel
3,41981830
asked Jan 19 at 21:02
Stef1611Stef1611
32
$endgroup$
add a comment |
$begingroup$
The Fourier transform of right handed exponential function $e^{-alpha t}$ is given by
$$mathcal{F}left[ e^{-alpha t}right] = frac{1}{alpha + j omega}.$$
Its derivative is
$$frac{d}{dt} e^{-alpha t} = -alpha e^{-alpha t}.$$
So the Fourier transform of derivative is
$$mathcal{F}left[ frac{d}{dt}e^{-alpha t}right] = -frac{alpha}{alpha + j omega}.$$
But if I use the Fourier transform derivative formula
$$j omega F(omega) = frac{j omega}{alpha + j omega}.$$
These results are different. Where is my mistake?
fourier-transform
edited Jan 19 at 21:21
Glorfindel
3,41981830
asked Jan 19 at 21:02
Stef1611Stef1611
32
$endgroup$
add a comment |
$begingroup$
The Fourier transform of right handed exponential function $e^{-alpha t}$ is given by
$$mathcal{F}left[ e^{-alpha t}right] = frac{1}{alpha + j omega}.$$
Its derivative is
$$frac{d}{dt} e^{-alpha t} = -alpha e^{-alpha t}.$$
So the Fourier transform of derivative is
$$mathcal{F}left[ frac{d}{dt}e^{-alpha t}right] = -frac{alpha}{alpha + j omega}.$$
But if I use the Fourier transform derivative formula
$$j omega F(omega) = frac{j omega}{alpha + j omega}.$$
These results are different. Where is my mistake?
fourier-transform
edited Jan 19 at 21:21
Glorfindel
3,41981830
asked Jan 19 at 21:02
Stef1611Stef1611
32
$endgroup$
The Fourier transform of right handed exponential function $e^{-alpha t}$ is given by
$$mathcal{F}left[ e^{-alpha t}right] = frac{1}{alpha + j omega}.$$
Its derivative is
$$frac{d}{dt} e^{-alpha t} = -alpha e^{-alpha t}.$$
So the Fourier transform of derivative is
$$mathcal{F}left[ frac{d}{dt}e^{-alpha t}right] = -frac{alpha}{alpha + j omega}.$$
But if I use the Fourier transform derivative formula
$$j omega F(omega) = frac{j omega}{alpha + j omega}.$$
These results are different. Where is my mistake?
fourier-transform
fourier-transform
edited Jan 19 at 21:21
Glorfindel
3,41981830
asked Jan 19 at 21:02
Stef1611Stef1611
32
edited Jan 19 at 21:21
Glorfindel
3,41981830
asked Jan 19 at 21:02
Stef1611Stef1611
32
edited Jan 19 at 21:21
Glorfindel
3,41981830
edited Jan 19 at 21:21
Glorfindel
3,41981830
edited Jan 19 at 21:21
Glorfindel
3,41981830
3,41981830
asked Jan 19 at 21:02
Stef1611Stef1611
32
asked Jan 19 at 21:02
Stef1611Stef1611
32
asked Jan 19 at 21:02
Stef1611Stef1611
32
32
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your mistake is that the derviative is
$$dfrac{d}{dt}e^{-alpha t}u(t) = e^{-alpha t}delta(t) -alpha e^{-alpha t}u(t)$$
From there, the rest follows and agrees with the derivative theorem of the Fourier transform.
answered Jan 20 at 0:18
Andy WallsAndy Walls
1,754139
$endgroup$
$begingroup$
thanks a lot for your answer. Now, everything is clear. I am sorry but I could not vote up because my reputation is too low. So, I could only accept the answer.
$endgroup$
– Stef1611
Jan 20 at 9:46
$begingroup$
@Stef1611: you're welcome.
$endgroup$
– Andy Walls
Jan 20 at 10:48
add a comment |
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%2f3079826%2fexponential-fourier-transform-and-derivative%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your mistake is that the derviative is
$$dfrac{d}{dt}e^{-alpha t}u(t) = e^{-alpha t}delta(t) -alpha e^{-alpha t}u(t)$$
From there, the rest follows and agrees with the derivative theorem of the Fourier transform.
answered Jan 20 at 0:18
Andy WallsAndy Walls
1,754139
$endgroup$
$begingroup$
thanks a lot for your answer. Now, everything is clear. I am sorry but I could not vote up because my reputation is too low. So, I could only accept the answer.
$endgroup$
– Stef1611
Jan 20 at 9:46
$begingroup$
@Stef1611: you're welcome.
$endgroup$
– Andy Walls
Jan 20 at 10:48
add a comment |
$begingroup$
Your mistake is that the derviative is
$$dfrac{d}{dt}e^{-alpha t}u(t) = e^{-alpha t}delta(t) -alpha e^{-alpha t}u(t)$$
From there, the rest follows and agrees with the derivative theorem of the Fourier transform.
answered Jan 20 at 0:18
Andy WallsAndy Walls
1,754139
$endgroup$
$begingroup$
thanks a lot for your answer. Now, everything is clear. I am sorry but I could not vote up because my reputation is too low. So, I could only accept the answer.
$endgroup$
– Stef1611
Jan 20 at 9:46
$begingroup$
@Stef1611: you're welcome.
$endgroup$
– Andy Walls
Jan 20 at 10:48
add a comment |
$begingroup$
Your mistake is that the derviative is
$$dfrac{d}{dt}e^{-alpha t}u(t) = e^{-alpha t}delta(t) -alpha e^{-alpha t}u(t)$$
From there, the rest follows and agrees with the derivative theorem of the Fourier transform.
answered Jan 20 at 0:18
Andy WallsAndy Walls
1,754139
$endgroup$
Your mistake is that the derviative is
$$dfrac{d}{dt}e^{-alpha t}u(t) = e^{-alpha t}delta(t) -alpha e^{-alpha t}u(t)$$
From there, the rest follows and agrees with the derivative theorem of the Fourier transform.
answered Jan 20 at 0:18
Andy WallsAndy Walls
1,754139
answered Jan 20 at 0:18
Andy WallsAndy Walls
1,754139
answered Jan 20 at 0:18
Andy WallsAndy Walls
1,754139
answered Jan 20 at 0:18
Andy WallsAndy Walls
1,754139
1,754139
$begingroup$
thanks a lot for your answer. Now, everything is clear. I am sorry but I could not vote up because my reputation is too low. So, I could only accept the answer.
$endgroup$
– Stef1611
Jan 20 at 9:46
$begingroup$
@Stef1611: you're welcome.
$endgroup$
– Andy Walls
Jan 20 at 10:48
add a comment |
$begingroup$
thanks a lot for your answer. Now, everything is clear. I am sorry but I could not vote up because my reputation is too low. So, I could only accept the answer.
$endgroup$
– Stef1611
Jan 20 at 9:46
$begingroup$
@Stef1611: you're welcome.
$endgroup$
– Andy Walls
Jan 20 at 10:48
$begingroup$
thanks a lot for your answer. Now, everything is clear. I am sorry but I could not vote up because my reputation is too low. So, I could only accept the answer.
$endgroup$
– Stef1611
Jan 20 at 9:46
$begingroup$
thanks a lot for your answer. Now, everything is clear. I am sorry but I could not vote up because my reputation is too low. So, I could only accept the answer.
$endgroup$
– Stef1611
Jan 20 at 9:46
$begingroup$
@Stef1611: you're welcome.
$endgroup$
– Andy Walls
Jan 20 at 10:48
$begingroup$
@Stef1611: you're welcome.
$endgroup$
– Andy Walls
Jan 20 at 10:48
add a comment |
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.
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%2f3079826%2fexponential-fourier-transform-and-derivative%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
