Mixing
Estimating time in harmonic signal
$begingroup$
I hope someone can help me with the following problem:
Assume a periodic signal of the form
$$begin{align}
s(t) &= sumlimits_{p=1}^P sin(pOmega_0t)\
&= sumlimits_{p=1}^P sin(theta_p(t)),
end{align}$$
with $0 leq t < frac{2pi}{Omega_0}$ (only one period of the fundamental frequency).
We take measurements $theta^*_p(t_0)$ of the phases $theta_p(t_0)$ at a certain (unknown) time $t_0$ for $p in [2,ldots,P]$ (i.e. for all harmonics except the fundamental). The measurements are phase wrapped, i.e. projected to the interval $[-pi,pi[$:
$$
theta^*_p(t_0) = theta_p(t_0) + n_p cdot 2pi
$$
How can we estimate/determine the time $t_0$ at which the measurements have been taken?
Here is a diagram to illustrate the problem:
From the diagram it becomes clear that for $p=2$, the phase value $theta^*_2(t_0)$ can occur at two points in time (see red marker). For $p=3$, the phase value $theta^*_3(t_0)$ can occur at three points in time, etc. But only at $t_0$ they all fall together. Can we use this property to estimate $t_0$?
signal-processing harmonic-analysis estimation-theory
edited Jan 19 at 9:25
Glorfindel
3,41981830
asked Jul 3 '13 at 9:38
kofferkoffer
194
$endgroup$
add a comment |
$begingroup$
I hope someone can help me with the following problem:
Assume a periodic signal of the form
$$begin{align}
s(t) &= sumlimits_{p=1}^P sin(pOmega_0t)\
&= sumlimits_{p=1}^P sin(theta_p(t)),
end{align}$$
with $0 leq t < frac{2pi}{Omega_0}$ (only one period of the fundamental frequency).
We take measurements $theta^*_p(t_0)$ of the phases $theta_p(t_0)$ at a certain (unknown) time $t_0$ for $p in [2,ldots,P]$ (i.e. for all harmonics except the fundamental). The measurements are phase wrapped, i.e. projected to the interval $[-pi,pi[$:
$$
theta^*_p(t_0) = theta_p(t_0) + n_p cdot 2pi
$$
How can we estimate/determine the time $t_0$ at which the measurements have been taken?
Here is a diagram to illustrate the problem:
From the diagram it becomes clear that for $p=2$, the phase value $theta^*_2(t_0)$ can occur at two points in time (see red marker). For $p=3$, the phase value $theta^*_3(t_0)$ can occur at three points in time, etc. But only at $t_0$ they all fall together. Can we use this property to estimate $t_0$?
signal-processing harmonic-analysis estimation-theory
edited Jan 19 at 9:25
Glorfindel
3,41981830
asked Jul 3 '13 at 9:38
kofferkoffer
194
$endgroup$
$begingroup$
just try to undestand the problem correctly, given a time $t_0$ and the phase $p=2$ respectively $theta_p(t_0)$, why shouldn't it be possible to find the fundamental that is half of phase?
$endgroup$
– al-Hwarizmi
Jul 4 '13 at 13:06
$begingroup$
Oh, maybe my wording didn't make that clear. The time $t_0$ is not known, only the phase values $theta_p(t_0)$ at that time are known. And as you can see from the diagram, a given phase $theta_p$ of a harmonic can occur at several time positions. I will edit the question to make that clearer.
$endgroup$
– koffer
Jul 4 '13 at 14:01
$begingroup$
Still dont get it, sorry. If you have $2 Omega_0 t_0 = theta_2$ and $3 Omega_0 t_0 = theta_3$ for instance, do you want to calculate $Omega_0$ and $t_0$?
$endgroup$
– al-Hwarizmi
Jul 4 '13 at 14:32
$begingroup$
No, $Omega_0$ is known. However, I can only measure the phase in the interval $[-pi,pi[$ (see my comment to Ron Gordon's answer).
$endgroup$
– koffer
Jul 4 '13 at 14:42
add a comment |
$begingroup$
I hope someone can help me with the following problem:
Assume a periodic signal of the form
$$begin{align}
s(t) &= sumlimits_{p=1}^P sin(pOmega_0t)\
&= sumlimits_{p=1}^P sin(theta_p(t)),
end{align}$$
with $0 leq t < frac{2pi}{Omega_0}$ (only one period of the fundamental frequency).
We take measurements $theta^*_p(t_0)$ of the phases $theta_p(t_0)$ at a certain (unknown) time $t_0$ for $p in [2,ldots,P]$ (i.e. for all harmonics except the fundamental). The measurements are phase wrapped, i.e. projected to the interval $[-pi,pi[$:
$$
theta^*_p(t_0) = theta_p(t_0) + n_p cdot 2pi
$$
How can we estimate/determine the time $t_0$ at which the measurements have been taken?
Here is a diagram to illustrate the problem:
From the diagram it becomes clear that for $p=2$, the phase value $theta^*_2(t_0)$ can occur at two points in time (see red marker). For $p=3$, the phase value $theta^*_3(t_0)$ can occur at three points in time, etc. But only at $t_0$ they all fall together. Can we use this property to estimate $t_0$?
signal-processing harmonic-analysis estimation-theory
edited Jan 19 at 9:25
Glorfindel
3,41981830
asked Jul 3 '13 at 9:38
kofferkoffer
194
$endgroup$
I hope someone can help me with the following problem:
Assume a periodic signal of the form
$$begin{align}
s(t) &= sumlimits_{p=1}^P sin(pOmega_0t)\
&= sumlimits_{p=1}^P sin(theta_p(t)),
end{align}$$
with $0 leq t < frac{2pi}{Omega_0}$ (only one period of the fundamental frequency).
We take measurements $theta^*_p(t_0)$ of the phases $theta_p(t_0)$ at a certain (unknown) time $t_0$ for $p in [2,ldots,P]$ (i.e. for all harmonics except the fundamental). The measurements are phase wrapped, i.e. projected to the interval $[-pi,pi[$:
$$
theta^*_p(t_0) = theta_p(t_0) + n_p cdot 2pi
$$
How can we estimate/determine the time $t_0$ at which the measurements have been taken?
Here is a diagram to illustrate the problem:
From the diagram it becomes clear that for $p=2$, the phase value $theta^*_2(t_0)$ can occur at two points in time (see red marker). For $p=3$, the phase value $theta^*_3(t_0)$ can occur at three points in time, etc. But only at $t_0$ they all fall together. Can we use this property to estimate $t_0$?
signal-processing harmonic-analysis estimation-theory
signal-processing harmonic-analysis estimation-theory
edited Jan 19 at 9:25
Glorfindel
3,41981830
asked Jul 3 '13 at 9:38
kofferkoffer
194
edited Jan 19 at 9:25
Glorfindel
3,41981830
asked Jul 3 '13 at 9:38
kofferkoffer
194
edited Jan 19 at 9:25
Glorfindel
3,41981830
edited Jan 19 at 9:25
Glorfindel
3,41981830
edited Jan 19 at 9:25
Glorfindel
3,41981830
3,41981830
asked Jul 3 '13 at 9:38
kofferkoffer
194
asked Jul 3 '13 at 9:38
kofferkoffer
194
asked Jul 3 '13 at 9:38
kofferkoffer
194
194
$begingroup$
just try to undestand the problem correctly, given a time $t_0$ and the phase $p=2$ respectively $theta_p(t_0)$, why shouldn't it be possible to find the fundamental that is half of phase?
$endgroup$
– al-Hwarizmi
Jul 4 '13 at 13:06
$begingroup$
Oh, maybe my wording didn't make that clear. The time $t_0$ is not known, only the phase values $theta_p(t_0)$ at that time are known. And as you can see from the diagram, a given phase $theta_p$ of a harmonic can occur at several time positions. I will edit the question to make that clearer.
$endgroup$
– koffer
Jul 4 '13 at 14:01
$begingroup$
Still dont get it, sorry. If you have $2 Omega_0 t_0 = theta_2$ and $3 Omega_0 t_0 = theta_3$ for instance, do you want to calculate $Omega_0$ and $t_0$?
$endgroup$
– al-Hwarizmi
Jul 4 '13 at 14:32
$begingroup$
No, $Omega_0$ is known. However, I can only measure the phase in the interval $[-pi,pi[$ (see my comment to Ron Gordon's answer).
$endgroup$
– koffer
Jul 4 '13 at 14:42
add a comment |
$begingroup$
just try to undestand the problem correctly, given a time $t_0$ and the phase $p=2$ respectively $theta_p(t_0)$, why shouldn't it be possible to find the fundamental that is half of phase?
$endgroup$
– al-Hwarizmi
Jul 4 '13 at 13:06
$begingroup$
Oh, maybe my wording didn't make that clear. The time $t_0$ is not known, only the phase values $theta_p(t_0)$ at that time are known. And as you can see from the diagram, a given phase $theta_p$ of a harmonic can occur at several time positions. I will edit the question to make that clearer.
$endgroup$
– koffer
Jul 4 '13 at 14:01
$begingroup$
Still dont get it, sorry. If you have $2 Omega_0 t_0 = theta_2$ and $3 Omega_0 t_0 = theta_3$ for instance, do you want to calculate $Omega_0$ and $t_0$?
$endgroup$
– al-Hwarizmi
Jul 4 '13 at 14:32
$begingroup$
No, $Omega_0$ is known. However, I can only measure the phase in the interval $[-pi,pi[$ (see my comment to Ron Gordon's answer).
$endgroup$
– koffer
Jul 4 '13 at 14:42
$begingroup$
just try to undestand the problem correctly, given a time $t_0$ and the phase $p=2$ respectively $theta_p(t_0)$, why shouldn't it be possible to find the fundamental that is half of phase?
$endgroup$
– al-Hwarizmi
Jul 4 '13 at 13:06
$begingroup$
just try to undestand the problem correctly, given a time $t_0$ and the phase $p=2$ respectively $theta_p(t_0)$, why shouldn't it be possible to find the fundamental that is half of phase?
$endgroup$
– al-Hwarizmi
Jul 4 '13 at 13:06
$begingroup$
Oh, maybe my wording didn't make that clear. The time $t_0$ is not known, only the phase values $theta_p(t_0)$ at that time are known. And as you can see from the diagram, a given phase $theta_p$ of a harmonic can occur at several time positions. I will edit the question to make that clearer.
$endgroup$
– koffer
Jul 4 '13 at 14:01
$begingroup$
Oh, maybe my wording didn't make that clear. The time $t_0$ is not known, only the phase values $theta_p(t_0)$ at that time are known. And as you can see from the diagram, a given phase $theta_p$ of a harmonic can occur at several time positions. I will edit the question to make that clearer.
$endgroup$
– koffer
Jul 4 '13 at 14:01
$begingroup$
Still dont get it, sorry. If you have $2 Omega_0 t_0 = theta_2$ and $3 Omega_0 t_0 = theta_3$ for instance, do you want to calculate $Omega_0$ and $t_0$?
$endgroup$
– al-Hwarizmi
Jul 4 '13 at 14:32
$begingroup$
Still dont get it, sorry. If you have $2 Omega_0 t_0 = theta_2$ and $3 Omega_0 t_0 = theta_3$ for instance, do you want to calculate $Omega_0$ and $t_0$?
$endgroup$
– al-Hwarizmi
Jul 4 '13 at 14:32
$begingroup$
No, $Omega_0$ is known. However, I can only measure the phase in the interval $[-pi,pi[$ (see my comment to Ron Gordon's answer).
$endgroup$
– koffer
Jul 4 '13 at 14:42
$begingroup$
No, $Omega_0$ is known. However, I can only measure the phase in the interval $[-pi,pi[$ (see my comment to Ron Gordon's answer).
$endgroup$
– koffer
Jul 4 '13 at 14:42
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If I may have understood your case then
given:
$$2 Omega_0 t_0 = theta_2$$
$$3 Omega_0 t_0 = theta_3$$
$$dots$$
and $theta_2$ and $theta_3$ known, then
$$Omega_0=frac{theta_2}{2 t_0}=frac{theta_3}{3 t_0}=dots=frac{theta_p}{p t_0}$$
So you see from your given constraints, $t_0$ can be any time, (it will change through all harmonics) and the system is under-determined .
The answer under the given constraints is that we can not specify a specific $t_0$ or ($Omega_0$).
For the new information you brought in your question we have now
$$theta^*_p(t_0) - n_p cdot 2pi = theta_p(t_0) $$
hence:
$$Omega_0=frac{theta^*_2(t_0) - n_2 cdot 2pi }{2 t_0}=dots=frac{theta^*_p(t_0) - n_p cdot 2pi }{p t_0}$$
It does not look better for getting $t_0$.
answered Jul 4 '13 at 14:44
al-Hwarizmial-Hwarizmi
2,64511131
$endgroup$
$begingroup$
Sorry, al-Hwarizmi, in my initial question I missed the point that the observed phase values $theta_p$ are actually phase-wrapped to the interval $[-pi,pi[$. I changed that now in the question and denoted the observed (wrapped) phase values by $theta^*_p$ to indicate that they are different from the original phase values $theta_p$.
$endgroup$
– koffer
Jul 4 '13 at 15:17
$begingroup$
no problem. I just tried to extend the answer. In any case I think you can translate the problem in a set of equations and see how many unknown variables versus number of equations or constraints required.
$endgroup$
– al-Hwarizmi
Jul 4 '13 at 15:46
add a comment |
$begingroup$
OK, the phase is obviously linear in time here, so we should assume that the phase measurements, when unwrapped, want to fall on a line. So the idea, albeit a bit crude, it to go through the measurements for increasing $p$, and increase the $p$th phase in increments of $2 pi$ until that phase is larger than the $p-1$th phase. As an example, allow me to glean off some values from your graphs:
$$left ( begin{array} \2 t_0 & 0.33 pi \ 3 t_0 & -0.55 pi \ 4 t_0 & 0.55 pi \ 5 t_0 & -0.39 piend{array} right )$$
Begin by taking the value of the phase at $t=2 t_0$ as is. For the phase at $t=3 t_0$, see that it is less than the phase at $t=2 t_0$, so increase by $2 pi$ to get $1.45 pi$, which is the smallest phase greater than the previous phase; we then move on to $4 t_0$, and you hopefully get the point. When you finish, you get something like this
$$left ( begin{array} \2 t_0 & 0.33 pi \ 3 t_0 & 1.45 pi \ 4 t_0 & 2.55 pi \ 5 t_0 & 3.61 piend{array} right )$$
A best fit line takes the form $theta(p) = -1.844 pi+1.094 pi p$, so plugging in $p=1$ corresponding to $t=t_0$, you get a phase of about $-0.75 pi$ (which makes sense on your graph), so that knowing $Omega_0$, you may then determine the value of $t_0$. I hope that helps.
answered Jul 4 '13 at 13:12
Ron GordonRon Gordon
122k14156266
$endgroup$
$begingroup$
Sorry, apparently I've missed an important detail in my question: I can only measure the phases $theta_p(t_0)$ in the interval $[-pi,pi[$. So the measured phase can be denoted by $theta^*_p(t_0) = theta_p(t_0) + n_p cdot 2pi$. So given a $theta^*_p(t_0)$ we have several possible times depending on $n_p$. In the problem above, the $p$-th harmonic can occur at exactly $p$ possible times. And I assume that the fact that they all fall together at $t_0$ could be used to determine $n_p$ for all harmonics. I will make that clearer in the original question.
$endgroup$
– koffer
Jul 4 '13 at 14:36
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%2f435188%2festimating-time-in-harmonic-signal%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If I may have understood your case then
given:
$$2 Omega_0 t_0 = theta_2$$
$$3 Omega_0 t_0 = theta_3$$
$$dots$$
and $theta_2$ and $theta_3$ known, then
$$Omega_0=frac{theta_2}{2 t_0}=frac{theta_3}{3 t_0}=dots=frac{theta_p}{p t_0}$$
So you see from your given constraints, $t_0$ can be any time, (it will change through all harmonics) and the system is under-determined .
The answer under the given constraints is that we can not specify a specific $t_0$ or ($Omega_0$).
For the new information you brought in your question we have now
$$theta^*_p(t_0) - n_p cdot 2pi = theta_p(t_0) $$
hence:
$$Omega_0=frac{theta^*_2(t_0) - n_2 cdot 2pi }{2 t_0}=dots=frac{theta^*_p(t_0) - n_p cdot 2pi }{p t_0}$$
It does not look better for getting $t_0$.
answered Jul 4 '13 at 14:44
al-Hwarizmial-Hwarizmi
2,64511131
$endgroup$
$begingroup$
Sorry, al-Hwarizmi, in my initial question I missed the point that the observed phase values $theta_p$ are actually phase-wrapped to the interval $[-pi,pi[$. I changed that now in the question and denoted the observed (wrapped) phase values by $theta^*_p$ to indicate that they are different from the original phase values $theta_p$.
$endgroup$
– koffer
Jul 4 '13 at 15:17
$begingroup$
no problem. I just tried to extend the answer. In any case I think you can translate the problem in a set of equations and see how many unknown variables versus number of equations or constraints required.
$endgroup$
– al-Hwarizmi
Jul 4 '13 at 15:46
add a comment |
$begingroup$
If I may have understood your case then
given:
$$2 Omega_0 t_0 = theta_2$$
$$3 Omega_0 t_0 = theta_3$$
$$dots$$
and $theta_2$ and $theta_3$ known, then
$$Omega_0=frac{theta_2}{2 t_0}=frac{theta_3}{3 t_0}=dots=frac{theta_p}{p t_0}$$
So you see from your given constraints, $t_0$ can be any time, (it will change through all harmonics) and the system is under-determined .
The answer under the given constraints is that we can not specify a specific $t_0$ or ($Omega_0$).
For the new information you brought in your question we have now
$$theta^*_p(t_0) - n_p cdot 2pi = theta_p(t_0) $$
hence:
$$Omega_0=frac{theta^*_2(t_0) - n_2 cdot 2pi }{2 t_0}=dots=frac{theta^*_p(t_0) - n_p cdot 2pi }{p t_0}$$
It does not look better for getting $t_0$.
answered Jul 4 '13 at 14:44
al-Hwarizmial-Hwarizmi
2,64511131
$endgroup$
$begingroup$
Sorry, al-Hwarizmi, in my initial question I missed the point that the observed phase values $theta_p$ are actually phase-wrapped to the interval $[-pi,pi[$. I changed that now in the question and denoted the observed (wrapped) phase values by $theta^*_p$ to indicate that they are different from the original phase values $theta_p$.
$endgroup$
– koffer
Jul 4 '13 at 15:17
$begingroup$
no problem. I just tried to extend the answer. In any case I think you can translate the problem in a set of equations and see how many unknown variables versus number of equations or constraints required.
$endgroup$
– al-Hwarizmi
Jul 4 '13 at 15:46
add a comment |
$begingroup$
If I may have understood your case then
given:
$$2 Omega_0 t_0 = theta_2$$
$$3 Omega_0 t_0 = theta_3$$
$$dots$$
and $theta_2$ and $theta_3$ known, then
$$Omega_0=frac{theta_2}{2 t_0}=frac{theta_3}{3 t_0}=dots=frac{theta_p}{p t_0}$$
So you see from your given constraints, $t_0$ can be any time, (it will change through all harmonics) and the system is under-determined .
The answer under the given constraints is that we can not specify a specific $t_0$ or ($Omega_0$).
For the new information you brought in your question we have now
$$theta^*_p(t_0) - n_p cdot 2pi = theta_p(t_0) $$
hence:
$$Omega_0=frac{theta^*_2(t_0) - n_2 cdot 2pi }{2 t_0}=dots=frac{theta^*_p(t_0) - n_p cdot 2pi }{p t_0}$$
It does not look better for getting $t_0$.
answered Jul 4 '13 at 14:44
al-Hwarizmial-Hwarizmi
2,64511131
$endgroup$
If I may have understood your case then
given:
$$2 Omega_0 t_0 = theta_2$$
$$3 Omega_0 t_0 = theta_3$$
$$dots$$
and $theta_2$ and $theta_3$ known, then
$$Omega_0=frac{theta_2}{2 t_0}=frac{theta_3}{3 t_0}=dots=frac{theta_p}{p t_0}$$
So you see from your given constraints, $t_0$ can be any time, (it will change through all harmonics) and the system is under-determined .
The answer under the given constraints is that we can not specify a specific $t_0$ or ($Omega_0$).
For the new information you brought in your question we have now
$$theta^*_p(t_0) - n_p cdot 2pi = theta_p(t_0) $$
hence:
$$Omega_0=frac{theta^*_2(t_0) - n_2 cdot 2pi }{2 t_0}=dots=frac{theta^*_p(t_0) - n_p cdot 2pi }{p t_0}$$
It does not look better for getting $t_0$.
answered Jul 4 '13 at 14:44
al-Hwarizmial-Hwarizmi
2,64511131
edited Jul 4 '13 at 15:38
answered Jul 4 '13 at 14:44
al-Hwarizmial-Hwarizmi
2,64511131
answered Jul 4 '13 at 14:44
al-Hwarizmial-Hwarizmi
2,64511131
answered Jul 4 '13 at 14:44
al-Hwarizmial-Hwarizmi
2,64511131
2,64511131
$begingroup$
Sorry, al-Hwarizmi, in my initial question I missed the point that the observed phase values $theta_p$ are actually phase-wrapped to the interval $[-pi,pi[$. I changed that now in the question and denoted the observed (wrapped) phase values by $theta^*_p$ to indicate that they are different from the original phase values $theta_p$.
$endgroup$
– koffer
Jul 4 '13 at 15:17
$begingroup$
no problem. I just tried to extend the answer. In any case I think you can translate the problem in a set of equations and see how many unknown variables versus number of equations or constraints required.
$endgroup$
– al-Hwarizmi
Jul 4 '13 at 15:46
add a comment |
$begingroup$
Sorry, al-Hwarizmi, in my initial question I missed the point that the observed phase values $theta_p$ are actually phase-wrapped to the interval $[-pi,pi[$. I changed that now in the question and denoted the observed (wrapped) phase values by $theta^*_p$ to indicate that they are different from the original phase values $theta_p$.
$endgroup$
– koffer
Jul 4 '13 at 15:17
$begingroup$
no problem. I just tried to extend the answer. In any case I think you can translate the problem in a set of equations and see how many unknown variables versus number of equations or constraints required.
$endgroup$
– al-Hwarizmi
Jul 4 '13 at 15:46
$begingroup$
Sorry, al-Hwarizmi, in my initial question I missed the point that the observed phase values $theta_p$ are actually phase-wrapped to the interval $[-pi,pi[$. I changed that now in the question and denoted the observed (wrapped) phase values by $theta^*_p$ to indicate that they are different from the original phase values $theta_p$.
$endgroup$
– koffer
Jul 4 '13 at 15:17
$begingroup$
Sorry, al-Hwarizmi, in my initial question I missed the point that the observed phase values $theta_p$ are actually phase-wrapped to the interval $[-pi,pi[$. I changed that now in the question and denoted the observed (wrapped) phase values by $theta^*_p$ to indicate that they are different from the original phase values $theta_p$.
$endgroup$
– koffer
Jul 4 '13 at 15:17
$begingroup$
no problem. I just tried to extend the answer. In any case I think you can translate the problem in a set of equations and see how many unknown variables versus number of equations or constraints required.
$endgroup$
– al-Hwarizmi
Jul 4 '13 at 15:46
$begingroup$
no problem. I just tried to extend the answer. In any case I think you can translate the problem in a set of equations and see how many unknown variables versus number of equations or constraints required.
$endgroup$
– al-Hwarizmi
Jul 4 '13 at 15:46
add a comment |
$begingroup$
OK, the phase is obviously linear in time here, so we should assume that the phase measurements, when unwrapped, want to fall on a line. So the idea, albeit a bit crude, it to go through the measurements for increasing $p$, and increase the $p$th phase in increments of $2 pi$ until that phase is larger than the $p-1$th phase. As an example, allow me to glean off some values from your graphs:
$$left ( begin{array} \2 t_0 & 0.33 pi \ 3 t_0 & -0.55 pi \ 4 t_0 & 0.55 pi \ 5 t_0 & -0.39 piend{array} right )$$
Begin by taking the value of the phase at $t=2 t_0$ as is. For the phase at $t=3 t_0$, see that it is less than the phase at $t=2 t_0$, so increase by $2 pi$ to get $1.45 pi$, which is the smallest phase greater than the previous phase; we then move on to $4 t_0$, and you hopefully get the point. When you finish, you get something like this
$$left ( begin{array} \2 t_0 & 0.33 pi \ 3 t_0 & 1.45 pi \ 4 t_0 & 2.55 pi \ 5 t_0 & 3.61 piend{array} right )$$
A best fit line takes the form $theta(p) = -1.844 pi+1.094 pi p$, so plugging in $p=1$ corresponding to $t=t_0$, you get a phase of about $-0.75 pi$ (which makes sense on your graph), so that knowing $Omega_0$, you may then determine the value of $t_0$. I hope that helps.
answered Jul 4 '13 at 13:12
Ron GordonRon Gordon
122k14156266
$endgroup$
$begingroup$
Sorry, apparently I've missed an important detail in my question: I can only measure the phases $theta_p(t_0)$ in the interval $[-pi,pi[$. So the measured phase can be denoted by $theta^*_p(t_0) = theta_p(t_0) + n_p cdot 2pi$. So given a $theta^*_p(t_0)$ we have several possible times depending on $n_p$. In the problem above, the $p$-th harmonic can occur at exactly $p$ possible times. And I assume that the fact that they all fall together at $t_0$ could be used to determine $n_p$ for all harmonics. I will make that clearer in the original question.
$endgroup$
– koffer
Jul 4 '13 at 14:36
add a comment |
$begingroup$
OK, the phase is obviously linear in time here, so we should assume that the phase measurements, when unwrapped, want to fall on a line. So the idea, albeit a bit crude, it to go through the measurements for increasing $p$, and increase the $p$th phase in increments of $2 pi$ until that phase is larger than the $p-1$th phase. As an example, allow me to glean off some values from your graphs:
$$left ( begin{array} \2 t_0 & 0.33 pi \ 3 t_0 & -0.55 pi \ 4 t_0 & 0.55 pi \ 5 t_0 & -0.39 piend{array} right )$$
Begin by taking the value of the phase at $t=2 t_0$ as is. For the phase at $t=3 t_0$, see that it is less than the phase at $t=2 t_0$, so increase by $2 pi$ to get $1.45 pi$, which is the smallest phase greater than the previous phase; we then move on to $4 t_0$, and you hopefully get the point. When you finish, you get something like this
$$left ( begin{array} \2 t_0 & 0.33 pi \ 3 t_0 & 1.45 pi \ 4 t_0 & 2.55 pi \ 5 t_0 & 3.61 piend{array} right )$$
A best fit line takes the form $theta(p) = -1.844 pi+1.094 pi p$, so plugging in $p=1$ corresponding to $t=t_0$, you get a phase of about $-0.75 pi$ (which makes sense on your graph), so that knowing $Omega_0$, you may then determine the value of $t_0$. I hope that helps.
answered Jul 4 '13 at 13:12
Ron GordonRon Gordon
122k14156266
$endgroup$
$begingroup$
Sorry, apparently I've missed an important detail in my question: I can only measure the phases $theta_p(t_0)$ in the interval $[-pi,pi[$. So the measured phase can be denoted by $theta^*_p(t_0) = theta_p(t_0) + n_p cdot 2pi$. So given a $theta^*_p(t_0)$ we have several possible times depending on $n_p$. In the problem above, the $p$-th harmonic can occur at exactly $p$ possible times. And I assume that the fact that they all fall together at $t_0$ could be used to determine $n_p$ for all harmonics. I will make that clearer in the original question.
$endgroup$
– koffer
Jul 4 '13 at 14:36
add a comment |
$begingroup$
OK, the phase is obviously linear in time here, so we should assume that the phase measurements, when unwrapped, want to fall on a line. So the idea, albeit a bit crude, it to go through the measurements for increasing $p$, and increase the $p$th phase in increments of $2 pi$ until that phase is larger than the $p-1$th phase. As an example, allow me to glean off some values from your graphs:
$$left ( begin{array} \2 t_0 & 0.33 pi \ 3 t_0 & -0.55 pi \ 4 t_0 & 0.55 pi \ 5 t_0 & -0.39 piend{array} right )$$
Begin by taking the value of the phase at $t=2 t_0$ as is. For the phase at $t=3 t_0$, see that it is less than the phase at $t=2 t_0$, so increase by $2 pi$ to get $1.45 pi$, which is the smallest phase greater than the previous phase; we then move on to $4 t_0$, and you hopefully get the point. When you finish, you get something like this
$$left ( begin{array} \2 t_0 & 0.33 pi \ 3 t_0 & 1.45 pi \ 4 t_0 & 2.55 pi \ 5 t_0 & 3.61 piend{array} right )$$
A best fit line takes the form $theta(p) = -1.844 pi+1.094 pi p$, so plugging in $p=1$ corresponding to $t=t_0$, you get a phase of about $-0.75 pi$ (which makes sense on your graph), so that knowing $Omega_0$, you may then determine the value of $t_0$. I hope that helps.
answered Jul 4 '13 at 13:12
Ron GordonRon Gordon
122k14156266
$endgroup$
OK, the phase is obviously linear in time here, so we should assume that the phase measurements, when unwrapped, want to fall on a line. So the idea, albeit a bit crude, it to go through the measurements for increasing $p$, and increase the $p$th phase in increments of $2 pi$ until that phase is larger than the $p-1$th phase. As an example, allow me to glean off some values from your graphs:
$$left ( begin{array} \2 t_0 & 0.33 pi \ 3 t_0 & -0.55 pi \ 4 t_0 & 0.55 pi \ 5 t_0 & -0.39 piend{array} right )$$
Begin by taking the value of the phase at $t=2 t_0$ as is. For the phase at $t=3 t_0$, see that it is less than the phase at $t=2 t_0$, so increase by $2 pi$ to get $1.45 pi$, which is the smallest phase greater than the previous phase; we then move on to $4 t_0$, and you hopefully get the point. When you finish, you get something like this
$$left ( begin{array} \2 t_0 & 0.33 pi \ 3 t_0 & 1.45 pi \ 4 t_0 & 2.55 pi \ 5 t_0 & 3.61 piend{array} right )$$
A best fit line takes the form $theta(p) = -1.844 pi+1.094 pi p$, so plugging in $p=1$ corresponding to $t=t_0$, you get a phase of about $-0.75 pi$ (which makes sense on your graph), so that knowing $Omega_0$, you may then determine the value of $t_0$. I hope that helps.
answered Jul 4 '13 at 13:12
Ron GordonRon Gordon
122k14156266
edited Jul 4 '13 at 16:15
answered Jul 4 '13 at 13:12
Ron GordonRon Gordon
122k14156266
answered Jul 4 '13 at 13:12
Ron GordonRon Gordon
122k14156266
answered Jul 4 '13 at 13:12
Ron GordonRon Gordon
122k14156266
122k14156266
$begingroup$
Sorry, apparently I've missed an important detail in my question: I can only measure the phases $theta_p(t_0)$ in the interval $[-pi,pi[$. So the measured phase can be denoted by $theta^*_p(t_0) = theta_p(t_0) + n_p cdot 2pi$. So given a $theta^*_p(t_0)$ we have several possible times depending on $n_p$. In the problem above, the $p$-th harmonic can occur at exactly $p$ possible times. And I assume that the fact that they all fall together at $t_0$ could be used to determine $n_p$ for all harmonics. I will make that clearer in the original question.
$endgroup$
– koffer
Jul 4 '13 at 14:36
add a comment |
$begingroup$
Sorry, apparently I've missed an important detail in my question: I can only measure the phases $theta_p(t_0)$ in the interval $[-pi,pi[$. So the measured phase can be denoted by $theta^*_p(t_0) = theta_p(t_0) + n_p cdot 2pi$. So given a $theta^*_p(t_0)$ we have several possible times depending on $n_p$. In the problem above, the $p$-th harmonic can occur at exactly $p$ possible times. And I assume that the fact that they all fall together at $t_0$ could be used to determine $n_p$ for all harmonics. I will make that clearer in the original question.
$endgroup$
– koffer
Jul 4 '13 at 14:36
$begingroup$
Sorry, apparently I've missed an important detail in my question: I can only measure the phases $theta_p(t_0)$ in the interval $[-pi,pi[$. So the measured phase can be denoted by $theta^*_p(t_0) = theta_p(t_0) + n_p cdot 2pi$. So given a $theta^*_p(t_0)$ we have several possible times depending on $n_p$. In the problem above, the $p$-th harmonic can occur at exactly $p$ possible times. And I assume that the fact that they all fall together at $t_0$ could be used to determine $n_p$ for all harmonics. I will make that clearer in the original question.
$endgroup$
– koffer
Jul 4 '13 at 14:36
$begingroup$
Sorry, apparently I've missed an important detail in my question: I can only measure the phases $theta_p(t_0)$ in the interval $[-pi,pi[$. So the measured phase can be denoted by $theta^*_p(t_0) = theta_p(t_0) + n_p cdot 2pi$. So given a $theta^*_p(t_0)$ we have several possible times depending on $n_p$. In the problem above, the $p$-th harmonic can occur at exactly $p$ possible times. And I assume that the fact that they all fall together at $t_0$ could be used to determine $n_p$ for all harmonics. I will make that clearer in the original question.
$endgroup$
– koffer
Jul 4 '13 at 14:36
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%2f435188%2festimating-time-in-harmonic-signal%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
$begingroup$
just try to undestand the problem correctly, given a time $t_0$ and the phase $p=2$ respectively $theta_p(t_0)$, why shouldn't it be possible to find the fundamental that is half of phase?
$endgroup$
– al-Hwarizmi
Jul 4 '13 at 13:06
$begingroup$
Oh, maybe my wording didn't make that clear. The time $t_0$ is not known, only the phase values $theta_p(t_0)$ at that time are known. And as you can see from the diagram, a given phase $theta_p$ of a harmonic can occur at several time positions. I will edit the question to make that clearer.
$endgroup$
– koffer
Jul 4 '13 at 14:01
$begingroup$
Still dont get it, sorry. If you have $2 Omega_0 t_0 = theta_2$ and $3 Omega_0 t_0 = theta_3$ for instance, do you want to calculate $Omega_0$ and $t_0$?
$endgroup$
– al-Hwarizmi
Jul 4 '13 at 14:32
$begingroup$
No, $Omega_0$ is known. However, I can only measure the phase in the interval $[-pi,pi[$ (see my comment to Ron Gordon's answer).
$endgroup$
– koffer
Jul 4 '13 at 14:42