Do square wave exist? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Why isn't the Fourier transform of a single sine wave cycle a single bar?How electromagnetic waves works, how they are created and how thay transform to electric signal?Can an alternating magnetic field be “informationally” shielded?square and triangle wave as a sum of sine waveEnergies carried by electric and magnetic fields in plane wavesEnergies of the fields in a plane wave, attenuated and unattenuatedAntenna transmission when a generic signal is appliedAbility of a 100km power transmission line to radiate 50Hz EM waves?Can antennas be viewed as light sources?Impedance calculation from voltage and current (exponential) waveforms
What do you call the main part of a joke?
Are sorcerers unable to use the Careful Spell metamagic option on themselves?
Random body shuffle every night—can we still function?
Did Mueller's report provide an evidentiary basis for the claim of Russian govt election interference via social media?
One-one communication
How does Belgium enforce obligatory attendance in elections?
How do I find out the mythology and history of my Fortress?
Drawing spherical mirrors
How to compare two different files line by line in unix?
Why are my pictures showing a dark band on one edge?
What does this say in Elvish?
Semigroups with no morphisms between them
What does it mean that physics no longer uses mechanical models to describe phenomena?
How to report t statistic from R
How do living politicians protect their readily obtainable signatures from misuse?
Tannaka duality for semisimple groups
AppleTVs create a chatty alternate WiFi network
Why are vacuum tubes still used in amateur radios?
What's the difference between the capability remove_users and delete_users?
Should a wizard buy fine inks every time he want to copy spells into his spellbook?
How did Fremen produce and carry enough thumpers to use Sandworms as de facto Ubers?
What is the chair depicted in Cesare Maccari's 1889 painting "Cicerone denuncia Catilina"?
What does 丫 mean? 丫是什么意思？
What is best way to wire a ceiling receptacle in this situation?
Do square wave exist?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Why isn't the Fourier transform of a single sine wave cycle a single bar?How electromagnetic waves works, how they are created and how thay transform to electric signal?Can an alternating magnetic field be “informationally” shielded?square and triangle wave as a sum of sine waveEnergies carried by electric and magnetic fields in plane wavesEnergies of the fields in a plane wave, attenuated and unattenuatedAntenna transmission when a generic signal is appliedAbility of a 100km power transmission line to radiate 50Hz EM waves?Can antennas be viewed as light sources?Impedance calculation from voltage and current (exponential) waveforms
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
If we send a square waveform through an antenna, will we get square electromagnetic waves with electric and magnetic fields looking like squares?
Also, since there is an abrupt/almost jump in amplitude, will we get very high frequency sine waves like predicted by the Fourier transform?
antenna electromagnetic fourier math-notation
edited 2 days ago
evildemonic
2,7811023
asked 2 days ago
user163416user163416
12339
$endgroup$
add a comment |
$begingroup$
If we send a square waveform through an antenna, will we get square electromagnetic waves with electric and magnetic fields looking like squares?
Also, since there is an abrupt/almost jump in amplitude, will we get very high frequency sine waves like predicted by the Fourier transform?
antenna electromagnetic fourier math-notation
edited 2 days ago
evildemonic
2,7811023
asked 2 days ago
user163416user163416
12339
$endgroup$
8
$begingroup$
A perfect square wave (0 rise / fall times) does not exist because it would require infinite bandwidth.
$endgroup$
– Peter Smith
2 days ago
2
$begingroup$
antennas have finite bandwidth
$endgroup$
– analogsystemsrf
2 days ago
2
$begingroup$
Infinite bandwidth and zero impedance
$endgroup$
– JonRB
2 days ago
$begingroup$
If the electric field is a close-to-ideal square wave, won't the magnetic field be more like a series of positive and negative spikes?
$endgroup$
– immibis
2 days ago
add a comment |
$begingroup$
If we send a square waveform through an antenna, will we get square electromagnetic waves with electric and magnetic fields looking like squares?
Also, since there is an abrupt/almost jump in amplitude, will we get very high frequency sine waves like predicted by the Fourier transform?
antenna electromagnetic fourier math-notation
edited 2 days ago
evildemonic
2,7811023
asked 2 days ago
user163416user163416
12339
$endgroup$
If we send a square waveform through an antenna, will we get square electromagnetic waves with electric and magnetic fields looking like squares?
Also, since there is an abrupt/almost jump in amplitude, will we get very high frequency sine waves like predicted by the Fourier transform?
antenna electromagnetic fourier math-notation
antenna electromagnetic fourier math-notation
edited 2 days ago
evildemonic
2,7811023
asked 2 days ago
user163416user163416
12339
edited 2 days ago
evildemonic
2,7811023
asked 2 days ago
user163416user163416
12339
edited 2 days ago
evildemonic
2,7811023
edited 2 days ago
evildemonic
2,7811023
edited 2 days ago
evildemonic
2,7811023
2,7811023
asked 2 days ago
user163416user163416
12339
asked 2 days ago
user163416user163416
12339
asked 2 days ago
user163416user163416
12339
12339
8
$begingroup$
A perfect square wave (0 rise / fall times) does not exist because it would require infinite bandwidth.
$endgroup$
– Peter Smith
2 days ago
2
$begingroup$
antennas have finite bandwidth
$endgroup$
– analogsystemsrf
2 days ago
2
$begingroup$
Infinite bandwidth and zero impedance
$endgroup$
– JonRB
2 days ago
$begingroup$
If the electric field is a close-to-ideal square wave, won't the magnetic field be more like a series of positive and negative spikes?
$endgroup$
– immibis
2 days ago
add a comment |
8
$begingroup$
A perfect square wave (0 rise / fall times) does not exist because it would require infinite bandwidth.
$endgroup$
– Peter Smith
2 days ago
2
$begingroup$
antennas have finite bandwidth
$endgroup$
– analogsystemsrf
2 days ago
2
$begingroup$
Infinite bandwidth and zero impedance
$endgroup$
– JonRB
2 days ago
$begingroup$
If the electric field is a close-to-ideal square wave, won't the magnetic field be more like a series of positive and negative spikes?
$endgroup$
– immibis
2 days ago
8
8
$begingroup$
A perfect square wave (0 rise / fall times) does not exist because it would require infinite bandwidth.
$endgroup$
– Peter Smith
2 days ago
$begingroup$
A perfect square wave (0 rise / fall times) does not exist because it would require infinite bandwidth.
$endgroup$
– Peter Smith
2 days ago
2
2
$begingroup$
antennas have finite bandwidth
$endgroup$
– analogsystemsrf
2 days ago
$begingroup$
antennas have finite bandwidth
$endgroup$
– analogsystemsrf
2 days ago
2
2
$begingroup$
Infinite bandwidth and zero impedance
$endgroup$
– JonRB
2 days ago
$begingroup$
Infinite bandwidth and zero impedance
$endgroup$
– JonRB
2 days ago
$begingroup$
If the electric field is a close-to-ideal square wave, won't the magnetic field be more like a series of positive and negative spikes?
$endgroup$
– immibis
2 days ago
$begingroup$
If the electric field is a close-to-ideal square wave, won't the magnetic field be more like a series of positive and negative spikes?
$endgroup$
– immibis
2 days ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
As you know (since you mentioned the Fourier transform), a square wave can be represented (well, almost -- see below) as the sum of an infinite series of sine waves. But it would not be possible to send a true square wave through any real physical antenna: As you move along the infinite series, the frequencies get higher and higher, and eventually you'll reach frequencies your antenna can't transmit, for various reasons. If you look at a chart of the electromagnetic spectrum, you will find that radio waves above a certain frequency are called "light", and your antenna probably can't reach those frequencies no matter how good it is.
(But, indeed, if you have an antenna that is capable of transmitting over a wide bandwidth -- that is, from very low to very high frequencies -- and you send some approximation of a square wave over it, you will see very high frequencies appear, just as predicted by the Fourier transform.)
There's also another problem: You can't quite actually approach a true square wave shape from any finite sum of sine waves, no matter how many. This problem is much more theoretical, and unlikely to actually come up in practice, but it's called the Gibbs phenomenon. It turns out that no matter how high in frequency you go, your approximation of a square wave will always overshoot at the big jumps from low to high and high to low. The overshoot will get shorter and shorter in time, the better your approximation (the higher in frequency you go.) But it will never go down in magnitude; it converges to about 9% of the size of the jump.
answered 2 days ago
Glenn WillenGlenn Willen
885211
$endgroup$
14
$begingroup$
You should rather say you can't actually make a true square wave from a finite sum of sine waves. From an infinite sum, you can. If you take the limit, the overshoot vanishes as you can see with an epsilon-delta-argument.
$endgroup$
– DerManu
2 days ago
3
$begingroup$
The Fourier series for a square wave does converge to a square wave, but it fails to converge uniformly to a square wave, because if you take finitely many (say, a trillion) terms of the series, it will still overshoot by about 9%. (Actually, no series of continuous functions converges uniformly to a square wave, since a square wave is not continuous. Still, the Fourier series is particularly problematic; there are other series which don't overshoot like that.)
$endgroup$
– Tanner Swett
2 days ago
4
$begingroup$
The sum converges pointwise to the square wave everywhere except at the transitions where it converges to the average of the left & right limits. The overshoot never vanishes, as the convergence is not uniform.
$endgroup$
– copper.hat
2 days ago
1
$begingroup$
@copper.hat: I recall reading that Foorier himself was rather unhappy with the fact that the amplitude of overshoot did not asymptotically approach zero as the number of terms increased. The fraction of the domain for which the function isn't within any particular epsilon of the correct value, however, does approach zero as the number of terms increases.
$endgroup$
– supercat
2 days ago
3
$begingroup$
Technically, any antenna will emit light if you get it hot enough
$endgroup$
– Nate Strickland
2 days ago
|
show 3 more comments
$begingroup$
No, perfect mathematical square waves don't exist in real world because square wave is not a continuous function (it does not have a derivative at the step). Therefore you can only approximate a square wave and the approximation does have very high frequencies, and at some point the antenna would not be able to send these so it would be a low-pass filter.
answered 2 days ago
JustmeJustme
2,5411413
$endgroup$
2
$begingroup$
Continuous functions don't exist in the real world either, because of quantum effects.
$endgroup$
– supercat
2 days ago
2
$begingroup$
Also the logic that "it does not have a derivative at the step" does not mean that the function is not continuous. Not differentiable does not imply not continuous. That being said, the function is not continuous since the one sided limits at the step don't agree.
$endgroup$
– Sean Haight
2 days ago
add a comment |
$begingroup$
In a more general case compared to the answers above, nothing can be stoped or started in zero time ie instantly. To do so would imply an infinitely high frequency component which would translate to infinite energy. The constraining factors are speed of light limitation of Special Relativity and Quantum Mechanics Uncertainty Principle.
The sharper the transition you want, the more energy you have to pump into the system
answered 2 days ago
Dirk BruereDirk Bruere
5,96853164
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("schematics", function ()
StackExchange.schematics.init();
);
, "cicuitlab");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "135"
;
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: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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
,
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%2felectronics.stackexchange.com%2fquestions%2f432994%2fdo-square-wave-exist%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As you know (since you mentioned the Fourier transform), a square wave can be represented (well, almost -- see below) as the sum of an infinite series of sine waves. But it would not be possible to send a true square wave through any real physical antenna: As you move along the infinite series, the frequencies get higher and higher, and eventually you'll reach frequencies your antenna can't transmit, for various reasons. If you look at a chart of the electromagnetic spectrum, you will find that radio waves above a certain frequency are called "light", and your antenna probably can't reach those frequencies no matter how good it is.
(But, indeed, if you have an antenna that is capable of transmitting over a wide bandwidth -- that is, from very low to very high frequencies -- and you send some approximation of a square wave over it, you will see very high frequencies appear, just as predicted by the Fourier transform.)
There's also another problem: You can't quite actually approach a true square wave shape from any finite sum of sine waves, no matter how many. This problem is much more theoretical, and unlikely to actually come up in practice, but it's called the Gibbs phenomenon. It turns out that no matter how high in frequency you go, your approximation of a square wave will always overshoot at the big jumps from low to high and high to low. The overshoot will get shorter and shorter in time, the better your approximation (the higher in frequency you go.) But it will never go down in magnitude; it converges to about 9% of the size of the jump.
answered 2 days ago
Glenn WillenGlenn Willen
885211
$endgroup$
14
$begingroup$
You should rather say you can't actually make a true square wave from a finite sum of sine waves. From an infinite sum, you can. If you take the limit, the overshoot vanishes as you can see with an epsilon-delta-argument.
$endgroup$
– DerManu
2 days ago
3
$begingroup$
The Fourier series for a square wave does converge to a square wave, but it fails to converge uniformly to a square wave, because if you take finitely many (say, a trillion) terms of the series, it will still overshoot by about 9%. (Actually, no series of continuous functions converges uniformly to a square wave, since a square wave is not continuous. Still, the Fourier series is particularly problematic; there are other series which don't overshoot like that.)
$endgroup$
– Tanner Swett
2 days ago
4
$begingroup$
The sum converges pointwise to the square wave everywhere except at the transitions where it converges to the average of the left & right limits. The overshoot never vanishes, as the convergence is not uniform.
$endgroup$
– copper.hat
2 days ago
1
$begingroup$
@copper.hat: I recall reading that Foorier himself was rather unhappy with the fact that the amplitude of overshoot did not asymptotically approach zero as the number of terms increased. The fraction of the domain for which the function isn't within any particular epsilon of the correct value, however, does approach zero as the number of terms increases.
$endgroup$
– supercat
2 days ago
3
$begingroup$
Technically, any antenna will emit light if you get it hot enough
$endgroup$
– Nate Strickland
2 days ago
|
show 3 more comments
$begingroup$
As you know (since you mentioned the Fourier transform), a square wave can be represented (well, almost -- see below) as the sum of an infinite series of sine waves. But it would not be possible to send a true square wave through any real physical antenna: As you move along the infinite series, the frequencies get higher and higher, and eventually you'll reach frequencies your antenna can't transmit, for various reasons. If you look at a chart of the electromagnetic spectrum, you will find that radio waves above a certain frequency are called "light", and your antenna probably can't reach those frequencies no matter how good it is.
(But, indeed, if you have an antenna that is capable of transmitting over a wide bandwidth -- that is, from very low to very high frequencies -- and you send some approximation of a square wave over it, you will see very high frequencies appear, just as predicted by the Fourier transform.)
There's also another problem: You can't quite actually approach a true square wave shape from any finite sum of sine waves, no matter how many. This problem is much more theoretical, and unlikely to actually come up in practice, but it's called the Gibbs phenomenon. It turns out that no matter how high in frequency you go, your approximation of a square wave will always overshoot at the big jumps from low to high and high to low. The overshoot will get shorter and shorter in time, the better your approximation (the higher in frequency you go.) But it will never go down in magnitude; it converges to about 9% of the size of the jump.
answered 2 days ago
Glenn WillenGlenn Willen
885211
$endgroup$
14
$begingroup$
You should rather say you can't actually make a true square wave from a finite sum of sine waves. From an infinite sum, you can. If you take the limit, the overshoot vanishes as you can see with an epsilon-delta-argument.
$endgroup$
– DerManu
2 days ago
3
$begingroup$
The Fourier series for a square wave does converge to a square wave, but it fails to converge uniformly to a square wave, because if you take finitely many (say, a trillion) terms of the series, it will still overshoot by about 9%. (Actually, no series of continuous functions converges uniformly to a square wave, since a square wave is not continuous. Still, the Fourier series is particularly problematic; there are other series which don't overshoot like that.)
$endgroup$
– Tanner Swett
2 days ago
4
$begingroup$
The sum converges pointwise to the square wave everywhere except at the transitions where it converges to the average of the left & right limits. The overshoot never vanishes, as the convergence is not uniform.
$endgroup$
– copper.hat
2 days ago
1
$begingroup$
@copper.hat: I recall reading that Foorier himself was rather unhappy with the fact that the amplitude of overshoot did not asymptotically approach zero as the number of terms increased. The fraction of the domain for which the function isn't within any particular epsilon of the correct value, however, does approach zero as the number of terms increases.
$endgroup$
– supercat
2 days ago
3
$begingroup$
Technically, any antenna will emit light if you get it hot enough
$endgroup$
– Nate Strickland
2 days ago
|
show 3 more comments
$begingroup$
As you know (since you mentioned the Fourier transform), a square wave can be represented (well, almost -- see below) as the sum of an infinite series of sine waves. But it would not be possible to send a true square wave through any real physical antenna: As you move along the infinite series, the frequencies get higher and higher, and eventually you'll reach frequencies your antenna can't transmit, for various reasons. If you look at a chart of the electromagnetic spectrum, you will find that radio waves above a certain frequency are called "light", and your antenna probably can't reach those frequencies no matter how good it is.
(But, indeed, if you have an antenna that is capable of transmitting over a wide bandwidth -- that is, from very low to very high frequencies -- and you send some approximation of a square wave over it, you will see very high frequencies appear, just as predicted by the Fourier transform.)
There's also another problem: You can't quite actually approach a true square wave shape from any finite sum of sine waves, no matter how many. This problem is much more theoretical, and unlikely to actually come up in practice, but it's called the Gibbs phenomenon. It turns out that no matter how high in frequency you go, your approximation of a square wave will always overshoot at the big jumps from low to high and high to low. The overshoot will get shorter and shorter in time, the better your approximation (the higher in frequency you go.) But it will never go down in magnitude; it converges to about 9% of the size of the jump.
answered 2 days ago
Glenn WillenGlenn Willen
885211
$endgroup$
As you know (since you mentioned the Fourier transform), a square wave can be represented (well, almost -- see below) as the sum of an infinite series of sine waves. But it would not be possible to send a true square wave through any real physical antenna: As you move along the infinite series, the frequencies get higher and higher, and eventually you'll reach frequencies your antenna can't transmit, for various reasons. If you look at a chart of the electromagnetic spectrum, you will find that radio waves above a certain frequency are called "light", and your antenna probably can't reach those frequencies no matter how good it is.
(But, indeed, if you have an antenna that is capable of transmitting over a wide bandwidth -- that is, from very low to very high frequencies -- and you send some approximation of a square wave over it, you will see very high frequencies appear, just as predicted by the Fourier transform.)
There's also another problem: You can't quite actually approach a true square wave shape from any finite sum of sine waves, no matter how many. This problem is much more theoretical, and unlikely to actually come up in practice, but it's called the Gibbs phenomenon. It turns out that no matter how high in frequency you go, your approximation of a square wave will always overshoot at the big jumps from low to high and high to low. The overshoot will get shorter and shorter in time, the better your approximation (the higher in frequency you go.) But it will never go down in magnitude; it converges to about 9% of the size of the jump.
answered 2 days ago
Glenn WillenGlenn Willen
885211
edited 2 days ago
answered 2 days ago
Glenn WillenGlenn Willen
885211
answered 2 days ago
Glenn WillenGlenn Willen
885211
answered 2 days ago
Glenn WillenGlenn Willen
885211
885211
14
$begingroup$
You should rather say you can't actually make a true square wave from a finite sum of sine waves. From an infinite sum, you can. If you take the limit, the overshoot vanishes as you can see with an epsilon-delta-argument.
$endgroup$
– DerManu
2 days ago
3
$begingroup$
The Fourier series for a square wave does converge to a square wave, but it fails to converge uniformly to a square wave, because if you take finitely many (say, a trillion) terms of the series, it will still overshoot by about 9%. (Actually, no series of continuous functions converges uniformly to a square wave, since a square wave is not continuous. Still, the Fourier series is particularly problematic; there are other series which don't overshoot like that.)
$endgroup$
– Tanner Swett
2 days ago
4
$begingroup$
The sum converges pointwise to the square wave everywhere except at the transitions where it converges to the average of the left & right limits. The overshoot never vanishes, as the convergence is not uniform.
$endgroup$
– copper.hat
2 days ago
1
$begingroup$
@copper.hat: I recall reading that Foorier himself was rather unhappy with the fact that the amplitude of overshoot did not asymptotically approach zero as the number of terms increased. The fraction of the domain for which the function isn't within any particular epsilon of the correct value, however, does approach zero as the number of terms increases.
$endgroup$
– supercat
2 days ago
3
$begingroup$
Technically, any antenna will emit light if you get it hot enough
$endgroup$
– Nate Strickland
2 days ago
|
show 3 more comments
14
$begingroup$
You should rather say you can't actually make a true square wave from a finite sum of sine waves. From an infinite sum, you can. If you take the limit, the overshoot vanishes as you can see with an epsilon-delta-argument.
$endgroup$
– DerManu
2 days ago
3
$begingroup$
The Fourier series for a square wave does converge to a square wave, but it fails to converge uniformly to a square wave, because if you take finitely many (say, a trillion) terms of the series, it will still overshoot by about 9%. (Actually, no series of continuous functions converges uniformly to a square wave, since a square wave is not continuous. Still, the Fourier series is particularly problematic; there are other series which don't overshoot like that.)
$endgroup$
– Tanner Swett
2 days ago
4
$begingroup$
The sum converges pointwise to the square wave everywhere except at the transitions where it converges to the average of the left & right limits. The overshoot never vanishes, as the convergence is not uniform.
$endgroup$
– copper.hat
2 days ago
1
$begingroup$
@copper.hat: I recall reading that Foorier himself was rather unhappy with the fact that the amplitude of overshoot did not asymptotically approach zero as the number of terms increased. The fraction of the domain for which the function isn't within any particular epsilon of the correct value, however, does approach zero as the number of terms increases.
$endgroup$
– supercat
2 days ago
3
$begingroup$
Technically, any antenna will emit light if you get it hot enough
$endgroup$
– Nate Strickland
2 days ago
14
14
$begingroup$
You should rather say you can't actually make a true square wave from a finite sum of sine waves. From an infinite sum, you can. If you take the limit, the overshoot vanishes as you can see with an epsilon-delta-argument.
$endgroup$
– DerManu
2 days ago
$begingroup$
You should rather say you can't actually make a true square wave from a finite sum of sine waves. From an infinite sum, you can. If you take the limit, the overshoot vanishes as you can see with an epsilon-delta-argument.
$endgroup$
– DerManu
2 days ago
3
3
$begingroup$
The Fourier series for a square wave does converge to a square wave, but it fails to converge uniformly to a square wave, because if you take finitely many (say, a trillion) terms of the series, it will still overshoot by about 9%. (Actually, no series of continuous functions converges uniformly to a square wave, since a square wave is not continuous. Still, the Fourier series is particularly problematic; there are other series which don't overshoot like that.)
$endgroup$
– Tanner Swett
2 days ago
$begingroup$
The Fourier series for a square wave does converge to a square wave, but it fails to converge uniformly to a square wave, because if you take finitely many (say, a trillion) terms of the series, it will still overshoot by about 9%. (Actually, no series of continuous functions converges uniformly to a square wave, since a square wave is not continuous. Still, the Fourier series is particularly problematic; there are other series which don't overshoot like that.)
$endgroup$
– Tanner Swett
2 days ago
4
4
$begingroup$
The sum converges pointwise to the square wave everywhere except at the transitions where it converges to the average of the left & right limits. The overshoot never vanishes, as the convergence is not uniform.
$endgroup$
– copper.hat
2 days ago
$begingroup$
The sum converges pointwise to the square wave everywhere except at the transitions where it converges to the average of the left & right limits. The overshoot never vanishes, as the convergence is not uniform.
$endgroup$
– copper.hat
2 days ago
1
1
$begingroup$
@copper.hat: I recall reading that Foorier himself was rather unhappy with the fact that the amplitude of overshoot did not asymptotically approach zero as the number of terms increased. The fraction of the domain for which the function isn't within any particular epsilon of the correct value, however, does approach zero as the number of terms increases.
$endgroup$
– supercat
2 days ago
$begingroup$
@copper.hat: I recall reading that Foorier himself was rather unhappy with the fact that the amplitude of overshoot did not asymptotically approach zero as the number of terms increased. The fraction of the domain for which the function isn't within any particular epsilon of the correct value, however, does approach zero as the number of terms increases.
$endgroup$
– supercat
2 days ago
3
3
$begingroup$
Technically, any antenna will emit light if you get it hot enough
$endgroup$
– Nate Strickland
2 days ago
$begingroup$
Technically, any antenna will emit light if you get it hot enough
$endgroup$
– Nate Strickland
2 days ago
|
show 3 more comments
$begingroup$
No, perfect mathematical square waves don't exist in real world because square wave is not a continuous function (it does not have a derivative at the step). Therefore you can only approximate a square wave and the approximation does have very high frequencies, and at some point the antenna would not be able to send these so it would be a low-pass filter.
answered 2 days ago
JustmeJustme
2,5411413
$endgroup$
2
$begingroup$
Continuous functions don't exist in the real world either, because of quantum effects.
$endgroup$
– supercat
2 days ago
2
$begingroup$
Also the logic that "it does not have a derivative at the step" does not mean that the function is not continuous. Not differentiable does not imply not continuous. That being said, the function is not continuous since the one sided limits at the step don't agree.
$endgroup$
– Sean Haight
2 days ago
add a comment |
$begingroup$
No, perfect mathematical square waves don't exist in real world because square wave is not a continuous function (it does not have a derivative at the step). Therefore you can only approximate a square wave and the approximation does have very high frequencies, and at some point the antenna would not be able to send these so it would be a low-pass filter.
answered 2 days ago
JustmeJustme
2,5411413
$endgroup$
2
$begingroup$
Continuous functions don't exist in the real world either, because of quantum effects.
$endgroup$
– supercat
2 days ago
2
$begingroup$
Also the logic that "it does not have a derivative at the step" does not mean that the function is not continuous. Not differentiable does not imply not continuous. That being said, the function is not continuous since the one sided limits at the step don't agree.
$endgroup$
– Sean Haight
2 days ago
add a comment |
$begingroup$
No, perfect mathematical square waves don't exist in real world because square wave is not a continuous function (it does not have a derivative at the step). Therefore you can only approximate a square wave and the approximation does have very high frequencies, and at some point the antenna would not be able to send these so it would be a low-pass filter.
answered 2 days ago
JustmeJustme
2,5411413
$endgroup$
No, perfect mathematical square waves don't exist in real world because square wave is not a continuous function (it does not have a derivative at the step). Therefore you can only approximate a square wave and the approximation does have very high frequencies, and at some point the antenna would not be able to send these so it would be a low-pass filter.
answered 2 days ago
JustmeJustme
2,5411413
answered 2 days ago
JustmeJustme
2,5411413
answered 2 days ago
JustmeJustme
2,5411413
answered 2 days ago
JustmeJustme
2,5411413
2,5411413
2
$begingroup$
Continuous functions don't exist in the real world either, because of quantum effects.
$endgroup$
– supercat
2 days ago
2
$begingroup$
Also the logic that "it does not have a derivative at the step" does not mean that the function is not continuous. Not differentiable does not imply not continuous. That being said, the function is not continuous since the one sided limits at the step don't agree.
$endgroup$
– Sean Haight
2 days ago
add a comment |
2
$begingroup$
Continuous functions don't exist in the real world either, because of quantum effects.
$endgroup$
– supercat
2 days ago
2
$begingroup$
Also the logic that "it does not have a derivative at the step" does not mean that the function is not continuous. Not differentiable does not imply not continuous. That being said, the function is not continuous since the one sided limits at the step don't agree.
$endgroup$
– Sean Haight
2 days ago
2
2
$begingroup$
Continuous functions don't exist in the real world either, because of quantum effects.
$endgroup$
– supercat
2 days ago
$begingroup$
Continuous functions don't exist in the real world either, because of quantum effects.
$endgroup$
– supercat
2 days ago
2
2
$begingroup$
Also the logic that "it does not have a derivative at the step" does not mean that the function is not continuous. Not differentiable does not imply not continuous. That being said, the function is not continuous since the one sided limits at the step don't agree.
$endgroup$
– Sean Haight
2 days ago
$begingroup$
Also the logic that "it does not have a derivative at the step" does not mean that the function is not continuous. Not differentiable does not imply not continuous. That being said, the function is not continuous since the one sided limits at the step don't agree.
$endgroup$
– Sean Haight
2 days ago
add a comment |
$begingroup$
In a more general case compared to the answers above, nothing can be stoped or started in zero time ie instantly. To do so would imply an infinitely high frequency component which would translate to infinite energy. The constraining factors are speed of light limitation of Special Relativity and Quantum Mechanics Uncertainty Principle.
The sharper the transition you want, the more energy you have to pump into the system
answered 2 days ago
Dirk BruereDirk Bruere
5,96853164
$endgroup$
add a comment |
$begingroup$
In a more general case compared to the answers above, nothing can be stoped or started in zero time ie instantly. To do so would imply an infinitely high frequency component which would translate to infinite energy. The constraining factors are speed of light limitation of Special Relativity and Quantum Mechanics Uncertainty Principle.
The sharper the transition you want, the more energy you have to pump into the system
answered 2 days ago
Dirk BruereDirk Bruere
5,96853164
$endgroup$
add a comment |
$begingroup$
In a more general case compared to the answers above, nothing can be stoped or started in zero time ie instantly. To do so would imply an infinitely high frequency component which would translate to infinite energy. The constraining factors are speed of light limitation of Special Relativity and Quantum Mechanics Uncertainty Principle.
The sharper the transition you want, the more energy you have to pump into the system
answered 2 days ago
Dirk BruereDirk Bruere
5,96853164
$endgroup$
In a more general case compared to the answers above, nothing can be stoped or started in zero time ie instantly. To do so would imply an infinitely high frequency component which would translate to infinite energy. The constraining factors are speed of light limitation of Special Relativity and Quantum Mechanics Uncertainty Principle.
The sharper the transition you want, the more energy you have to pump into the system
answered 2 days ago
Dirk BruereDirk Bruere
5,96853164
answered 2 days ago
Dirk BruereDirk Bruere
5,96853164
answered 2 days ago
Dirk BruereDirk Bruere
5,96853164
answered 2 days ago
Dirk BruereDirk Bruere
5,96853164
5,96853164
add a comment |
add a comment |
Thanks for contributing an answer to Electrical Engineering 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%2felectronics.stackexchange.com%2fquestions%2f432994%2fdo-square-wave-exist%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
8
$begingroup$
A perfect square wave (0 rise / fall times) does not exist because it would require infinite bandwidth.
$endgroup$
– Peter Smith
2 days ago
2
$begingroup$
antennas have finite bandwidth
$endgroup$
– analogsystemsrf
2 days ago
2
$begingroup$
Infinite bandwidth and zero impedance
$endgroup$
– JonRB
2 days ago
$begingroup$
If the electric field is a close-to-ideal square wave, won't the magnetic field be more like a series of positive and negative spikes?
$endgroup$
– immibis
2 days ago