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
Is it OK to use the testing sample to compare algorithms?
Sally's older brother
An isoperimetric-type inequality inside a cube
Is a copyright notice with a non-existent name be invalid?
NIntegrate on a solution of a matrix ODE
The Nth Gryphon Number
What did Turing mean when saying that "machines cannot give rise to surprises" is due to a fallacy?
How could a hydrazine and N2O4 cloud (or it's reactants) show up in weather radar?
How to resize main filesystem
By what mechanism was the 2017 UK General Election called?
What criticisms of Wittgenstein's philosophy of language have been offered?
How to ask rejected full-time candidates to apply to teach individual courses?
In musical terms, what properties are varied by the human voice to produce different words / syllables?
As a dual citizen, my US passport will expire one day after traveling to the US. Will this work?
Inverse square law not accurate for non-point masses?
How does the body cool itself in a stillsuit?
Problem with display of presentation
Where did Ptolemy compare the Earth to the distance of fixed stars?
How can I list files in reverse time order by a command and pass them as arguments to another command?
How do Java 8 default methods hеlp with lambdas?
Any stored/leased 737s that could substitute for grounded MAXs?
One-one communication
Did any compiler fully use 80-bit floating point?
Weaponising the Grasp-at-a-Distance spell
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 Apr 17 at 22:26
evildemonic
2,7811023
asked Apr 17 at 7:22
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 Apr 17 at 22:26
evildemonic
2,7811023
asked Apr 17 at 7:22
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
Apr 17 at 7:50
2
$begingroup$
antennas have finite bandwidth
$endgroup$
– analogsystemsrf
Apr 17 at 13:00
2
$begingroup$
Infinite bandwidth and zero impedance
$endgroup$
– JonRB
Apr 17 at 20:35
$begingroup$
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
Apr 17 at 22:25
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 Apr 17 at 22:26
evildemonic
2,7811023
asked Apr 17 at 7:22
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 Apr 17 at 22:26
evildemonic
2,7811023
asked Apr 17 at 7:22
user163416user163416
12339
edited Apr 17 at 22:26
evildemonic
2,7811023
asked Apr 17 at 7:22
user163416user163416
12339
edited Apr 17 at 22:26
evildemonic
2,7811023
edited Apr 17 at 22:26
evildemonic
2,7811023
edited Apr 17 at 22:26
evildemonic
2,7811023
2,7811023
asked Apr 17 at 7:22
user163416user163416
12339
asked Apr 17 at 7:22
user163416user163416
12339
asked Apr 17 at 7:22
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
Apr 17 at 7:50
2
$begingroup$
antennas have finite bandwidth
$endgroup$
– analogsystemsrf
Apr 17 at 13:00
2
$begingroup$
Infinite bandwidth and zero impedance
$endgroup$
– JonRB
Apr 17 at 20:35
$begingroup$
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
Apr 17 at 22:25
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
Apr 17 at 7:50
2
$begingroup$
antennas have finite bandwidth
$endgroup$
– analogsystemsrf
Apr 17 at 13:00
2
$begingroup$
Infinite bandwidth and zero impedance
$endgroup$
– JonRB
Apr 17 at 20:35
$begingroup$
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
Apr 17 at 22:25
8
8
$begingroup$
A perfect square wave (0 rise / fall times) does not exist because it would require infinite bandwidth.
$endgroup$
– Peter Smith
Apr 17 at 7:50
$begingroup$
A perfect square wave (0 rise / fall times) does not exist because it would require infinite bandwidth.
$endgroup$
– Peter Smith
Apr 17 at 7:50
2
2
$begingroup$
antennas have finite bandwidth
$endgroup$
– analogsystemsrf
Apr 17 at 13:00
$begingroup$
antennas have finite bandwidth
$endgroup$
– analogsystemsrf
Apr 17 at 13:00
2
2
$begingroup$
Infinite bandwidth and zero impedance
$endgroup$
– JonRB
Apr 17 at 20:35
$begingroup$
Infinite bandwidth and zero impedance
$endgroup$
– JonRB
Apr 17 at 20:35
$begingroup$
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
Apr 17 at 22:25
$begingroup$
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
Apr 17 at 22:25
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 Apr 17 at 8:00
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
Apr 17 at 12:50
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
Apr 17 at 13:53
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
Apr 17 at 13:53
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
Apr 17 at 18:36
3
$begingroup$
Technically, any antenna will emit light if you get it hot enough
$endgroup$
– Nate Strickland
Apr 17 at 22:48
|
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 Apr 17 at 8:46
JustmeJustme
2,5511413
$endgroup$
2
$begingroup$
Continuous functions don't exist in the real world either, because of quantum effects.
$endgroup$
– supercat
Apr 17 at 18:37
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
Apr 17 at 21:57
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 Apr 17 at 10:05
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 Apr 17 at 8:00
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
Apr 17 at 12:50
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
Apr 17 at 13:53
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
Apr 17 at 13:53
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
Apr 17 at 18:36
3
$begingroup$
Technically, any antenna will emit light if you get it hot enough
$endgroup$
– Nate Strickland
Apr 17 at 22:48
|
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 Apr 17 at 8:00
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
Apr 17 at 12:50
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
Apr 17 at 13:53
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
Apr 17 at 13:53
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
Apr 17 at 18:36
3
$begingroup$
Technically, any antenna will emit light if you get it hot enough
$endgroup$
– Nate Strickland
Apr 17 at 22:48
|
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 Apr 17 at 8:00
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 Apr 17 at 8:00
Glenn WillenGlenn Willen
885211
edited Apr 17 at 22:44
answered Apr 17 at 8:00
Glenn WillenGlenn Willen
885211
answered Apr 17 at 8:00
Glenn WillenGlenn Willen
885211
answered Apr 17 at 8:00
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
Apr 17 at 12:50
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
Apr 17 at 13:53
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
Apr 17 at 13:53
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
Apr 17 at 18:36
3
$begingroup$
Technically, any antenna will emit light if you get it hot enough
$endgroup$
– Nate Strickland
Apr 17 at 22:48
|
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
Apr 17 at 12:50
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
Apr 17 at 13:53
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
Apr 17 at 13:53
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
Apr 17 at 18:36
3
$begingroup$
Technically, any antenna will emit light if you get it hot enough
$endgroup$
– Nate Strickland
Apr 17 at 22:48
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
Apr 17 at 12:50
$begingroup$
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
Apr 17 at 12:50
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
Apr 17 at 13:53
$begingroup$
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
Apr 17 at 13:53
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
Apr 17 at 13:53
$begingroup$
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
Apr 17 at 13:53
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
Apr 17 at 18:36
$begingroup$
@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
Apr 17 at 18:36
3
3
$begingroup$
Technically, any antenna will emit light if you get it hot enough
$endgroup$
– Nate Strickland
Apr 17 at 22:48
$begingroup$
Technically, any antenna will emit light if you get it hot enough
$endgroup$
– Nate Strickland
Apr 17 at 22:48
|
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 Apr 17 at 8:46
JustmeJustme
2,5511413
$endgroup$
2
$begingroup$
Continuous functions don't exist in the real world either, because of quantum effects.
$endgroup$
– supercat
Apr 17 at 18:37
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
Apr 17 at 21:57
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 Apr 17 at 8:46
JustmeJustme
2,5511413
$endgroup$
2
$begingroup$
Continuous functions don't exist in the real world either, because of quantum effects.
$endgroup$
– supercat
Apr 17 at 18:37
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
Apr 17 at 21:57
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 Apr 17 at 8:46
JustmeJustme
2,5511413
$endgroup$
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 Apr 17 at 8:46
JustmeJustme
2,5511413
answered Apr 17 at 8:46
JustmeJustme
2,5511413
answered Apr 17 at 8:46
JustmeJustme
2,5511413
answered Apr 17 at 8:46
JustmeJustme
2,5511413
2,5511413
2
$begingroup$
Continuous functions don't exist in the real world either, because of quantum effects.
$endgroup$
– supercat
Apr 17 at 18:37
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
Apr 17 at 21:57
add a comment |
2
$begingroup$
Continuous functions don't exist in the real world either, because of quantum effects.
$endgroup$
– supercat
Apr 17 at 18:37
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
Apr 17 at 21:57
2
2
$begingroup$
Continuous functions don't exist in the real world either, because of quantum effects.
$endgroup$
– supercat
Apr 17 at 18:37
$begingroup$
Continuous functions don't exist in the real world either, because of quantum effects.
$endgroup$
– supercat
Apr 17 at 18:37
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
Apr 17 at 21:57
$begingroup$
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
Apr 17 at 21:57
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 Apr 17 at 10:05
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 Apr 17 at 10:05
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 Apr 17 at 10:05
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 Apr 17 at 10:05
Dirk BruereDirk Bruere
5,96853164
answered Apr 17 at 10:05
Dirk BruereDirk Bruere
5,96853164
answered Apr 17 at 10:05
Dirk BruereDirk Bruere
5,96853164
answered Apr 17 at 10:05
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
Apr 17 at 7:50
2
$begingroup$
antennas have finite bandwidth
$endgroup$
– analogsystemsrf
Apr 17 at 13:00
2
$begingroup$
Infinite bandwidth and zero impedance
$endgroup$
– JonRB
Apr 17 at 20:35
$begingroup$
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
Apr 17 at 22:25