If one sends short bursts of electric current that abruptly turn on and off, why does the receiver receive a...
$begingroup$
An Introduction to Information Theory: Symbols, Signals and Noise, by John R. Pierce, says the following:
The difficulty which Morse encountered with his underground wire remained an important problem. Different circuits which conduct a steady electric current equally well are not necessarily equally suited to electrical communication. If one sends dots and dashes too fast over an underground or undersea circuit, they are run together at the receiving end. As indicated in Figure II-1, when we send a short burst of current which turns abruptly on and off, we receive at the far end of the circuit a longer, smoothed-out rise and fall of current. This longer flow of current may overlap the current of another symbol sent, for instance, as an absence of current. Thus, as shown in Figure II-2, when a clear and distinct signal is transmitted it may be received as a vaguely wandering rise and fall of current which is difficult to interpret.
Of course, if we make our dots, spaces, and dashes long enough, the current at the far end will follow the current at the sending end better, but this slows the rate of transmission. It is clear that there is somehow associated with a given transmission circuit a limiting speed of transmission for dots and spaces. For submarine cables this speed is so slow as to trouble telegraphers; for wires on poles it is so fast as not to bother telegraphers. Early telegraphists were aware of this limitation, and it, too, lies at the heart of communication theory.
As someone who does not have an electrical engineering background, I find the described phenomenon perplexing. If one sends short bursts of electric current that abruptly turn on and off, why is it true that, depending on the type of circuit, the receiver may receive a smoothed-out current, rather than the discrete currents that were sent? One would expect, naively perhaps, that the signals received would be identical to the signals sent?
I would greatly appreciate it if people could please take the time to answer this using language that can be understood by someone without an electrical engineering background.
current signal
asked Dec 20 '18 at 14:09
The PointerThe Pointer
30919
$endgroup$
|
show 4 more comments
$begingroup$
An Introduction to Information Theory: Symbols, Signals and Noise, by John R. Pierce, says the following:
The difficulty which Morse encountered with his underground wire remained an important problem. Different circuits which conduct a steady electric current equally well are not necessarily equally suited to electrical communication. If one sends dots and dashes too fast over an underground or undersea circuit, they are run together at the receiving end. As indicated in Figure II-1, when we send a short burst of current which turns abruptly on and off, we receive at the far end of the circuit a longer, smoothed-out rise and fall of current. This longer flow of current may overlap the current of another symbol sent, for instance, as an absence of current. Thus, as shown in Figure II-2, when a clear and distinct signal is transmitted it may be received as a vaguely wandering rise and fall of current which is difficult to interpret.
Of course, if we make our dots, spaces, and dashes long enough, the current at the far end will follow the current at the sending end better, but this slows the rate of transmission. It is clear that there is somehow associated with a given transmission circuit a limiting speed of transmission for dots and spaces. For submarine cables this speed is so slow as to trouble telegraphers; for wires on poles it is so fast as not to bother telegraphers. Early telegraphists were aware of this limitation, and it, too, lies at the heart of communication theory.
As someone who does not have an electrical engineering background, I find the described phenomenon perplexing. If one sends short bursts of electric current that abruptly turn on and off, why is it true that, depending on the type of circuit, the receiver may receive a smoothed-out current, rather than the discrete currents that were sent? One would expect, naively perhaps, that the signals received would be identical to the signals sent?
I would greatly appreciate it if people could please take the time to answer this using language that can be understood by someone without an electrical engineering background.
current signal
asked Dec 20 '18 at 14:09
The PointerThe Pointer
30919
$endgroup$
14
$begingroup$
RC low pass filter effect.
$endgroup$
– panic attack
Dec 20 '18 at 14:12
7
$begingroup$
@panicattack I don't understand what that means?
$endgroup$
– The Pointer
Dec 20 '18 at 14:12
9
$begingroup$
@panicattack This probably needs a less technical explanation, as per questioner's last paragraph.
$endgroup$
– pjc50
Dec 20 '18 at 14:19
8
$begingroup$
And do not write answers in comments...
$endgroup$
– Arsenal
Dec 20 '18 at 14:20
3
$begingroup$
@ElliotAlderson jusaca’s answer is the type of explanation that I was looking for. Although I don’t know what a “low pass filter” is, the provided explanation was enough to clarify what is going on.
$endgroup$
– The Pointer
Dec 20 '18 at 14:34
|
show 4 more comments
$begingroup$
An Introduction to Information Theory: Symbols, Signals and Noise, by John R. Pierce, says the following:
The difficulty which Morse encountered with his underground wire remained an important problem. Different circuits which conduct a steady electric current equally well are not necessarily equally suited to electrical communication. If one sends dots and dashes too fast over an underground or undersea circuit, they are run together at the receiving end. As indicated in Figure II-1, when we send a short burst of current which turns abruptly on and off, we receive at the far end of the circuit a longer, smoothed-out rise and fall of current. This longer flow of current may overlap the current of another symbol sent, for instance, as an absence of current. Thus, as shown in Figure II-2, when a clear and distinct signal is transmitted it may be received as a vaguely wandering rise and fall of current which is difficult to interpret.
Of course, if we make our dots, spaces, and dashes long enough, the current at the far end will follow the current at the sending end better, but this slows the rate of transmission. It is clear that there is somehow associated with a given transmission circuit a limiting speed of transmission for dots and spaces. For submarine cables this speed is so slow as to trouble telegraphers; for wires on poles it is so fast as not to bother telegraphers. Early telegraphists were aware of this limitation, and it, too, lies at the heart of communication theory.
As someone who does not have an electrical engineering background, I find the described phenomenon perplexing. If one sends short bursts of electric current that abruptly turn on and off, why is it true that, depending on the type of circuit, the receiver may receive a smoothed-out current, rather than the discrete currents that were sent? One would expect, naively perhaps, that the signals received would be identical to the signals sent?
I would greatly appreciate it if people could please take the time to answer this using language that can be understood by someone without an electrical engineering background.
current signal
asked Dec 20 '18 at 14:09
The PointerThe Pointer
30919
$endgroup$
An Introduction to Information Theory: Symbols, Signals and Noise, by John R. Pierce, says the following:
The difficulty which Morse encountered with his underground wire remained an important problem. Different circuits which conduct a steady electric current equally well are not necessarily equally suited to electrical communication. If one sends dots and dashes too fast over an underground or undersea circuit, they are run together at the receiving end. As indicated in Figure II-1, when we send a short burst of current which turns abruptly on and off, we receive at the far end of the circuit a longer, smoothed-out rise and fall of current. This longer flow of current may overlap the current of another symbol sent, for instance, as an absence of current. Thus, as shown in Figure II-2, when a clear and distinct signal is transmitted it may be received as a vaguely wandering rise and fall of current which is difficult to interpret.
Of course, if we make our dots, spaces, and dashes long enough, the current at the far end will follow the current at the sending end better, but this slows the rate of transmission. It is clear that there is somehow associated with a given transmission circuit a limiting speed of transmission for dots and spaces. For submarine cables this speed is so slow as to trouble telegraphers; for wires on poles it is so fast as not to bother telegraphers. Early telegraphists were aware of this limitation, and it, too, lies at the heart of communication theory.
As someone who does not have an electrical engineering background, I find the described phenomenon perplexing. If one sends short bursts of electric current that abruptly turn on and off, why is it true that, depending on the type of circuit, the receiver may receive a smoothed-out current, rather than the discrete currents that were sent? One would expect, naively perhaps, that the signals received would be identical to the signals sent?
I would greatly appreciate it if people could please take the time to answer this using language that can be understood by someone without an electrical engineering background.
current signal
current signal
asked Dec 20 '18 at 14:09
The PointerThe Pointer
30919
asked Dec 20 '18 at 14:09
The PointerThe Pointer
30919
edited Dec 22 '18 at 13:58
The Pointer
asked Dec 20 '18 at 14:09
The PointerThe Pointer
30919
asked Dec 20 '18 at 14:09
The PointerThe Pointer
30919
asked Dec 20 '18 at 14:09
The PointerThe Pointer
30919
30919
14
$begingroup$
RC low pass filter effect.
$endgroup$
– panic attack
Dec 20 '18 at 14:12
7
$begingroup$
@panicattack I don't understand what that means?
$endgroup$
– The Pointer
Dec 20 '18 at 14:12
9
$begingroup$
@panicattack This probably needs a less technical explanation, as per questioner's last paragraph.
$endgroup$
– pjc50
Dec 20 '18 at 14:19
8
$begingroup$
And do not write answers in comments...
$endgroup$
– Arsenal
Dec 20 '18 at 14:20
3
$begingroup$
@ElliotAlderson jusaca’s answer is the type of explanation that I was looking for. Although I don’t know what a “low pass filter” is, the provided explanation was enough to clarify what is going on.
$endgroup$
– The Pointer
Dec 20 '18 at 14:34
|
show 4 more comments
14
$begingroup$
RC low pass filter effect.
$endgroup$
– panic attack
Dec 20 '18 at 14:12
7
$begingroup$
@panicattack I don't understand what that means?
$endgroup$
– The Pointer
Dec 20 '18 at 14:12
9
$begingroup$
@panicattack This probably needs a less technical explanation, as per questioner's last paragraph.
$endgroup$
– pjc50
Dec 20 '18 at 14:19
8
$begingroup$
And do not write answers in comments...
$endgroup$
– Arsenal
Dec 20 '18 at 14:20
3
$begingroup$
@ElliotAlderson jusaca’s answer is the type of explanation that I was looking for. Although I don’t know what a “low pass filter” is, the provided explanation was enough to clarify what is going on.
$endgroup$
– The Pointer
Dec 20 '18 at 14:34
14
14
$begingroup$
RC low pass filter effect.
$endgroup$
– panic attack
Dec 20 '18 at 14:12
$begingroup$
RC low pass filter effect.
$endgroup$
– panic attack
Dec 20 '18 at 14:12
7
7
$begingroup$
@panicattack I don't understand what that means?
$endgroup$
– The Pointer
Dec 20 '18 at 14:12
$begingroup$
@panicattack I don't understand what that means?
$endgroup$
– The Pointer
Dec 20 '18 at 14:12
9
9
$begingroup$
@panicattack This probably needs a less technical explanation, as per questioner's last paragraph.
$endgroup$
– pjc50
Dec 20 '18 at 14:19
$begingroup$
@panicattack This probably needs a less technical explanation, as per questioner's last paragraph.
$endgroup$
– pjc50
Dec 20 '18 at 14:19
8
8
$begingroup$
And do not write answers in comments...
$endgroup$
– Arsenal
Dec 20 '18 at 14:20
$begingroup$
And do not write answers in comments...
$endgroup$
– Arsenal
Dec 20 '18 at 14:20
3
3
$begingroup$
@ElliotAlderson jusaca’s answer is the type of explanation that I was looking for. Although I don’t know what a “low pass filter” is, the provided explanation was enough to clarify what is going on.
$endgroup$
– The Pointer
Dec 20 '18 at 14:34
$begingroup$
@ElliotAlderson jusaca’s answer is the type of explanation that I was looking for. Although I don’t know what a “low pass filter” is, the provided explanation was enough to clarify what is going on.
$endgroup$
– The Pointer
Dec 20 '18 at 14:34
|
show 4 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Your cable behaves like a low pass filter, which means that high frequencies are getting damped. The longer the cable, the stronger is this effect.
Impulses have, due to their fast rise and fall, pretty fast frequency components.
If these high frequencies are damped, your impulse "smears" over time and you get the desired result you posted in your question.
edited Dec 21 '18 at 12:48
Community♦
1
answered Dec 20 '18 at 14:28
jusacajusaca
54337
$endgroup$
1
$begingroup$
This is the type of explanation I was looking for. Thanks!
$endgroup$
– The Pointer
Dec 20 '18 at 14:35
19
$begingroup$
This comes from the inductance and capacitance distributed along the cable. You can shove current in one end and some of it will go into charging up the capacitor formed by the cable. When you stop shoving current in one end, the inductance keeps the current flowing and discharges the capacitor. That is where the electrons go.
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– Ross Millikan
Dec 20 '18 at 16:02
5
$begingroup$
I like to think of it like how a fly-wheel smooths out the rotation of a car engine. Every second stroke is a small explosion (so many abrupt 'shoves' or pulses), but because the fly-wheel is heavy it won't react instantaneously to the pulse, and as it's heavy, it will continue spinning after the pulse. Similar things happen in a signal cable. The cable itself is first charged up slightly (which takes a certain time), and then that charge is released (which also takes a certain time)
$endgroup$
– AkselA
Dec 20 '18 at 18:29
5
$begingroup$
Interesting to add: This actually makes high voltage AC power transmission less efficient over long distances. Despite the high up-front cost and inefficiency of switching DC voltages, the cost is justified over very long stretches of high voltage DC transmission.
$endgroup$
– Alexander
Dec 20 '18 at 20:53
add a comment |
$begingroup$
Perhaps it would be useful to think of it differently. Instead of shoving electricity around, pretend that you have a very long pipe, some water (at pressure) and a valve.
If you turn on the valve at one end of the pipe, it will take a certain amount of time to pressurize the pipe and force the water through. At the other end, the water will eventually come out, but as a slow increase in flow, followed by a slow decrease.
If you turn the water on and off quickly enough, it will simply appear at the other and as a moderate flow.
As mentioned elsewhere, in electrical terms, this is due to the capacitance, and inductance of the long cable. The longer the cable/pipe, the greater the apparent smoothing effect. The reasons (physics) and math may be different, but the results rhyme, and hopefully that will help you understand.
answered Dec 20 '18 at 18:06
CommodoreCommodore
1512
$endgroup$
3
$begingroup$
Right! Slowed by the inductance, the charge literally does "pile up" in the capacitance and then "dribble out".
$endgroup$
– AaronD
Dec 20 '18 at 18:51
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To extend the water analogy, walk to your local swimming pool, grab a kickboard, and try to make a nice wall of water ala the movie "Ten Commandments". Not only can you not make a stationary vertical wall, you can't make a moving sharp wavefront.
$endgroup$
– Ben Voigt
Dec 21 '18 at 16:30
add a comment |
$begingroup$
Infinitely sharp edges in the signal need infinite bandwidth to transmit which is not possible with real world wires. Taken a long enough wire, it passes only low frequencies and high frequencies are attenuated, so the sharp fast edges get smeared into slower wavy edges and thus you need to send longer pulses to see the voltage to slowly rise up to proper detection level at the receiving end. The signals degrade as each unit length of the copper wire can be thought as a series resistance with inductance and it also has stray parallel capacitance and leakage resistance to its surroundings, otherwise this model is known as a transmission line. The series resistance with parallel capacitance is a low-pass filter. The capacitance and inductance forms what is known as the characteristic impedance of the transmission line, and if the transmitting and receiving ends do not match the transmission line impedance, a voltage pulse will reflect back to wire to some extent so it ping-pongs back and forth in the wire until it settles. As the voltage pulse travels in the wire at roughly two thirds of the speed of light, all this how much the signal degrades when it travels through the transmission line or keeps bouncing between the ends of transmission line determines how fast signals you can transmit and how far you can transmit them without being too degraded.
answered Dec 20 '18 at 17:55
JustmeJustme
85518
$endgroup$
4
$begingroup$
"Taken a long enough wire, it passes only low frequencies" -- I think the question was exactly why that happens. It's also not even about just infinitely sharp edges.
$endgroup$
– ilkkachu
Dec 20 '18 at 18:00
$begingroup$
Thanks, I added description why that happens.
$endgroup$
– Justme
Dec 20 '18 at 19:02
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your cable behaves like a low pass filter, which means that high frequencies are getting damped. The longer the cable, the stronger is this effect.
Impulses have, due to their fast rise and fall, pretty fast frequency components.
If these high frequencies are damped, your impulse "smears" over time and you get the desired result you posted in your question.
edited Dec 21 '18 at 12:48
Community♦
1
answered Dec 20 '18 at 14:28
jusacajusaca
54337
$endgroup$
1
$begingroup$
This is the type of explanation I was looking for. Thanks!
$endgroup$
– The Pointer
Dec 20 '18 at 14:35
19
$begingroup$
This comes from the inductance and capacitance distributed along the cable. You can shove current in one end and some of it will go into charging up the capacitor formed by the cable. When you stop shoving current in one end, the inductance keeps the current flowing and discharges the capacitor. That is where the electrons go.
$endgroup$
– Ross Millikan
Dec 20 '18 at 16:02
5
$begingroup$
I like to think of it like how a fly-wheel smooths out the rotation of a car engine. Every second stroke is a small explosion (so many abrupt 'shoves' or pulses), but because the fly-wheel is heavy it won't react instantaneously to the pulse, and as it's heavy, it will continue spinning after the pulse. Similar things happen in a signal cable. The cable itself is first charged up slightly (which takes a certain time), and then that charge is released (which also takes a certain time)
$endgroup$
– AkselA
Dec 20 '18 at 18:29
5
$begingroup$
Interesting to add: This actually makes high voltage AC power transmission less efficient over long distances. Despite the high up-front cost and inefficiency of switching DC voltages, the cost is justified over very long stretches of high voltage DC transmission.
$endgroup$
– Alexander
Dec 20 '18 at 20:53
add a comment |
$begingroup$
Your cable behaves like a low pass filter, which means that high frequencies are getting damped. The longer the cable, the stronger is this effect.
Impulses have, due to their fast rise and fall, pretty fast frequency components.
If these high frequencies are damped, your impulse "smears" over time and you get the desired result you posted in your question.
edited Dec 21 '18 at 12:48
Community♦
1
answered Dec 20 '18 at 14:28
jusacajusaca
54337
$endgroup$
1
$begingroup$
This is the type of explanation I was looking for. Thanks!
$endgroup$
– The Pointer
Dec 20 '18 at 14:35
19
$begingroup$
This comes from the inductance and capacitance distributed along the cable. You can shove current in one end and some of it will go into charging up the capacitor formed by the cable. When you stop shoving current in one end, the inductance keeps the current flowing and discharges the capacitor. That is where the electrons go.
$endgroup$
– Ross Millikan
Dec 20 '18 at 16:02
5
$begingroup$
I like to think of it like how a fly-wheel smooths out the rotation of a car engine. Every second stroke is a small explosion (so many abrupt 'shoves' or pulses), but because the fly-wheel is heavy it won't react instantaneously to the pulse, and as it's heavy, it will continue spinning after the pulse. Similar things happen in a signal cable. The cable itself is first charged up slightly (which takes a certain time), and then that charge is released (which also takes a certain time)
$endgroup$
– AkselA
Dec 20 '18 at 18:29
5
$begingroup$
Interesting to add: This actually makes high voltage AC power transmission less efficient over long distances. Despite the high up-front cost and inefficiency of switching DC voltages, the cost is justified over very long stretches of high voltage DC transmission.
$endgroup$
– Alexander
Dec 20 '18 at 20:53
add a comment |
$begingroup$
Your cable behaves like a low pass filter, which means that high frequencies are getting damped. The longer the cable, the stronger is this effect.
Impulses have, due to their fast rise and fall, pretty fast frequency components.
If these high frequencies are damped, your impulse "smears" over time and you get the desired result you posted in your question.
edited Dec 21 '18 at 12:48
Community♦
1
answered Dec 20 '18 at 14:28
jusacajusaca
54337
$endgroup$
Your cable behaves like a low pass filter, which means that high frequencies are getting damped. The longer the cable, the stronger is this effect.
Impulses have, due to their fast rise and fall, pretty fast frequency components.
If these high frequencies are damped, your impulse "smears" over time and you get the desired result you posted in your question.
edited Dec 21 '18 at 12:48
Community♦
1
answered Dec 20 '18 at 14:28
jusacajusaca
54337
edited Dec 21 '18 at 12:48
Community♦
1
edited Dec 21 '18 at 12:48
Community♦
1
edited Dec 21 '18 at 12:48
Community♦
1
1
answered Dec 20 '18 at 14:28
jusacajusaca
54337
answered Dec 20 '18 at 14:28
jusacajusaca
54337
answered Dec 20 '18 at 14:28
jusacajusaca
54337
54337
1
$begingroup$
This is the type of explanation I was looking for. Thanks!
$endgroup$
– The Pointer
Dec 20 '18 at 14:35
19
$begingroup$
This comes from the inductance and capacitance distributed along the cable. You can shove current in one end and some of it will go into charging up the capacitor formed by the cable. When you stop shoving current in one end, the inductance keeps the current flowing and discharges the capacitor. That is where the electrons go.
$endgroup$
– Ross Millikan
Dec 20 '18 at 16:02
5
$begingroup$
I like to think of it like how a fly-wheel smooths out the rotation of a car engine. Every second stroke is a small explosion (so many abrupt 'shoves' or pulses), but because the fly-wheel is heavy it won't react instantaneously to the pulse, and as it's heavy, it will continue spinning after the pulse. Similar things happen in a signal cable. The cable itself is first charged up slightly (which takes a certain time), and then that charge is released (which also takes a certain time)
$endgroup$
– AkselA
Dec 20 '18 at 18:29
5
$begingroup$
Interesting to add: This actually makes high voltage AC power transmission less efficient over long distances. Despite the high up-front cost and inefficiency of switching DC voltages, the cost is justified over very long stretches of high voltage DC transmission.
$endgroup$
– Alexander
Dec 20 '18 at 20:53
add a comment |
1
$begingroup$
This is the type of explanation I was looking for. Thanks!
$endgroup$
– The Pointer
Dec 20 '18 at 14:35
19
$begingroup$
This comes from the inductance and capacitance distributed along the cable. You can shove current in one end and some of it will go into charging up the capacitor formed by the cable. When you stop shoving current in one end, the inductance keeps the current flowing and discharges the capacitor. That is where the electrons go.
$endgroup$
– Ross Millikan
Dec 20 '18 at 16:02
5
$begingroup$
I like to think of it like how a fly-wheel smooths out the rotation of a car engine. Every second stroke is a small explosion (so many abrupt 'shoves' or pulses), but because the fly-wheel is heavy it won't react instantaneously to the pulse, and as it's heavy, it will continue spinning after the pulse. Similar things happen in a signal cable. The cable itself is first charged up slightly (which takes a certain time), and then that charge is released (which also takes a certain time)
$endgroup$
– AkselA
Dec 20 '18 at 18:29
5
$begingroup$
Interesting to add: This actually makes high voltage AC power transmission less efficient over long distances. Despite the high up-front cost and inefficiency of switching DC voltages, the cost is justified over very long stretches of high voltage DC transmission.
$endgroup$
– Alexander
Dec 20 '18 at 20:53
1
1
$begingroup$
This is the type of explanation I was looking for. Thanks!
$endgroup$
– The Pointer
Dec 20 '18 at 14:35
$begingroup$
This is the type of explanation I was looking for. Thanks!
$endgroup$
– The Pointer
Dec 20 '18 at 14:35
19
19
$begingroup$
This comes from the inductance and capacitance distributed along the cable. You can shove current in one end and some of it will go into charging up the capacitor formed by the cable. When you stop shoving current in one end, the inductance keeps the current flowing and discharges the capacitor. That is where the electrons go.
$endgroup$
– Ross Millikan
Dec 20 '18 at 16:02
$begingroup$
This comes from the inductance and capacitance distributed along the cable. You can shove current in one end and some of it will go into charging up the capacitor formed by the cable. When you stop shoving current in one end, the inductance keeps the current flowing and discharges the capacitor. That is where the electrons go.
$endgroup$
– Ross Millikan
Dec 20 '18 at 16:02
5
5
$begingroup$
I like to think of it like how a fly-wheel smooths out the rotation of a car engine. Every second stroke is a small explosion (so many abrupt 'shoves' or pulses), but because the fly-wheel is heavy it won't react instantaneously to the pulse, and as it's heavy, it will continue spinning after the pulse. Similar things happen in a signal cable. The cable itself is first charged up slightly (which takes a certain time), and then that charge is released (which also takes a certain time)
$endgroup$
– AkselA
Dec 20 '18 at 18:29
$begingroup$
I like to think of it like how a fly-wheel smooths out the rotation of a car engine. Every second stroke is a small explosion (so many abrupt 'shoves' or pulses), but because the fly-wheel is heavy it won't react instantaneously to the pulse, and as it's heavy, it will continue spinning after the pulse. Similar things happen in a signal cable. The cable itself is first charged up slightly (which takes a certain time), and then that charge is released (which also takes a certain time)
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– AkselA
Dec 20 '18 at 18:29
5
5
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Interesting to add: This actually makes high voltage AC power transmission less efficient over long distances. Despite the high up-front cost and inefficiency of switching DC voltages, the cost is justified over very long stretches of high voltage DC transmission.
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– Alexander
Dec 20 '18 at 20:53
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Interesting to add: This actually makes high voltage AC power transmission less efficient over long distances. Despite the high up-front cost and inefficiency of switching DC voltages, the cost is justified over very long stretches of high voltage DC transmission.
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– Alexander
Dec 20 '18 at 20:53
add a comment |
$begingroup$
Perhaps it would be useful to think of it differently. Instead of shoving electricity around, pretend that you have a very long pipe, some water (at pressure) and a valve.
If you turn on the valve at one end of the pipe, it will take a certain amount of time to pressurize the pipe and force the water through. At the other end, the water will eventually come out, but as a slow increase in flow, followed by a slow decrease.
If you turn the water on and off quickly enough, it will simply appear at the other and as a moderate flow.
As mentioned elsewhere, in electrical terms, this is due to the capacitance, and inductance of the long cable. The longer the cable/pipe, the greater the apparent smoothing effect. The reasons (physics) and math may be different, but the results rhyme, and hopefully that will help you understand.
answered Dec 20 '18 at 18:06
CommodoreCommodore
1512
$endgroup$
3
$begingroup$
Right! Slowed by the inductance, the charge literally does "pile up" in the capacitance and then "dribble out".
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– AaronD
Dec 20 '18 at 18:51
$begingroup$
To extend the water analogy, walk to your local swimming pool, grab a kickboard, and try to make a nice wall of water ala the movie "Ten Commandments". Not only can you not make a stationary vertical wall, you can't make a moving sharp wavefront.
$endgroup$
– Ben Voigt
Dec 21 '18 at 16:30
add a comment |
$begingroup$
Perhaps it would be useful to think of it differently. Instead of shoving electricity around, pretend that you have a very long pipe, some water (at pressure) and a valve.
If you turn on the valve at one end of the pipe, it will take a certain amount of time to pressurize the pipe and force the water through. At the other end, the water will eventually come out, but as a slow increase in flow, followed by a slow decrease.
If you turn the water on and off quickly enough, it will simply appear at the other and as a moderate flow.
As mentioned elsewhere, in electrical terms, this is due to the capacitance, and inductance of the long cable. The longer the cable/pipe, the greater the apparent smoothing effect. The reasons (physics) and math may be different, but the results rhyme, and hopefully that will help you understand.
answered Dec 20 '18 at 18:06
CommodoreCommodore
1512
$endgroup$
3
$begingroup$
Right! Slowed by the inductance, the charge literally does "pile up" in the capacitance and then "dribble out".
$endgroup$
– AaronD
Dec 20 '18 at 18:51
$begingroup$
To extend the water analogy, walk to your local swimming pool, grab a kickboard, and try to make a nice wall of water ala the movie "Ten Commandments". Not only can you not make a stationary vertical wall, you can't make a moving sharp wavefront.
$endgroup$
– Ben Voigt
Dec 21 '18 at 16:30
add a comment |
$begingroup$
Perhaps it would be useful to think of it differently. Instead of shoving electricity around, pretend that you have a very long pipe, some water (at pressure) and a valve.
If you turn on the valve at one end of the pipe, it will take a certain amount of time to pressurize the pipe and force the water through. At the other end, the water will eventually come out, but as a slow increase in flow, followed by a slow decrease.
If you turn the water on and off quickly enough, it will simply appear at the other and as a moderate flow.
As mentioned elsewhere, in electrical terms, this is due to the capacitance, and inductance of the long cable. The longer the cable/pipe, the greater the apparent smoothing effect. The reasons (physics) and math may be different, but the results rhyme, and hopefully that will help you understand.
answered Dec 20 '18 at 18:06
CommodoreCommodore
1512
$endgroup$
Perhaps it would be useful to think of it differently. Instead of shoving electricity around, pretend that you have a very long pipe, some water (at pressure) and a valve.
If you turn on the valve at one end of the pipe, it will take a certain amount of time to pressurize the pipe and force the water through. At the other end, the water will eventually come out, but as a slow increase in flow, followed by a slow decrease.
If you turn the water on and off quickly enough, it will simply appear at the other and as a moderate flow.
As mentioned elsewhere, in electrical terms, this is due to the capacitance, and inductance of the long cable. The longer the cable/pipe, the greater the apparent smoothing effect. The reasons (physics) and math may be different, but the results rhyme, and hopefully that will help you understand.
answered Dec 20 '18 at 18:06
CommodoreCommodore
1512
answered Dec 20 '18 at 18:06
CommodoreCommodore
1512
answered Dec 20 '18 at 18:06
CommodoreCommodore
1512
answered Dec 20 '18 at 18:06
CommodoreCommodore
1512
1512
3
$begingroup$
Right! Slowed by the inductance, the charge literally does "pile up" in the capacitance and then "dribble out".
$endgroup$
– AaronD
Dec 20 '18 at 18:51
$begingroup$
To extend the water analogy, walk to your local swimming pool, grab a kickboard, and try to make a nice wall of water ala the movie "Ten Commandments". Not only can you not make a stationary vertical wall, you can't make a moving sharp wavefront.
$endgroup$
– Ben Voigt
Dec 21 '18 at 16:30
add a comment |
3
$begingroup$
Right! Slowed by the inductance, the charge literally does "pile up" in the capacitance and then "dribble out".
$endgroup$
– AaronD
Dec 20 '18 at 18:51
$begingroup$
To extend the water analogy, walk to your local swimming pool, grab a kickboard, and try to make a nice wall of water ala the movie "Ten Commandments". Not only can you not make a stationary vertical wall, you can't make a moving sharp wavefront.
$endgroup$
– Ben Voigt
Dec 21 '18 at 16:30
3
3
$begingroup$
Right! Slowed by the inductance, the charge literally does "pile up" in the capacitance and then "dribble out".
$endgroup$
– AaronD
Dec 20 '18 at 18:51
$begingroup$
Right! Slowed by the inductance, the charge literally does "pile up" in the capacitance and then "dribble out".
$endgroup$
– AaronD
Dec 20 '18 at 18:51
$begingroup$
To extend the water analogy, walk to your local swimming pool, grab a kickboard, and try to make a nice wall of water ala the movie "Ten Commandments". Not only can you not make a stationary vertical wall, you can't make a moving sharp wavefront.
$endgroup$
– Ben Voigt
Dec 21 '18 at 16:30
$begingroup$
To extend the water analogy, walk to your local swimming pool, grab a kickboard, and try to make a nice wall of water ala the movie "Ten Commandments". Not only can you not make a stationary vertical wall, you can't make a moving sharp wavefront.
$endgroup$
– Ben Voigt
Dec 21 '18 at 16:30
add a comment |
$begingroup$
Infinitely sharp edges in the signal need infinite bandwidth to transmit which is not possible with real world wires. Taken a long enough wire, it passes only low frequencies and high frequencies are attenuated, so the sharp fast edges get smeared into slower wavy edges and thus you need to send longer pulses to see the voltage to slowly rise up to proper detection level at the receiving end. The signals degrade as each unit length of the copper wire can be thought as a series resistance with inductance and it also has stray parallel capacitance and leakage resistance to its surroundings, otherwise this model is known as a transmission line. The series resistance with parallel capacitance is a low-pass filter. The capacitance and inductance forms what is known as the characteristic impedance of the transmission line, and if the transmitting and receiving ends do not match the transmission line impedance, a voltage pulse will reflect back to wire to some extent so it ping-pongs back and forth in the wire until it settles. As the voltage pulse travels in the wire at roughly two thirds of the speed of light, all this how much the signal degrades when it travels through the transmission line or keeps bouncing between the ends of transmission line determines how fast signals you can transmit and how far you can transmit them without being too degraded.
answered Dec 20 '18 at 17:55
JustmeJustme
85518
$endgroup$
4
$begingroup$
"Taken a long enough wire, it passes only low frequencies" -- I think the question was exactly why that happens. It's also not even about just infinitely sharp edges.
$endgroup$
– ilkkachu
Dec 20 '18 at 18:00
$begingroup$
Thanks, I added description why that happens.
$endgroup$
– Justme
Dec 20 '18 at 19:02
add a comment |
$begingroup$
Infinitely sharp edges in the signal need infinite bandwidth to transmit which is not possible with real world wires. Taken a long enough wire, it passes only low frequencies and high frequencies are attenuated, so the sharp fast edges get smeared into slower wavy edges and thus you need to send longer pulses to see the voltage to slowly rise up to proper detection level at the receiving end. The signals degrade as each unit length of the copper wire can be thought as a series resistance with inductance and it also has stray parallel capacitance and leakage resistance to its surroundings, otherwise this model is known as a transmission line. The series resistance with parallel capacitance is a low-pass filter. The capacitance and inductance forms what is known as the characteristic impedance of the transmission line, and if the transmitting and receiving ends do not match the transmission line impedance, a voltage pulse will reflect back to wire to some extent so it ping-pongs back and forth in the wire until it settles. As the voltage pulse travels in the wire at roughly two thirds of the speed of light, all this how much the signal degrades when it travels through the transmission line or keeps bouncing between the ends of transmission line determines how fast signals you can transmit and how far you can transmit them without being too degraded.
answered Dec 20 '18 at 17:55
JustmeJustme
85518
$endgroup$
4
$begingroup$
"Taken a long enough wire, it passes only low frequencies" -- I think the question was exactly why that happens. It's also not even about just infinitely sharp edges.
$endgroup$
– ilkkachu
Dec 20 '18 at 18:00
$begingroup$
Thanks, I added description why that happens.
$endgroup$
– Justme
Dec 20 '18 at 19:02
add a comment |
$begingroup$
Infinitely sharp edges in the signal need infinite bandwidth to transmit which is not possible with real world wires. Taken a long enough wire, it passes only low frequencies and high frequencies are attenuated, so the sharp fast edges get smeared into slower wavy edges and thus you need to send longer pulses to see the voltage to slowly rise up to proper detection level at the receiving end. The signals degrade as each unit length of the copper wire can be thought as a series resistance with inductance and it also has stray parallel capacitance and leakage resistance to its surroundings, otherwise this model is known as a transmission line. The series resistance with parallel capacitance is a low-pass filter. The capacitance and inductance forms what is known as the characteristic impedance of the transmission line, and if the transmitting and receiving ends do not match the transmission line impedance, a voltage pulse will reflect back to wire to some extent so it ping-pongs back and forth in the wire until it settles. As the voltage pulse travels in the wire at roughly two thirds of the speed of light, all this how much the signal degrades when it travels through the transmission line or keeps bouncing between the ends of transmission line determines how fast signals you can transmit and how far you can transmit them without being too degraded.
answered Dec 20 '18 at 17:55
JustmeJustme
85518
$endgroup$
Infinitely sharp edges in the signal need infinite bandwidth to transmit which is not possible with real world wires. Taken a long enough wire, it passes only low frequencies and high frequencies are attenuated, so the sharp fast edges get smeared into slower wavy edges and thus you need to send longer pulses to see the voltage to slowly rise up to proper detection level at the receiving end. The signals degrade as each unit length of the copper wire can be thought as a series resistance with inductance and it also has stray parallel capacitance and leakage resistance to its surroundings, otherwise this model is known as a transmission line. The series resistance with parallel capacitance is a low-pass filter. The capacitance and inductance forms what is known as the characteristic impedance of the transmission line, and if the transmitting and receiving ends do not match the transmission line impedance, a voltage pulse will reflect back to wire to some extent so it ping-pongs back and forth in the wire until it settles. As the voltage pulse travels in the wire at roughly two thirds of the speed of light, all this how much the signal degrades when it travels through the transmission line or keeps bouncing between the ends of transmission line determines how fast signals you can transmit and how far you can transmit them without being too degraded.
answered Dec 20 '18 at 17:55
JustmeJustme
85518
edited Dec 20 '18 at 18:47
answered Dec 20 '18 at 17:55
JustmeJustme
85518
answered Dec 20 '18 at 17:55
JustmeJustme
85518
answered Dec 20 '18 at 17:55
JustmeJustme
85518
85518
4
$begingroup$
"Taken a long enough wire, it passes only low frequencies" -- I think the question was exactly why that happens. It's also not even about just infinitely sharp edges.
$endgroup$
– ilkkachu
Dec 20 '18 at 18:00
$begingroup$
Thanks, I added description why that happens.
$endgroup$
– Justme
Dec 20 '18 at 19:02
add a comment |
4
$begingroup$
"Taken a long enough wire, it passes only low frequencies" -- I think the question was exactly why that happens. It's also not even about just infinitely sharp edges.
$endgroup$
– ilkkachu
Dec 20 '18 at 18:00
$begingroup$
Thanks, I added description why that happens.
$endgroup$
– Justme
Dec 20 '18 at 19:02
4
4
$begingroup$
"Taken a long enough wire, it passes only low frequencies" -- I think the question was exactly why that happens. It's also not even about just infinitely sharp edges.
$endgroup$
– ilkkachu
Dec 20 '18 at 18:00
$begingroup$
"Taken a long enough wire, it passes only low frequencies" -- I think the question was exactly why that happens. It's also not even about just infinitely sharp edges.
$endgroup$
– ilkkachu
Dec 20 '18 at 18:00
$begingroup$
Thanks, I added description why that happens.
$endgroup$
– Justme
Dec 20 '18 at 19:02
$begingroup$
Thanks, I added description why that happens.
$endgroup$
– Justme
Dec 20 '18 at 19:02
add a comment |
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14
$begingroup$
RC low pass filter effect.
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– panic attack
Dec 20 '18 at 14:12
7
$begingroup$
@panicattack I don't understand what that means?
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– The Pointer
Dec 20 '18 at 14:12
9
$begingroup$
@panicattack This probably needs a less technical explanation, as per questioner's last paragraph.
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– pjc50
Dec 20 '18 at 14:19
8
$begingroup$
And do not write answers in comments...
$endgroup$
– Arsenal
Dec 20 '18 at 14:20
3
$begingroup$
@ElliotAlderson jusaca’s answer is the type of explanation that I was looking for. Although I don’t know what a “low pass filter” is, the provided explanation was enough to clarify what is going on.
$endgroup$
– The Pointer
Dec 20 '18 at 14:34