# Signal propogating in a wire

gursimran
While studying the type of signal propogation in telephone wires, I read that it has a bandwidth of 400-3000 Hz and the signal travels with the speed of light(as in case of all electric signals). So electronics at the two ends communicate atonce. One changes the voltage at one end and the other detects, it changes again and the other samples again. So the signal is transfered.

But if suppose we have a very long wire and the time of signal tranfer is finite(say 1s) and the electronics on one side change the signal before the voltage on the other side has actually changed. Does we have a so called voltage wave that travells?? What happens in that scenario???

yungman
The wires has to go in pair to form a guided structure. The signal that travel through the structure is actually a TEM wave. It is not the voltage that travel through the wire. TEM wave travelling through the conductive guided structure cause surface current and that is the current that the recieving end detected and become voltage. It is almost like voltage travel at close to light speed. But actually it is the electromagnetic wave. I have a post currently here about TEM wave. Read that and you'll understand more.

Read the last few posts when I intentionally make the circuit into a non-guided structure. I am not convinced one way or the other, but I believe voltage is travelling very slow. It it is the TEM wave that travel at a fraction of the light speed.

BTW, a normal electrical wire with two or three conductors you buy in store is considered as two wire transmission line. Together with coax lines, striplines are all guided structure. So without knowing it, you already put in a guided structure and TEM wave travel in high speed.

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gursimran
The wires has to go in pair to form a guided structure. The signal that travel through the structure is actually a TEM wave. It is not the voltage that travel through the wire. TEM wave travelling through the conductive guided structure cause surface current and that is the current that the recieving end detected and become voltage. It is almost like voltage travel at close to light speed. But actually it is the electromagnetic wave. I have a post currently here about TEM wave. Read that and you'll understand more.

Read the last few posts when I intentionally make the circuit into a non-guided structure. I am not convinced one way or the other, but I believe voltage is travelling very slow. It it is the TEM wave that travel at a fraction of the light speed.

BTW, a normal electrical wire with two or three conductors you buy in store is considered as two wire transmission line. Together with coax lines, striplines are all guided structure. So without knowing it, you already put in a guided structure and TEM wave travel in high speed.

Thanks for replying. I understood what you intended to say. But as the electromagnetic wave travels carrying information(say amplitude as a function of time) to long distances and the information is not lost(even the time difference between the two signals (say 'a') remain same). Do you believe that the signal will get transmitted even in case of a guided structure, just like a EM wave do.

I have instinct about normal wires that, if I apply a voltage(say 5v)(amplitude) and it does reach the other end and recorded or acted upon by the instrument, before the next signal(voltage say 1) is applied, the previous signal will be lost. This is due to the natural instinct that only one voltage can remain in a wire at any moment of time. But if the information flows as a wave the surely the medium can carry information without losing any..

I will prefer to study the TEM wave more and will consult in this post if I get something useful. Anyways once again thanks for helping and keep me going..

Phrak
The wires has to go in pair to form a guided structure. The signal that travel through the structure is actually a TEM wave. It is not the voltage that travel through the wire. TEM wave travelling through the conductive guided structure cause surface current and that is the current that the recieving end detected and become voltage. It is almost like voltage travel at close to light speed. But actually it is the electromagnetic wave. I have a post currently here about TEM wave. Read that and you'll understand more.

Read the last few posts when I intentionally make the circuit into a non-guided structure. I am not convinced one way or the other, but I believe voltage is travelling very slow. It it is the TEM wave that travel at a fraction of the light speed.

Why do you thing the voltage, or a potential difference between conductors, doesn't propagate along the pair as well? The electric field lines between conductors would seem to indicate this.

yungman
Why do you thing the voltage, or a potential difference between conductors, doesn't propagate along the pair as well? The electric field lines between conductors would seem to indicate this.

That was what I thought of for the longest time. In all the engineering EM books, they use voltage and current phasors to show propagation. But if you look at the the TEM transmission lines, it is the TEM wave that propagate down. The consequence of the H field case surface current that you see as current and voltage coming out on the other end of the phone wire. Because it is the EM wave that propagate down in this case the telephone wires which is a two parallel wire wave guide. Read the other post I posted about this subject, if you think otherwise please join in, the more the marrier!!!

Antiphon
Yungman, you are incorrect on both counts.

You do not need a pair of wires to form a guided wave structure. You do not even need a conductor. An insulating dielectric works just fine.

Second, you do not need to resort to the field theory explanation and confuse the OP when he was right in the first place. You do have a voltage wave and a current wave which travels down the wires exactly as the OP suggested. Your best answer to him would have been "yes". Please stop giving excessively complicated answers to obviously simple questions. You are confusing, not helping the people with the questions.

yungman
Thanks for replying. I understood what you intended to say. But as the electromagnetic wave travels carrying information(say amplitude as a function of time) to long distances and the information is not lost(even the time difference between the two signals (say 'a') remain same). Do you believe that the signal will get transmitted even in case of a guided structure, just like a EM wave do.

I have instinct about normal wires that, if I apply a voltage(say 5v)(amplitude) and it does reach the other end and recorded or acted upon by the instrument, before the next signal(voltage say 1) is applied, the previous signal will be lost. This is due to the natural instinct that only one voltage can remain in a wire at any moment of time. But if the information flows as a wave the surely the medium can carry information without losing any..

I will prefer to study the TEM wave more and will consult in this post if I get something useful. Anyways once again thanks for helping and keep me going..

What you are refer to is the case the wire is much much shorter than the wave length of the signal. If you have a 1GHz signal, the wave length is about 8" or something, if you have a wire more than 1", you cannot assume the signal is the same at every point, that is the whole thing about transmission lines.

Gold Member
There is no essential difference between the way a simple battery, switch and bulb circuit on the bench or a 1000km transmission line behave. In both cases, an EM wave is involved. It just happens that, at low frequencies, the complications of time delay and propagation can be ignored. There are many circuit designs where the propagation times even around a single circuit board are very relevant. It's all a matter of degree.

yungman
Yungman, you are incorrect on both counts.

You do not need a pair of wires to form a guided wave structure. You do not even need a conductor. An insulating dielectric works just fine.

Second, you do not need to resort to the field theory explanation and confuse the OP when he was right in the first place. You do have a voltage wave and a current wave which travels down the wires exactly as the OP suggested. Your best answer to him would have been "yes". Please stop giving excessively complicated answers to obviously simple questions. You are confusing, not helping the people with the questions.

He specifically asked about telephone wire which is two wire. We are not talking about other medium.

Please explain to me how wave travelling down as voltage wave and current wave instead of EM wave. And no, I don't think it is a simple answer because if it is just simple current travel down a conducting wire, it would be very slow because of the velocity of the electrons in good conductor. How do you explain the signal travel in fraction of light speed and how the dielectric medium play a role in the speed if the current just travel inside the conductor.

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Antiphon
He specifically asked about telephone wire which is two wire. We are not talking about other medium.

Please explain to me how wave travelling down as voltage wave and current wave instead of EM wave.

I know that's what he asked; you gave a misleading answer that it requires two wires to form a guided wave structure. It requires no wires at all.

The field solution yields the simple voltage and current representations when the relevant integrals are carried out. To get the voltage between the wires you integrate E.dl along a line between the wires. This gives the voltage at any point along the wires. The voltage wave corresponds exactly and indentically to the field solution result for engineering purposes. He can visualize the voltage wave without trying to digest field theory in 3D to arrive at the correct result.

The current is obtained by a closed line integral of H.dl around a surface that encloses one of the wires.

An answer on these forums should be directed to match the level of understanding of the question for maximum learning effectiveness.

yungman
I know that's what he asked; you gave a misleading answer that it requires two wires to form a guided wave structure. It requires no wires at all.

The field solution yields the simple voltage and current representations when the relevant integrals are carried out. To get the voltage between the wires you integrate E.dl along a line between the wires. This gives the voltage at any point along the wires. The voltage wave corresponds exactly and indentically to the field solution result for engineering purposes. He can visualize the voltage wave without trying to digest field theory in 3D to arrive at the correct result.

The current is obtained by a closed line integral of H.dl around a surface that encloses one of the wires.

An answer on these forums should be directed to match the level of understanding of the question for maximum learning effectiveness.

WRONG!!!

The voltage and current is the consequence of the EM wave.....exactly like you said you V and I phasor derived from E and H. People need to know it is not that simple.

Until you can explain why voltage and current can travel at this great speed and why it is affected by the medium the conductor is in, you better keep your opinion to yourself.

It is a very important concept that people in this field should understand.

Antiphon
Now you bring up phasors which are a frequency-domain construct and not helpful in understanding wavefronts and speeds of propagation.

You confused someone asking a simple question about transformers with a detour about the current-carrying capacity of conductors.

In this thread you are completely confused yourself about how a current pulse travels down a pair of wires. Yes, the fields govern the propagation. But the line voltages and currents are defined by the integrals I gave above, and it is exactly that simple.

It is you who should be keeping you inaccurate opinions to yourself. I will nevertheless indulge your childish admonition and give the answers to your questions since you did not.

The voltage and current waves travel along the wire at the propagation speed of the fields around the wire. This is a function of the dielectric constant and magnetic permeability of the medium around the wire. The wave velocity is given by v=1/sqrt(LC) where L and C are the inductance and capacitance per unit length. This is always c or less. Yet again, you incorrectly told the poster that on a TEM line like this the velocity is c. It is not. It depends on the wire diameter and spacing as well as the medium the wires are embedded in.

Youngman, it is very clear that you know just enough to be dangerous but not enough to be helpful. You also appear to be more interested in flaunting your incomplete knowledge of fields than you are in helping the readers get helpful answers. Do yourself a favor and only give clear answers when you know them, not when you feel like showing how "much" you know.

yungman
Now you bring up phasors which are a frequency-domain construct and not helpful in understanding wavefronts and speeds of propagation.
You confused someone asking a simple question about transformers with a detour about the current-carrying capacity of conductors.

In this thread you are completely confused yourself about how a current pulse travels down a pair of wires. Yes, the fields govern the propagation. But the line voltages and currents are defined by the integrals I gave above, and it is exactly that simple.
I think the only one that is confused is you, coming up with the notion that it is the voltage tranvelling down the wire. Which is WRONG.
It is you who should be keeping you inaccurate opinions to yourself. I will nevertheless indulge your childish admonition and give the answers to your questions since you did not.
I think you are here to disrupt rather than answer the question. Speaking of childish, Why don't you explain to me again how the voltage and current travel down the wire in close to light speed again. Something about $V=-\int_c \vec E \cdot d \vec l$!!!
The voltage and current waves travel along the wire at the propagation speed of the fields around the wire.
Read the transmission line again, it is the H that cause surface current and the result current is the consequence of the H field. Which again is the EM wave again!!!
This is a function of the dielectric constant and magnetic permeability of the medium around the wire. The wave velocity is given by v=1/sqrt(LC) where L and C are the inductance and capacitance per unit length. This is always c or less. Yet again, you incorrectly told the poster that on a TEM line like this the velocity is c. It is not. It depends on the wire diameter and spacing as well as the medium the wires are embedded in.
So what do you call this? Voltage travelling down?
Youngman, it is very clear that you know just enough to be dangerous but not enough to be helpful. You also appear to be more interested in flaunting your incomplete knowledge of fields than you are in helping the readers get helpful answers. Do yourself a favor and only give clear answers when you know them, not when you feel like showing how "much" you know.

Carm down!!!This is a highly educated forum, save your insult to yourself. There is no place for this kind of childish comment. Debate the validity of the content and cut the insult. If I am wrong, state your case..........which so far you seem to back up what I said about EM travelling down.

I report this to the administrator already, I am not going to reply to any of the insult, just the valid debate on the theory.

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Antiphon
My mistake, yungman, you didn't say the wave travels at c.

Sometimes my patience wears thin with people who dispense bad advice over and over. I'm sure you mean well and I can tell you are trying. I apologize.

Phrak
Born2bwire is very well versed in this, worth listening to, and would certainly do a better job at this than I, but here goes...

In a pair of transmission wires, as in a coaxial cable, the signal will propagate at somewhat less than the speed of light. The signal components will consist of changes in: potential and current directed along the length of the wire, and both transverse electric and magnetic fields.

The simplest geometry to consider (other than a rectangular waveguide) is a coaxial cable without dielectric or magnetic material. The solution requires solving the differential equations for the full time dependent Maxwell equations.

For instance $\nabla \cross H = J_f + \partial D / \partial t$ with the displacement current included.

Although the individual charges may travel through the wire at a few centimeters per hour, depending upon current density, the modulations in current keep up with the rest of the signal.

yungman
Born2bwire is very well versed in this, worth listening to, and would certainly do a better job at this than I, but here goes...

In a pair of transmission wires, as in a coaxial cable, the signal will propagate at somewhat less than the speed of light. The signal components will consist of changes in: potential and current directed along the length of the wire, and both transverse electric and magnetic fields.

The simplest geometry to consider (other than a rectangular waveguide) is a coaxial cable without dielectric or magnetic material. The solution requires solving the differential equations for the full time dependent Maxwell equations.

For instance $\nabla \cross H = J_f + \partial D / \partial t$ with the displacement current included.

Although the individual charges may travel through the wire at a few centimeters per hour, depending upon current density, the modulations in current keep up with the rest of the signal.

Also I have to mention BobS for helping me in the other post. He is of big help.

gursimran
There is a lot of technical stuff and debate going on here. I really appreciate the effort of yungman for helping me on my doubt although I'm not that technical sound in physics (I'm a undergraduate student in computer science) to understand and comment in any debate going on here. Still I want to learn physics more and plan to do my further study in physics. Presently I'm studying electromagnetics from 'Fundamentals of electromagnetics with Matlab'. Please suggest me a better book if it exist.

One thing that I have concluded is that I need not worry a lot about the signal delay when working on electronics. It happens quite fast (close to speed of light).

What I intended to ask was in case of a 'very long wire' take large enough for light to travel. Suppose light(EM wave signal) take 10s to travel along the wire. When I apply a voltage of 5V at one end at t=0, a voltage of 4V at t=1, 3V at t=2, 2V at t=3, 1V at t=4 and finally 0V at t=5 what is the voltage as different times at the other end.

Some points that I would like to get answers for specifically are ::
1. Once the TEM wave reaches the other end, does it take neglegible time(no time) for the voltage on the other end to rise to the voltage information that TEM wave carries.(voltage applied on the first end)
2. When the wire is long enough if TEM wave is in the way to the other end and I change the voltage on the first end. what is the voltage response as a function of time on the other end..

Gold Member
If there is no dispersion and the line has been terminated with its characteristic impedance*, then the shape of the pulse / step should be the same as when it was launched (merely delayed by about 4ns per metre). If there is a different load on the end of the line then the pulse shape will change - maybe with some ringing or with the edge 'rounded off'. To ensure that a signal is not mangled by a transmission line, it needs to be of uniform construction and with no 'discontinuities' along its length or at the end.

Actually, if you are dealing with high data rate digital signals, the propagation time can be surprisingly relevant around a circuit board. But, of course, other components (soggy amplifiers, for instance) may have more of an effect. You would need to check in each case.

* A uniform transmission line has a quantity associated with it that is called its characteristic impedance, Z0. For a lossless line,
Z0 = √(L/C)
where L is the Inductance per unit length and C is the Capacity per unit length. Typical values would be 50Ω for coaxial line and 200Ω for twisted pairs.

yungman
There is a lot of technical stuff and debate going on here. I really appreciate the effort of yungman for helping me on my doubt although I'm not that technical sound in physics (I'm a undergraduate student in computer science) to understand and comment in any debate going on here. Still I want to learn physics more and plan to do my further study in physics. Presently I'm studying electromagnetics from 'Fundamentals of electromagnetics with Matlab'. Please suggest me a better book if it exist.
I have 8 books on this subject. For serious engineering EM, the best would be "Field and Wave Electromagnetics" by David K Cheng. And I would suppliment with "Introduction to Electrodynamics" by David Griffiths. Something about Davids that they write good EM books!!!:rofl: Not to mention "Classical Electrodynamics" by J David Jackson!!! Forget this book, this is the holy graile of EM.:rofl:
One thing that I have concluded is that I need not worry a lot about the signal delay when working on electronics. It happens quite fast (close to speed of light).
For now, I don't think you need to worry. I've gone much deeper than what you were asking because I so wish someone would tell me about EM instead of voltage that travel on the wires. I had been confused about this for a long time.
What I intended to ask was in case of a 'very long wire' take large enough for light to travel. Suppose light(EM wave signal) take 10s to travel along the wire. When I apply a voltage of 5V at one end at t=0, a voltage of 4V at t=1, 3V at t=2, 2V at t=3, 1V at t=4 and finally 0V at t=5 what is the voltage as different times at the other end.

Some points that I would like to get answers for specifically are ::
1. Once the TEM wave reaches the other end, does it take neglegible time(no time) for the voltage on the other end to rise to the voltage information that TEM wave carries.(voltage applied on the first end)
2. When the wire is long enough if TEM wave is in the way to the other end and I change the voltage on the first end. what is the voltage response as a function of time on the other end..

Sophiecentaur gave a very detail description already and I don't think you need to over worry. In real life, they put repeaters along the way, they never let a signal travel for 4 second, not even 0.4 second on a transmission line. Remember speed of light goes around the earth something like 8 times( don't quote me on this) per second. So in dielectric medium, it is still going to have speed to go around at least one time per second!!! 4 second make it a very very long line!!! They use repeater at regular interval to re-shape the signal. For now, don't worry too much.

Gold Member
Remember, the delay and dispersion (delay inequality) only becomes relevant when you are trying to send pulses (or analogue signals) which are very fast and close together. If you are sending a data stream at 1Gb/s then a 1ns difference in delay time (15cms of length) would be enough to be out of step by 1 whole bit. Much less delay could constitute a problem in a high speed computer circuit. With speeds of 3GHz, it's a different ballgame from 'old fashioned' machines with clock speeds of 3Mhz.

Phrak
As best I can tell, dispersion is a result of impedence variation with frequency. This could be a greater factor in digital transmission as the signal also consists of higher frequencies about the base signal.

When the frequency between transitions is high enough (shorter than half the wire length/signal velocity) the termination load doesn't have an effect on the transmitted signal, but there will be a reflected signal back from the load due to any mismatch between load impedance and characteristic impedance of the line. This reflected energy is not absorbed by the load so the signal is degraded in amplitude. Kinks in the line are also reflection points. So are vias on a PC board that send one trace to another layer. (Changing layers also has the effect of changing the characteristic impedance with any associated ground planes as well as coupling to other traces.)

yungman
I am no expert by my understanding is the velocity vary a little with frequency.

$$\vec u \;=\; \frac 1 {\sqrt{\mu \epsilon_c}} \;\hbox { where } \;\epsilon_c \;=\; \epsilon + j\frac {\sigma}{\omega}$$

There is the $\omega$ in the equation and that will change with frequency.

For example in a 2.4G data rate serial transmission, each data period is about 200pS. If you have a square wave that contain all odd frequency harmonics, only take about 20 to 30pS change in arrival time at the reciever to distort the shape. This invole only very very slight change of velocity of the harmonic frequency to do that. Even high quality coax or other transmission medium will have a slight conductance and will affect the velocity slightly.

At least this is my basic understanding.

Bob S
As best I can tell, dispersion is a result of impedence variation with frequency. This could be a greater factor in digital transmission as the signal also consists of higher frequencies about the base signal.
As you point out, the dispersion [in a TEM line] is a result of the impedance variation with frequency, specifically the inductance per unit length L. The characteristic impedance is Z = √(L/C), and the velocity of propagation is v = 1/√(LC), where C is the capacitance per unit length. The change in L is due to the fact that at lower frequencies, the larger skin depth (which is proportional to 1/√f) of the conductor (like copper) permits the transverse magnetic field to penetrate into the conductor, thus increasing the inductance L per unit length and decreasing the propagation velocity. See for example

http://www.ece.vt.edu/swe/lwa/memo/lwa0136.pdf

The effect on the risetime of a square wave is shown in post # 11 of

Bob S

gursimran
Sophiecentaur gave a very detail description already and I don't think you need to over worry. In real life, they put repeaters along the way, they never let a signal travel for 4 second, not even 0.4 second on a transmission line. Remember speed of light goes around the earth something like 8 times( don't quote me on this) per second. So in dielectric medium, it is still going to have speed to go around at least one time per second!!! 4 second make it a very very long line!!! They use repeater at regular interval to re-shape the signal. For now, don't worry too much.

Ya I know, I need not worry a lot of this because it is a practically impossible situation but its a point of theoretical importance ot understand what type of signals travel across the wire. In case of simple EM waves travelling across the air, they may travel many seconds without losing aby information because the wavefront remains the same.. Nevertheless I got the stuff about repeaters and got the right direction to move further in this subject. Thanks a lot for that. Still I would appreciate if someone could tell voltage as a function of time on the other end, that I asked in my previous post..

Phrak
As you point out, the dispersion [in a TEM line] is a result of the impedance variation with frequency, specifically the inductance per unit length L. The characteristic impedance is Z = √(L/C), and the velocity of propagation is v = 1/√(LC), where C is the capacitance per unit length. The change in L is due to the fact that at lower frequencies, the larger skin depth (which is proportional to 1/√f) of the conductor (like copper) permits the transverse magnetic field to penetrate into the conductor, thus increasing the inductance L per unit length and decreasing the propagation velocity. See for example

http://www.ece.vt.edu/swe/lwa/memo/lwa0136.pdf

The effect on the risetime of a square wave is shown in post # 11 of

Bob S

A square wave contains only odd harmonics. I think you loop step needs adjusting using For n = 1 to 500 step 2.

Bob S
A square wave contains only odd harmonics. I think you loop step needs adjusting using For n = 1 to 500 step 2.
Thanks for taking the time to read my code. You are absolutely correct, for a 50% duty cycle square wave. For an arbitrary duty cycle square wave, like my example, both even and odd harmonics are required.

Bob S

Phrak
Thanks for taking the time to read my code. You are absolutely correct, for a 50% duty cycle square wave. For an arbitrary duty cycle square wave, like my example, both even and odd harmonics are required.

Bob S

No problem. I haven't been able to find a decent graph on the web of the distortion caused by time delay of harmonics so I'm curious to see what you can come up with.

What does your graph look like with the correction?

Bob S
The first thumbnail shows the calculated and measured dispersion of 10-ns square pulses in RG-58 and RG-213 coac cables, taken from

http://www.jlab.org/Hall-B/notes/clas_notes91/note91-007.pdf

The second thumbnail shows the Fourier series calculation of the dispersion of 10-ns pulses in various coax cables, using parameter c2 on page 3 of above pdf.

Bob S

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