Question on TEM mode transmission lines.

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  • #1
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I want to verify the theory of TEM mode tx lines how the signal travel. Please correct me if I am wrong. This mainly refer to stripline, microstrip( assume TEM) etc.

1) The signal is transmited as EM wave through the dielectric between the two lines, not actually current through the conductors.

2) Charge and current density on the surface of the conductor only as the consequence of the Maxwell equations:

[tex] \nabla X \vec H = \vec J_{free} \;\;\hbox { on the surface of the conductor plates.}[/tex]

[tex]\nabla X \vec E = \rho_{free} \;\;\hbox { on the surface of the conductor plates.}[/tex]

3) The electrical signal is launghed by injection of current at the input end, the current density create the TEM wave according to the Maxwell equation:

[tex] \nabla X \vec H = \vec J_{free} \;\;\hbox { and} \;\;\nabla X \vec E = \rho_{free} [/tex]

4) At the end of the transmission line, the electrical signal is the consequence of the surface current density and charges according to the same Maxwell's equation. This electrical signal is to drive any electrical circuit as if the electrical signal is driven directly through the transmission line.


Bottom line, in the middle of the transmission, it is TEM electromagnetic wave. The electrical signal is the consequence of the surface charge and surface current density while the EM wave propagate in between the plates of the stripline or microstrip.

I alway thought it was the electrical signal travelling through the conductor before but I don't think that is true because the velocity of electrons are too slow.

Please help me by commenting on this.
 

Answers and Replies

  • #2
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Exactly. The velocity of the signal is v = sqrt(1/LC) in meters/second, and the impedance of the TEM signal is Z = sqrt(L/C) in ohms, where L is the inductance per unit length (Henrys per meter), and C is the capacitance per unit length (Farads per meter).

Skin effect losses in conductors force the currents to the surface at high frequencies. The signal power is given by the Poynting vector; P dA = ∫E x H dA.

This electrical signal is to drive any electrical circuit as if the electrical signal is driven directly through the transmission line.
True, if the impedance of the electrical circuit is equal to the impedance of the transmission line.

Bob S
 
  • #3
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Thanks

All this time, I kept thinking it is the voltage and current that travel throught the stripline or microstrip because it is always voltage and current phasor being used. Something kept bugging me because the speed of electrons moving in the conductor don't match up and also if the signal is really travelling in the conductor, why the [itex] \epsilon [/itex] of the dielectric affect the characteristic impedance and velocity. Finally I read the whole thing over and realize it is the TEM wave that travel through the tx line and there is a transistion from electrical signal to TEM wave at the input and TEM wave back to electrical signal at the output.

Take me the third time studying over to realize this!!!!:redface:

Thanks

Alan
 
  • #4
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I have another question. Say in a dc circuit, if I have a very long wire from the battery with a switch connecting to a light far away. Why the light still turn on as fast as EM wave travelling the same distance?

I can see when voltage first apply, EM wave generated and travel down fast due to step in voltage. But after that, it has to be the conduction current in the wire that keep it going. If the current velocity is slow and take a few seconds to travel from one end to the other, why is the light turn on and stay on continuously? Or is it an effect like water pipe, even it is very long, the same water molecule take say 60 seconds to travel the pipe, but the output turn on instantaneously because the water at the input push the whole pipeful of water forward? If so, you can go back and argue about the tx line that electrons is being pushed also!!!!
 
  • #5
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Yes, the pipe analogy is correct. Also, the current and voltage on a TEM waveguide are valid variables to describe the wave. The energy is in the space between the wires but the currents are right there and track with the fields.

Your equations aren't correct. The curl of E is not equal to the charge density but to the magnetic current density. The charge density is related to the divergence of E.
 
  • #6
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Yes, the pipe analogy is correct. Also, the current and voltage on a TEM waveguide are valid variables to describe the wave. The energy is in the space between the wires but the currents are right there and track with the fields.

Your equations aren't correct. The curl of E is not equal to the charge density but to the magnetic current density. The charge density is related to the divergence of E.
My bad. I was rushing to post it. I actually meant

[tex] \nabla \cdot \vec D \;=\; \rho_{free}[/tex]

Bottom line is the free current density and charge that are in action.
 
  • #7
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I have another question. Say in a dc circuit, if I have a very long wire from the battery with a switch connecting to a light far away. Why the light still turn on as fast as EM wave travelling the same distance?

I can see when voltage first apply, EM wave generated and travel down fast due to step in voltage. But after that, it has to be the conduction current in the wire that keep it going. If the current velocity is slow and take a few seconds to travel from one end to the other, why is the light turn on and stay on continuously? Or is it an effect like water pipe, even it is very long, the same water molecule take say 60 seconds to travel the pipe, but the output turn on instantaneously because the water at the input push the whole pipeful of water forward? If so, you can go back and argue about the tx line that electrons is being pushed also!!!!
Anyone can help?
 
  • #8
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Anyone can help?
Put a battery and a switch at one end of a long transmission line. Put a matched termination at the other end. Close the switch at t=0, so the voltage on the transmission line at the switch is a step function V(t) = V0 for t>0. Do a Laplace transform from the time domain to the frequency domain. Now you can use the telegraphers equation for each individual frequency. See

http://en.wikipedia.org/wiki/Telegrapher's_equations

So now the signal arriving at the termination is the sum over all frequencies. The signal for each frequency is of the form

V(w,x) = e-αx·V0·eiωt - ikx

Put in values for attenuation α and propagation factor k at every frequency. According to the telegrapher's equations, the signal propagation velocity v(ω) = sqrt(1/LC) at every frequency. This velocity is dependent only on the capacitance C and inductance L per unit length. Now transform back to the time domain.This is the voltage step function at the termination. How does this differ from the input step function?

Besides the attenuation, the major change at low frequencies is the small increase of the transmission line inductance, when the skin depth penetration is greater. So Z(ω) = sqrt(L/C) is slightly larger, and v(ω) = sqrt(1/LC) is slightly smaller at low frequencies.

Bob S
 
  • #9
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Put a battery and a switch at one end of a long transmission line. Put a matched termination at the other end. Close the switch at t=0, so the voltage on the transmission line at the switch is a step function V(t) = V0 for t>0. Do a Laplace transform from the time domain to the frequency domain. Now you can use the telegraphers equation for each individual frequency. See

http://en.wikipedia.org/wiki/Telegrapher's_equations

So now the signal arriving at the termination is the sum over all frequencies. The signal for each frequency is of the form

V(w,x) = e-αx·V0·eiωt - ikx

Put in values for attenuation α and propagation factor k at every frequency. According to the telegrapher's equations, the signal propagation velocity v(ω) = sqrt(1/LC) at every frequency. This velocity is dependent only on the capacitance C and inductance L per unit length. Now transform back to the time domain.This is the voltage step function at the termination. How does this differ from the input step function?

Besides the attenuation, the major change at low frequencies is the small increase of the transmission line inductance, when the skin depth penetration is greater. So Z(ω) = sqrt(L/C) is slightly larger, and v(ω) = sqrt(1/LC) is slightly smaller at low frequencies.

Bob S
I see, thanks for your help. So everything is EM!!! Is it Fourier Transform you refer to instead of Laplace?
 
  • #10
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Put a battery and a switch at one end of a long transmission line. Put a matched termination at the other end. Close the switch at t=0, so the voltage on the transmission line at the switch is a step function V(t) = V0 for t>0. Do a Laplace transform from the time domain to the frequency domain. Now you can use the telegraphers equation for each individual frequency. See

http://en.wikipedia.org/wiki/Telegrapher's_equations

So now the signal arriving at the termination is the sum over all frequencies. The signal for each frequency is of the form

V(w,x) = e-αx·V0·eiωt - ikx

Put in values for attenuation α and propagation factor k at every frequency. According to the telegrapher's equations, the signal propagation velocity v(ω) = sqrt(1/LC) at every frequency. This velocity is dependent only on the capacitance C and inductance L per unit length. Now transform back to the time domain.This is the voltage step function at the termination. How does this differ from the input step function?

Besides the attenuation, the major change at low frequencies is the small increase of the transmission line inductance, when the skin depth penetration is greater. So Z(ω) = sqrt(L/C) is slightly larger, and v(ω) = sqrt(1/LC) is slightly smaller at low frequencies.

Bob S
OK, I am going to push it a little more:

What if I hook up a light bulb with a loonnnng long wire half way across the world to connect to a switch and then to the +ve side of teh battery. Then I connect the -ve side of the battery with a long long wire and keep going around the world ( not the same path as the original wire to the switch.) and come back to the light bulb. So the circuit is one big loop around the world. There is no wave guide to speak of, assume neither wire is connect to the ground ( no microstrip effect using the earth). There should not be TEM wave travelling along the wire. The only TEM wave is pointing towards the center of the wire according to Poynting vector where E is parallel to the wire, B is circling the wire and E X B = -r pointing to the center of the wire.

Now how come the light still turn on so fast?
 
  • #11
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5
I see, thanks for your help. So everything is EM!!! Is it Fourier Transform you refer to instead of Laplace?
The book Transmission Lines and Wave Propoagation (4th edition) by Philip Magnusson is the best book I know of that discusses transmission lines. Few books discuss skin effect losses properly.

Chapter 5 discusses the Laplace transform of step functions in transmission lines, using the skin loss approximation shown in the first thumbnail. Because √(jω) = (1 + j)√ω, the skin effect loss causes both an attenuation (measured in nepers) and a phase delay (measured in radians). This √f frequency dependence of losses is rather difficult to use in Laplace transforms however.

For this particular case, it is easier to use a Fourier series expansion. A square wave is represented by the series sin(nα/(nα). In thumbnail 3, a 5-ns pulse on a coax with skin effect losses is shown. The code for this pulse is shown on thumbnail 2.

Bob S
 

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  • #12
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The book Transmission Lines and Wave Propoagation (4th edition) by Philip Magnusson is the best book I know of that discusses transmission lines. Few books discuss skin effect losses properly.

Chapter 5 discusses the Laplace transform of step functions in transmission lines, using the skin loss approximation shown in the first thumbnail. Because √(jω) = (1 + j)√ω, the skin effect loss causes both an attenuation (measured in nepers) and a phase delay (measured in radians). This √f frequency dependence of losses is rather difficult to use in Laplace transforms however.

For this particular case, it is easier to use a Fourier series expansion. A square wave is represented by the series sin(nα/(nα). In thumbnail 3, a 5-ns pulse on a coax with skin effect losses is shown. The code for this pulse is shown on thumbnail 2.

Bob S
Actually I have the book, I'll take a look. Thanks.

How about my last question. What I am trying is to guarante not to have a wave guide so the current has to travel the wire. What will happen?

Thanks, you are of big help.

Alan
 
  • #13
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Anyone?
 
  • #14
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The only TEM wave is pointing towards the center of the wire according to Poynting vector where E is parallel to the wire, B is circling the wire and E X B = -r pointing to the center of the wire.
Before your question is answered, we need to understand the meaning and effect of the above statement.

In a standard coaxial cable, the signal (mostly) travels as a TEM signal between the inner and outer conductors (If the conductors were superconducting, the TEM approximation is exact). If the center conductor has resistance, then there is a longitudinal voltage drop along the conductor, and an electric field parallel to the conductor, in addition to the radial electric field in the TEM wave. So the azimuthal H x the longitudinal E indicates a radial component of E x H. In this case, there is a small component of the Poynting vector that points into the center conductor, and indicates therefore that a component of the power in the TEM wave is flowing into the wire.

In your case, you are in essence stating that all the power is flowing into the wire, and no power is flowing along the wire. Is this correct?

Why do you believe this is true?

Bob S
 
  • #15
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Before your question is answered, we need to understand the meaning and effect of the above statement.

In a standard coaxial cable, the signal (mostly) travels as a TEM signal between the inner and outer conductors (If the conductors were superconducting, the TEM approximation is exact). If the center conductor has resistance, then there is a longitudinal voltage drop along the conductor, and an electric field parallel to the conductor, in addition to the radial electric field in the TEM wave. So the azimuthal H x the longitudinal E indicates a radial component of E x H. In this case, there is a small component of the Poynting vector that points into the center conductor, and indicates therefore that a component of the power in the TEM wave is flowing into the wire.

In your case, you are in essence stating that all the power is flowing into the wire, and no power is flowing along the wire. Is this correct?

Why do you believe this is true?

Bob S
I am refering to post #10.

I understand from your answer before that with a guided structure like the coax, parallel line, stripline etc., both H and E are perpendicular to the line and poynting vector is parallel to the line. So the TEM wave travel along the line with great speed of [itex] \frac 1 {\sqrt {\mu \epsilon}} [/itex].

But in #10, I am trying to make it so there is no guided structure by wrapping around the earth so there is no return path along the signal line. With no guide structure, you cannot set up an TEM wave. The current has to travel inside the wire. And according to the calculation, the speed of electron travelling inside the wire is very slow. So why the light still turn on fast.

I know this is rediculous, usually you power anything with a two wire cable so you always have a parallel wire as a guided structure so the TEM wave can propagate. But in my example, I only have one line only so it is not a guided structure.
 
  • #16
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There should not be TEM wave travelling along the wire. The only TEM wave is pointing towards the center of the wire according to Poynting vector where E is parallel to the wire, B is circling the wire and E X B = -r pointing to the center of the wire.
Try using a superconducting wire, which cannot support an electric field E along the surface of the wire. Is there still a Poynting vector?

Bob S
 
  • #17
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Try using a superconducting wire, which cannot support an electric field E along the surface of the wire. Is there still a Poynting vector?

Bob S
No, if it is super conductor, there won't be any voltage drop along the wire and no E, so there won't be Poynting vector into the wire.

How about my question in #10? I cannot think of a way the EM flow along the wire and the light should come on after a long delay!!! But that just sounds wrong. I don't know what to think.
 
  • #18
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Here is one solution

Before I answer the question, let's set the following baseline:
1) No radiation
2) No wire resistance, so no Poynting vector pointing into wire, hence no electric field parallel to wire.
x) No magnetic fields inside wire.
3) Wire radius a = 0.01 meters, Earth radius b = 6.4 x 10^6 meters
4) No dielectrics anywhere
5) No magnetic materials anywhere
6) Light bulb resistance R = 12 ohms
8) Battery voltage 12 volts

The self inductance of single-turn wire loop around the Earth is Lloop = μ0b[Ln(8b/a) -2] = 164 Henrys

So the L/R time constant is 164/12 = 14 seconds.

Bob S
 
  • #19
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Here is one solution

Before I answer the question, let's set the following baseline:
1) No radiation
2) No wire resistance, so no Poynting vector pointing into wire, hence no electric field parallel to wire.
x) No magnetic fields inside wire.
3) Wire radius a = 0.01 meters, Earth radius b = 6.4 x 10^6 meters
4) No dielectrics anywhere
5) No magnetic materials anywhere
6) Light bulb resistance R = 12 ohms
8) Battery voltage 12 volts

The self inductance of single-turn wire loop around the Earth is Lloop = μ0b[Ln(8b/a) -2] = 164 Henrys

So the L/R time constant is 164/12 = 14 seconds.

Bob S
Thanks for the reply. So you do agree in this way, I am not going to have a guided structure? The thing still puzzle me is that the velocity of the electrons are so slow in good conductor, 14 seconds still sound quite fast!!! What you are doing is assuming quasi static ( for lack of better words ) or lump element representation. Is this valid?
 

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