Z0 of a line with unequal wires

[modmix]

In case of a transmission line consisting of two wires in parallel (aka Balanced line) the characteristic impedance Z0 is given by this equation:

where l is half the distance between the wire centres, R is the wire radius and \mu, \epsilon are respectively the permeability and permittivity of the surrounding medium.

I have two questions I cannot answer myself:

• what is the characteristic impedance in case of wires of different diameter ?

• In case of a tube rather than a solid wire: does the characteristic impedance depend soley on the outer diameter or does the inner diameter play a role as well?

Thanks in advance for any hint.
Ulli

 

[davenn]

Hi modmix
welcome to PF

the formula is very similar but it also takes into account the ratios of the diameters of the inner and outer conductors along with the dielectric constant.
I don't know how to do latex to do the formula here so will aim you at one of the best sources on the net ...

http://microwaves101.com/encyclopedia/coax.cfm#Z0

have a read down that page ... it should answer your questions

hope that helps you :)

cheers
Dave

 

[modmix]

Thanks Dave for both, the welcome and your answer.
Nice to be back at what I did more than 30 years ago at university ,-)

I guess I wasn't precise enough - the page you pointed at talks about coax.

I'm looking for a formula to describe a Twin Line where the diameter '2a' of both wires is different ( eg. 0.8 mm and 2 mm).

from: Transmission Line Parameter Calculator

cheers
Ulli

 

[davenn]

ahhh ok

In all my years of working with radio or other transmission lines, a twin cable or open wire lines are always the same diameter. Am not sure of why you would want or how you would end up with them being different diameters ?
Coax is the only instance where I am familiar with differences in conductor sizes

Just curious, is there a specific situation where this is applicable for you ?

cheers
Dave

 

[modmix]

Sure, in almost 100% the case wires do have the same diameter.

A while ago I made an SPDIF cable with two wires in parallel. Lucky me, the cable plays music better than all commercial cables I've tried so far. Don't ask me why (but, of course, I would like to know how SPDIF transmission properties do affect sound quality - might be a matter of jitter...).

As SPDIF signal ground is normally connected to the case of the connected components and, as ground loops due to mains is quite common in such a situation, the idea is to reduce the gound loop impact on the signal by reducing the ground wire resistance (similar to the concept of parallel Earth connector (PEC).

To be able to that, knowing the required geometrie to achieve 75 Ω would be great.

Just to thicker wire was the first idea. It turned out that the distance between thicker wires for a 75 Ω line are very very small and impedance changes far to much in case of small derivations from the needed geometrie.

The hope is, to get a less sensitive solution by using just one thick wire.

cheers
Ulli

 

[davenn]

OK

usually SPDIF uses either coax or optical fibre and maximum cable length is recommended to be ~ 10 metres

Because its a digital signal, I don't see that there would be any affect on the audio quality unless there is significant degradation to the digital signal. That is if the bit stream leaving the transmitter can be correctly identified and processed at the receiver then the audio won't be affected

Dave

 

[modmix]

usually SPDIF uses either coax ...

Sure.
Apart from the shielding, I guess, this is mainly for the simple reason that controlling the impedance for 75 Ω is a bit tricky with wires in parallel.

Because its a digital signal, I don't see that there would be any affect on the audio quality unless there is significant degradation to the digital signal. That is if the bit stream leaving the transmitter can be correctly identified and processed at the receiver then the audio won't be affected

Sure.
At least in theory.
In the real world, different digital cables do produce different sound quality - I guess mainly do to different jitter they invoke.
I did experimental physics - theory comes second ,-)

That's why I asked about impedance of parallel wires of different diameter - I simply do not know how to get the formula derived from Maxwell's equations ,-)

cheers
Ulli

 

[Baluncore]

It is the outside diameter that matters with RF conductors. They are modeled as thin walled conductive cylinders. The current flows only in the surface. With a coaxial line the signal propagates through the dielectric between the outside of the inner and the inside of the outer conductors.

Audio quality is not degraded when transmitted over a digital link unless there are bit errors. That is better fixed by using a higher quality coaxial line rather than an exposed parallel line. The exposed parallel line will pick up external signals that will on occasion cause more significant bit errors than the original coaxial line. Parallel lines have less dielectric and so are characterised by significantly lower losses than coaxial cables, maybe your line is too long and you need a radio link.

A parallel transmission line in an audio environment will need to be twisted or it will generate EMI. When it is twisted it will need some dielectric insulation for support. That will again increase the dielectric losses. The distribution of the dielectric will defeat any simple Zo formula and so suggests the use of FEA.

The characteristic impedance of a transmission line, Zo = Sqrt(L/C); where L and C are the inductance and capacitance per unit length. So to calculate the impedance you must calculate the self and mutual inductance of the two parallel conductors along with their capacitance in the presence of a distribution of dielectric that distorts the electric field. Those are both very involved computations.

Now the problem is that you cannot make transmission lines with impedances below 200 ohms without significant capacitance which requires a dielectric. That results in losses which in turn increases the bit errors.

 

[modmix]

It is the outside diameter that matters with RF conductors. They are modeled as thin walled conductive cylinders. The current flows only in the surface. With a coaxial line the signal propagates through the dielectric between the outside of the inner and the inside of the outer conductors.

That is true for lossless lines at high frequencies, I guess. Think about skin effect, for example.
Here graph from a german manufactuar:

http://www.tmr-audio.de/homemain/wissenswertes-ueber-hifi/faq/leitungen/111-wellenwiderstand-bei-nf-leitungen

Audio quality is not degraded when transmitted over a digital link unless there are bit errors.

And bits are bits. Sorry, that's a too simple model. It doesn't fit to my experience.

and so suggests the use of FEA.

What is FEA, pls.

Now the problem is that you cannot make transmission lines with impedances below 200 ohms without significant capacitance which requires a dielectric.

I'm sorry to tell you that I did both, 75 Ω and 110 Ω using wires instead of coax. 75 Ω isn't twisted. Nevertheless, music is best using this cable.

cheers
Ulli

 

[davenn]

sorry, perceived personal experience isn't really a scientific answer.
you would need to provide measurable and repeatable data to support your claims :)

you seem to be falling into that category of those that believe that OFC cables produce better music quality
their problem is mainly ... that after spending $100's on the cables , they are not likely to admit that it makes no difference, so the myth is perpetuated

both baluncore and I have told you that unless the bit stream incurs errors there is going to be NO change in the information that is transmitted and received. that is easily measured and can be shown under repeatable tests

I would love to see how you are getting your low Zo's. show us your test setups and results proving such

cheers
Dave

 

[Baluncore]

In case of a tube rather than a solid wire: does the characteristic impedance depend soley on the outer diameter or does the inner diameter play a role as well?

The Zo of a transmission line is only important when the length of the line is greater than about one 10th of a wavelength. For a 30m long line that equates to a 300m wavelength which is equivalent to 1MHz. So by the time your frequency rises high enough for the cable to be treated as a transmission line the skin effect depth in copper will be only 66.um = 0.066 mm = 2.6 thousandths of an inch. The digital data rate on your line is above 1MHz so on the German manufacturers marketing graph the horizontal black line Zo = Sqrt(L/C) is the one that counts. If you were not transmitting digital data but analogue audio then the line would be treated differently, as a lump rather than a line. That graph is for analogue signals not for digital data streams. It confuses matching the line with driving and terminating the lump.

And bits are bits. Sorry, that's a too simple model. It doesn't fit to my experience.

You must differentiate between the transmission of the bits and the initial “analogue to digital” and final “digital to analogue” conversion processes. The transmission will not degrade the audio if there are no bit errors over the link. That is a simple fact. There is jitter in every data stream but the D to A converter at the far end is being clocked with data at a much lower rate than the jitter. 32 bit serial data with 10% jitter is only 1/320 of the sample rate which is 640 times the highest frequency in the audio. If there is loss of audio quality it is because the converters are poorly designed.

FEA is finite element analysis. It solves for the capacitance and the inductance of a discrete model from which you must compute Zo = Sqrt(L/C). The general transmission line impedance equations use exactly the same Zo = Sqrt(L/C).

I'm sorry to tell you that I did both, 75 Ω and 110 Ω using wires instead of coax.

How do you know you “did” those characteristic impedances? Do you really have a time domain reflectometer? I expect you used insulation which has losses so limits the range, just like a coaxial line. Because your signal works once does not mean it will always work since interference is spasmodic, people walk around with mobile electronics. Coaxial lines are used to avoid interference.

Nevertheless, music is best using this cable.

You will believe what you want to believe. I can only give you a guide to understanding the physics of transmission lines. Belief systems are outside the remit of the Physics Forum's peer reviewed science.

 

[modmix]

Hi Dave,

sorry, perceived personal experience isn't really a scientific answer.

I agree completely
With one remark: personal experience might be a reason for further investitagtion.

you would need to provide measurable and repeatable data to support your claims :)

I'm working on that, too.
It's not so easy to find affordable measurment tools with high enough resolution.

In a german forum Gert reported some measurements on how jitter depends on an SPDIF cable (the vco voltage at the receiver is recorded).

Apogee Big Ben als Jittermonitor
Welches Koaxkabel für S/PDIF? p.5 gives some spectrograms like this

The like to the recorded files doesn't work anymore.
Give me your email address in case you would like to get the files.​

If you agree, I'ld like to come back to the thread question (formular for Z0 of a lecher wire with wires with unequal diameter).

Thanks.
Ulli

 

[modmix]

FEA is finite element analysis. It solves for the capacitance and the inductance of a discrete model from which you must compute Zo = Sqrt(L/C). The general transmission line impedance equations use exactly the same Zo = Sqrt(L/C).

Thanks.

How do you know you “did” those characteristic impedances? Do you really have a time domain reflectometer?

Yes. Sort of: an ATtiny13 on a LT Demo Circuit board gives pulses watched with a 400 MHz scope:

You will believe what you want to believe. I can only give you a guide to understanding the physics of transmission lines. Belief systems are outside the remit of the Physics Forum's peer reviewed science.

100% agreeded.
That's why I asked in my previous post to come back to my question, which is pure physics, I'ld say.

Thanks for your comments.
Ulli

 

[modmix]

solved

see Bild 4a and equation (7) in mgk1.pdf.

cheers
Ulli

 

[sophiecentaur]

A while ago I made an SPDIF cable with two wires in parallel. Lucky me, the cable plays music better than all commercial cables I've tried so far. Don't ask me why (but, of course, I would like to know how SPDIF transmission properties do affect sound quality - might be a matter of jitter...).

Unless you were getting a significant / measurable difference in error rate then how would Jitter, Noise or frequency response have any effect? Any subtle difference you could be getting, using your connection cable, could very likely be down to some other difference between your system and the comparison system. Did you actually do a Blind A/B comparison between the two systems? If not, then your observation is questionable - along with a large number of other opinions that we read from HiFi enthusiasts.

 

[modmix]

Unless you were getting a significant / measurable difference in error rate then how would Jitter, Noise or frequency response have any effect?

That is exactly the question I'ld like to get an answer for ,-)
When CD was introduced, I didn't like what I heard as some other did. All CD fans told me: "That's digital.That's just 0 and 1. That must be right! You are wrong by not accepting the truth."

Later it turned out that jitter is important and that first CD players didn't do it very well.
I'm used to being told being wrong when I do believe in what I (and other people) do listen ,-)

Any subtle difference you could be getting, using your connection cable, could very likely be down to some other difference between your system and the comparison system.

What if the difference can be heard in my system just by changing one cable?
What if the same happens in other systems doing the same change?

Did you actually do a Blind A/B comparison between the two systems? If not, then your observation is questionable - along with a large number of other opinions that we read from HiFi enthusiasts.

I only have one eye ,-)

To be serious:

Once a person building quite good analog stuff got a digital cable - just to get his opion. He didn't like digital audio much, was just starting to get a setup which did well enough for his eras. He put the cable into his system. Later, his wife went into call him for lunch. She asked quite irritated: "What did you do? All the treble is gone!" - blind enough?

Not the treble but the artefacts due to jitter were almost gone - something one must in many cases learn to appreciate ,-)
​

Once a friend played some warm and frozen ripped CDs. CDs I never had heard before. After an hour or so, I had a kind of signature related with either warm or frozen. Many hours later we where listing to the reproduction by a prototype amplifier - the focus was clearly on the amp. Within the middle of that session, I asked for confirmation that the played track is from a frozen rip. Was confirmed. - blind enough?

We did a check with a β-version of the Mutec MC-3+ smart clock used as a re-clocker. I had two different verisions at that time. I simply wanted to get a second opinion on that gear. In a first run, the result was reverse to mine. Then we flipped the digital cables (the said better cable now between re-clocker and dac). Now the result was 'correct' ,-)
The friend left the room. I stayed for further listening.
As it wasn't the best heard configuration, I stood up and changed to the best configuration. Well, it wasn't as good, as I've heard this before - may be I'm simply tiered, I thought.
The friend came back and immediately started complaining loudly about the bad sound. After a while, I asked him: "Do you know that you are telling me, that the so far best heard configuration doesn't play well?" "No! that cannot be the case!". He went to see what configuration we were listenig to. Yes, seemed to be the best one. He went back for further listening. As he was still sure that that wasn't right, he went again to see what's going on. "What did you do?" he suddenly asked. "Nothing except for going back to the best configuration..." - "Hee, hee, you made a mistake! the digital cable is not connected in the right direction!"
After having changed that, sound was indeed better.
My friend didn't know that I had changed the setup.
I didn't know about the direction of the digital cable.
- blind enough?

I do stop here.

And I do admit that the listening capability does vary among persons.

cheers
Ulli

 

[sophiecentaur]

When you digitally encode an analogue signal, you build in inherent distortion by the sampling process and, of course there will be imperfections introduced in the implementation of the ADC. They use so many bits and sample at the frequency they do in order to reduce the impairments to a level that is imperceptible / acceptable (based on groups of 'experts' with golden ears).The distortions and noise which are associated with this process can be made measurably less than any analogue tape recording or with vinyl pressing - factors that are always ignored by the anti-digital brigade. If you do nothing with the resulting digital signal except reproduce it with no errors then the information will not, by definition, be changed in any way. The channel is indistinguishable from a short piece of copper wire.
If you play out that digital signal with a high enough quality DAC, then there will be no additional impairment. The digital part of the channel is totally transparent unless errors have been introduced by some fault.
It is a fact that the distortions due to tape and the characteristics of vinyl recording are very audible and identifiable and people have got used to them enough to like them. This is along the same lines as the fondness some people have for valve amplifiers and their 'sound'. It's subjective and perfectly ok. But the quality of such systems is due to the original program signal having been modified. It's not the same as when it started out.
Once you take a 'CD quality' digital signal and try to compress it into a channel with a lower data rate, there is a risk (certainty in most cases) that the quality will be impaired. 'Digital' is often sold as 'perfect' even when it's an mp3 signal at 48kbps. Look at all those dodgy adverts.

@modmix Do you have any way of measuring the error rates in your digital system? If the error rates are not the same then you might expect the programme to sound different. (Fair testing and all that.) There are good and bad systems for masking and correcting digital errors.

Probably best just to stick to the Physics or Engineering on PF; things that you can actually measure. There are plenty of other forums where the subjective aspects of media are discussed without wet blankets like me.

 

[modmix]

Probably best just to stick to the Physics or Engineering on PF; things that you can actually measure.

As far as I remember, I simply asked for a certain formula - quite a matter of "Physics or Engineering", right?
May be it was not a good idea to answer off topic questions... I just wanted to be polite.
I offline here.
Ulli

 

[sophiecentaur]

As far as I remember, I simply asked for a certain formular - quite a matter of "Physics or Engineering", right?
May be it was not a good idea to answer off topic questions... I just wanted to be polite.
I offline here.
Ulli

You made a statement about performance, too, with an implied claim. That was what I was responding to. (Post 5)

 

[modmix]

#4 asked an off topic question:

Just curious, is there a specific situation where this is applicable for you ?

That was answered by me in #5 - sorry for trying to be polite,

#1 has my original question.
#3 some clarification

EOT

 

[davenn]

wasn't really off topic as we were still trying to discover what it was you were on about
it took some time for you to finally tell us about what you were really doing
and it was then that we found out you were into twilight zone science

not to worry
hopefully you now understand the physics of it

Dave

 

[modmix]

Why should the question for a certain formula being justified ?
#5 and the discussion following is regraded being off topic by me. It neither gives an answer nor does it give any contribution to the thread question. Instead of giving an answer there was a discussion of why I want to do something not common - at least from my point of you.

No, I still do not understand the physics of it - the given reference seem to be incorrect (formular given for wires with different diameter does give the same values for the special case of equal diameter as obtained by the normally used formula).

But, I do not expect to get some advice or inside from those people living in the dark age where it is known that men can not be trusted and the truth is written in the ancient books.

Regards from the twilight zone - where we do have at least a bit of light possibly giving us some enlightenment at the end.

Ulli


PS:
Still very much keen to get a correct formula (see posting #1).

 

[sophiecentaur]

PS:
Still very much keen to get a correct formula (see posting #1).

Here's a suggested way of getting what you want.
The answer you get should be somewhere between the Z₀ for a pair of the thick and a pair of thin wires, as a check. Before you go to the trouble of all that follows, just find what it would be for equal diameters. If there's not much difference then it may well not be worth bothering. (How accurate can your match be and what is the impedance in the rest of the circuit? There will be an acceptable VSWR that isn't 1.000)

You need to know the C per unit length and the L per unit length:
(Z₀ = √(L/C).
To find the C.
This is certainly going to be a pretty good approximation for finding the capacitance between the lines.
Treat the two lines separately and take a perpendicular line, half way between the two centres. This line will be a line of symmetry and the distance to use for your calculations- the fields (each wire on its own) would be shown by those radial lines. Of course, they will be curves when there are two lines but you can still expect symmetry (or at least near symmetry). The capacitance between the two lines will be the equivalent to two capacitances in series, each one, corresponding to the capacitance of one line and the plane of symmetry. You can google the formula for C of an unbalanced line.
To find the L, find the inductance of two parallel wires of a equal diameter and spacing (easily googled). Then take the mean for the two different diameters.

P.S. PF does try to deliver when it can!

 

[Baluncore]

I will attempt to restrict my analysis to transmission line construction, modelling and practical use. Since transmission lines are multidimensional, I will try to separate out those dimensions and avoid the post-modernist confusion that can arise when making soup. There are really only a few concepts that must be appreciated to understand why things are still done the way they are.


The original modmix hypothesis was something along the lines that; an asymmetric parallel transmission line would in some way provide a better quality transmission environment than say, a coaxial cable. While I would really love to agree with the modmix hypothesis, whenever I try to find a practical way to make it happen, I come across a practical roadblock that prevents advance in that direction.

It has been said that “the grass always appears greener on the other side of the fence”, but when you finally get there it is no better. The conclusion I often reach is that, if modmix could see what his hypothesis really implied, then modmix would not want to go there. There is nothing wrong with “wishful thinking”, indeed it is necessary for innovation, but when it can be shown to be a delusion through a careful analysis, then it must be recognised as a “fiction” rather than a “reality”. Marketing hinges on wishful thinking, reality does not. Now I will try to justify my assertion that asymmetric parallel lines are not as good as coaxial lines in this application. I will also attempt to explain why a simple equation relating Z to practical parallel line dimensions is not possible.

1. How many universes are there ?
Consider two coaxial cables lying next to each other. The signals in one are not mixed with the signals in the other because the centre conductors cannot “see” each other. They can only see their own reflection from the inside of their outer braid. That means the bandwidth employed can be reused in every cable without interference. With a parallel line on the other hand, the external fields are free to interfere and be interfered with by all other signals in our universe. This can only be minimised by twisting the pair of conductors, keeping them symmetrical in size and driving them as a differential pair. Twisted pairs have significant analogue “cross-talk”, something that is a real problem in the telephone - audio world. The last thing I would want in an audio environment is an exposed digital signal that would couple into all low level analogue signal cables.

2. When is a general equation available for Z. Or why is the electric field diagram useful, (Bilt 2, in the reference).
Since Z = Sqr(L / C) any equation for Z must solve for the inductance and the capacitance.

Conductors.
It is very rare for there to be any magnetic material present in a line, mainly because it leads ultimately to very high losses, so inductance is determined simply by the external surfaces of the conductors alone. Those surfaces also determine the resistive losses. The bigger the conductors the lower those ohmic losses.

Insulators.
Capacitance is the real problem here. Since some form of insulation is needed to keep the two conductors apart there will be a distribution of dielectric in the electric field. That dielectric has two effects. (Here I will ignore single frequency tuned RF lines and metal insulators). The first effect is that it contributes to losses, especially at higher frequencies. The second problem is that to generate a simple equation the dielectric must be positioned so it's boundaries follow equipotentials of the electric field as shown in Bild 2 of the earlier reference. Now, because the equipotentials are all circles about different centres, it is not possible to write a simple equation for Z where the insulation about a wire is symmetrical. It is clear that an equation for any particular sized cylindrical conductors can be written as all the equipotentials are circular.

Equation(7) is written for an asymmetric pair of circular conductors immersed in an ocean of dielectric. It can not therefore be rewritten for a specific coaxial distribution of dielectric about one conductor, like an insulated wire.

3. Losses due to dimensional sensitivity.
Examining eqn(7) we see the Arcosh() term gives a most stable impedance when d1 = d2. At that point, for Z = 75 ohms, we find that the air-gap between the two conductors is only about 20% of their diameter. As the two conductors differ more in their diameter the gap needed becomes proportionally narrower and much more critical. This means that for asymmetric conductors any slight change in the gap width along the line will lead to big changes in Z. Each local change in Z will generate a partial signal reflection accompanied by frequency sensitive standing waves, digital signal distortion and high losses. When the air-gap is small it will need to be replaced with an insulated spacer. That insulator must be very carefully controlled dimensionally since it is critically placed where the electric field is steepest. Any variation in dielectric characteristics will again lead to ZL variation and therefore line losses. For the same Z the insulation will need to be thinner than the original computed gap for air.

4. Better quality cable. There is always better grade coaxial cable available at exponentially higher prices. Using lines designed for microwave applications would reduce losses. Consider for example Heliax LDF4-75 or LMR600-75. (The 75 is important as it is the impedance).

Conclusion.
Knowing an equation for the impedance of an asymmetric parallel line does not ultimately help. Instead it highlights the folly of using asymmetrical parallel conductors for digital signals.

 

[f95toli]

If we ignore the application for a moment...

The answer to the original question is that I am pretty sure there are no closed form formulas; you would need numerical calculations (finite elements); it is a trivial calculation for any solver that can do electrostatics (and even simple if the solver gives you the capacitance per unit length).

See e.g.

http://www.pdesolutions.com/reprints/impedance.pdf

for a description of how to calculate Z0 for an arbitrary cross section.

I use COMSOL for things like this, but there are free PDE solvers out there that presumably could solve a problem as simple as this (all you need to do is to calculate the static E field around a PEC in a dielectric).

The closest thing I can think of where they ARE analytical formulas would be an asymmetric coplanar stripline. I've only used these a couple of times. and each time I gave up on using the formulas; simply because they still require some decent software to actually calculate something (they involve lots of elliptic integrals). That said, asymmetric lines are nevertheless sometimes useful for RF applications.

 

[sophiecentaur]

I have been wondering what this thread is all about, actually. The OP appears to be dealing with a small-scale (home?) setup in which there is enough loss / dispersion / crosstalk to be causing bit errors - hence the reported impairment to the music. The problem seems to relate to the connecting cables if there really is a difference in perceived sound quality for different arrangements. That implies something is operating marginally and that signal levels are not what they should be. How often is overall bit rate affected by short connecting cables that are connected correctly to properly functioning equipment? Afaiaa, it's not a major problem even on long digital links; a link budget and system spec will ensure the right signal levels are presented to receiving terminals and the coding will take care of expected error rates.
Billions of Pounds are at stake where high capacity digital links are involved and it seems to me highly unlikely that the standard systems are very far short of optimum, using the available equipment. There would be no earthly reason not to use asymmetrical 'balanced' feeder if it will truly perform better. (Something about the inherent imbalance would suggest greater susceptibility to crosstalk, for a start; was it not the improvement in the balance of modern twisted pairs that allowed such an enormous jump in data capacity?)
I find it all a bit hard to take seriously without a bit more help in the shape of some more reference sources about the merits of asymmetrical feeder.

 

[jasonRF]

You can do this exactly either with conformal mapping or by images. The images approach is easiest, in my opinion. Consider the potential due to 2 parallel line charges of the same density \rho_\ell but opposite sign, separated by some distance. The equipotentials are circles, though not centered on each of the line charges. You essentially just construct the solution to satisfy your geometry. Let D be the separation of the two axes, and R_1, R_2 be the two radii. Then I believe that the capacitance per unit length ends up being
<br /> C_0 = \frac{2 \pi \epsilon}{\cosh^{-1}\left[\frac{D^2 - R_1^2 - R_2^2}{2 R_1 R_2} \right]}<br />
To get the inductance, you use the fact that we know that C_0 L_0 = \mu \epsilon. Finally,
<br /> Z_0 = \sqrt\frac{L_0}{C_0}.<br />
This does reduce to the correct formula for R_1=R_2. You just need to use the fact that \cosh^{-1}(2 x^2 - 1) = 2\cosh^{-1}(x)

jason

 

[Baluncore]

JasonRF.
That is OK, apart from a missing factor of 4, but it does not permit the modelling of insulation that is symmetrical about the smaller conductor since the equipotentials are all circles about different centres. Any spacer insulation between the two conductors crosses the line of symmetry so there cannot be a closed equation.

The reference in post #14 has equation(7) and the diagram in Bild 4a, and the conformable electric field in Bild 2.
ZL = 60 / Sqr(Er) * Arcosh( (4*s*s - d1*d1 - d2*d2) / (2*d1*d2) )

 

[jasonRF]

JasonRF.
That is OK, apart from a missing factor of 4, but it does not permit the modelling of insulation that is symmetrical about the smaller conductor since the equipotentials are all circles about different centres. Any spacer insulation between the two conductors crosses the line of symmetry so there cannot be a closed equation.

The reference in post #14 has equation(7) and the diagram in Bild 4a, and the conformable electric field in Bild 2.
ZL = 60 / Sqr(Er) * Arcosh( (4*s*s - d1*d1 - d2*d2) / (2*d1*d2) )

Somehow I missed that post - thanks! That document looks great - are you aware of an english translation? I have forgotten what little German I learned in high school. Anyway, it looks like i did not miss a factor of 4, since my R_1 is their d_1/2. The formula does assume a dielectric medium of infinite extent, of course, but I thought that was the approximation that the OP was asking for.

jason

 

[Baluncore]

are you aware of an english translation

I looked but I did not find one.
If anyone knows of one, please let us know.