Question about ohms in a coaxial cable. - Page 1
Forum section: electronics / rf-microwave

author: Ben321
author_rank: Frequent Contributor
timestamp: March 19, 2025, 08:30:50 pm
content: What exactly does it mean when a coaxial cable is called a "75ohm cable" or a "50ohm cable"? Wouldn't it make more sense to say the impedance of the cable is "75 ohms per foot" or "75 ohms per meter" or something like that? The longer a cable is, the higher its impedance is. Wires exhibit resistance and also inductance (since current flowing in them produces a magnetic field). But the resistance of resistors add together when placed in series, as well is the inductance of inductors. And using a longer cable is equivalent to adding more resistors and inductors in series. So if a cable that's 3 feet long is 75 ohms, then a longer run of that same cable (lets say 10 feet) should have a much higher impedance. I don't know why you see coaxial cables of all different lengths all labeled as 75ohms or 50ohms. It doesn't seem to make any sense.

author: coppercone2
author_rank: Super Contributor
timestamp: March 19, 2025, 08:38:15 pm
content: no, the longer a cable the higher its loss increases but impedance stays the same.Capacitance, resistance and inductance all increase but the impedance stays the same.Its a fundamental 'ratio' that can be calculated based on the diameter of the different parts of the coaxial line

author: Kleinstein
author_rank: Super Contributor
timestamp: March 19, 2025, 08:46:14 pm
content: The impedance is the ratio of voltage to current for an infinite long cable, or a cable terminated the correct (same as cable impedance) impedance to get no reflection.With no lesses the impedance is also the same as the ratio of inductance to capacitance for a short piece of the cable.So it is not ohms per meter or km, but ohms.

author: Ben321
author_rank: Frequent Contributor
timestamp: March 19, 2025, 08:49:25 pm
content: Quote from: coppercone2 on March 19, 2025, 08:38:15 pmSo does that mean that a longer run of coax also will use a larger diameter of cable to compensate? Because I've seen some offerings on websites where they have a reel of cable and you buy it by the foot, so you can choose your own length, from the same cable, so that means you aren't selecting a larger diameter to compensate for longer lengths.I'm just curious how the cable's impedance is figured out. I realize that capacitance also increases (adding more capacitors in parallel) because there's capacitance between the core and the metal sheath of the cable. And yes I know that capacitance and inductance (or more specifically capacitive reactance and inductive reactance) will cancel out if matched correctly. But this also depends on frequency. Capacitive reactance and inductive reactance are functions of frequency. higher frequency increases inductive reactance, and decreases capacitive reactance. And at one specific frequency those cancel out (resonance). So then you only need to deal with the resistance of the cable, which can't be canceled out. But even that (a purely resistive cable) only happens at one specific frequency. But that frequency is never explicitly stated on the coaxial cables you can buy. It only gives an impedance, never the frequency at which this impedance was measured.So again, I'm not sure how to actually interpret the ohms measurement of a cable. What does it actually mean, and how does it affect actual usage of the cable (the practical importance of the ohms measurement of the cable)?

author: coppercone2
author_rank: Super Contributor
timestamp: March 19, 2025, 08:51:02 pm
content: Fun question : what happens to the impedance of the coaxial cable if you yank out the wire and replace it with a piece of graphite from a pencil?

author: Ben321
author_rank: Frequent Contributor
timestamp: March 19, 2025, 08:55:56 pm
content: Quote from: Kleinstein on March 19, 2025, 08:46:14 pmInfinitely long cable? Actually an infinitely long cable will have infinite resistance and therefore infinite impedance. Even silver (the best conductor, many times better than copper) still has a non-zero resistivity. Regardless of the reactance, infinite resistance will produce infinite impedance (Z = sqrt(R^2 + X^2)). So I don't think that saying an infinite stretch of coaxial cable has 75ohms impedance is in any way accurate. Unless of course they are making this statement about the cable while completely ignoring the resistance of the conductors, and exclusively describing the reactance.

author: coppercone2
author_rank: Super Contributor
timestamp: March 19, 2025, 08:57:27 pm
content: what you might be doing is ignoring the speed of lightwill the electron refuse to go towards a nearby capacitor just because the tunnel is infinitely long?When you think about infinities you need to think about what you can see, not what is at the endThe infinity is useful because it halts the problem in a sort of transient state so it can be scrutinizedand silver is like 10% better then copper. we wish it was many times

author: 2N3055
author_rank: Super Contributor
timestamp: March 19, 2025, 09:09:47 pm
content: https://helloworld922.blogspot.com/2013/04/online-transmission-line-simulation.htmlCoaxial cable is transmission line.Characteristic impedance is a measure how much current will cable pull from the source when you apply certain voltage in a process of charging the cable, while pulse front propagates towards the other end. On the other end of the cable you can have short, but electricity takes time to reach the other end. Until it does it will consume only some limited current, defined as U/Characteristic impedance (50/75Ω or whatever cable says it is).Once pulse reaches other end of the cable where you have short and after all the reflections die out, current source will get loaded with large current that is only defined by DC copper resistance.If other side of coax is unloaded (nothing connected) once pulse front reaches other side, and again, reflections die out, cable will stop pulling current. It's internal capacitance is charged and no current flows anymore.Infinitely long cable will forever pull only current defined as voltage applied divided by it's characteristic impedance, regardless of actual metal DC resistance. Because it will be charging forever..That is what 50 Ω or 75Ω is.You can easily measure it if you have signal source with known source impedance.Like signal generator with 50 Ω source impedance.You enable signal on output of siggen, and measure voltage unloaded.Than connect the cable and again measure voltage of signal when loaded.Ratio of how much voltage dropped is because source and load created voltage divider..Little math and voila ..

author: radiolistener
author_rank: Super Contributor
timestamp: March 19, 2025, 09:23:37 pm
content: Quote from: Ben321 on March 19, 2025, 08:30:50 pmA coaxial cable's impedance, such as 75 Ω or 50 Ω refers to its characteristic impedance, which is a fundamental property of the cable determined by its geometry and materials.This impedance represents the resistance observed at the cable's input, which would be equivalent to a resistor of the same value. Even if the cable is terminated with a different impedance at its output, the characteristic impedance is what you will measure at the input - at least until a current pulse travels through the cable, reflects off the end, and returns to the input. These reflections occur if the termination impedance does not match the cable’s characteristic impedance, leading to signal degradation, standing waves, or power loss in high-frequency applications.Quote from: Ben321 on March 19, 2025, 08:30:50 pmNo, this is incorrect because characteristic impedance is not a simple resistive loss along the cable’s length. Instead, it is a property of the transmission line that arises due to the propagation delay of the electromagnetic wave traveling through the cable.Unlike resistive losses, which convert electrical energy into heat, characteristic impedance is determined by the way energy is stored and transferred within the cable. Since the cable has a nonzero length, the wave takes time to travel through it. During this time, energy is not instantly delivered to the load but is temporarily stored in the cable as a combination of electric and magnetic fields due to its capacitance and inductance per unit length. Essentially, the cable acts as an energy reservoir for the waves it transmits. This stored energy defines the impedance that the signal "sees" as it propagates, independent of the cable’s length.Quote from: Ben321 on March 19, 2025, 08:30:50 pmNo. The characteristic impedance of a coaxial cable (e.g., 50Ω or 75Ω) is a constant property of the cable and does not change with length. It is determined by the cable’s geometry (conductor sizes and spacing) and the dielectric material, affecting how an electromagnetic wave propagates through the cable.If you take a coaxial cable of infinite length, you will observe that its input impedance is equal to its characteristic impedance. This happens because, in an infinite cable, the signal never encounters a termination or discontinuity, meaning no reflections occur. As a result, the wave propagates indefinitely, and the input impedance remains constant at the characteristic impedance for an unlimited duration.For a finite-length cable, the characteristic impedance is observed only for a limited time—specifically, until the signal reaches the cable's end, reflects (if the termination impedance does not match), and returns to the input. Once the reflected wave reaches the input, the system transitions to a steady-state condition, where the combined effect of the incident and reflected waves determines the observed input impedance. At this point, any mismatch between the load impedance and the characteristic impedance of the cable becomes apparent.

author: TimFox
author_rank: Super Contributor
timestamp: March 19, 2025, 10:10:05 pm
content: A simple way to understand characteristic impedance of a transmission line:If you “terminate” the far end of the line with an impedance equal to that value, regardless of length, you will see that value of impedance at the driven “near end”.This is due to the absence of reflections at the properly-terminated far end.Neglecting loss (due to resistance and dielectric loss), the value is resistive given byZ0= (L’ / C’)1/2,whereL’andC’are the inductance and capacitance per unit length (ft, in, m, etc.), which depend on cross-section dimensions and the dielectric constant of the insulation.If the termination is other thanZ0, the impedance at the near end will be reactive (inductance or capacitance) that depends on both length and frequency, including resonance conditions on length and frequency.

author: bdunham7
author_rank: Super Contributor
timestamp: March 19, 2025, 10:48:25 pm
content: Quote from: Ben321 on March 19, 2025, 08:30:50 pmYou can either understand it mathematically by learning the physical principles or you can acquire a sort of intuitive understanding by knowing the behavior. In any case, you've already gotten several good explanations so I'll just provide some simple observations on coax cables.First, the cables have small amount of ohmic resistance which is not an important factor here, and the have some capacitance (commonly specified per foot) and inductance. They also have physical dimensions, including length, and exist in a universe where we have time. One important characteristic here is that the signal in a coax cable travels somewhat slower than light, about 2/3c or 1ns/foot or ~4ns/meter. That last part is important.If you had a 50-ohm coax cable that was 250 million meters long it would take about a billion nanoseconds, or 1 second for the signal to reach the other end. If you had the cable open at both ends and then measured at one end with an ohmmeter, you would get a reading of 50 ohms--for about two seconds. That's how long it would take for the signal to reach the other end, be reflected and travel back. Once two seconds had elapsed, the ohmmeter would read an open circuit. There are models you can read about where the coax cable is described as a series of capacitors that charge up as your signal propagates down the line, but I'll just stick to the behavior.Now realistically, a 250 megameter coax line is impractical and parasitic losses and interference would likely make the experiment fail. However, on a shorter timescale you can understand the operation of a transmission line this way. If you have a 2-meter coax, for something like an ohmmeter at DC it is just two wires with some capacitance between the, typically some 10s of pF per foot or so. At low frequencies, mains power or audio or even a bit higher, the 8ns transit time means that the coax is almost entirely like a pair of open wires. If that transit time is much less than the period of the signal, then it doesn't behave like a transmission line, or IOW the transmission line properties have negligible effect on the system.Now if you put a short pulse into a 50R coax, say 1ns, then that pulse will travel down the coax until it reaches the end. If the end is open, the signal will be reflected back. If it is terminated by a resistor that happens to have the an ohmic resistance that corresponds to the coax characteristic impedance, then the pulse will be absorbed and not reflected. If you have a capacitor or different value resistor, then other things happen. The coax "looks like" a 50R resistor to the device producing the pulse for at least 16ns, then after that things change unless the coax was terminated on the other end with a 50R resistor in which case it will always look like 50R.

author: G0HZU
author_rank: Super Contributor
timestamp: March 20, 2025, 12:35:09 am
content: It's probably worth mentioning that if you do low frequency experiments with coaxial cable or if you send a very long pulse down a very long coax cable you won't get the results described because the characteristic impedance of (real world) coaxial cable isn't flat vs frequency as you explore lower and lower frequencies. This assumes coax cable made with copper (or similar) conductors at room temperature where the coax will have some resistance in the conductors.The characteristic impedance of (say) RG174 coax is only about 50 ohms up at RF frequencies. Eg greater than 500 kHz as a ballpark estimate. As the frequency is reduced below this, the characteristic impedance rises (and becomes increasingly complex). Once you get down to audio frequencies the characteristic impedance could have a magnitude of 1000 ohms for example.

author: ahbushnell
author_rank: Frequent Contributor
timestamp: March 20, 2025, 01:11:06 am
content: Or you could say before the reflection return on a finite cable.

author: JonMorton
author_rank: Contributor
timestamp: March 20, 2025, 01:33:36 am
content: Quote from: Ben321 on March 19, 2025, 08:30:50 pmCheck this video:

author: Ben321
author_rank: Frequent Contributor
timestamp: March 20, 2025, 04:24:51 am
content: Quote from: bdunham7 on March 19, 2025, 10:48:25 pmActually if the cable was 250 million meters long, then the resistance of the copper would become important, and couldn't just be ignored as a matter of a few milliohms as it is on reasonable real-world lengths of coax. And at that point, its resistance would be more important than the characteristic impedance of the cable. SO much so, that your signal would get completely absorbed by the cable (dissipated as thermal losses) before it reached the other end. The whole cable would just be a huge resistor at that point.

author: Andy Chee
author_rank: Super Contributor
timestamp: March 20, 2025, 04:48:38 am
content: Quote from: Ben321 on March 20, 2025, 04:24:51 amAgain, DO NOT confuse the characteristic impedance with DC resistance. They are two very different properties of transmission lines. They are NOT synonymous.However, you are correct. RG-174 will have greater attenuation than RG-213 for the same length, despite both sharing a characteristic impedance of 50 ohms. A 100 metre coiled drum of RG-174 can be used as a 50 ohm dummy load for UHF and above, with the RF energy being dissipated along its length.

author: bdunham7
author_rank: Super Contributor
timestamp: March 20, 2025, 05:04:39 am
content: Quote from: Ben321 on March 20, 2025, 04:24:51 amWhich is why I saidQuoteIt's a concept to aid in understanding, not a call to make a YouTube video with millions of meters of coax. However, the ohmic resistance of the long cable wouldn't add to the characteristic impedance in that way. In reality, pieces of coax appear to be capacitors to low frequency signals. A really long piece of coax will not look like a resistor, it will charge up and then appear to be an open circuit unless there's enough dielectric leakage to allow some current through.As I said, you can try to understand it intuitively or you can go with the math, Z0 = (L’ / C’)1/2. But since you won't easily obtain L' and C', you can just use a calculator where someone else has done the work. Note that it isn't just coax cables that have characteristic impedances, connectors/tees/PCB traces/etc all have them as well. You can get both 50 ohm and 75 ohm BNC connectors, for example.https://www.pasternack.com/t-calculator-coax-cutoff.aspxThen if you want to think about really long cables, you can look up Olver Heaviside's telegrapher's equations or learn about how oscilloscope probe cables work to control dispersion.

author: Tation
author_rank: Regular Contributor
timestamp: March 20, 2025, 07:04:45 am
content: Quote from: Ben321 on March 20, 2025, 04:24:51 amThe DC resistance of the cable, measured from here to 2.5e8 m from here will be huge, but the propagating wave doesn't see such resistance, the wavefront only sees the local geometry of the cable, such geometry determines the ratio between the propagating E and H fields at that section of the cable and that ratio also corresponds to the impedance seen at cable input (and is the characteristic impedance of the cable). When the generator drives the line, it does not know if the line is short, long or infinitely long, nor what is the impedance connected at cable end, the "instantaneous" impedance it sees only depends on how is the cable right here. Only after the wave has reached cable end, reflected and arrived again to the generator, will it see a different impedance. If there's no reflection at the load, the generator will see the same (characteristic) impedance forever.Of course the DC resistance of the wires will produce losses, as well as the imperfect dielectric between conductors, these losses will manifest in the characteristic impedance of the cable no longer being real, but complex, but still not being dependent on cable length. Losses will depend on the imaginary part of the characteristic impedance and length. Note that the impedance seen at the input of a sufficiently long, lossy, cable will be equal to its characteristic impedance, no matter the load, as the wave reaching the load will be small against the wave at generator output (due to losses) and the reflected wave reaching the generator will be also attenuated in the same amount, so it will be negligible against the incident wave, so the generator does not see it.A 250 million meter RG-58A cable will have an attenuation of about 100 million dB!! (about 13 dB/100 ft), so the reflected wave reaching the generator will be, at least, 200 million dB below the wave output by the generator... well, that's small enough for it to be negligible(the path loss from Earth to Voyager I is around 308 dB). Such reflected wave would "reach" the generator approximately 1.1 s after the generator started to generate it. About 215 m of RG-58A are enough for 308 dB attenuation.The Wikipedia entry for the telegrapher's equationshttps://en.m.wikipedia.org/wiki/Telegrapher%27s_equationsmay be of interest. Here the relation between the "per meter" parameters of the cable (resistance along and between wires, capacitance between wires and inductance along wires, all of them "per meter") and the characteristic impedance (ohms) and losses (dB/m) are shown.

author: gf
author_rank: Super Contributor
timestamp: March 20, 2025, 07:31:35 am
content: Quote from: Ben321 on March 20, 2025, 04:24:51 amIt seems you are only considering DC. Feed a (say) 100MHz AC signal into the long cable, and the AC voltage source will see the characteristic impedance as load.Btw, the characteristic impedance Z0 is only constant for an ideal, lossless transmission line. For a lossy transmission line (and a real-world cable is lossy), Z0 is a function of frequency. At high frequencies, Z0 approaches an ohmic value of say 75 ohms, or 50 ohms or whatever (this is the number which is usually specified), but at low frequencies, Z0 becomes complex and |Z0| is much higher, increasing with decreasing frequency.

author: ftg
author_rank: Frequent Contributor
timestamp: March 20, 2025, 08:59:57 am
content: What made me grasp this back in the day was an exercise where the coax impedance was calculated via the lumped element representation.In a sense coax is "just" a long inductor and capacitor.As parasitics and DC resistance has also been mentioned, I attached also one example that includes those.

author: G0HZU
author_rank: Super Contributor
timestamp: March 20, 2025, 10:58:14 am
content: If you were to feed a step waveform (a really long pulse) from a 50R pulse generator into that really long coax to try and demonstrate the cable ZO is always 50 ohms then this is a poor test in reality because it would be seen to begin to go wrong within a few hundred nanoseconds if the aim was to monitor the voltage level (or the current) at the feed point and then convert this to a resistance measurement. This assumes the test is done at normal room temperature (and not close to absolute zero).If something like RG174 was used the voltage at the feed point would be seen to slowly ramp up and it would be quite obvious by about 1 microsecond that the demonstration was flawed. You wouldn't have to use a very long run of coax to see this either.Even a few hundred metres of RG174 would show the same problem by 1 microsecond into the test. The voltage at the feed point would be seen to be ramping up over time because of the resistive losses in the cable. The voltage at the feed point might ramp up to be 20% higher after about 1 microsecond for example. This means that the current into the coax is slowly falling over time.The voltage ramp up by 20% at the feed point (or the decline in input current) after 1 microsecond will happen if the cable is 200 metres long or 2km long or 200 km long.

author: Zero999
author_rank: Super Contributor
timestamp: March 20, 2025, 11:15:10 am
content: Quote from: Ben321 on March 19, 2025, 08:55:56 pmNo it won't, because the insulator is imperfect, so there will be some leakage and the longer the cable, the more leakage. More importantly, the cable also has inductance and capacitance, as mentioned by many above, which will also become infinite.The characteristic impedance is the impedance of a theoretically infinite length of cable. In practical terms, it's the impedance seen by the source, when a signal is initially applied. For example, suppose the cable is 50 Ohm and the length of the cable is such that it takes 1µs for the signal to propagate along it. If you connect one end of the cable to a voltage source, it will initially see an impedance of 50 Ohm, irrespective of whatever is connected to the other end, until 2µs, which is the time for the signal to travel down the cable and be reflected of whatever is connected to the other end, assuming it's not a 50 Ohm resistor.In order to prevent reflections, the impedance on the load side of the cable needs to match the cable's characteristic impedance, otherwise not all of the energy will be absorbed and it will be reflected back to the transmitter.

author: radiolistener
author_rank: Super Contributor
timestamp: March 20, 2025, 01:56:18 pm
content: Quote from: Zero999 on March 20, 2025, 11:15:10 amYes, that's correct - characteristic impedance is what the source "sees" before the signal has had time to travel to the end of the cable and reflect back. During this initial period, you cannot determine whether a load is connected at the far end or not. Instead, the cable behaves as if it were a resistor with a value equal to its characteristic impedance.This behavior is determined by the cable’s distributed capacitance and inductance, which are uniformly distributed along its entire length. The two conductors of the coaxial cable act as the plates of a capacitor, while the conductors themselves exhibit inductance. These parameters are typically expressed as capacitance per unit length (pF/m) and inductance per unit length (µH/m).A key property of characteristic impedance is that it remains constant regardless of cable length. This is because when you increase the length of the cable, both its capacitance and inductance increase proportionally. Since characteristic impedance is defined as:\$Z = \sqrt{\frac{L'}{C'}}\$where L' is inductance per unit length and C' is capacitance per unit length, any increase in length affects both parameters equally, keeping their ratio - and thus the characteristic impedance - constant. This fundamental property ensures that a coaxial cable maintains the same impedance whether it is one meter or one kilometer long.

author: bdunham7
author_rank: Super Contributor
timestamp: March 20, 2025, 02:42:44 pm
content: Quote from: Ben321 on March 19, 2025, 08:55:56 pmI missed this and perhaps we haven't realized what it is that you don't understand. Do you understand that the 50 ohms (or whatever impedance) of a coax is measured from core to shield on one end of an open cable, not from end-to-end? I mentioned that in my first reply but did't emphasize it at all. Rereading some of your questions I'm wondering if that is where you are hung up.

author: TimFox
author_rank: Super Contributor
timestamp: March 20, 2025, 04:28:57 pm
content: Quote from: G0HZU on March 20, 2025, 12:35:09 amI disagree about low-frequency operation of coaxial transmission lines.There are substantial changes incharacteristic impedanceat high frequencies, but the behavior at low frequencies follows the normal description of transmission lines.Thought experiment:Connect a sine-wave generator (with constant source impedance of arbitrary value) to three meters of RG-58/U coax terminated at the far end with a good 50\$\Omega\$ resistor.Then measure the AC voltage at both ends as you lower the frequency from, say, 1 MHz down to 100 Hz. You should find the same value on the two voltmeters (with a small difference due to cable loss) over this range.Now, go to an extreme mismatch by removing the termination resistor.Since at these frequencies the line is “short” (about 15 ns for velocity factor 2/3 and 3 m length) compared with the period (>1000 ns) the cable looks like a capacitor of constant value (about 246 pF at 3 m) and the frequency response (over this range of low frequencies) is determined by the source impedance and this capacitance.Stuff does happen at very high frequencies due to the dielectric, skin effect, cross-sectional dimensions, etc.It is important to note that "characteristic impedance" is a parameter of the cable (with some frequency dependence), and the measured impedance depends on your circuit and frequency.

author: G0HZU
author_rank: Super Contributor
timestamp: March 20, 2025, 05:40:17 pm
content: Quote from: TimFox on March 20, 2025, 04:28:57 pmOK but your experiment doesn't demonstrate the characteristic impedance of the cable. The characteristic impedance (Zo) of typical coax cable (eg RG174, RG316 RG58 etc) definitely does change at lower frequencies and there are lots of ways to measure it.A while back I measured the Zo of some RG316 cable. See the results below for real and imaginary and magnitude across 100 Hz to nearly 1 MHz. You can see the Zo increases as the frequency reduces. I did it by measuring the impedance with the far end shorted and then with the far end open. I then calculate Zo as Zo = sqrt(Zopen*Zshort) at every frequency point.It's important to use a sensible length of coax cable when doing these tests (or it becomes difficult to measure the open and short impedances at some frequencies) and for this frequency range I used a 22 metre run of RG316 coax cable.The result for Zo is given below. This result should be independent of the cable length.

author: radiolistener
author_rank: Super Contributor
timestamp: March 20, 2025, 05:48:44 pm
content: Quote from: TimFox on March 20, 2025, 04:28:57 pmIt is a well-known characteristic of coaxial cables that at low frequencies, their measured impedance deviates from the specified characteristic impedance. This occurs because the definition of characteristic impedance assumes ideal transmission line behavior, which is valid primarily at high frequencies.Usually it occurs at frequencies below approximately 100 kHz and becomes increasingly significant below approximately 10 kHz:At ultra-high frequencies (above 10-100 GHz), the impedance of coax cable also becomes unstable

author: Tation
author_rank: Regular Contributor
timestamp: March 20, 2025, 05:53:44 pm
content: It is not unexpected at all for Zo to be different near DC than at higher frequencies. The usual equation for it: \$Z_0=60\sqrt{\mu_r \over\epsilon_r}\ln {b\over a}\$, is derived assuming that the skin effect is well established and the currents circulate only in the surfaces of the conductors. Near DC that's not the case and current fills the conductors, thus different E and H and different Zo.On the other hand, if the frequency is high enough, it is possible to excite non-TEM modes in the line (TEM = Transverse Electro Magnetic → both E and H are orthogonal to the cable axis). These modes, designated as TEnm and TMnm (TE = Transversal Electric —there's H field along axis—, TM = Transverse Magnetic —there's E field along axis. n and m are integers describing how many cycles does show the E, H fields across the cable section) have the property that they cannot exist below a given frequency, called the cut-off frequency of the mode. In a coaxial cable, the mode having the lowest cut-off frequency (apart from the "usual" TEM mode, with a cut-off of 0 Hz) is the TE11 mode. The TE11 cut-off freq. for a RG58 cable is around 30 GHz. Above that frequency the power into the cable is distributed among the different available modes (at least TEM + TE11), being difficult to control the exact way it distributes, so this regime is to be avoided (in any case, at such frequencies, losses are too high for a RG58 cable).Hollow waveguides do not support TEM modes, only TE and TM. Usually they work above the cutoff frequency of the lowest mode (TE10 por rectangular guides, TM11 for cylindrical ones) and below the cut-off freq. of the next mode. Thus only one mode is excited. Also, hollow waveguides do not work at low frequencies.

author: bdunham7
author_rank: Super Contributor
timestamp: March 20, 2025, 06:05:01 pm
content: Quote from: G0HZU on March 20, 2025, 05:40:17 pmCan you tell us exactly how you measured the impedance of the cable at those low frequencies? Did your measurements take phase into account?

author: Wallace Gasiewicz
author_rank: Super Contributor
timestamp: March 20, 2025, 06:23:45 pm
content: Agree with radiolistener.There are other effects in wire or in coax at low freq, say a few KHz. There are things like eddy flow in transmission lines which describes current flow at low freq. Rather than skin effect which is used to explain transmission at higher freq. The characteristic impedance of coax is different at below about 100 KHz. At a few KHz the impedance is much higher. Here is an article from Belden:https://www.belden.com/blogs/broadcast/understanding-skin-effect-and-frequencyI really do not think there is a coax that has a characteristic impedance at audio frequencies.You cannot use coax as long AC transmission line because it is just a big capacitor.

author: G0HZU
author_rank: Super Contributor
timestamp: March 20, 2025, 06:26:47 pm
content: Quote from: bdunham7 on March 20, 2025, 06:05:01 pmI'm pretty sure I used an Analog Discovery 2 and a sense resistor. This does take phase into account. The results I got were very close to what I was expecting based on a simulation of a basic coax model of RG316 cable.

author: G0HZU
author_rank: Super Contributor
timestamp: March 20, 2025, 06:39:11 pm
content: If it helps, I also measured an 86 metre drum of Sucoform 141 cable with the Analog Discovery 2. This is a very nice RF cable with similar dimensions to RG58. I was able to measure down to 10 Hz this time.I also did the same thing on a simulator using an independent model of the coax cable (provided by the simulator).The two results are shown below. The AD2 does struggle a bit up above 100 kHz because I'm using a basic test fixture but the agreement was still very good.I've used a log scale for the impedance this time and I've taken the absolute value of the imaginary part (shown in blue) to allow it to become positive and it overlays nicer on the graph. But the blue imaginary trace is really a negative reactance.

author: bdunham7
author_rank: Super Contributor
timestamp: March 20, 2025, 07:11:24 pm
content: Quote from: G0HZU on March 20, 2025, 06:26:47 pmDid you record the actual individual current measurements of the open and shorted tests? If I understand the result, the characteristic impedance is about 2k at 100Hz. So if you terminate the cable with 2k and you supply a 100Hz signal to the cable at say 5VRMSand you get a current of about 5/2000 or 2.5mARMS, whichreflectstmatchesagrees with the 2k impedance. But how is that result any different than if the characteristic impedance of the cable were infinity, 50R or near-zero? What is the signficance of saying that a 20M cable has a characteristic impedance of 2k @ 100Hz?

author: G0HZU
author_rank: Super Contributor
timestamp: March 20, 2025, 08:16:24 pm
content: Quote from: bdunham7 on March 20, 2025, 07:11:24 pmI'm not sure what you mean here. I just measured the complex impedance with a short at the far end and then with an open at the far end and I do this at all the test frequencies. I then compute Zo from this.The input impedance of a transmission line is a function of the attenuation factor (alpha), the phase factor (beta), the characteristic impedance (Zo) and the length of the line and what the termination impedance is at the far end. By placing an open at the far end and measuring the impedance (Zoc) and then placing a short at the far end and measuring the impedance (Zsc), the two unknown factors (phase and attenuation) can be cancelled out in the equation and so can the (known) line length and this just leaves the unknown Zo.So armed with the input impedance measured with a short termination (Zsc) and the input impedance with an open termination (Zoc), it's possible to reverse engineer Zo from Zo = sqrt(Zoc*Zsc). I did this for about 50 test frequencies spaced across 100 Hz to about 400 kHz and plotted the results.That's all I've done and I believe that this is a valid way to work out Zo. In theory at least, this Zo is independent of cable length so don't pay too much attention to the fact that I used about 22 metres of cable to do this test.

author: bdunham7
author_rank: Super Contributor
timestamp: March 20, 2025, 10:06:54 pm
content: Quote from: G0HZU on March 20, 2025, 08:16:24 pmI'm going to have to think about this some more. Could you explain how you modeled the expected impedances of the cables? What formula, calculator or assumptions are involved?I did a quick test on a coil of 75-ohm cable some installer left behind, it is about 3-4 meters long (I didn't uncoil it) with my LCR tester. It uses a 1VAC stimulus at 100, 1000, 10k and 100k Hz. I don't have time right now to post the entire set of results, but I got somthing similar to what you did. I also measured the capacitance (open) and the inductance (shorted) as well as the DCR of the inner, outer and combination.For 100Hz, I got 4.55M reactance, Θ=-89.98 (open) and 0.77R, Θ=0.48 (short). That results in a calculated Z0of 1871R. However, I measured the capacitance and resistance directly. I got 345pF (which varied very little over the frequency range, it was 348pF @ 100kHz) and 0.765R in the DCR range. So my Zopenappears to be entirely due to the capacitive reactance and my Zshortappear to be entirely due to DCR. The issue there is that the impedance calculated this way depends entirely on the DCR, L is nowhere to be found. The measured 10µH of inductance would result in 6mR of reactance, essentially nothing. Of course if you double the cable length, the DCR and capacitance both double, so that appears consistent. However, if you replaced the core of the coax with aluminum and raised the DCR, your resulting calculated Z0would change dramatically whereas the characteristic impedance at 100MHz should not.So tell me what I'm missing or where I'm going wrong. I think the issue here is that at these frequencies parasitic losses and other characteristics dominate over any transmission line effects.

author: RoV
author_rank: Regular Contributor
timestamp: March 20, 2025, 10:34:42 pm
content: The fact that the impedance of a real transmission line goes high and complex at low frequency is treated on any electromagnetics textbook. See e.g. chapter 2 of Pozar's "Microwave Engineering" 2nd ed.With reference to telegrapher's equations that model most 2-wire transmission lines including coax, in a real line there are non only C and L (capacitance and inductance per unit length), but also G and R (shunt conductance and series resistance per unit length). Impedance is derived as:Z0 = sqrt((R+j*omega*L)/(G+j*omega*C)), where omega = 2*pi*f.G is normally negligible, except for very high frequencies, but R is not.Above 0.1-1 MHz, for normal cables it happens that R << omega*L, so Z0 ~= sqrt(L/C) that is the usual equation.But at low frequencies Z0 ~= sqrt((R+j*omega*L)/(j*omega*C)) and this explains the odd behavior.It doesn't happen only to coax cables: for example, it is well known that typical phone twisted pairs have an impedance close to ~600 ohm at audio frequency, while it is ~100 ohm at larger frequencies (e.g. for ADSL usage).

author: bdunham7
author_rank: Super Contributor
timestamp: March 20, 2025, 10:46:59 pm
content: Quote from: RoV on March 20, 2025, 10:34:42 pmYes, it seems obvious simply because it becomes a capacitor, thus as the frequency drops the reactive impedance goes higher and Θ goes to -90. The additional terms you have supplied have the answers as does my experimental observation that the DCR term completely dominates at 100Hz. I think the issue I have is whether you call the resulting calculated impedance Z0= (Zopen*Zshort)1/2a "characteristic impedance". I'm not sure what designating this LF impedance result as the "characteristic impedance" means as a practical matter. IOW, I don't see the point in designating an impedance value to the cable alone.

author: TimFox
author_rank: Super Contributor
timestamp: March 20, 2025, 11:03:53 pm
content: I stand corrected about characteristic impedance at low frequencies: I neglected the loss components of series resistance and shunt conductance.In the equation forZ0quoted by RoV that includes losses, Z0 = sqrt((R+j*omega*L)/(G+j*omega*C)), where omega = 2*pi*f. we see that at low frequencies the resistance per unit lengthR' will not be small compared with the tinyy value ofwL'at very low frequencyw.Reasonable approximations: at low frequency, where skin effect is not important,R' will be roughly constant, andL' is constant (although its reactance falls with frequency).At 100 kHz, the skin depth in copper is approximatly 0.2 mm.For the capacitanceC' and conductanceG' per unit length,C' is constant and its susceptance falls with frequency. However, for a good dielectric we expect a constant loss factorD=G/wCon the order of 10-3, which is only a small correction to the denominator, while the effect of constantR' is a large correction, leading to a complex characteristic impedance, at low frequencies.Consider RG-58C/U. Its RF specification isZ0= 50\$\Omega\$ andC' = 28.5 pF/ft = 93.5 pF/m. Therefore,L' =Z02xC' = 234 nH/m.Approximating the resistance by that of the center conductor, which is 19 x #33 (0.0071) in tinned copper, or (0.207/19)/ft = 0.036\$\Omega\$/m.Atf= 24.5 kHz, the inductive reactancewL' will equal the series resistance, and the correction to the numerator becomes large.

author: G0HZU
author_rank: Super Contributor
timestamp: March 20, 2025, 11:22:15 pm
content: Quote from: bdunham7 on March 20, 2025, 10:46:59 pmI can suggest a thought experiment if that helps... Repeat your open and short impedance measurements (in your head) on a longer cable. Then do it on a much longer cable. Keep making the cable longer and longer and you should notice that Zoc and Zsc start to look quite similar when the cable becomes really long. Eventually they will essentially measure the same if the cable is long enough. When this happens you have discovered the characteristic impedance of the cable as both Zsc and Zoc will be equal to the characteristic impedance Zo.Zo = sqrt(Zoc*Zsc) becomes Zo = sqrt(Zo*Zo)

author: coppercone2
author_rank: Super Contributor
timestamp: March 20, 2025, 11:28:30 pm
content: need large diameter coaxial cable for low frequency work. The skin depth is only 8.5mm at 60Hz, so a foot diameter inner conductor might work.Maybe someone can make a casting to make a low frequency 50Hz coaxial cable? A few tons of aluminum might resolve this argumentThe other issue might be plating, I assume a solid copper core LMR400 cable would do better at lower frequencies then some plated one. There must be a available skin depth ratio that can be determined for good transmission behavior (the ratio of 'dead material' in the center of the inner conductor to the 'available material', so the thickness of a 'conductive pipe'), and the same thing for the outer pipe. Then it should get more complicated for stranded stuff and also braidsIs that what this is in VLF station cuttler in maine? I wonder how thick that plating is.

author: coppercone2
author_rank: Super Contributor
timestamp: March 20, 2025, 11:54:41 pm
content: So how are you supposed to think about it when the skin depth is deeper then the material?To increase the impedance of a normal functional coaxial cable it would need a thinner inner conductor.What are those electrons doing. It would be like they are not on the surface and only flowing in a tiny path in the middle of the conductor (i.e. instead of having the ID be 1mm, the ID is 0.1mm, which would give you a higher impedance using the formulaThe rule with those cables seems to be that the higher the ratio the bigger the impedance.Like to get a coaxial cable with 1kOhm impedance, for a 10mm OD, you would need a 0.0000000001mm inner conductorFor 0.001mm, its only 470 ohms.The diameter of a electron is 2.8 femtometers. are they going single file in the middle?and that is assuming they are going through a big tube that has a normally functioning outer surface. Then you get to some maximum impedance for a given size for a single file electron flowwhat is going on in thereWhen the coaxial cable impedance is 2kOhm there is something very strange happening!

author: TimFox
author_rank: Super Contributor
timestamp: March 21, 2025, 02:51:01 am
content: When the skin depth is several times greater than the material thickness (i.e., greater than the radius of a cylinder), that means that the current flows throughout the whole interior of the cylinder at that frequency.(The normal definition of skin depth is the distance from the surface where the current falls off by 1/e, or about 37%.)

author: coppercone2
author_rank: Super Contributor
timestamp: March 21, 2025, 05:29:25 am
content: but why does the impedance rise by so much? The 'ratio' of diameters needed for the impedance measured (1K Ohm+) is absurd.The explaination is a bit hard to follow, but it sounds like that whole calculation used for conventional impedance determination by ratio of the diameters becomes largely irrelevant.Like are the electrons not near the surface doing something funny?

author: coppercone2
author_rank: Super Contributor
timestamp: March 21, 2025, 05:57:44 am
content: https://i.sstatic.net/Iqb7F.pngThis is a graph from beldon from a stack exchanger threadConfusing because inductance seems to be dropped. But the capacitance is the part that seems weird to me. Since its not on the surface there is like copper in the 'way' . The inductance part does not bother me, from the concept of a "short at dc"seems like their 'burried' in the depths of the copperthe wave just going through the copper like its fiber line comes to mind too, its translucent lolLike ok, resistance is staying similar, capacitance is going down, inductance is going down because of the 2piF. Since the denominator is going smaller and the numerator adder term is going smaller, the resistor is heavy and it starts making a heavy fraction. But as to what is actually going on in there amazes meif the frequency is way too high you need to increase the diameter just the same as you need to increase the diameter when its too low because the shell it has is too small

author: gf
author_rank: Super Contributor
timestamp: March 21, 2025, 06:08:59 am
content: At low frequencies, R is the DC resistance. R becomes only larger at *high* frequencies due to skin effect.

author: coppercone2
author_rank: Super Contributor
timestamp: March 21, 2025, 06:51:02 am
content: actually that brings another question what is the highest frequency you can get through a metal coaxial cable made to arbitrary specs, like if you can get it mechanically to whatever dimensions you pleasedoes it come down to something related to the thickness of an atom? or like the minimum thickness for something to be a 'sea of electrons' definition?but even if there is a lower limit to conductivity (I think its the 3d matrix), you could just make it wider and wider.but above a certain frequency it would not interact with the minimum silver or copper thickness for something to be a conductive metal. I wonder what that number is, it certainly seems like a conventional materials limit of some kind.Does anyone know what that number is, what is the.. minimum skin depth of a material? 10 atoms stack? There should be another regime there where any known metal making up a coaxial cable would be a semiconductor only?So if a silver atom is 0.144nm, 10 atoms would be ~1.4nm. At roughly 2,000,000 GHz (2000 THz), the skin depth is 1.4nm. If the skin depth get's any smaller, its like not interacting with a conductive material anymore? Like what is going on there? metal is like broken above Ultraviolet C ? brain overloadedis this why x-ray reflectors are shoddy?

author: gf
author_rank: Super Contributor
timestamp: March 21, 2025, 07:59:14 am
content: Quote from: bdunham7 on March 20, 2025, 10:46:59 pmThat's not how the characteristic impedance is primarly defined.It just happens that you can prove mathematically that (Zopen*Zshort)1/2of a finite-length transmission line is equal to the load-invariant input impedance of an infinite-length TL (with the same RLGC parameters), offering an opportunity to measure/calculate Z0of a finite-length TL by measuring its Zopenand Zshort

author: Zero999
author_rank: Super Contributor
timestamp: March 21, 2025, 08:29:06 am
content: Quote from: coppercone2 on March 21, 2025, 06:51:02 amI'm not sure what you're asking? Are you talking about coaxial cables or conductors in general?The coaxial cable needs to be much thinner than the shortest wavelength being transmitted, otherwise it no longer behaves as a lumped transmission line. Other modes such as its diameter, the distance between the outer and inner conductors and the diameter of the inner conductor come into play.In the case of metals, it depends on the material itself. Metals conduct electricity because they have free electrons which are not bound to atoms, as is the case with insulators. A metal is a lattice of positive ions surrounded by free electrons which have similar properties to a plasma. When an electromagnetic wave hits the metal, these free electrons oscillate at the same frequency of the wave, like ripples on the surface of water, until they're damped. At high frequencies (above the plasma frequency) the electrons can no longer keep up, so the metal effectively stops being an electrical conductor.. According to Wikipedia, the plasma frequency of most metals is in the ultraviolet region, which is why they're silver, more reddish of goldish metals, have lower plasma frequencies, causing them to absorb blue and violet wavelengths.https://en.wikipedia.org/wiki/Plasmon#Explanation

author: coppercone2
author_rank: Super Contributor
timestamp: March 21, 2025, 08:46:04 am
content: I noticed that it seems like that number would put a upper limit on the function of a coaxial line made of metal at 2000THz because the skin depth is less then the "conductive metal behavior' depth of the metal (which I guessed to be 10 atoms).I guess it kinda puts a limit on all metallic conductorsnot sure if that helpedI wonder if its related to this plasma frequency. I just did a calculation to see how thick 10 atoms of silver would be stacked up, and what the frequency would need to be to have a smaller skin depth then this, and I got 2000THz,which seems to agree with this plasma frequency.Because if you got 10 atoms, and the skin depth is 9 atoms, it seems to be not working correctly anymore.It's weird that I got a similar result to some complex sounding behavior involving plasma like things. Can it be related? 2000THz is something in the UV-C frequency.I used no other information then the diameter of a silver atom and a skin-depth calculator online and a guess that 10 atoms would be enough to have 'sea like behavior' (from the old explanation that a metal has a sea of electrons, because when you have a 10 atom thick silver pipe, it seems like it has enough electrons to behave 'fludically' ). Maybe the plasmon is just a advanced explanation of 'sea of electrons'?So this would also put a UV-frequency limit for twin leads and twisted pairs too, I guess. Well not sure, because then you also have proximity effect, so maybe twin lead would need to be a little thicker and thus maybe only get to visible operation instead of UV? I am having trouble to think about how proximity effect could interact with this,. but my brain is smokingThe end goal is just to have upper limits of frequency for twin leads and coaxial connectors (and I guess waveguide) made out of metal. I think that is a interesting 'barrier'. And it allows you to start thinking how you could bypass it. Skin depth of neutronium? and I need to look up super conductors, but they are made of atoms so it might not be any better. Metallic hydrogen comes to mind too. Hmm. Or graphene.I don't think its as absurd to think about it as it sounds because I have been reading that all the vacuum oscillator (like a magnatron) can be scaled down quite a bit to make even optical frequencies. I wonder if those too have a frequency limit of generating signals based on the skin depth of metal. I wonder if metallic hydrogen does something to fix this, because its so small. But its also kind of deficient in valance electrons.

author: A.Z.
author_rank: Super Contributor
timestamp: March 21, 2025, 09:13:57 am
content: Quote from: Ben321 on March 19, 2025, 08:30:50 pmhttps://en.m.wikipedia.org/wiki/Characteristic_impedanceHTH

author: Zero999
author_rank: Super Contributor
timestamp: March 21, 2025, 09:15:48 am
content: Quote from: coppercone2 on March 21, 2025, 08:46:04 amPlasma frequency isn't related to skin depth. It's the same reason why lower frequency radio waves bounce off the ionosphere and higher frequencies pass straight through into space.Transmission lines fail way below the plasma frequency. The maximum frequency depends on the physical dimensions. You want a 2000THz transmission line, then you need to think about something with a diameter much smaller than the wavelength, say 20nm.

author: Tation
author_rank: Regular Contributor
timestamp: March 21, 2025, 09:37:53 am
content: Quote from: bdunham7 on March 20, 2025, 10:06:54 pmAs explained by G0HZU, this time with equations, from the formula for the impedance at a transmission line input:$$Z_{in}(l)=Z_0{Z_L+Z_0\tanh(l(\alpha+j\beta)) \over Z_0+Z_L\tanh(l(\alpha+j\beta))}$$being \$Z_L\$ the load impedance at line end, \$Z_0\$ the characteristic impedance of the line, α its attenuation constant, β its phase constant and \$l\$ its length. If \$Z_L = 0\$ one gets \$Z_{SC}(l)=Z_0\tanh(l(\alpha+j\beta))\$. If \$Z_L\to\infty\$ then \$Z_{OC}(l)=Z_0/\tanh(l(\alpha+j\beta))\$. Multiplying both one gets \$Z_{SC}(l)Z_{OC}(l)=Z_0^2\$ and thus the characteristic impedance of the line can be computed, no matter line lenght, α and β as \$Z_0=\sqrt{Z_{SC}Z_{OC}}\$. Not the definition of \$Z_0\$, but a means to compute it from measurements.All these comes from the telegrapher's equations, that come from Maxwell's equations, so generic enough. Also assume that there's only a TEM mode propagating thru the line. \$Z_0\$, α and β may depend on frequency, but in the previous derivations that's not relevant and experiment shows that \$Z_0\$ does indeed vary with frequency.

author: Tation
author_rank: Regular Contributor
timestamp: March 21, 2025, 10:47:46 am
content: Quote from: bdunham7 on March 20, 2025, 10:06:54 pmFrom the datasheet of a 75 Ω cable (an RG-59 BU): conductor resistance 0.159 Ω/m, capacitance 67 pF/m. Supposing that the cable length is 5 m (instead of your 3~4 m assumption): its (OC) capacitance should be 67 pF/m * 5 m = 335 pF. Its (SC) resistance 0.157 Ω * 5 = 0.785 Ω. At 100 Hz these translate to reactances of -4.75j MΩ and 0.785 Ω. The characteristic impedance of the cable, thus, must be \$\sqrt{-4.75j\text{ M}\Omega\cdot0.785\text{ }\Omega}\$ giving ~1931 Ω with -45º phase. All agreeing, more or less, with your measurements.There's nothing wrong here. Simply put, \$Z_0\$ varies with frequency. Its expression is:$$Z_0=\sqrt{R'+jωL'\over G'+jωC'}$$being R', L', G' and C' the resistance (of wires), inductance (of wires), conductance (between wires, due to dielectric and geometry) and capacitance (between wires, also due to dielectric and geometry) per unit length. At high frequencies: \$Z_{0\text{ HF}}=\sqrt{L'/C'}\$, which is real and more or less constant. At low frequencies, with a good enough dielectric, \$Z_{0\text{ LF}}=\sqrt{R'/jωC'}\$, which increases with lower frequencies, has a phase of -45º and depends on \$\sqrt{R_{DC}}\$.But, in any case, at 100 Hz this 5 m transmission line has a length negligible compared to the wavelength, so it is not necessary to use all this, simply consider it as a series resistance of 0.785 Ω with a shunt capacitance of 335 pF and all is OK. Yes, at LF both G' and L' do not matter, only R' and C' do, but the transmission line approach is still valid, although not necessary.

author: G0HZU
author_rank: Super Contributor
timestamp: March 21, 2025, 12:21:23 pm
content: I did a load of measurements about a year ago to demonstrate Zo and I've dug out some more old results. There's a version of the E5071C VNA at work that goes down to just 9 kHz and I measured the 22 metres of RG316 cable on it across 9 kHz through to about 10 MHz. I measured the cable as a full two port device (s11, s21, s12, s22) rather than measuring just one end as s11. I then exported this as an s2p file and this can be used as a model of the coax.I can now do lots of things with itThe first thing I did was to calculate the characteristic impedance Zo when terminated in a short and open using Zo = sqrt(Zoc*Zsc). See the first plot below. This shows Zo to be just under 250 ohms at 9kHz (green trace).I used the Genesys simulator to calculate sqrt(Zsc*Zoc) and it then produced the plot for me.The other neat thing I can do is daisy-chain dozens of copies of the model in series to effectively make up a much longer cable. I put together 136 copies in a cascade to make up a cable that would be about 3km long. This is long enough to be so lossy that the input impedance should just show the characteristic impedance of the cable as there should only be a forward wave at the input. Any reflections will never make it back to the source because of the losses in the cable even at just 9kHz. See the second plot below. This agrees very closely with the Zo = sqrt(Zoc*Zsc) equation in the plot below.To show the agreement better, I've overlaid the Zo traces for the equation (green trace) and the 3km long cascade (pink trace) in this second plot and the agreement is impressively close! They overlay almost perfectly! This shows how the equation agrees with reality.

author: G0HZU
author_rank: Super Contributor
timestamp: March 21, 2025, 12:40:23 pm
content: At work, I have access to the Cadence/AWR Microwave Office RF simulator and this has a fairly decent model for coax cable. It allows the dimensions of the coax to be defined as well as the resistivity of the conductors and the dielectric properties can also be defined.It's therefore really easy to do experiments on crazy line lengths and look at Zo at low frequencies. At the same time I did the VNA experiments with the E5071C VNA I put together a 3km long model of RG316 cable in AWR MWO and left the far end unterminated. This is effectively the same as the cascaded experiment using the VNA data from the E5071C VNA.The result for the AWR simulation of the 3km run of RG316 coax is shown below. This also shows that the Zo is just under 250 ohms at 9 kHz. So the AWR model agrees very closely with the previous experiment with the VNA. Normally, I can access AWR Microwave Office from home remotely but this is down at the moment so I can't do any fresh experiments with the AWR coax model.

author: coppercone2
author_rank: Super Contributor
timestamp: March 21, 2025, 11:28:08 pm
content: Quote from: Zero999 on March 21, 2025, 09:15:48 amright, I totally forgot about the other requirements. But I still think thats a interesting calculation or observation, even if it does not answer the question. (minimum thickness of a conductor). It still seems like it means something that you can have a frequency that can't use the minimum thickness of a metal conductor, its like its just not there.. it like can't interact normally anymoredoes this have a name?and it's just coincidence that my answer is also ultraviolet?

author: Ben321
author_rank: Frequent Contributor
timestamp: March 24, 2025, 08:10:13 am
content: Quote from: gf on March 20, 2025, 07:31:35 amWhile the source may only see the signal as a 75ohms (or 50ohms) load when driven at an RF frequency, you also need to ask what the actual LOAD sees at the other end of the coax. It will see a hugely weakened (to the point it's well below the noise floor of the inputs of all but the best top-of-the-line receivers) signal, that has been mostly absorbed (dissipated as heat) by the resistance of the coax, when the coax is literally millions of meters long. Remember that Z = sqrt(X^2 + R^2). Z is total impedance. X is reactance. And R is resistance. As you can see, even in the BEST case scenario, where X = 0, you will still have Z=R. This means that for a very long length of coax, the total current flowing through the load will be VERY LOW. And this also means the total voltage as seen by the load will be very low, due to voltage divider effect.Remember that the load (such as a radio receiver) will have a terminating resistor internally at 75ohms between the signal pin input BNC connector and the ground, for the input to the antenna. So if you have a test signal generator with its output impedance being 75ohms (it usually will have a 75ohm resistor in series with the signal pin of the BNC output connector to force it to have 75ohms output resistance at all frequencies) and 250 million meters of 75ohm coax between the test generator, and lets say its DC resistance is 1ohm per 25 meters, that's 10 million ohms of DC resistance between the generator and the receiver. Even at ideal operating conditions that RF signal is going to going through a huge resistance of 10 million ohms. So the current (albeit an RF AC current instead of DC current) flowing through the load will be I = V/R because in an ideal setup all reactances have been removed (as they are in the case of a properly tuned coax).R = R1+R2+R3 = 10000150(R1 = 75 ohms output resistor in sig-gen, R2 = 10000000 ohms coax, R3 = 75 ohms terminating resistor in load)Assuming the open circuit voltage from the sig-gen is 1v amplitude (peak voltage, not peak-to-peak voltage, and not RMS voltage) sinewave, then the current through the whole circuit, including the load, will beI =1/10000150 = 0.0000000999985 A (peak current, not peak-to-peak current, and not RMS current)Now you have that much current flowing through the load (radio receiver, or alternatively a 75ohms dummy load resistor) which has 75ohms input impedance, which means the voltage seen by the load isV = I*R3 = 0.0000000999985 * 75 = 0.0000074998875 Volts (which is about 7.5 uV)Note that a sinewave with a 7.5 uV amplitude (peak voltage, not peak-to-peak voltage, and not RMS voltage). This would be undetectably weak in most receivers.Remember, there's no magic way of bypassing the DC resistance of the metal in the coax. The voltage of the signal as seen by the receiver will always decrease with longer lengths of the same coax cable. In fact, the resistance will INCREASE with frequency. The higher the frequency, the more the skin effect takes place, and the less current will flow through the center of the copper wire core of the coax, and the more it will flow o the SURFACE of that same copper wire. In effect, the cross sectional area of the copper wire, as seen by the electric current, will be reduced at higher frequencies, and thus the resistance the current sees will be higher. If it's 10million ohms at DC for that 250 million meters length, maybe the resistance will be 1 billion ohms at a frequency of 10 MHz. And correspondingly the voltage as seem by the load will decrease by a factor of 100, such that the voltage as seen by the load in the above example would be only 75nV. That is so low, that really only things like radio telescopes, designed to pick up the weakest signals from space, would even be able to detect a signal this weak.

author: Tation
author_rank: Regular Contributor
timestamp: March 24, 2025, 08:57:57 am
content: So, at 10 MHz, you expect around 143 dB losses in the 250 million meter coax (20*log(75 nv/1 V))Well, you are off by around 100 million dB, that's about 5 million orders of magnitude. So insert 5000000 zeros into your numbers and you will be closer to the correct values.You insist on consider the cable as a lumped element device. That's plainly wrong, your analysis is not applicable. The signal does not see 1 GOhm at all, and the current does not see 1 GOhm.Transmission line theory exists because lumped element analysis cannot cope with all these.

author: Tation
author_rank: Regular Contributor
timestamp: March 24, 2025, 02:03:05 pm
content: If you put an AC ammeter in series with the inner conductor of the coaxial, you will see that its measurements decay exponentially with distance from generator. And that's even in the case of a perfect dielectric (vacuum).

author: MarkT
author_rank: Frequent Contributor
timestamp: March 26, 2025, 09:20:13 pm
content: Forget all this DC resistance of a long cable, you should be thinking in terms of milliohms/metre, picofarads/metre and nanohenries/metre, and thinking of waves propagating along the line. Take a walk on the Heaviside (!!)