dumb coax question - Page 1
Forum section: electronics / rf-microwave

author: A.Z.
author_rank: Super Contributor
timestamp: April 21, 2023, 01:17:54 pm
content: probably my poor old brain is just farting again, whatever, I can't just find an answer, so I'm asking hereLet's say we have 100ft of 50 Ohm coax, the manufacturer states that with a length of 100ft at a given frequency, that coax will have a 4dB lossNow, we measure the loss and find it's correct, that is 4dB as stated in the cable datasheet, till now, everything's fineBut now, let's say that we change the impedance at one end (or both ends) of the cable to. something other than its 50 Ohm characteristic impedanceWhat happens to the cable loss now ? Will those 4dB still stand true and sum up to the losses due to reflections or will those 4dB change ?

author: yl3akb
author_rank: Regular Contributor
timestamp: April 21, 2023, 05:01:58 pm
content: The available power at the end of the cable where mismatch load is connected would always be 4 dB less than injected power at cable input.In practice though You would be measuring transmission coeffcient (S21) which by definition is ratio of absorbed and incident waves, but if cable impedance is different than port impedance of measurement system, there would be less absorbed power, and measured loss would be larger than 4 dB due to mismatch loss.

author: A.Z.
author_rank: Super Contributor
timestamp: April 21, 2023, 07:24:15 pm
content: yep, got around it right after hitting the post button, as usual...

author: RoGeorge
author_rank: Super Contributor
timestamp: April 21, 2023, 08:38:38 pm
content: The problem with energy transfer (or power) is that part of it gets reflected back to the source, at each point where the impedance changes (along the traveling path).The impedance concept has a more general meaning in physics. It can be mechanical impedance, electrical impedance, acoustical impedance, etc. The broader concept of impedance is defined asthe ratio between a cause and its effect. No matter which kind of impedance it is, mechanical, or electrical, or the impedance of public while listening to the news, the matching of impedance is still necessary if we want the maximum transfer to happen. If not, only a fraction of the "cause" stimuli will enter and produce an "effect" into the load, while the rest of the stimuli will "backfire".For the cable, if not well matched with the source, a part of the power will be reflected back to the source, and another part will enter into the cable. Once entered, that fraction that got into the cable will propagate to the other end, and will lose 4dB over those 100m of cable. Now the entered power less the lost 4dB will have to be transferred further, to the load.Same principle applies at the exit point from the cable. If the load perfectly matches the cable, then all power that managed to arrive at the other end of the cable will be transferred to the load. If the cable impedance doesn't perfectly match the load's impedance, then a fraction of the power will be reflected back into the cable, and only a fraction from the power that arrived to the other end will be transferred to the load.The funniest thing is that the fraction that has been reflected back into the cable, will travel back 100m to the entry point, will lose 4dB on the way back, then hit the entering point, which was not well matched, so it will reflect again into the cable, and so on, back and forth, just like the water waves are hitting the walls of a bathtub and reflect back and forth.All the direct waves and the reflected waves add up together, so it will mess the amplitude seen by the load. Sometimes the reflected wave can enter back into the transmitter, and add up with the transmitted wave at that moment, and together burn the power amplifier.

author: A.Z.
author_rank: Super Contributor
timestamp: April 22, 2023, 05:26:20 am
content: thanks to all people who replied, but I see I've been unclear, I know about stationary regime and its effects, what I was wondering was about the coax losses, I mean, if the coax materials (center insulator, braid...) could/would somewhat have a different/higher loss by themselves when the coax isn't properly matched, aside from losses due to reflections

author: yl3akb
author_rank: Regular Contributor
timestamp: April 22, 2023, 07:31:38 am
content: If You operate the coax near its voltage breakdown, then mismatch (standing waves) could cause pemanent damage to the cable and of course increased loss

author: RoGeorge
author_rank: Super Contributor
timestamp: April 22, 2023, 07:45:09 am
content: Quote from: A.Z. on April 22, 2023, 05:26:20 amPlease don't mind my long post above, I was only trying to arrange thoughts in my had, I'm a very beginner in RF practice.About the losses of the cable itself, yes, they will stay the same no matter what's connected at its ends. The losses are a property of the cable alone. The cable doesn't "know" what's connected at its ends. Even more, the wave that propagates through the cable doesn't "know" what impedance will met at the other end.While traveling inside the cable, the wave will only "see" the point inside the cable where the wave is at that moment (I picture the wave more like a front of a wave, or a single "sample" from the wave if you want - though there is no sampling happening, just that it help me picturing a single point). Point by point, the wave will advance through the cable until the other end. At any given moment, the wavefront cannot "see" what is in front of it, or behind. Only when the wavefront of the wave arrives at the other end, the wave will suddenly "bump" into a different impedance, the load's impedance.Because (at any given moment) the wavefront can only see the point where the wave is (at that snapshot in time), while inside the cable the wave will interact only with the cable, and lose the same 4dB, no matter what's connected to the cable.

author: gf
author_rank: Super Contributor
timestamp: April 23, 2023, 08:35:33 am
content: Quote from: A.Z. on April 22, 2023, 05:26:20 amThe loss of (say) 4dB refers to a one-time travel of the wave through the cable, in one direction. If a fraction of the signal arriving at the end happens to be reflected, then the wave traveling in reverse direction is again attenuated 4dB. And so on... That's how it is defined. Maybe your subjective notion of "cable loss" is different from this definition?

author: gf
author_rank: Super Contributor
timestamp: April 23, 2023, 08:45:15 am
content: Quote from: yl3akb on April 21, 2023, 05:01:58 pmAnd since the characteristic impedance of alossycable is not real-valued, but always complex, it cannevermatch the 50 Ohm port impedance of a VNA exactly.

author: radiolistener
author_rank: Super Contributor
timestamp: April 23, 2023, 09:53:51 am
content: Quote from: A.Z. on April 21, 2023, 01:17:54 pmin that case coax cable turns into LC circuit. The RF wave flows through cable and have 4 dB loss, then part of the wave energy is reflected back and flows back to the input of cable. This reflected wave also has 4 dB loss. Next the rest of reflected wave leads to a change of input cable impedance.So, the answer on your question is yes, 4 dB loss is still here, but now loss is multiplied by Q due to reflections and the total loss will be more than 4 dB. The actual loss will depends on Q-factor of that LC circuit and depends on the coax cable electrical length (phase delay), loss and VSWR on the cable output.

author: A.Z.
author_rank: Super Contributor
timestamp: April 23, 2023, 12:21:18 pm
content: Quote from: gf on April 23, 2023, 08:45:15 amwell, no... imagine terminating a 50 Ohm coax with a totally resistive load of (say) 1481 Ohm or, if you prefer add an inductance and make it complex but far away from the coax characteristic impedance, now at a given frequency F, will the dielectric/braid still show the characteristic loss as for datasheet ? See my doubt now ?

author: RoGeorge
author_rank: Super Contributor
timestamp: April 23, 2023, 12:48:26 pm
content: Quote from: A.Z. on April 23, 2023, 12:21:18 pmYes, because the wave doesn't know what is connected at the other end, until the wave will hit the end, and only at that moment the wave will find out the terminator's impedance.

author: A.Z.
author_rank: Super Contributor
timestamp: April 23, 2023, 12:56:28 pm
content: Quote from: RoGeorge on April 23, 2023, 12:48:26 pmok but what if the characteristics of (say) the dielectric material, under those circumstances, are different from the standard ?

author: radiolistener
author_rank: Super Contributor
timestamp: April 23, 2023, 12:59:50 pm
content: Quote from: A.Z. on April 23, 2023, 12:21:18 pmcoax cable loss is still the same, but now you're using the cable in different mode - it is used to transfer direct waves and reflected waves simultaneously and both path have loss according to the cable specification, so the total loss will be more than that which is specified for a single way wave travel.

author: RoGeorge
author_rank: Super Contributor
timestamp: April 23, 2023, 01:09:18 pm
content: From a typical datasheet like this, for examplehttps://www.tme.eu/Document/151b7fc68e82e16d70ee1f3bf3c18100/M_004-005-006-007_RG-coaxial_cables.pdfit seems only the temperature or the frequency can change the losses/meter, because the losses in the table are given a a specified temperature and at a specified frequency. At least on paper, it looks like the terminator's impedance doesn't affect the dB/m losses.I don't know how it really is in practice.

author: radiolistener
author_rank: Super Contributor
timestamp: April 23, 2023, 01:21:04 pm
content: Quote from: A.Z. on April 23, 2023, 12:56:28 pmCharacteristic resistance is not loss resistance. These are different resistances.Characteristic resistance appears here because distributed inductance and distributed capacitance prevents short circuit for AC current. Since inductance and capacitance compensate each other, the cable reactance is zero. But it still has active component in the input impedance, because coax cable can accumulate RF energy due to time delay required to travel wave forward and back. So, when you push some DC pulse with U voltage, the current will be limited with the cable until the moment when the wave travels forward and then back in the cable.If you measure current consumed by the cable input before reflected wave will be returned back, you can calculate the cable characteristic impedance with R = U / I, according to the Ohm law. But this is not heat loss resistance, because it happens due to phase delay in the cable. In other words, this resistance appears here because some RF energy is required to fill the cable before it appears on the cable output. It didn't lost, its just stored in the cable in a form of electromagnetic waves which will still present on the cable output for some time even if you disconnect cable from RF wave source.Cable loss affects power loss on the cable output, but not it's characteristic resistance on the cable input.Different dielectric material properties requires different coax cable geometry to keep characteristic resistance. Also it leads to a different wave propagation speed, because the wave energy flows within dielectric between central conductor and coax shield.

author: gf
author_rank: Super Contributor
timestamp: April 23, 2023, 02:40:47 pm
content: A RLGC model for a lossy transmission line is usually characterized by two complex numbers:Propagation ConstantCharacteristic ImpeadanceBoth quantities are frequency-dependent (and likely also temperature-dependent). For a given frequency (and temperature) they are cable-specific constants, which do not depend on the source or load connected to the cable. The "loss per length unit" from the datasheet is basically just the real part of this propagation constant (but with a different scaling).Maybe you should not try to associate a particular observable meaning with this loss constant. If you connect a lossy cable to an arbitrary source and load, you cannot observe the cable's intrinsic loss constant directly. But if you measure both, the cable's S11 and S21, then you can calculate its propagation constant (and also its characteristic impedance) from the measured S11 and S21. And once you know the cable's propagation constant, you also know its loss constant.[ And as already mentioned, don't assume that the characteristic impedance of alossy"50 Ohm cable" is really 50 Ohms. It isn't. Only lossless transmission lines have a purely resistive characteristic impedance. ]See alsohttp://weewave.mer.utexas.edu/MED_files/Former_Students/thesis_dssrtns/Friar_R_diss/ch_2.pdfhttps://tsapps.nist.gov/publication/get_pdf.cfm?pub_id=924664

author: radiolistener
author_rank: Super Contributor
timestamp: April 23, 2023, 03:17:14 pm
content: Quote from: gf on April 23, 2023, 02:40:47 pm50 Ω coax cable has fixed 50 Ω impedance and it doesn't depends on cable loss. Ideal cable with zero loss and a real coax cable have the same impedance, because it depends on coax cable geometry and magnetic and electric properties of isolator and coductor.Of course impedance can have some tolerance, but usually error is small enough to not be taken into account. If you don't use some fake Chinese coax cableYou can measure cable parameters by measure open end cable capacitance C and short circuit end cable inductance L:Z = sqrt( L / C ) V = 1 / sqrt( L * C ) [m/s]where Z is characteristic impedance and V is a wave propagation speed.To get more precise, use more longest cable, 10 meters is good enough length to measure Z and V with low error.As you can see resistance R for heat loss is missing from the impedance formula, so both parameters - impedance and velocity factor doesn't depends on cable loss.PS: velocity factor VF = V / c, where c is speed of light in the vacuum constant

author: gf
author_rank: Super Contributor
timestamp: April 23, 2023, 03:47:10 pm
content: Quote from: radiolistener on April 23, 2023, 03:17:14 pmThis only applies to alosslesstransmission line (LC model).For alossytransmission line (-> RLCG model) it isZ = sqrt( (R + j * omega * L) / (G + j * omega * C) )Lossy means that either R != 0 or G != 0 (or both != 0).And then the characteristic impedance Z is no longer real-valued, but has even a reactive component.See equation (2.2) inhttp://weewave.mer.utexas.edu/MED_files/Former_Students/thesis_dssrtns/Friar_R_diss/ch_2.pdfFor R = G = 0, the equation degrades to Z = sqrt( L / C )

author: switchabl
author_rank: Frequent Contributor
timestamp: April 23, 2023, 03:57:46 pm
content: If the Heaviside condition (G/C = R/L) is satisfied, the reactive component cancels out and the characteristic impedance reduces to Z_0 = sqrt(L/C) as in the lossless case.

author: radiolistener
author_rank: Super Contributor
timestamp: April 23, 2023, 03:59:16 pm
content: gfI think its too complicated to use RLCG model and doesn't have sense here, just because difference will be too small to be taken into account. The error of L and C measurement will affect Z and V values much more.Also coax cable has not ideal geometry and material properties and as result it has some Z and V value variations over it's length. If you bend coax cable a little or push it at some point, it also leads to some change of Z and V value at that point of cable. All it leads to a some degree of tolerance. So, you can just assume that Z = sqrt( L / C ) and impedance doesn't have reactive part.

author: A.Z.
author_rank: Super Contributor
timestamp: April 23, 2023, 04:19:12 pm
content: temperature dependence is something to consider, if a cable is mismatched, temperature will raise (depending on power and mismatch) so impedance will vary...Also, if the dielectric has a given loss, it won't remain the same as temperature raises