Correct model for interference from an unterminated transmission line branch - Page 1
Forum section: electronics / rf-microwave

author: oschonrock
author_rank: Regular Contributor
timestamp: February 26, 2021, 04:46:59 pm
content: Question about the correct mathematical model for a branched transmission lineBaseline SetupWaveform generator: Output 10-60MHz sine wave, 8Vp-p, Output impedance 50OhmsRG58 50Ohm coax: propagation speed 0.659c (length ~0.5m, but that should not be relevant)50Ohm "feed through" terminator connected to200Mhz digital oscilloscopeResult: 4Vp-p sine wave. Amplitude flat across frequency range. 50% attenuation, as expected, due to "voltage divider" of the 50Ohm source with 50Ohm terminatorInterference SetupAdd a coax-T before 50Ohm feed through terminatorConnect additional ~1.6m piece of RG58 coax to the 3rd leg of the T.Leave the end of the ~1.6m piece unterminated.Result: As expected, we get reflection at unterminated end. This results in interference being seen by oscilloscope. Amplitude measured by oscilloscope collapses to near zero (about 200mVp-p in reality) at about 31Mhz. Given prop-speed of 1.98e8m/sthis is at λ=6.38m, which is a good match for λ/4=1.63m. At 10Mhz λ≈20m interference almost perfectly "constructive" resulting in an observed amplitude of ~4Vp-p, very similar to the baseline. (note there is no "doubling" due to constructive interference, see below)QuestionWhat is the correct relationship for observed amplitude in the interference setup as a function of frequency?My attempt was to just model sinusoidal addition from here:https://dspguru.com/files/Sum_of_Two_Sinusoids.pdfgives us:Acos(ωt+α)+Bcos(ωt)=sqrt(A^2+B^2+2ABcos(α)) * cos(ωt+arctan(Asin(α)Acos(α)+B))And if we assume near perfect reflection (we nearly have that), then A and B are equal. Just focusing on the amplitude of resultant we have:ResultantAmplitude=sqrt(2A^2+2A^2cos(α))This does indeed give the rough shape of the frequency response however the scale is wrong. Expression above gives 2A at α=0which is clearly wrong by factor of 2 (no doubling). It correctly shrinks to zero at α=π. The amplitudes at the intermediate frequencies are also too high.A simple linear scale factor to push the low frequency amplitudes down to the correct, observed value, fails to align the frequencies around 20Mhz.I suspect this is because we are not modelling the actual network of impedances here, and the reflected wave is not a "second source" (hence no doubling).There is also the potential issue that the reflected wave may(!) see two 50Ohm loads in parallel, ie 25Ohms and that there will therefore be a secondary reflection back to the open ended stub?I am ideally looking for a simple electrical equivalent circuit, for example see attached equvcct.pngZ in that diagram will be changing from an "open circuit" at low frequencies to a "short circuit" ~ 31Mhz.So, in summary, what I am looking for is an expression for the signal seen by the oscilloscope across the 50Ohm feed through terminator which incorporates a good model for the changing Z.

author: yl3akb
author_rank: Regular Contributor
timestamp: February 26, 2021, 05:23:43 pm
content: If I understand correctly, you need an expression of voltage across the feed-through termination resistor?First thought would be to calculate combined impedance of 'feed-through' R paralleled with input Z of 'transmission line + unterminated Z'. Formula here could be used for Zinhttps://en.wikipedia.org/wiki/Transmission_line#Input_impedance_of_lossless_transmission_lineAs a result you would obtain simple voltage divider formed by output Z of your generator and calculated combined Z allowing to calculate the voltages. I you are lazy like me, this might help:http://hyperphysics.phy-astr.gsu.edu/hbase/electric/vdivac.htmlP.S You should use numbering of components, when showing the schematics, otherwise it is pain to discuss it.

author: oschonrock
author_rank: Regular Contributor
timestamp: February 26, 2021, 06:33:24 pm
content: Yup! top marks for that approach. Thank you.basically this link, does ithttps://en.wikipedia.org/wiki/Transmission_line#OpenThe rest is just complex number, but simply circuit analysis.My model vs actual now looks like the image below:(note that that data is for a 52cm coax piece where the destructive interference is just above 90Mhz. I was trying to keep it simpler in above example. The plot also uses wavelength for x-axis rather than frequency, because it directly relates to the length of the unterminated stub)answer spelled out on stackexchange which has better formatting for code and equations:https://electronics.stackexchange.com/a/550274/213363

author: oschonrock
author_rank: Regular Contributor
timestamp: February 26, 2021, 06:41:25 pm
content: BTW, Was my concern about the secondary reflection correct? ie is possible that the reflected wave sees 25Ohms?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 26, 2021, 06:57:30 pm
content: It does, yes. This test would usually be done with the source terminator only, no need for more.The reflected impedance (from the very high end termination) is still very low at the feedpoint, giving a deep null.Tim

author: oschonrock
author_rank: Regular Contributor
timestamp: February 26, 2021, 08:29:43 pm
content: Thanks for confirming that. makes perfect sense.I'll run some tests both ways and compare.

author: oschonrock
author_rank: Regular Contributor
timestamp: February 27, 2021, 06:32:56 pm
content: I ran some figures both ways, ie with and without terminator at the scope.Firstly note that I am actually using a ~50cm stub, which means I need ~90Mhz for interference. My signal generator doesn't go that high (I only have a Siglent SDG 1062X).So what I am doing is using the special square wave circuitry in that SDG1062 to generate 10-12Mhz square waves. These have sinusoidal components well over 200Mhz. The I was inspecting the resulting waveform using the FFT function on my SDS 1204X-E and comparing amplitudes of the top 8 harmonics (up to 150Mhz). To get more data I am then changing the square wave base frequency in 0.2Mhz steps from 10-12Mhz, which multiplies up.Automation. Because all this gets tedious, I started downloading the waveform data onto a USB and doing the FFT in python (results are nearly indentical BTW). Then to save more time, I am now talking to the AWG and the scope over a LAN connection using sockets from python. Changing the frequency on the AWG and then downloading the waveform data from the scope. Twice for all frequencies once without the stub in place and once with it to get the interference. So the entire process is now automated and takes about 30seconds to do.So I did 10-12Mhz with/without interference stub TWICE.. once with the terminator and once without it.Here are the results: Attached are saved result curve_fits for "with the terminator" (first graph) and "without the terminator" (second graph).I don't want to judge, but the "with the terminator" (first graph) has a better data/model fit.Any comments / ideas?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 27, 2021, 07:48:31 pm
content: Hmm, neat. Wonder if that has to do with reflections, the scope or signal source being nonideal, that sort of thing.I guess it would be good to compare with the no-stub case, straight into the scope with/out pass-thru terminator.Tim

author: cdev
author_rank: Super Contributor
timestamp: February 27, 2021, 07:52:31 pm
content: The free Windows program SimSmith as well as some similar programs for Linux (GSMC) make it fairly straightforward to figure problems like this out.

author: oschonrock
author_rank: Regular Contributor
timestamp: February 28, 2021, 12:32:54 am
content: I have taken a slightly different approach in order to remove some potential sources of error.I am now using a longer stub ~1.6m which puts the destructive interference frequency at ~ 31Mhz and therefore in the range of my AWG. So I tried without the Square Waves and FFT, and just used sine waves. 3Mhz => 60Mhz in 1Mhz steps. All still controlled using SCPI over sockets from python.On the scope side I am now just measuring the Pk-Pk so I decided to run the built in stats, for 30 samples and get the mean and std-dev so I have "better" error estimate. I had to write code for a sort of "auto-ranging" to maintain decent resolution of the scope throughout the destructive trough. It goes right down to 50mV per division now and you can see that in the sharp and low trough in the graph. I believe that the FFT type analysis I was doing earlier was too inaccurate when signals got small, just because the raw data lacked resolution.I am also combining the errors from normal/interference measurements statistically correctly now (Before I was just using a 3% guestimate)Result? Both with and without terminator are remarkably close to their respective models now (which are of course different due to a different voltage divider). The With-terminator is perhaps still "slightly"(!!) better?? It's certainly not worse, which is interesting given we thought it should suffer from secondary reflections? Perhaps at higher frequencies / longer lengths?The final error estimate on the length shown on the graphs is clearly wrong/too small. But the fit is impressive.Comments?

author: oschonrock
author_rank: Regular Contributor
timestamp: February 28, 2021, 04:00:45 pm
content: Thanks for the Tip on SimSmith. Spent some hours watching the excellent intro videos and came up with this cct and chart to model my situation. .ssx file also attached (inside a .zip to make forum happy).I am very much a rookie at SimSmith so bear with me.Comments on SimSmith model setup:I modelled the setup as 2 pieces of RG58 TX line with the actual lengths into a 1G (open circuit) Load. The 2 pieces are separated by either a 1M shunt resistor (the input impedance of the oscilloscope) or a 50Ohm Shunt resistor (the feed through terminator).I used one of those isolating blocks to separate the 2 setups. The left hand one is the one without the terminator and R1 is the scope with 1M input resistance. I also tried putting a 15pF capacitance in parallel to the 1M to more accurately model the osc input, but it made no material difference.The generators are not "cloned" because I adjusted them to give the same voltage at the oscilloscope/terminator at low frequency. I did the in practice as well. I means the circuit with the terminator has 4x the output power of the one without the terminator (halving of the voltage divider squared). ie 80mW and 20mW. This produces 2V amplitudes at low frequencies which matches my experimental conditions.Here is my attempt at interpreting the result of the SimSmith chart.I have plotted voltage in R1 and R2 (ie at the oscilloscope input)Also plotted SWR for interestThe voltage plot makes perfect sense to me with the expected collapse just above 30Mhz. Both R1 & R2 are the same with R2 in pink being a sharper notch as also observed in my experiment.I understand the R2 SWR plot, but Ido notunderstand what is going on with the SWR at R1 (in the cct without the terminator)If I expand the frequency range to 120Mhz (easy now!) I can see the expected repeated destructive interference cycles.I was unable to find any evidence (or way of measuring really?) the suspected secondary reflection in the right hand case with the terminator, because it sees 50||50=25Ohm. Does it exist? Am I not looking correctly?Any coments on this simulation setup or the results?Many thanks

author: oschonrock
author_rank: Regular Contributor
timestamp: February 28, 2021, 07:09:57 pm
content: I found the secondary reflection. Had to be hiding there somewhere.I had to move to a Time domain reflectometry approach. First I followed this video from ae6ty:to set up Sim Smith for a TDR pulse. and without terminator (measuring at the oscilloscope as before) is gives just a single reflection off the unterminated end.Sim-SingleReflectionWithoutTerminator.png belowThen change the impedance at oscilloscope to 50Ohms (inserting the terminator) and we get 2 distinct relfections. The second one weaker and inverted. Because it has been inverted during refelection off the lower 50||50=25 impedance back at the osc/scource end - the travelled the length of unterminated stub (andback again) a second time, before arriving at the oscilloscope at twice the delay of the first reflection.Sim-DoubleReflectionWithTerminator.png belowThe circuit used for simulation was:Sim-PulseFindTDRCct.png belowand the .ssx file in a .zip here:PulseFind.zipNot quite sure about the amplitudes...but the timings and polarities all tie up.RealityCan we measure this second reflection is real life? Yes we can, just. The shortest pulse I can make out of the AWG is ~ 50ns.And without the terminator the single additive reflection looks like this:SingleReflectionWithoutTerminator.png belowAfter adding the terminator we can see a clear, extrasubtrativereflection ~35ns after the start of the pulse:DoubleReflectionWithTerminator.pngAgain, not quite sure about the sizes of the relfections vs the simulations, but the polarities and timings line up well and within the margin of error.Feel like I have (possibly over) analysed this situation to death now, and feel like I understand it all (except that SWR curve in post above). I suspect the "frequency response" / Bode Plot type approach I was using early was not sensitive enough to pick up the secondary reflection.Any mistakes?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 28, 2021, 08:11:31 pm
content: There ya goTim

author: oschonrock
author_rank: Regular Contributor
timestamp: March 02, 2021, 01:14:22 pm
content: I also worked out why I was completely unable to detect the secondary reflection using my "frequency response" technique.That is because the secondary reflection has to travel exactly 2x more the length of the stub, L.So the primary reflection (off the unterminated end) travels 2L, and the secondary reflection travels an extra 2L, ie 4L behind the initial wave from the driver. If we take the case of interest of L = lambda / 4, then the primary reflection is entirely destructive - no inversion due to open end, and phase difference of lambda /2.The secondary reflection however, is inverted when it is reflected off the 50||50=25 Ohms it sees at the oscilloscope. it then travels an extra lambda / 2, so a total phase shift of lamdawithone inversion.ie the secondary relfectionalsointerferes destructively at L = lambda / 4.So all it does is make a "sharper notch", which is exactly what my "complex impedance" type analysis shows for the case with the terminator.QED?So it all ties up. The secondary reflection is "hidden" within the frequency respsonse notch of the primary refection. It adds to it.

author: cdev
author_rank: Super Contributor
timestamp: March 05, 2021, 08:09:26 pm
content: "antenna mode" (Seen in a time nuts post yesterday.)