Strange(ish) behaviour of FRA/Bode -- help needed - Page 1
Forum section: electronics / rf-microwave

author: Professzore
author_rank: Contributor
timestamp: September 27, 2023, 05:16:12 pm
content: Dear OMs!I'm in a deep testing session of a Keysight DSOX1204G. One key factor to buy is atrustable FRA capability, as@nctnicosuggested in a reply:https://www.eevblog.com/forum/testgear/waveform-generator-recommendations-needed-(based-on-verbose-explanation)/msg4892759/#msg4892759.The following four examples drove me a little frustrated, and I feel I miss something and/orI do not know somethingwhich is mandatory for the understanding.DUT: YU1LM's BPF for 20 m, like this one in a different setup:https://wiki.batc.org.uk/images/b/be/2m_BP_FILTER-YU1LM.pdfScope: Keysight DSOX1204G, AWG: GW Instek MFG-2230M.Case 1:Setup as described in this Youtube video:SINE output, 200 mVpp, sweep from 10 MHz to 20 MHz, marker output at 17 MHz (blue trace).The result is a clear BPF response with great attenuation out of the desired 14 MHz center.Seescope_41Case 2:Setup: scope Gen output split to DUT-in and to Channel 2, DUT-out to Channel 1Noise output, 200 mVpp, FFT from 10 MHz to 20 MHz (source CH1), cursor at 18.4 MHz and 14.4 MHz.Peaks as expected.Seescope_48Case 3:Setup: as in Case 2.Gen output set to SINE, 600 mVpp, frequency manually set to the highest response curve on Channel 1 (DUT output), which is around 540 mVpp, 13,97 MHz. aANice curve, negligible attenuation, no distortion.Seescope_45Case 4:Setup: as in Case 2.FRA from 10 MHz to 20 MHz, 200 mVpp. Peaks are around 12 MHz and 16 MHz, with serious (nearly 20 dB) attenuation at 14 MHz.This is themoment.Seescope_44M'kay... The problem is -- and at this point I beg someone to give me a rope -- cases 1, 2, and 3 are all absolute measurements (with different scales, but this doesn't make any sense here), all show a fairly good response curve. In case 4, FRA (Bode) analysis is about a ratio between an input signal and an output signal. It does mean that we can see two peaks on FRA because the attenuation is the smallest on these points (output relative to input), which peaks are possibly produced by the input (/reverse) attenuation of the BPF. In other words, the input signal (channel 2, green) is attenuated, so the ratio between input and output became much higher than near the "peak" frequency of the BPF. Just see the same screenshot of Case 3 with a different frequency:Seescope_46So, there must be something I don't know, even can't think about. So I can bet the the possible cause of this frustrating situation (I can't make measurements properly on a BPF), but I don't know why, and I have no any idea how to make it properly.Please,describe this phenomenon, and suggest a new setup/measurement procedure to reach my goal.Thanks in advance!73, DX de HA5CSTP.s.: sorry for the terrible attachments...

author: RFDx
author_rank: Regular Contributor
timestamp: September 28, 2023, 03:12:56 am
content: Quote from: Professzore on September 27, 2023, 05:16:12 pmA lot of relevant information is missing here. The ripple in the passband of the filter is very high that's why I wonder if the 20m BPF is terminated in 50 Ohm? What kind of probes (1:1, 10:1) are you using? Schematic of the BPF? FRA configuration (Input CH2, output CH1) on the scope OK?A simple, manual sweep between 10 and 20MHz is straightforward, do that first: Waveform generator, 50 Ohm cable, BPF, 50 Ohm cable and 50 Ohm termination at the input of the scope. The amplitude response will be your reference for comparison with the automated FRA.

author: Professzore
author_rank: Contributor
timestamp: September 28, 2023, 06:50:37 pm
content: Quote from: RFDx on September 28, 2023, 03:12:56 amHi,Thanks for spending some precious time with my problem. I highly appreciate that.The second part isdone. The result is the same as seen inCase 1. There is no minor difference, though, response curve is less “peaky”.Missing information:Indeed there werenotermination, by the way, the “output load” was set to High-Z. Just trying with termination, see below.FRA setup is OK. Always checked twice.Probe setting is set to 1:10, if the probe is set to 1:10 (reference signal split from generator output), otherwise set correctly (1:1, as all other “probes” are 50 Ohm SMA-to-BNC coaxial cables).BPF schematics is slightly re-designed by K1FM, the original, very verbose documentation is available here:https://dokumen.tips/download/link/input-hf-6m-bp-lp-filters-for-sdr-transceivers-yu1lm.html. The actual board and the component values are like this:https://oshpark.com/profiles/k1fm(QRP BPF, on the bottom of the page).Actual schematics attached.Experiences with 50 Ohm termination.In short:no significant change.First peak is still around 12 MHz (+4 dB), the second peak (smaller) is around 16 MHz (–0.23 dB), 14 MHz "center" (–0.96 dB).Also tried with a completely different BPF (tunable, much wider response curve, less attenuation, also for 20 m). Now is way off (tuned improperly), though, it shows an easily recognizable peak on the “target” freq., on any of the above-mentioned measurement cases. In other words: the results are within the acceptable difference produced by the difference of the measuring protocols. So the "Case 4" setup seems to be valid, the only remaining question is:why?P.s.: as far as I know, Keysight is investigating this phenomenon (or soon will).Again, thanks for your reply.73 de HA5CST

author: Professzore
author_rank: Contributor
timestamp: September 28, 2023, 06:58:05 pm
content: Hi,And as an extra, two more screenshots.Input: 60 mVpp sweep. 10 to 20 MHz (_50.png) and 1 to 20 MHz (_51.png).Green: input (AWG output).Yellow: DUT output.Input signal is severely attenuated at around 12 MHz (relative gain to the output), output signal seems to be a “fair” BPF response curve.I am totally lost.

author: Professzore
author_rank: Contributor
timestamp: September 28, 2023, 08:46:11 pm
content: PossiblySOLVED.More details during the weekend.

author: RFDx
author_rank: Regular Contributor
timestamp: September 28, 2023, 11:56:56 pm
content: Quote from: Professzore on September 28, 2023, 06:50:37 pmI was asking if the output of the filter is terminated with 50 Ohm, not what the output of the generator is switched to (High-Z / 50 Ohm).Quote from: Professzore on September 28, 2023, 06:50:37 pmSee attachement"A - FRA Problem". What you have to do to get the real frequency response of the filter is to measure the ratio V_CH1 / V0. V0 would be yourconstantreference voltage. The problem is you don't have access to V0. What you are really doing is measuring the ratio V_CH1 / V_CH2. V_CH2, the voltage at the input of the filter, is now your new reference voltage. V_CH2 isnot constant, the amplitude changes with frequency which is why you see two peaks in the FRA.Because the filter doesn't behave like a steady 50 Ohm load, you need to isolate the reference voltage V_CH2 from the input of the BP-filter. Attachement"B- FRA Solution"shows a simple, passive solution with four additional resistors (50 Ohm Owen splitter) that achieves sufficient isolation. The FRA is now almost identical to the real frequency response of the filter. An (active) splitter with very high isolation and a scope with 50 Ohm inputs would improve the measurement setup further.

author: Professzore
author_rank: Contributor
timestamp: September 29, 2023, 05:28:35 pm
content: Hi,Thank you very muchfor your verbose explanation.On the other side, at a Hungarian forum,we found another possible reason. Possibly the sum of these two is the source of this “strange” result. The front of this BPF (L1 and C1) acts like an oscillator, and it seriously affects the apparent impedance of the filter. This change is shunting the incoming signal at around 13 MHz.All other part of your explanation is clear for the current level of understanding, andalso very motivating to learn even more about this topic.Anyway, both the filter and the measurement setup are far from final. Despite the quality, features, and possibilities of these instruments, the project is more like anexperiment, than a real working scenario.May I ask you which software/service did you use to run these simulations?Thanks again,73 de HA5CST

author: RFDx
author_rank: Regular Contributor
timestamp: September 30, 2023, 06:51:00 am
content: Quote from: Professzore on September 29, 2023, 05:28:35 pmI assure you there is no oscillator or any kind of oscillations involved. To make the filter oscillate at one port or the other you would have to undampen the filter by providing externally a negative resistance. That is not possible with the current measurement setup.The cause for the peaks in the FRA is clear and simple. The 50 Ohm input impedance of the filter is confined to the middle (15MHz) of the passband where return loss and SWR are excellent. At the corners (13.38MHz / 17MHz) of the passband the impedance already decreased to 24 Ohm (see attached screenshots). This means the load and therefore the output voltage of the AWG changes with frequency. Using this voltage directly as a reference voltage for CH2 is flawed and inevitably leads to the peaks (ratio CH1/CH2 exceeds 1) you observed.As shown, the solution is also simple: Stabilize the voltage into the reference channel CH2 by providing some degree of isolation between the input of the filter and CH2. With a dual channel AWG you could synchronize both channels, use the first channel to drive the filter and the second channel as a reference source for the FRA.Quote from: Professzore on September 29, 2023, 05:28:35 pmNI Multisim. LTspice is a similar tool, is free and very popular.

author: Professzore
author_rank: Contributor
timestamp: October 21, 2023, 04:33:05 pm
content: Thank you very much! I highly appreciate all your efforts to clarify thisAll clear. Also my short future for new learning is well planned.73!