Take a look at the antenna with a VNA and sweep with the Phase function.
Let’s do that!
There are lots of competing firmwares for the nanoVNA, and having tried many and found them wanting, I use the latest firmware from ttrftech, the ‘originator’ of the nanoVNA. So, my comments are in the context of that firmware.
Let’s look at display of the magic phase quantity with a very good load on the nanoVNA, you might think of this load as the ultimate goal of an antenna system.
Above is a screenshot of my nanoVNA where I have selected the ONLY display format labelled phase, and it can be seen that the yellow trace appears to be quite random.
Above is a screenshot of nanoVNA Saver which seems the preferred PC client of the masses, again the same good load is attached. The upper left plot is the ONLY phase plot derived from s11, again it is quite random. Also show are plots of impedance (which is very good), VSWR (which is very good), and Return Loss (which is very good). Return Loss might look noisy, and it is, but it is always greater than 65dB… excellent! The only plot that has NO VALUE in this case is the phase plot!
In fact, when the magnitude of s11 becomes very small, the phase of s11 becomes dominated my measurement noise and it worthless. Yes, the closer you approach the Nirvana of VSWR=1 (ReturnLoss very high, s11 very low), the less value in the phase of s11.
Let’s look at a sweep of a real antenna, a 5/8λ 144MHz vertical on my car, looking into 4m of RG58 feedline.
On this chart, the easiest curve to interpret for most hams is the VSWR curve (magenta). The markers show its minimum (1.09) and the VSWR=1.5 bandwidth (145.05150.05MHz)… this is a good antenna from that point of view… but it could be improved by lengthening a little to move the frequency for minimum VSWR down to 147MHz (… but there is no adjustment left).
So, look at the Return Loss blue curve. Return Loss is related to VSWR and you could make exactly the same conclusions. We should accept Return Loss > 15dB.
Look at the s11 phase curve in red. It does not cross the zero phase line (the middle of the chart, in this sweep, it is 44° at minimum VSWR even though it is sometimes less at higher VSWR. Can you make any rational conclusion from the phase curve, and does the fact it is not zero condemn the antenna system?
Look at the R and X curves, green and black. Can you draw any conclusions from them directly? Can you see where the phase of R+jX would be zero? Hint: it is where X is zero… but hey, that doesn’t happen with this antenna system.
After all that information overload, the VSWR curve is the key performance indicator, and I could have used an ordinary VSWR meter to come to the same conclusions pretty much.
Yet another example where the focus on s11 phase is so misguided.
]]>Reflection Bridge and Return Loss Bridge are somewhat synonymous, in practice to measure Return Loss one is interested in the magnitude of the response, and to measure the complex reflection coefficient or s11, both magnitude and phase are of interest.
Above is Oristopo’s graph.
We can create a model of a Return Loss Bridge or Reflection Bridge in Simsmith and plot its response for swept Zu.
Above is the Simsmith model with plot for R swept from 12500Ω and X=0Ω (for simplicity). When plotted on a log frequency scale, the s11 response is symmetric, and the markers at 50/10 and 50*10 both produce s11=1.73dB.
Note that this simulated bridge complies with ALL the requirements for correct response:
For convenience, the source power is defined as 16W so that it produces 1W or 0dBW at the detector when Zu=0 (or ∞).
The calculated s11 (or ReturnLoss) can be calculated easily to verify these two cases using a calculator, or good online calculator. For example using Calculate VSWR and Return Loss from Zload (or Yload or S11) and Zo.
From a point of digitising the s11 response, the challenge is as great for 5Ω as it is for 500Ω. Very low and very high impedances are sensitive to different aspects of the fixture, so it is easy to make a fixture that compromises high or low impedances.
When s11 becomes relatively large (ie approaching 1, or 0dB) as it does for measurement of very high and very low impedances, the ADC resolution becomes an issue, internal noise of the instrument becomes significant, and accurate phase measurement is more difficult, and as a result, measurement accuracy is compromised.
Several recent articles have used measurements of transmission line sections with SC and OC terminations.
Above is an example where at HF, s11 >0.05dB, which is the magnitude of s11 with a load of 17370+j0Ω, or 0.1439Ω. Sure, there is some noise, but the measurements are usable for the purpose at hand.
One wonders if some online experts have condemned high impedance measurements as grossly inaccurate based on their own experience, perhaps with flawed fixtures, maybe they are just quoting another online expert they have read.
Generalised assertions by online experts that VNAs cannot accurately measure impedance above a few hundred ohms are not borne out by careful measurement experience of known DUTs in appropriate fixtures… or they have unreal expectations about the accuracy required for common analyses.
]]>Oristopo gives a diagram and explanation.
Above is his diagram. He gives an expression that he states applies when R1=R3=R4=Rm: im = sqrt(Vf*(Rm – R2)/(12*Rm + 4*R2)).
This is deeply flawed, if R2>Rm the expression results in the square root of a ve number… which might be acceptable in a complex number scenario, but this is a DC circuit.
Nevertheless, let us calculate the current with R2=0 and R2=5 while R1=R3=R4=Rm=50.
We can calculate \(ReturnLoss=20 log10 \frac{0.28867513459481287}{0.26940795304016235}=0.6 \;dB\).
Wrong, the ReturnLoss of a 5Ω load on a 50Ω Return Loss Bridge should be 1.7dB.
So, the circuit / expression does not have the response of a Return Loss Bridge.
That is understandable, the schematic is not that of a Return Loss Bridge. If a Return Loss bridge uses R1=R3=R4=Rm, then its source MUST also have a source impedance Rs where Rs=R1=R3=R4=Rm for an accurate Return Loss response.
Analysis based on the schematic above with Zs=0 is not representative of the Return Loss Bridge used in accurate instruments, and conclusions are not soundly based.
The Return Loss Bridge is a deceptively simple thing… it does take careful attention to all details to obtain accurate results.
In the more general sense of a VNA, the reflection bridge must respond proportionally to the complex reflection coefficient.
Oristopo’s graph would imply that the nanoVNA can measure down to zero ohms but not above a few hundred ohms. The simple fact is that a reflection bridge calibrated for 50Ω returns the same magnitude voltage for 2.5+j0Ω load as for 1000+j0Ω, just the phase is opposite… so if measurement noise is a problem for one, it is likely to be much the same for the other.
Generalised assertions by online experts that VNAs cannot accurately measure impedance above a few hundred ohms are not borne out by careful measurement experience of known DUTs in appropriate fixtures… or they have unreal expectations about the accuracy required for common analyses.
]]>Some unexpected ‘bumps’ on the measured response of a short SC transmission line section were concerning, there was no apparent explanation.
The bump around 80MHz had no obvious explanation, and appeared to be an artifact of the measurement fixture, or the instrument. The s11 values from 70150MHz are suspect.
The expected s11 response can be gleaned from a Simsmith simulation.
The line section was then measured using a VNWA3E calibrated with the same fixture.
Above, no sign of the bump, the response is quite as expected.
My baseline config of ttrftech firmware v0.8 and nanoVNAmod client was calibrated with the same fixture and the same measurement made.
No sign of the bump there either. Of concern though is the low end results, s11 should not be greater than 0dB.
So, same VNA, same fixture, same DUT and different results for the oneofeleven suite, not just different but an unexpected / unexplained bump.
I did create an issue on github re a bunch of other errors with computed ‘info’ values in the application, no response yet, but it is early days.
Some further testing on this issue is detailed at https://github.com/OneOfEleven/NanoVNAH/issues/4 .
Update 15/03/2021: two issues listed at https://github.com/OneOfEleven/NanoVNAH have quietly disappeared and new issues cannot be added, perhaps a sign that the software has gone unsupported.
]]>Above is the schematic of the amplifier, analysis here is of the 25W configuration using a 2n5591.
The figure above shows the details of T1, a Ruthroff 1:4 unun.
The initial question was whether this would work as an air cored structure… but the question seemed motivated by difficulty in getting the amplifier to work properly.
So, let’s review the matching scheme. It is a combination of three components, T1, C4 and C5.
Consulting the datasheet, we see that the recommended load for the 2n5991 for 25W out on 12.5V at 175MHz is 4+j2Ω. That will be a little different at 144MHz due to the transistor capacitance having different susceptance at the lower frequency, but not greatly, it is a good place to start.
As mentioned there are three components in the matching network, but the operation of T1 is far from nominal 1:4, and for a transformation from 4Ω to 16Ω, you would choose a line with Zo=8Ω, that is not practicable, so there will be standing waves on that line section and therefore significant impedance transformation.
Since there is significant impedance transformation on the line, the characteristics of the line become important.
The originally specified #20 (0.81mm) was not on hand but some 0.71mm is available and will be used.
Minimum enamel thickness specified for 0.7mm wire ranges 3080µm, let’s assume the medium covering of 53µm. Average cover may be a little more. The wire measures 0.755mm overall, but that alone does not imply the enamel thickness.
Using TWLLC, we can get a ball part estimate of Zo using a guess of vf=0.7 based on experience.
0.071 ECW twisted pair 

Parameters  
Conductivity  5.800e+7 S/m 
Rel permeability  1.000 
Diameter  0.000710 m 
Spacing  0.000763 m 
Velocity factor  0.700 
Loss tangent  0.000e+0 
Frequency  146.000 MHz 
Twist rate  100 t/m 
Length  1.000 m 
Results  
Zo  33.50j0.68 Ω 
Velocity Factor  0.7000 
Twist factor  0.9725 
So, Zo in the range 3035Ω is likely.
A test section of 255m length was made and measured with SC and OC terminations using a VNWA3E.
Above are the s11 measurements for SC and OC.
From that dataset we can calculate Zo.
Calculation of Zo over most of this range looks ok, it has the typical turn up at low frequencies, and there is a problem measuring close to its quarter wave resonance. Around 150MHz, Zo is around 33Ω, quite close to expectation.
We can also calculate vf.
vf is 0.665 around 150MHz, so the earlier guess was not too far off the mark.
Let’s build a Simsmith model to find a matching solution and explore the sensitivity to component values.
Taking the target load impedance for the source to be 4+j2Ω, we can use Simsmith to model the network and tweak it for a match to a 4+j2Ω generator.
Element D models the Ruthroff 1:4 unun transmission line transformer. Lcm is a calculated value for the common mode inductance of the transmission line section, a two turn solenoid to accommodate the length of the transmission line section.
It is not very convenient to work with a Smith chart with complex reference impedance, as can be seen it warps the Z space.
Instead, let’s add an element so that we can use a purely real Zo.
Above, Z1 offsets the j2 component of the desired network input impedance so that the Zo is 4Ω for a ‘normal’ Smith chart scaling. Z1 is not part of the actual network, but purely a fixup.
Element D models the Ruthroff 1:4 unun transmission line transformer. Lcm is a calculated value for the common mode inductance of the transmission line section, a two turn solenoid to accommodate the length of the transmission line section.
If you follow the impedance changes at each element of the Simsmith model, C2 and D are the most significant (excluding from the dummy Z1). Impedance transformation in D is mostly due to transmission line effects.
Not surprisingly, matching is very sensitive to C2 and length, vf and Zo of the transformer D. As it turns out, the common mode inductance Lcm is not very critical, hence no need for a magnetic core.
ARRL. 1977. The radio amateurs handbook. ARRL p453
]]>This article reports nanoVNA measurement of a two wire line where no common mode countermeasures were taken.
Above is a Smith chart of the complex reflection coefficient Γ (s11) looking into a length of nominally 142Ω transmission line of similar type to that in the reference article, the chart is normalised to Zref=142+j0Ω. Note the locus is a spiral, clockwise with increasing frequency, and centred on the chart prime centre Zref. More correctly it is centred on transmission line Zo, and the keen observer might note that the spirals are offset very slightly downwards, actual Zo is not exactly 142Ω, but 142jXΩ where X is small and frequency dependent, a property of practical lines with loss.
Above is a Smith chart of the same data, but normalised to Zref=50+j0Ω, ie the prime centre is 50+j0Ω. The spirals are offset, but again they are centred on Zo. It might not look centred on Zo, but note that the spiral inwards is not symmetric on this scale, and if the line was long enough, the spirals converge on Zo.
So what does this spiral with Zref=50+j0Ω translate to in a cartesian plot of s11 vs f?
Whilst the value of s11 wrt actual Zo, when mapped under Zref=50+j0Ω (as in the VNA reports), it has a cyclic variation with f.
Now that we know the phenomena to expect of a normal practical (lossy) transmission line, we can review a measurement data set.
The DUT is the same test line documented at Measure transmission line Zo – nanoVNA – PVC speaker twin. It is 1m in length and that article documented nominal Zo≈142Ω and VF≈0.667.
The DUT is directly connected to the nanoVNA Port 1 jack (CH0 in nano speak), no measures have been taken to reduce potential for common mode excitation of the system, and the nanoVNA is connected to the desk computer by a USB cable… all a bit nondescript because the only detail that is important is no common mode counter measures used.
Above is a Smith chart of a scan from 1100MHz. The form is broadly a clockwise spiral from low to high f, but notable deviations at the high and low f end. Beware, it might look like 80% of the locus is a spiral, but it represents more like just 30% of the frequency range.
Let’s look at s11 for a better perspective.
Above is a plot of s11 vs f. Recall that we might expect some smooth cyclic variation in s11 with f, and that accounts for the dip at Marker 1, but the shenanigans around 30MHz are an anomaly cause by common mode excitation. Likewise the range 66100MHz is affected.
This dataset is seriously compromised by common mode excitation of the DUT, the data is worthless and no reliable conclusions can be drawn about the pure differential mode characteristics of the DUT.
Measure transmission line Zo – nanoVNA – PVC speaker twin reported measurements and analysis of the same line section, and it described the common mode countermeasures.
Common mode current gives rise to unquantified radiation loss and affects the input impedance of the line section.
If countermeasures are apparently needed and are not described, the experimenter may be naive and none were used, and the data is questionable, possibly worthless… certainly suspect until the common mode issue can be clarified.
]]>Well, as the article showed, it is not quite the nobrainer but with care, it can give good results. This article documents such a measurement of a 0.314mm cable.
The nanoVNA was carefully SOLT calibrated from 1 to 201MHz. Care includes that connectors are torqued to specification torque… no room here for hand tight, whether or not with some kind of handwheel adapter or surgical rubber tube etc.
Above is the Smith chart view over the frequency range from a little under λ/8 to a little over λ/8. It is as expected, a quite circular arc with no anomalies. Since the DUT is coax, and the connector is tightened to specification torque, we would expected nothing less. The situation may be different with two wire lines if great care is not taken to minimise common mode excitation. The sotware does not show Marker 2 properly, it should be between ‘c’ and ‘i’ of the word Capacitive.
Above is the R,X scan, Marker 2 is at the λ/4 resonance (X≈0) and Marker 1 is exactly half of that frequency, so λ/8 electrical length at which point X=Zo.
From the λ/8 measurement, we can calculate VF=0.314/(1.767384/4)=0.710 which reconciles with the datasheet.
]]>This article shows the use of the Smith chart to look for departures from pure transmission line behavior in that test, or any other that depends on measuring purely Zin of a length of line in purely differential mode with short circuit or open circuit termination.
Above is a Smith chart plot of what we should see looking into a line of similar characteristic swept from 1 to 20MHz. There is no magic there, this is basic transmission lines and Smith chart.
Above is the Smith chart plot of a scan of the NEC model with the USB cable included. A significant anomaly can be seen and the measurement frequency is in the seriously disturbed area, little wonder the test gave wrong results. The whole curve is wrong, and will give erroneous results.
Any significant common mode current gives rise to power lost to radiation, and the measured impedance is not that of the transmission line alone. The Smith chart might not have strong hints as in the above, but it is a first place to look for departure from expected pure transmission line behavior.
You would not see this huge hint of an invalid experimental setup if you focus on the narrow band of frequencies where the electrical length is λ/8.
Above is the NEC Smith chart plot with the USB cable removed, quite as expected.
]]>Apparent gross failures are often wrongly attributed to factors like manufacturing tolerances, polluted line surface, other esoteric factors etc that might imply a knowledgeable author… but that is social media, an unreliable source of information.
Let’s explore an estimate using measurements with a nanoVNA using the popular eighth wavelength (λ/8) method.
The λ/8 method relies upon the property of a lossless line terminated in an open circuit that differential impedance \(Z_d=\jmath X= \jmath \left Z_0 \right cot \left(\pi/4\right)= \jmath\left Z_0 \right\). So, if you measure the reactance looking into the λ/8 (\(\frac{\piᶜ}{4} \:or\: 45°\)), you can estimate Zo as equal to the magnitude of the reactance.
A similar expression can be written for the case of a short circuit termination and it leads to the same result that you can estimate Zo as equal to the magnitude of the reactance (an exercise for the reader).
The fact that the two cases lead to the same result can be used to verify that the line length is in fact λ/8 (they will not be equal if the length is a little different to λ/8)… though writeups rarely mention this, or perform the test.
So, the method depends critically on:
Most online articles do not include details of the measurement setup, perhaps thinking that it not all that relevant. Of course, one of the greatest failings in experiments is to ignore some factor that is in fact relevant.
The nanoVNA is such a limited device without a computer attached, lets model a scenario that might well be used by the naive.
Above is a graphic of the scenario. It comprises a vertical section of the modelled transmission line from height 2 to 7m, so it is 5m in length and comprises 1mm diameter copper conductors spaced 20mm, dielectric is a vacuum. Impedance is measured in the horizontal segment bonding both sides of the transmission line.
We might expect the quarter wave resonance of this part of the scenario alone would be approximately 15MHz, but the NEC model gives a slightly lower frequency of 14.97MHz and therefore λ/8 is 7.485MHz.
Also included is a connection from one side of the transmission line to real ground to represent the ground connection of the nanoVNA via its USB cable to a computer used to capture the measurement results.
Above is the impedance plot from the model, it looks well behaved and we might suggest that we would expect that the nanoVNA would measure Zin=Rin+jXin=92.5j195Ω or Z=215Ω.
If you not into the ‘j value’ stuff, you might then say that Zo=215Ω. If you were a little more savvy, you might say that Zo=X=195Ω. Not a whole lot of difference… but the difference should be concerning. In fact, the relatively large value of R in Z should sound a warning that the naive might overlook.
Let’s look at a simple model of the transmission line using TWLLC.
Parameters  
Conductivity  5.800e+7 S/m 
Rel permeability  1.000 
Diameter  0.001000 m 
Spacing  0.020000 m 
Velocity factor  1.000 
Loss tangent  0.000e+0 
Frequency  7.485 MHz 
Twist rate  0 t/m 
Length  0.125 wl 
Zload  1.000e+100+j0.000e+0 Ω 
Yload  0.000000+j0.000000 S 
Results  
Zo  444.04j1.48 Ω 
Velocity Factor  1.0000 
Length  45.000 °, 0.125000 λ, 5.006554 m, 1.670e+4 ps 
Line Loss (matched)  2.28e2 dB 
Line Loss  >100 dB 
Efficiency  ~0 % 
Zin  8.486e1j4.418e+2 Ω 
Above is an extract of the output.
Key calculated results are:
Note that:
By comparison with the TLLC prediction, a key point of reconciliation failure is that in the measurement Rin is relatively quite large and quite inconsistent with even a low loss line, and so the premise for applying the λ/8 method vanishes.
Let’s model a better DUT,
Above is a graphic of the scenario. It comprises a vertical section of the modelled transmission line from height 2 to 7m, so it is 5m in length and comprises 1mm diameter copper conductors spaced 20mm, dielectric is a vacuum. Impedance is measured in the horizontal segment bonding both sides of the transmission line.
We might expect the quarter wave resonance of this part of the scenario alone would be approximately 15MHz, but the NEC model gives a slightly lower frequency of 14.97MHz and therefore λ/8 is 7.485MHz.
There is no connection to ground, the instrument and the transmission line section are relatively isolated from ground.
Above is the impedance plot from the model, it looks well behaved and we might suggest that we would expect that the nanoVNA would measure Zin=Zoc=Rin+jXin=0.879j458Ω.
As a simple check, Rin is relatively small, and we might accept that Zo=Xin=458Ω… but this value is sensitive to the electrical length.
Running the same model with a short circuit termination, Zsc=3.98234+j463.85Ω.
We can calculate Zo=Ro+jXo from Zoc and Zsc.
As you see, Ro lies between the Xoc and Xsc.
This calculation if of higher accuracy than basing Zo on Xoc or Xsc alone of an assumed λ/8 section.
A simpler approximation for low loss line is \(Z_o\approx \sqrt{X_{oc} X_{sc}}=460.8\) which reduces sensitivity to actual length.
Calculated Zo based on the second NEC model does not reconcile exactly with the TWLLC calculation as they use different methods of modelling, but the results are reasonably close.
What’s the problem was asked. The problem is that the λ/8 method depends on a valid differential mode impedance measurement, and that did not happen in the first example, the common mode current path prevented pure differential mode excitation of the DUT.
Relationship between angle of reflection coefficient and angle of impedance
It was stated above that the angle (or phase) of s11 or Γ is not the same as the angle (or phase) of Z.
Given Zo and Γ, we can find θ, the angle of Z.
\(
Z=Z_0\frac{1+\Gamma}{1\Gamma}\)Zo and Γ are complex values, so we will separate them into the modulus and angle.
\(
\left  Z \right  \angle \theta =\left  Z_0 \right  \angle \psi \frac{1+\left \Gamma \right  \angle \phi}{1\left \Gamma \right  \angle \phi} \\
\theta =arg \left ( \left  Z_0 \right  \angle \psi \frac{1+\left \Gamma \right  \angle \phi}{1\left \Gamma \right  \angle \phi} \right )\)We can see that the θ, the angle of Z, is not simply equal to φ, the angle of Γ, but is a function of four variables: \(\left  Z_0 \right , \psi , \left \Gamma \right , \& \: \phi\) .
It is true that if ψ=0 and φ=0 that θ=0, but that does not imply a wider simple equality. This particular combination is sometimes convenient, particularly when ψ=0 as if often the case with a VNA.
This article offers a simulation of a load similar to a 7MHz half wave dipole.
The load comprises L, L1, and C1 and the phase of s11 (or Γ) and phase of Z (seen at the source G) are plotted, along with VSWR.
Firstly, note that the two phase plots are very different, but in this case they cross over at phase=0 at 7.077MHz.
Secondly, note that even thought both phases are zero at 7.077MHz, the VSWR is 1.20. Neither phase demonstrates the best conditions for least feed line loss, minimum VSWR is slightly lower at 7.099MHz.
Maximum power in the load coincides with minimum VSWR at 7.099MHz.
Beware of claims that phase (of something) is the optimisation target, the author probably doesn’t really understand this stuff.
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