Line loss under standing waves – recommendation of dodgy tool on eHam

In a discussion about using a 40m centre fed half wave dipole on 80m, the matter of feed line loss came up and online expert KM1H offered:

Use this to help make up your mind. Add it to the normal coax loss. http://www.csgnetwork.com/vswrlosscalc.html

This is to suggest that the feed line loss under standing waves can be calculated with that calculator.

He then berates and demeans a participant for commenting on his recommendation, bluster is par for the course in these venues.

Calculator analysis

The calculator in question states this calculator is designed to give the efficiency loss of a given antenna, based on the input of VSWR (voltage standing wave ratio) and other subsequent factors.

This is a bit wishy washy, efficiency loss is not very clear. The usual meaning of efficiency is PowerOut/PowerIn, and the usual meaning of loss is PowerIn/PowerOut, both can be expresssed in dB: LossdB=10*log(Loss) and EfficiencydB=10*log(Efficiency). Continue reading Line loss under standing waves – recommendation of dodgy tool on eHam

The sign of reactance – challenge reality check

The sign of reactance – a challenge posed a problem, a set of R,|X| data taken with an analyser of a quite simple network and asked readers to solve the sign of X over the range, ie to transform R,|X| to  R,X.

The sign of reactance – challenge solution gave a solution to the challenge, and The sign of reactance – challenge discussion provided some discussion about the problem and solution.

Some correspondents have asserted that the challenge (see above Smith chart) contains a response that is contrived for the purpose and not representative of real world antenna systems. Continue reading The sign of reactance – challenge reality check

The sign of reactance – challenge discussion

The sign of reactance – a challenge posed a problem, a set of R,|X| data taken with an analyser of a quite simple network and asked readers to solve the sign of X over the range, ie to transform R,|X| to  R,X.

It is widely held that this is a trivial matter, and lots of software / firmware implement algorithms that fail on some scenarios. Though the scenario posed was designed to be a small set that provides a challenging problem, it is not purely theoretical, the characteristics of the data occur commonly in real world problems and the challenge data is derived from measurement of a real network.

Above is a Smith chart plot of the measured data that was transformed to the R,|X| for the challenge. Continue reading The sign of reactance – challenge discussion

The sign of reactance – challenge solution

The sign of reactance – a challenge posed a problem, a set of R,|X| data taken with an analyser of a quite simple network and asked readers to solve the sign of X over the range, ie to transform R,|X| to  R,X.

It is widely held that this is a trivial matter, and lots of software / firmware implement algorithms that fail on some scenarios. Though the scenario posed was designed to be a small set that provides a challenging problem, it is not purely theoretical, the characteristics of the data occur commonly in real world problems and the challenge data is derived from measurement of a real network.

Imported and rendered graphically in ZPlots we have:

The network measured is comprised from analyser, a 2.8m length of RG58/CU, a tee piece feeding a 50 resistor on one branch and on the other branch, another 2.8m length of RG58/CU with a 4.7Ω resistor termination.

The challenge is: what is the sign of X across the frequency range? Continue reading The sign of reactance – challenge solution

The sign of reactance – a challenge

Over time, readers of The sign of reactance have suggested that determining the sign of reactance with an antenna analyser that does not directly measure the sign is not all that difficult, even for beginners. The article shoots down some of the most common algorithms as failures on simple cases.

This article gives measurements made from a simple network of two identical lengths of 50Ω coax, a 50Ω resistor and a 4.7Ω resistor. It is a network designed to offer a challenge to the simple algorithms, and it IS solvable analytically… but not with most algorithms and software,

Here is the data from measurement made with an AA-600 and then all – signs removed, so in fact the Xs column is |Xs|.

"Zplots file generated by AntScope"
"Freq(MHz)","Rs","Xs"
9.000000,78.13,53.66
9.250000,82.12,51.10
9.500000,86.10,47.83
9.750000,89.46,44.00
10.000000,92.30,39.90
10.250000,94.53,35.39
10.500000,96.21,30.71
10.750000,97.17,26.14
11.000000,97.49,21.54
11.250000,97.30,17.12
11.500000,96.54,13.04
11.750000,95.47,9.14
12.000000,93.92,5.68
12.250000,92.16,2.70
12.500000,90.25,0.17
12.750000,88.13,2.50
13.000000,85.94,4.50
13.250000,83.67,6.15
13.500000,81.45,7.36
13.750000,79.29,8.38
14.000000,77.22,9.21
14.250000,75.21,9.78
14.500000,73.23,10.16
14.750000,71.44,10.37
15.000000,69.70,10.25
15.250000,67.99,10.23
15.500000,66.50,9.99
15.750000,65.10,9.68
16.000000,63.81,9.27
16.250000,62.65,8.72
16.500000,61.59,8.15
16.750000,60.55,7.54
17.000000,59.69,6.86
17.250000,58.97,6.20
17.500000,58.20,5.43
17.750000,57.66,4.68
18.000000,57.14,3.81
18.250000,56.77,2.98
18.500000,56.47,2.16
18.750000,56.22,1.22
19.000000,56.04,0.38
19.250000,56.07,0.50
19.500000,56.02,1.38
19.750000,56.12,2.29
20.000000,56.41,3.15
20.250000,56.68,4.03
20.500000,57.11,4.86
20.750000,57.51,5.72
21.000000,58.06,6.61
21.250000,58.77,7.45
21.500000,59.54,8.22
21.750000,60.47,8.95
22.000000,61.44,9.75
22.250000,62.52,10.34
22.500000,63.77,10.97
22.750000,65.11,11.55
23.000000,66.56,12.02
23.250000,68.11,12.38
23.500000,69.82,12.64
23.750000,71.75,12.82
24.000000,73.67,12.84
24.250000,75.96,12.67
24.500000,78.12,12.27
24.750000,80.40,11.72
25.000000,83.05,10.69
25.250000,85.56,9.68
25.500000,88.29,8.09
25.750000,90.92,6.21
26.000000,93.63,3.91
26.250000,96.17,1.13
26.500000,98.61,2.16
26.750000,100.68,5.92
27.000000,102.51,10.11
27.250000,103.87,14.90
27.500000,104.65,19.98
27.750000,104.71,25.32
28.000000,103.98,30.95
28.250000,102.58,36.48
28.500000,100.14,41.97
28.750000,97.08,47.32
29.000000,93.07,51.86

Imported and rendered graphically in ZPlots we have:

The challenge is what is the sign of X across the frequency range? Continue reading The sign of reactance – a challenge

Is a ham transmitter conjugate matched to its load?

Following on from KL7AJ on the Conjugate Match Theorem, KL7AJ on the Conjugate Match Theorem – analytical solution asked the question Is a ham transmitter conjugate matched to its load?

The answer speaks to the relevance of Walt Maxwell’s Conjugate Mirror proposition to ham stations. Continue reading Is a ham transmitter conjugate matched to its load?

KL7AJ on the Conjugate Match Theorem – analytical solution

KL7AJ on the Conjugate Match Theorem asked the question Should we have expected this outcome?

Let us solve a very similar problem analytically where measurement errors do not contribute to the outcome.

Taking the load impedance to be the same 10.1+j0.2Ω, and calculating for a T match similar to the MFJ-949E (assuming L=26µH, QL=200, and ideal capacitors) we can find a near perfect match.

The capacitors are 177.2 and 92.93pF for the match.

Now turning the network around by swapping the capacitors and changing the load to 50+j0Ω. Continue reading KL7AJ on the Conjugate Match Theorem – analytical solution

KL7AJ on the Conjugate Match Theorem

KL7AJ proposed a little test for his readers on QRZ:

One of the most useful (and sometimes astonishing) principles in radio is the Conjugate Match theorem. In the simplest terms, what this says is that the maximum power will be transferred between a source (like a transmitter) and a load (like an antenna), when the source impedance is the COMPLEX CONJUGATE of the load impedance (or vice versa).
Here’s a neat little experiment to prove the conjugate match theorem. You need four basic ingredients: an antenna analyzer like the MFJ259 (or an actual impedance bridge, if you know how to use one). A good low loss antenna tuner. A good 50 ohm resistor. And a good 200 ohm resistor. And some appropriate connecting hardware, namely some short bits of coax.

Step 1) connect the 50 ohm resistor to the OUTPUT of the antenna tuner. Connect the antenna analyzer to the INPUT of the antenna tuner.

Step 2) Adjust the antenna tuner to get precisely 50 ohms, zero reactance on the antenna analyzer. This step simply confirms everything is working.

Step 3) Replace the 50 ohm resistor with the 200 ohm resistor. Readjust the antenna tuner to get 50 ohms, zero reactance on the antenna analyzer. Do not disturb the antenna tuner adjustments after this point.

Step 4) Remove the 200 ohm resistor and insert the antenna analyzer in its place (at the OUTPUT of the antenna tuner).

Step 5) Insert the 50 ohm resistor at the INPUT of the antenna tuner.

Step 6) Take a careful reading of the antenna analyzer. (What do you think it will say?)

10 points for anyone who will correctly explain why this works.

Some clarifications

Jacobi maximum power transfer theorem

Jacobi published his maximum power transfer theorem in 1840. It states that maximum power is transferred from a (Thevenin) source to a load when the load resistance is equal to the (Thevenin equivalent) source resistance.

It was later adapted to apply to AC circuits with sinusoidal excitation, maximum power is transferred from a (Thevenin) source to a load when the load impedance is the complex conjugate of the (Thevenin equivalent) source impedance.

Walt Maxwell’s Conjugate Mirror

(Maxwell 2001 24.5) states

To expand on this definition, conjugate match means that if in one direction from a junction the impedance has the dimensions R + jX, then in the opposite direction the impedance will have the dimensions R − jX. Further paraphrasing of the theorem, when a conjugate match is accomplished at any of the junctions in the system, any reactance appearing at any junction is canceled by an equal and opposite reactance, which also includes any reactance appearing in the load, such as a non-resonant antenna. This reactance cancellation results in a net system reactance of zero, establishing resonance in the entire system. In this resonant condition the source delivers its maximum available power to the load. …(1)

Note that it states that if a conjugate match is established an any junction, then a conjugate match occurs in any (all) other junctions, simultaneously a conjugate match exists everywhere. Continue reading KL7AJ on the Conjugate Match Theorem

UHF series coaxial connector characteristic impedance

Measurements of Insertion VSWR of UHF series connectors consistently show increasing Insertion VSWR with frequency, an issue that often impacts measurement accuracy.

My own article Exploiting your antenna analyser #12 is but one of many.

Measurements consistently hint that the defect is that the characteristic impedance is typically somewhere between 30 and 40Ω.

Above is a dimensioned drawing from Amphenol (https://www.amphenolrf.com/connectors/uhf.html). Continue reading UHF series coaxial connector characteristic impedance

Review of the Amidon AB_200_10 balun

The Amidon AB_200_10 2-30MHz, 1KW balun and knock-offs have been around for a very long time, I recall Dick Smith selling them in the early 1970s in Australia.

They were regarded as the epitome of the art… but it was not a very well understood art.

Lets analyse the common implementation as a Ruthroff 4:1 voltage balun in a 50:200Ω scenario.

Ruthroff 4:1 voltage balun

In this implementation, Amidon’s instructions show 16 bifilar turns on a T200-2 core.

A very simple model is to consider the device as an ideal transformer with a shunt magnetising impedance equal to the impedance of the 16t winding that appears across the 50Ω terminals. This has its greatest effect at low frequencies and although it is specified from 2-30MHz, lets analyse it at 3.5MHz.

The powdered iron core has very low loss at 3.5MHz, sufficiently so that we can ignore the imaginary component of µr for this analysis and take µr to be 10+j0.

Above is a calculation of the magnetising impedance and admittance under those assumptions. The magnetising admittance (0.00-j0.0134S) appears in shunt with the transformed load admittance (0.02S) so we can simply add them to find the admittance seen by the transmitter (0.02-j0.0134S). Continue reading Review of the Amidon AB_200_10 balun