Antennas – Page 29 – owenduffy.net

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

Last update: 14th September, 2020, 8:05 PM

End fed half wave matching – voltage rating of compensation capacitors

The so-called End Fed Half Wave antenna system has become more popular, particularly in the form of a broadband matching transformer in combination with a wire operated harmonically over perhaps three octaves (eg 7, 14, 21, 28MHz).

The broadband transformer commonly uses a medium µ ferrite toroid core, and a turns ratio of around 8:1. Flux leakage results in less than the ideal n^2 impedance transformation, and a capacitor is often connected in parallel with the 50Ω winding to compensate the transformer response on the higher bands.

David, VK3IL posted EFHW matching unit in which he describes a ferrite cored transformer matching unit that is of a common / popular style.

My EFHW match box. 3:24 turns ration on a FT140-43 toroid with a 150pF capacitor across the input.

Above is David’s pic of his implementation. It is a FT140-43 toroid with 3 and 24t windings and note the 150pF capacitor in shunt with the coax connector.

The article End fed matching – analysis of VK3IL’s measurements gives the following graph showing the effects of compensation for various resistive loads. Continue reading End fed half wave matching – voltage rating of compensation capacitors

Last update: 14th April, 2024, 1:36 PM

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

Last update: 4th December, 2017, 7:02 AM

The sign of reactance – challenge discussion

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

Last update: 24th November, 2017, 9:18 AM

The sign of reactance – challenge solution

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

Last update: 23rd November, 2017, 8:32 PM

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

Last update: 21st November, 2017, 10:46 AM

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?

Last update: 4th November, 2018, 12:56 PM

KL7AJ on the Conjugate Match Theorem – analytical solution – Winsmith

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 – Winsmith

Last update: 9th January, 2021, 6:32 AM

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

Last update: 2nd November, 2017, 7:58 PM

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

Last update: 29th October, 2017, 11:16 AM