W0QE’s video #80: High Power Balun with #31 Ferrite Material gives some measurements and simulations of a FT240-31 inductor with 11 and 14 turns.

In the video he states:

It turns out that the heating effects in the coil are related to the voltage across the coil only, not the current through the it or anything else.

In fact, there is current flowing through the inductor and that develops a voltage difference across the ends. When we are talking about the self inductance properties, then we are talking about the voltage induced in the inductor as a direct result of the current flowing through the inductance.

Let’s look at his own figures to demonstrate,

Above is his Simsmith model. Let us focus on just the left hand two elements L and R1 (for the 11t inductor) as it is a quite complicated model. L was derived from a measurement of the inductor in a fixture, and to some extent the fixture is captured.

The voltage applied to both L and R is given as 272.2V, this is around the voltage that would be applied to a 50+j0Ω load at 1500W, so his experiment to to simulate the conditions if that inductor was the 50Ω winding of a 50Ω transformer at 1500W.

The impedance of R1 is given as 4134-j328.1Ω (interpolated possibly) from his measurement of Z.

From V and Z, we can calculate \(Y=\frac1Z\) and \(I=\frac{V}Z\), and powers \(P_i=I^2R\) and \(P_v=V^2G\) (R and G being the real parts of Z and Y respectively).

Let’s use Python as a complex number calculator to make the maths easy.

>>> import math >>> import cmath >>> z=4134-328.1j >>> y=1/z >>> y (0.00024038230052189223+1.907823725235434e-05j) >>> v=272.2 >>> v 272.2 >>> p50=v**2/50 >>> p50 1481.8568 >>> pv=v**2*y.real >>> pv 17.810607331400476 >>> i=v/z >>> i (0.06543206220205906+0.005193096180090851j) >>> pi=abs(i)**2*z.real >>> pi 17.81060733140048

We can see there is current flowing, and that calculated pv and pi reconcile (and they reconcile with Simsmith)… so the power dissipated (ie converted to heat) in the inductor can be calculated from current or voltage, his statement quoted earlier is plainly wrong.

So, if either current or voltage can be used, is one better?

This is more a question of whether voltage or current can be measured reasonably accurately and conveniently.

Whilst in W0QE’s test configuration it is fairly easy to make a valid voltage measurement of the inductor shunted by the 50Ω dummy load at the T (the test was designed for that), measurement of a ‘floating’ common mode choke is not so easy as the instrument may significantly disturb the thing being measured (it doesn’t have the 50Ω shunt load, circuit impedance is now much higher and the effect of the instrument leads etc is much greater).

A better approach for the common mode choke might be to use a clamp on RF ammeter and use measured Z of the choke to obtain \(P_{choke}=I^2 \mathbb{R}(Z_{choke})\) or \(P_{choke}=I^2 Real(Z_{choke})\).

I make the observation that many hams write about common mode current, but it is rare to see valid measurement of common mode current.

Another question that arises is the accuracy of measurement of Z. In this instance, the choke in fixture is very close to self resonant (evidenced by the smallish magnitude of X) and the real part of Z is quite sensitive to layout and shunt capacitance such as the fixture… giving rise to significant uncertainty or R around the resonant frequency. You might question whether the fixture actually represents the intended deployment scenario.

He also calculates the expected flux level as 50.7 gauss and mutters that they are rated for 57 gauss… yes, rated for 57 gauss

.

Above is from the Fair-rite datasheet for #31 material. It is easy to see that saturation flux density is of the order of 1000 gauss (0.1T), so operating at 50 gauss is way way below saturation. Like many if not most ferrite ham applications at HF, loss becomes a problem way before saturation.

Nothing in the Fair-rite #31 datasheet relates core loss to flux density.

I could not see where he used this magic 57 or 50.7 gauss in any power calculation, all of his power calculation is based on a small signal measurement of the inductor magnetising impedance. 57 gauss may have been sourced from Amidon… but keep in mind that Amidon is not a manufacturer, Fair-rite is a manufacturer. So much for the claim that loss is dependent on voltage and not current.

In this application, the power dissipated in an inductor can be calculated from Z and either current through it, or voltage across it.

If you want to estimate the core loss in a common mode choke, carefully measure the common mode impedance and the common mode current and calculate the core loss.

Uncertainty in Z can be significant, especially near the inductor’s self resonant frequency.

]]>Is this Segal’s law at play?

There are several common contributors including:

- faulty, dirty, or not properly mated connectors and cables;
- VSWR meters that are not accurate at low power levels; and
- influence of the common mode current path on VSWR.

The first and obvious question is, are all the cables and connectors sound and properly tightened? Some types of connector (eg UHF series, SMA) depend on tightness of the screw ring / nut for proper electrical contact of the outer conductor. An aerosol can of Isopropyl Alcohol (IPA) and some cleaning swabs / brushes are very handy to ensure connectors are clean.

Should you trust your VSWR meter – detector linearity discusses the second issue.

To the third issue mentioned, if you truly want to compare the two instruments, you must measure EXACTLY the same thing… and that means that when you disconnect the coax from the back of the radio and connect it to the analyser, you must RESTORE the common mode current path. I usually do this by holding the analyser coax connector (with antenna cable attached) firmly against the transceiver connector outer threads, and preferably isolating my fingers from the metal using an insulating sheet.

Above, an example of the nanoVNA with minimal adapters to UHF series, antenna patch lead attached and the shell of the connector held in good contact with the transceiver connector at far left. The white sheet is a silicone sheet to insulate my hand from the other stuff so that it is as close to the operating configuration as reasonably possible.

In this case, the minimum VSWR reconciles very well with the IC-7300 VSWR meter, provided the tx power is more than 60W.

]]>Section 1-36 states explicitly that it is applicable to Iron Powder and Ferrite, which is interesting because they are very different materials from a loss point of view.

Basically, their method depends on a maximum safe value for peak flux density.

They give an expression for peak flux density \(B_{max}=\frac{10^8E }{4.44 A_e N F}\) and the following table of design limits for Bmax.

Note that the table and formula are independent of ferrite mix type (though they do mention that “these figures may vary slightly according to the type of material being used.”

So, lets work some examples being a 3t primary at 7MHz on FT240 cores of #43 and #61 material. The two material types are very popular for HF RF inductors and transformers, and were chosen for that reason rather than designing an example that shows extreme difference.

Using VK1SV’s online calculator of Amidon’s method, we get…

By trial and error, it is found that 83V is just ‘safe’ for this transformer, irrespective of material.

At around 60gauss (0.006T), flux density is way way below saturation (>1000gauss (0.1T))

Now lets do a permeability based calculation of magnetising admittance and then power lost in the core Pcore at 83V applied for the #43 core.

\(P_{core}=V^2 G_{core}=83^2 \cdot 0.00232=16.0 \; W\) 16W is on the high side for this core in free air, more so if it is enclosed.

Note, this is a small signal analysis, but as explained, flux density is way way below saturation.

Now lets do a permeability based calculation of magnetising admittance and then power lost in the core Pcore at 83V applied for the #61 core. The Amidon method is probably unsafe.

\(P_{core}=V^2 G_{core}=83^2 \cdot 0.000147=1.013 \; W\) 1W is on the low side for this core in free air, it can probably dissipate more like 10W in free air, somewhat less enclosed. The Amidon method is unduly safe.

There is a huge difference, the core loss with #43 material is 1600% of that with #61.

Note, this is a small signal analysis, but as explained, flux density is way way below saturation.

The simplest way is to make a measurement of the magnetising admittance (or impedance from which you can calculate admittance).

One method is to put three turns onto each of the cores and sweep in with your calibrated nanoVNA. Find Z from s11 at 7MHz, and invert it to find G and B (Some hammy tools might give you Rp||XP, so G=1/Rp). Then calculate \(P_{core}=V^2 G_{core}=83^2 G_{core} \; W\).

Now keep in mind that the tolerance of ferrite are fairly wide, so your measured results won’t exactly reconcile with the calculations above, but you should find a wide disparity in the core loss for the two materials and question the usefulness of Amidon’s method.

So, before you accept online expert’s advice on either side of this argument… make some measurements and develop trust in the tools you use.

Amidon’s method is manifestly poor. People who recommend it have probably not tested its validity, and both damages their credibility.

For ferrites at RF (and zero DC current), operation is more likely to be limited by dissipation than magnetic saturation, and in that case, the flux density is not usually very interesting.

]]>Let’s work an example using Simsmith to do some of the calculations.

Scenario:

- 20m ground mounted vertical base fed against a 2.4m driven earth electrode @ 0.5MHz;
- 10m RG58A/U coax; and
- Receiver with 500+j0Ω ohms input impedance and Noise Figure 20dB.

An NEC-4.2 model of the antenna gives a feed point impedance of 146-j4714Ω and radiation efficiency of 0.043%, so radiation resistance \(Rr=146 \cdot 0.00043=0.0063\).

Above, the NEC antenna model summary.

Above is a Simsmith model of the system scenario.

R1 and G model the antenna, G uses Rr for Zo, and R1 contains the balance of the feed point impedance.

With the useZo source type, the source would deliver 1W or 0dBW to a conjugate matched load.

The next important figure is the power into the 500Ω load L. it is -58.3dBW. Simsmith has calculated the solution to the antenna loss elements, mismatches and coax loss under standing waves. Effectively, the average gain of the antenna system (everything to the right of L) is -58.3dB. Such an antenna is likely to have a Directivity of around 4dB, in fact the NEC model calculates 4.8dB. So the maximum gain is -58+4.8=-53.2dB.

The burning question is whether it is sufficiently good to hear most signals. Well, a better question is how much does it degrade off-air signal to noise ratio (S/N). All receivers degrade S/N, but how much degradation occurs in this scenario.

We need to think about the ambient noise. Lets use ITU-R P.372 for guidance on the expected median noise in a rural precint.

Above, ambient noise figure @ 0.5MHz is 75.54dB.

Now lets calculate the Signal to Noise Degradation (SND).

At 4.58 dB it is not wonderful, the weakest signals (ie those with low S/N) we be degraded significantly, stronger signals (those with high S/N) will be degraded by the SAME amount, but for instance reducing S/N from 20 to 15dB is not so significant.

Applying this to your own scenario

The information fed into the calculations included:

- Rr;
- feed point impedance;
- transmission line details;
- Rx input impedance and NF; and
- Ambient noise expectation.

To calculate your own scenario, you need to find these quantities with some accuracy.

Tools:

]]>The very popular nanoVNA-H will be used to make the measurements.

The scenario:

- nanoVNA fully calibrated from 1.5-1.8MHz using a 200mm length coax lead on Port 2 (nanoVNA CH1);
- 10m of RG58C/U; and
- f=1.65MHz.

Above is a block diagram of the test configuration. nanoVNA measurements are wrt 50Ω, so \(P=\frac{V^2}{50}\) and \(V=\sqrt{50P}\).

1650000 0.019864771 -5.590068373 0.986123401 -18.626393762 1.0 0.0 1.0 0.0

Above is a record from the .s2p file in MA format.

Above is calculation of Zin=52.07-j0.1722Ω from s11 from the .s2p file. Also calculated is MismatchLoss=0.002dB. This method of calculating MismatchLoss is only correct if either source or load impedance is purely real, which is true in this case.

Converting the magnitude of s21 from the .s2p file, we get |s21|=-0.12dB.

Let’s review some meanings of terms (in the 50Ω matched VNA context):

- \(TransmissionLoss=\frac{PowerIn}{PowerOut}=\frac{P_1}{P_2}\);
- \(s11=\frac{V_{1r}}{V_{1i}}\);
- \(InputMismatchLoss=\frac{P_{1i}}{P_{1}}=\frac1{(1-|s11|^2)}\);
- \(s21=\frac{V_{2i}}{V_{1i}}\); and
- \(InsertionLoss=\frac{P_{1}}{P_{2}} \approx \frac{P_{1i}}{P_{2i}} =\frac1{|s21|^2}\).

It is assumed that Zin of VNA Port 2 is 50+j0Ω, and that therefore P2r=0. Error in Zin of VNA Port 2 flows into the results. A 10dB attenuator is fitted to Port 2 prior to calibration to improve accuracy of Zin.

With the quantities expressed in dB, we can derive that \(TransmissionLoss=-|s21|-InputMismatchLoss\).

In the example given above, \(TransmissionLoss=-|s21|-InputMismatchLoss \\=0.120-0.002=0.12\: dB\).

In the example given above, \(InsertionLoss=-|s21|=0.12\: dB\).

Because this is a nominally matched scenario, reflected power at the input and output ports is very small and InsertionLoss≈TransmissionLoss.

The calculated figure is lower than might be expected from the datasheet, but there are issues with interpolation of loss figures in the transition region.

Zo of the nominally 50Ω cable is not exactly Zo at the test frequency. Indications are that it is around 50.95-j1.16Ω (see above). At this frequency, skin effect is not fully developed and internal impedance of the centre conductor becomes more significant, raising Zo.

- Smith, P. 1995. Electronic applications of the Smith chart 2nd ed. Noble Publishing Tucker.

The Transmission Loss of a line section may not be directly given by any measures displayed by a VNA, it may take some interpretation and some accounting for elements that can be measured.

]]>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) with Simsmith we can find a near perfect match.

The capacitors are 177.2 and 92.9pF for the match.

Also calculated is the impedance looking back from the load to the source shown here as L_revZ. The impedance looking back towards the 50Ω load is 17.28-j0.6216Ω, which is quite close to the value obtained by measurement, 18.0-j0.8Ω (which is dependent on the actual Q of the ATU elements).

Is there some smoke and mirrors in calculation of L_revZ? Lets turn the network around.

Now turning the network around by swapping the capacitors and changing the load to 50+j0Ω.

Above, the impedance looking back towards the 50Ω load is 17.28-j0.62Ω, which consistent with the L_revZ calculation and is quite close to the value obtained by measurement, 18.0-j0.8Ω (which is dependent on the actual Q of the ATU elements).

So, in answer to the question Should we have expected this outcome?

, the answer is yes, it is not surprising and quite similar to what we might expect from a network of this type.

Walt Maxwell’s Conjugate Mirror (Maxwell 2001 24.5) which imbues a magic system wide conjugate match with certain benefits, a utopia, which does not apply to systems that include any loss, it does not apply to real world systems. Maxwell does not state that limitation of his proposition.

Is a ham transmitter conjugate matched to its load? Watch for a follow up post.

- KL7AJ on the Conjugate Match Theorem
- Maxwell, Walter M. 2001. Reflections II. Sacramento: Worldradio books.
- Kl7ajConjugateMatch Simsmith models

]]>

The very popular nanoVNA-H will be used to make the measurements.

The scenario:

- nanoVNA fully calibrated from 1-5MHz using a 200mm length coax lead on Port 2 (nanoVNA CH1);
- 35m of CCS RG6/U (close to an electrical quarter wavelength);
- 75-50Ω Minimum Loss Pad (5.72dB); and
- f=1.65MHz (close to a quarter wavelength.

Above is a block diagram of the test configuration. nanoVNA measurements are wrt 50Ω, so \(P=\frac{V^2}{50}\) and \(V=\sqrt{50P}\).

1650000 0.252864092 -14.895982563 0.447711895 -85.899042128 1.0 0.0 1.0 0.0

Above is a record from the .s2p file in MA format.

Above is calculation of Zin=81.37-j11.30Ω from s11 from the .s2p file. Also calculated is MismatchLoss=0.287dB. This method of calculating MismatchLoss is only correct if either source or load impedance is purely real, which is true in this case.

Converting the magnitude of s21 from the .s2p file, we get |s21|=-6.98dB.

Let’s review some meanings of terms (in the 50Ω VNA context):

- \(TransmissionLoss=\frac{PowerIn}{PowerOut}=\frac{P_1}{P_2}\);
- \(s11=\frac{V_{1r}}{V_{1i}}\);
- \(InputMismatchLoss=\frac{P_{1i}}{P_{1}}=\frac1{(1-|s11|^2)}\); and
- \(s21=\frac{V_{2i}}{V_{1i}}\).

It is assumed that Zin of VNA Port 2 is 50+j0Ω, and that therefore P2r=0. Error in Zin of VNA Port 2 flows into the results. A 10dB attenuator is fitted to Port 2 prior to calibration to improve accuracy of Zin.

With the quantities expressed in dB, we can derive that \(TransmissionLoss=-|s21|-InputMismatchLoss\).

In the example given above, \(TransmissionLoss=-|s21|-InputMismatchLoss \\=6.98-0.286=6.69\: dB\).

The InsertionLoss of the Minimum Loss Pad is 5.7dB, and that must be deducted from the total TransmissionLoss to find the TransmissionLoss of the coax, \(TransmissionLoss_{coax}=TransmissionLoss_{total}-InsertionLoss_{MLP} \\=6.69-5.72=0.97\: dB\)

Note that this is not exactly the Matched Line Loss, the 75Ω line is not perfectly 75Ω, it was terminated by the nominal 75-50Ω Minimum Loss Pad and 50Ω VNA port so there is small standing wave. Nevertheless, MLL will be very close to 0.97dB or 0.028dB/m.

A measurement of Rin of a resonant section very slightly higher in frequency allows calculation of MLL.

Above, calculated MLL assuming Zo is around 78Ω at this frequency (a consequence of increased internal impedance of the centre conductor due to poor skin effect).

You might be tempted to apply (Smith 1995)’s expression for loss due to standing waves \(\frac{Loss_{mismatched}}{Loss_{matched}}=\frac{1+S^2}{2S}\) but this scenario does not satisfy his conditions for it to apply, a much misused expression.

- Smith, P. 1995. Electronic applications of the Smith chart 2nd ed. Noble Publishing Tucker.

The Transmission Loss of a line section may not be directly given by any measures displayed by a VNA, it may take some interpretation and some accounting for elements that can be measured.

]]>The very popular nanoVNA-H will be used to make the measurements.

The scenario:

- nanoVNA fully calibrated from 1-5MHz using a 200mm length coax lead on Port 2 (nanoVNA CH1);
- 35m of CCS RG6/U (close to an electrical quarter wavelength);
- three 50Ω terminations in shunt with VNA Port 2; and
- f=1.65MHz (close to a quarter wavelength.

The transmission line load is four 50Ω loads in parallel, one of them being VNA Port 2. Only one quarter of the output power is captured by the VNA, so there is effectively a loss of 6.02dB in that configuration. It also delivers a 12.5+j0Ω load the the transmission lines, VSWR is about 6. Note this power division is based on the assumption that Zin of Port 2 is 50+j0Ω, and error in Zin flows into the result. A 10dB attenuator is fitted to Port 2 prior to calibration to improve accuracy of Zin.

Above is a block diagram of the test configuration. nanoVNA measurements are wrt 50Ω, so \(P=\frac{V^2}{50}\) and \(V=\sqrt{50P}\).

1650000 0.710496228 1.456050447 0.255798404 -84.390357361 1.0 0.0 1.0 0.0

Above is a record from the .s2p file in MA format.

Above is calculation of Zin=293.8+j211.42Ω from s11 from the .s2p file. Also calculated is MismatchLoss=3.052dB. This method of calculating MismatchLoss is only correct if either source or load impedance is purely real, which is true in this case.

Converting the magnitude of s21 from the .s2p file, we get |s21|=-11.84dB.

Let’s review some meanings of terms (in the 50Ω VNA context):

- \(TransmissionLoss=\frac{PowerIn}{PowerOut}=\frac{P_1}{P_2}\);
- \(s11=\frac{V_{1r}}{V_{1i}}\);
- \(InputMismatchLoss=\frac{P_{1i}}{P_{1}}=\frac1{(1-|s11|^2)}\); and
- \(s21=\frac{V_{2i}}{V_{1i}}\).

It is assumed that Zin of VNA Port 2 is 50+j0Ω, and that therefore P2r=0. Error in Zin of VNA Port 2 flows into the results. A 10dB attenuator is fitted to Port 2 prior to calibration to improve accuracy of Zin.

With the quantities expressed in dB, we can derive that \(TransmissionLoss=-|s21|-InputMismatchLoss\).

In the example given above, \(TransmissionLoss=-|s21|-InputMismatchLoss \\=11.84-3.052=8.79\: dB\).

As explained, the 6.02dB loss of the power divider in the load needs to be deducted to obtain the coax TransmissionLoss of 2.77dB.

You might be tempted to apply (Smith 1995)’s expression for loss due to standing waves \(\frac{Loss_{mismatched}}{Loss_{matched}}=\frac{1+S^2}{2S}\) but this scenario does not satisfy his conditions for it to apply, a much misused expression.

- Smith, P. 1995. Electronic applications of the Smith chart 2nd ed. Noble Publishing Tucker.

The Transmission Loss of a line section may not be directly given by any measures displayed by a VNA, it may take some interpretation and some accounting for elements that can be measured.

]]>The very popular nanoVNA-H will be used to make the measurements.

The scenario:

- nanoVNA fully calibrated from 1-5MHz using a 200mm length coax lead on Port 2 (nanoVNA CH1);
- 35m of CCS RG6/U (close to an electrical quarter wavelength); and
- f=1.65MHz (close to a quarter wavelength.

The DUT was swept one way from 1.5-1.8MHz and saved as a .s2p file.

1656000 0.385966661 -6.813504102 0.815628707 -85.534776748 1.0 0.0 1.0 0.0

Above is a record from the .s2p file in MA format.

Above is calculation of Zin=111.2-j11.97Ω from s11 from the .s2p file. Also calculated MismatchLoss=0.701dB. This method of calculating MismatchLoss is only correct if either source or load impedance is purely real, which is true in this case. A more general calculation uses Kurokawa’s expression.

Above, Kurokawa’s 1/(1-|s|^2) is the correct MismatchLoss in the more general case. They reconcile, as we expect.

Converting the magnitude of s21 from the .s2p file, we get |s21|=-1.77dB.

Let’s review some meanings of terms (in the 50Ω VNA context):

- \(TransmissionLoss=\frac{PowerIn}{PowerOut}=\frac{P_1}{P_2}\);
- \(s11=\frac{V_{1r}}{V_{1i}}\);
- \(InputMismatchLoss=\frac{P_{1i}}{P_{1}}=\frac1{(1-|s11|^2)}\); and
- \(s21=\frac{V_{2i}}{V_{1i}}\).

In the example given above, \(TransmissionLoss=-|s21|-InputMismatchLoss \\=1.77-0.701=1.07\: dB\)

Note that this is not the Matched Line Loss, the 75Ω line was terminated by the 50Ω VNA port so there is a standing wave.

You might be tempted to apply (Smith 1995)’s expression for loss due to standing waves \(\frac{Loss_{mismatched}}{Loss_{matched}}=\frac{1+S^2}{2S}\) but this scenario does not satisfy his conditions for it to apply, a much misused expression.

- Smith, P. 1995. Electronic applications of the Smith chart 2nd ed. Noble Publishing Tucker.

This article discusses the quantity V2/V1 for a special case, a 180° transmission line section.

180° transmission line sections are often used as part of a balun for VHF/UHF antennas.

Above is an application of a 180° line, a ‘half wave balun’, the U shaped section is 180° in electrical length.

In this role, the intention is to deliver at the end of the line section a voltage equal in magnitude and opposite in phase to that at the input, fairly independently of the impedance at the end of the section (within reason). Readers will recognise that this is the behavior of a good Voltage Balun.

Let’s look at an example, a 4 element Yagi for 144MHz with a folded dipole feed. The feed point data is derived from a single ended feed, and multiplied by two in the Simsmith model to simulate the impedance of one side of the folded dipole connected to the end of the 180° line section.

Above is a Simsmith model of the feed components for the 180° feed branch. Note the wide variation in Yagi feed point impedance (the magenta curve). Note that 50Ω coax is used (similar to RG/58AU), so there is a significant standing wave on the balun section, this is common practice and it illustrates an important aspect of the line section… that impedance looking in (the blue curve) is fairly similar to the load impedance (the magenta curve). Recall that the line section is only exactly 180° at one frequency, 144.2 MHz in this case. If lower loss coax was used, the two curves would be closer.

The magnitude and phase of V2/V1 are calculated and plotted.

Above, a plot of magnitude and phase of V2/V1. The scales are expanded, the magnitude is almost 1 with little variation and phase approximately 180° with only small variation… especially in the few hundred kHz of typical operating range for SSB.

So, when the two branches are feed in parallel (ie, V1 is identical), the feed point voltages are almost equal in magnitude and opposite in phase even though there might be significant variation in feed point impedance.

This property could be thought of as the voltage forcing property of a half wave

.

We can develop an approximate solution for the value of V2/V1.

Considering the transmission line section to be a two port network, we can say that \(\frac{V_2}{V_1}=\frac1{cosh(\gamma l)+\frac{Z_0}{Z_2} sinh(\gamma l)}\). The result is dominated by \(cosh(\gamma l)\) when Z2 is of the order of Zo. Since in this case \(\gamma l \approx \jmath \pi\), \(cosh(\gamma l) \approx -1\) in the region of interest so we can simplify to \(\frac{V_2}{V_1} \approx -1=1 \angle -180 ^{\circ}

\).

Here is an example calc in Python for RG59A/U at 144MHz. gamma and length are specified in units of wavelength.

>>> import math >>> import cmath >>> z2=113+20j # >>> gamma=3.447e-2+6.283e+0j #from TLLC /wavelength >>> l=0.5 >>> zo=50 >>> gl=gamma*l >>> gl #approx j pi (0.017235+3.1415j) >>> v2v1=1/(cmath.cosh(gl)+zo/z2*cmath.sinh(gl)) >>> v2v1 #approx -1 (-0.992518269569061-0.0013300496437132702j) >>> cmath.polar(v2v1) (0.9925191607522839, -3.1402525786628877) >>> cmath.phase(v2v1)/cmath.pi*180 -179.92321936245702 >>> abs(v2v1) #note this is approximately 1 0.9925191607522839

A half wave of transmission line can be used as a phase inverter, and gives good results even in the presence of moderate standing waves on the line section.

]]>