# When simple explanations target a simple audience

Modern hams live busy lives and it is difficult to fit everything in to the available time / resources etc. So, there is an appetite for the skinny on some key topics, the inside info that took the wise a long time to learn.

This article discusses one of those articles containing the skinny on VSWR, What is VSWR: Voltage Standing Wave Ratio, it takes only a minute or two to read and there is a six minute video for those who prefer that.

The issues discussed here are common in the ham world explanation of VSWR and analysing them provides a learning opportunity. The video contains the issues mentioned below… and some.

Right up front, eager readers are given a take home message. If something prevents them finishing the article they have learned something they can repeat as pros. So satisfying!

In order to obtain the maximum power transfer from the source to the transmission line, or the transmission line to the load, be it a resistor, an input to another system, or an antenna, the impedance levels must match.

In other words for a 50Ω system the source or signal generator must have a source impedance of 50Ω, the transmission line must be 50Ω and so must the load

Ok, it states clearly and unequivocally that a necessary condition for maximum power transfer it that source must match line and line must match load.

We will test that proposition, but firstly the detailed explanation follows…

Issues arise when power is transferred into the transmission line or feeder and it travels towards the load. If there is a mismatch, i.e. the load impedance does not match that of the transmission line, then it is not possible for all the power to be transferred.

As power cannot disappear, the power that is not transferred into the load has to go somewhere and there it travels back along the transmission line back towards the source.

When this happens the voltages and currents of the forward and reflected waves in the feeder add or subtract at different points along the feeder according to the phases. In this way standing waves are set up.

The astute reader will have noticed that the source in the diagram above is a pure voltage source, no Thevenin source impedance element is shown schematically so it is implicitly zero. The source is NOT matched to the line. The following discussion and analysis works around that and applies for any source impedance.

Taking the first paragraph, then for for all the power to be transferred the line must be lossless, so the analysis below is for a lossless line to mimic the context of the quote.

## A simple test case

Now lets explore a simple case of the configuration in the diagram above where Zo=100Ω, Zload=50Ω, Vload=1V, and the length of the line is an electrical half wavelength.

We will work backwards from the load, because that is how steady state transmission lines are solved.

We can calculate the complex reflection coefficient Γ at the load end of the transmission line, it is -0.333+j0.

We can calculate the forward wave voltage at the load Vfl, Vfl=Vload/(1+Γ)=1/(1-0.333+j0)=1.5V.

We can calculate the reflected wave voltage at the load Vrl, Vrl=Vfl*Γ=1.5*(-0.333+j0)=-0.5V.

We can calculate the currents at the load by dividing Vfl and Vrl by Zo to get If and Ifl, If=0.015A and Irl=-0.005A.

The power at the load Pout=(Vfl+Vrl)*(Ifl-Irl)=(1.5+-0.5)*(0.015–0.005)=0.02W. (The Ir term is negative because Ir flows in the opposite direction to If.)

Now lets calculate the voltages, currents and power at the source end.

Because the line is lossless and exactly a half wave long, the source end voltages and currents are exactly 180° out of phase with the load end (forward wave leads, reflected wave lags, but the result is simply that the sign of all the components in the earlier formulas are the opposite.

The power at the source end Pin=(Vfs+Vrs)*(Ifs-Irs)=(-Vfl+-Vrl)*(-Ifl–Irl)=(-1.5+–0.5)*(-0.015+-0.005)=0.02W.

The efficiency is Pout/Pin=0.02/0.02*100=100%.

We can calculate the impedance seen by the source, it is (Vfs+Vrs)/(Ifs-Irs)=(-Vfl+-Vrl)/(-Ifl–Irl)=(-1.5+–0.5)/(-0.015—0.005)=50Ω. The load seen by the source is also 50Ω and the power delivered by the source to the load is exactly the same as if the load was connected directly to the source, proving the quoted statement is not true in all cases.

Under the specific scenario, there is zero different between the matched load and the case of a lossless half wave of line inserted.

I have used a lossless scenario only because the original scenario was implicitly lossless. Drawing wider conclusions from lossless cases is fraught with dangers.

The article refers to sub-articles for definitions of key quantities, and some of those are wrong or inconsistent. For example, his expressions $$\Gamma=\frac {Z_l-Z_0} {Z_l+Z_0}$$ and $$\Gamma=\sqrt{ \frac {P_r} {P_f}}$$ are inconsistent, the first returns a complex value and the second can only return a real value.

## The matter of Thevenin source impedance

The Jacobi Maximum Power Transfer Theorem (JMPT) does say that the power delivered by a linear source is maximised when the load impedance is the complex conjugate of the source impedance. Linear is the mean that all possible combinations of terminal voltage is linearly related to current by the expression that Vt=Vs-I*Zs.

The issue is that most ham transmitters are not well represented by a Thevenin equivalent circuit, and so the JMPT does not apply. Nevertheless the model is used without proof that it applies.

## Simple explanations

Simple explanations are appealing, but when explanations are appealing because they are simple, you are in dangerous territory.

One of the challenges of teaching is to give learners models that are a foundation, a valid foundation that does not need to be unlearned to enable further learning. The referenced article contains statements that may seem convenient to the purpose but are not soundly based and must be unlearned for the learner to progress further. Teaching convenient lies is a betrayal of the trust that learners should have in their teacher.