Surely there cannot be more forward power than the transmitter makes?

Let’s explore a simple numerical example of a practical line operating in Transverse Electro Magnetic (TEM) mode (the usual thing for practical coax lines at HF).

Let’s review the meaning of 50Ω line.

It means that the line geometry imposes a natural constraint on a wave travelling in the line that V/I=50… but remember that TEM waves are free to travel in (only) two directions. This natural ratio of V/I is called the characteristic impedance Zo.

Consider this…

Let’s say that at some point on a 50Ω line (it could be start, finish, or anywhere in between), V=100V and I=1A, therefore Z=V/I=100/1=100Ω. This situation could be created at the end of a transmission line by simply attaching a 100Ω load.

Zo is a complex value (having real and imaginary parts), but for simplicity we will assume that Zo is a purely real value, Zo=50+j0Ω. It makes the maths easier, it becomes arithmetic you can do in your head, and in fact practical transmission lines at 100MHz have Zo that is very very close to a purely real number.

So how can V/I at a point on a Zo=50Ω line be 100/1=100Ω?

The solution

So the voltage at the point of interest point is 100V and I=1A, Z=100Ω. (For simplicity, I am assuming I in phase with V, so the actual power is easily calculated P=V*I=100*1=100W. P is the energy flow past the point averaged over a full RF cycle.

This 100W is the power that the transmitter ‘makes’.

Because of the Zo constraint that a travelling wave MUST have V/I=50, then this 100V/1A can only be due to the sum of two waves travelling in opposite directions, there MUST be a forward wave and a non-zero reflected wave such that their sums deliver the observed voltage and current.

What would a Bird 43 50Ω directional wattmeter indicate?

A Bird 43 50Ω wattmeter ‘explains’ V and I at a point as being due to two waves travelling in opposite directions (taking into account their relative phase). The Bird wattmeter is calibrated for a purely real Zo, 50Ω in this example.

Now because there is 100W flowing, then the difference between the indicated forward wave Pf and reflected wave Pr MUST be 100W, ie \(P=P_f-P_r=100\). This does not tell us the value of Pf or Pr.

To find Pr and Pf, we need to calculate the complex reflection coefficient: \(\Gamma=\frac{Z-Z_0}{Z+Z_0}=\frac{100-50}{100+50}=0.33\).

Now \(V=V_f(1+\Gamma)\), which we can rearrange to \(V_f=\frac{V}{(1+\Gamma)}=\frac{100}{1+0.33}=75\) and therefore \(V_r=V-V_f=100-75=25\).

So, we know Vf and Vr, we can calculate \(P_f=\frac{{V_f}^2}{Z_0}=\frac{{75}^2}{50}=112.5\) and \(P_r=\frac{{V_r}^2}{Z_0}=\frac{{25}^2}{50}=12.5\) and \(P=P_f-P_r=100\).

Similarly we can say \(I=I_f(1-\Gamma)\) (-ve due to the wave travelling in the opposite direction), which we can rearrange to \(I_f=\frac{I}{(1-\Gamma)}=\frac{1}{1-0.33}=1.5\) and therefore \(I_r=I-I_f=1-1.5=-0.5\).

Forward power can be greater than the transmitter ‘makes’, you must deduct the reflected power to find the ‘net’ power, and that ‘net’ power cannot be greater than the actual transmitter output power under those load conditions (though it may be greater than the transmitter’s notional ‘rated’ power).

Surely there cannot be more forward power than the transmitter makes?

The transmission line supports TEM waves travelling in two possible directions, and you must consider both waves to calculate the power at a point on the line. Forward power is a notional quantity associated with one of two possible travelling waves.

The law of conservation of energy is not violated by any of the discussion above.