Nichols: The Two Bird Experiment

With the following introduction, (Nichols nd) tries to demonstrate some important principles. He says…

This really tests your understanding of transmission line theory.

Above is Nichols’ test setup, simple enough.

With the transmitter keyed, the transmatch is adjusted to show zero reflected power on Bird Wattmeter #1. Transmitter is then adjusted to generate exactly 100 watts of forward power indicated on Bird Wattmeter #1. Bird directional Wattmeter #2 indicates about 36 watts of REFLECTED power. (Charts are readily available to show that a 4:1 mismatch gives about 36% reflected power).

Wattmeter #1 reads Pfwd=100 and Pref=0 as stated in the referenced article, \(P_{fwd}-P_{ref}=100W\). I assume the Bird elements are 50Ω ones, ie the reference impedance is 50+j0Ω.

The Transmatch is stated to have negligible loss, let’s assume zero loss. On that basis, \(P_{fwd}-P_{ref}=100W\) on the load side of the Transmatch. We also know the because of the 200+jΩ load on wattmeter #2, the notional VSWR is 4.

Question: What does the FORWARD power of Directional Wattmeter #2 show. And why?

Let’s calculate the magnitude of the reflection coefficient.


We can also find Pfwd-Pref in terms of Pfwd-Pref and ρ.

\(P_{ref}=\rho^2 P_{fwd} \\
P_{fwd}-P_{ref}=P_{fwd} \left(1-\rho^2\right) \\
P_{fwd}=\frac{P_{fwd}-P_{ref}}{1-\rho^2} \\
P_{fwd}=\frac{100}{1-0.6^2} \\
P_{fwd}=156.25 \\
P_{ref}=0.6^2 \cdot 156.25 = 56.25 \\
P_{fwd}-P_{ref}=156.25-56.25=100 \\\)

So, the answer to Nichols’ question is Pfwd=156.25W, and his assertion that Bird directional wattmeter #2 indicates about 36 watts of
is plainly wrong, it should read 56.25W.

Nichols’ solution implies that Pfwd=100W on both sides of the Transmatch, whereas Pfwd-Pref=100W (by virtues of the lossless Transmatch).

Yes, it does test your transmission line theory, might be time for review?

It is hard to reconcile the theory of this scenario with his claim on this topic p.s.: I always base my claims on actual measurements.