A little challenge was posted online, a request to explain this nominal 50-25Ω transformer.
Don’t get tricked by the 2:1 impedance ratio, it is probably nominal.
Since this is an RF transformer, let’s assume that the coax line sections work in TEM mode. It is likely to be very low loss, let’s assume it is lossless for ease of analysis.
I interpret the dashed lines to mean that there is some form of coupling between the three coax sections, but without a picture we are led by explaining how it does what it is purported to do and from that inferring the construction.
Firstly, the thinner coax with no connections to the shield at any point behaves essentially as a plain conductor, the current that flows on its inner conductor also flows on the outer surface of its outer conductor.
Let’s redraw the circuit using a simplified transformer equivalence for a short transmission line.
The top pair of windings is the top coax section, the middle pair of windings is the bottom coax section, and the bottom winding is the thin / middle coax section with no connections to its shield.
I have shown all windings fully flux coupled because this is necessary for this thing to work, each winding needs to have the same voltage across it..
So starting with the right hand three series windings, let’s assume current I (1I) flows in this series circuit. Since the two windings at left are fully flux coupled, current 1I flows in each winding, 2I flows in from the 50 ohm input wire and 2I is contributed from the top end of these two windings to the 25 ohm output wire. The right hand windings contribute 1I to the 25 ohm output wire.
Device input current is 2I and output current is 3I, and impedance transformation ratio \(\frac{Z_{out}}{Z_{in}}=(\frac23)^2=0.444\), not quite the nominal 0.5.
So, having explained how it achieves what it is meant to do, and that explanation depending on all windings being tightly flux coupled, it would appear that the coax device has all three coaxes inside a shared toroidal core or equivalent.