I had a matching technique described to me by a professional broadcast engineer. Here's the problem: you have a vertical antenna with a low resonant impedance1, let's say \$38\Omega\$, but you need a better match to a \$50\Omega\$ feedline.
Solution: put a coil between the base of the aerial and ground. Then, you feed off a tap in the coil:
(image source: G3TSO on QSL.net)
How could this work? I can understand how adding a shunt inductor could improve the match if the antenna presents a capacitive load. But, why feed off a tap in the middle of the coil? If it's working as an autotransformer, wouldn't it be making the antenna impedance appear to be less to the feedline (the opposite of what we are trying to accomplish)? Also, if the antenna is already resonant, wouldn't the addition of shunt inductance make the match worse? Or is this why the feed is off a tap? Please help me understand how this works.
1: in my particular case, the impedance is even lower (I measure \$14\Omega\$) due to shortening of the antenna.
Best Answer
The only way I can see it working is if the lower impedance antenna (38 ohm) were connected to the "tap" of the auto transformer. A 38 ohm tap and a 50 ohm feedline means the tap is at 87% of the way up i.e. \$(0.87)^2 \times 50\Omega = \$ 37.8 ohms.
An 18 ohm tap is at 60% i.e. \$(0.60)^2 \times 50\Omega = \$ 18 ohms.
I can't see any other way round this other than the broadcast engineer got in a muddle or your ears need washing out LOL.
NB These calculations assume the auto transformer windings are 100% coupled to each other.