Electronic – Impedance Matching of Antenna and Amplifier

I only have control over what is in the dotted rectangle.

If that is true you'll need two 25 ohm resistors in series with each feed to the 50 ohm terminator to avoid reflections (placed local to the 50ohm). The sending end impedance can be zero ohms providing the receiving end is perfect (which at the moment it isn't without the two 25 ohm resistors previously mentioned). If the receiving end were not perfect, a reflection would occur but this would be swallowed by the sending end if it had exactly 100 ohms terminating it: -

So what would a 1st reflection cause at the receiving end - the waveform the receiver receives could be put "out of shape" (digital transmissions) or changed in amplitude (for CW type transmissions). Neither will be earth shattering and probably livable with but, the 2nd reflection could be a different story. The 2nd reflection arises from the first reflection bouncing off the incorrectly terminated sending end and by the time it reaches the receiving end it may be enough to turn a digital 1/0 signal into a digital 0/1 symbol i.e. it causes inter-symbol interference.

So, for digital transmissions, longer cables can make the 2nd reflection trespass the timeslot of "later" bit and shorter cables have less of a problem. On the other hand, longer cables are likely to attenuate a reflection more so it's a little difficult to be general about this problem.

If in doubt, as a first measure, ensure the receiving end is accurately terminated. I hardly ever feed out to a coax or twinax via a line termination impedance so this should be OK. If you can't fix the receiving end then hope that the reflections are not significant enough to cause inter-symbol corruptions. For a CW signal I don't think you'll see total blackouts with a 100 ohm line being terminated in 50 ohms.

Well, I don't know if the schematic you uploaded is what you have in your case, or it is only an example. Checking it, it has a mistake, and maybe it is the origin of the issue.

Let me explain it from the begining:

1. For maximum output power, you need to match the external components of the Source to be its conjugate. That means that the real part seen from the Source must be 10 ohm, and the imaginary part must be the oposite than the produced by the 2nF internal capacitance.

2. As you have 100ohm Rload, you need to convert it to an equivalent 10ohm. The only way to reduce the equivalent value of a resistor is adding another in parallel. That is why you add an inductor in parallel to R2. The L value is calculated to obtain 10 ohm as real part of this parallel equivalent reactance.

3. For obtaining a Requivalent of 10ohm, you use a inductor that introduce a complex inductance, so you need to add a capacitor in serie (that works as a negative imaginary value) to remove this complex part.

4. When you calculate the C needed to remove the imaginary part, you must take into account the existing C on the source.

To solve that, there is a few formulas used, that includes the Q value concept as explained in previous answers.

Calculations:

The 1,06nF required is not what you need to place outside the Source, this is the equivalent C needed in all the LC network. As you have an internal 2nF capacitance, you need to calculate the outside C to get an equivalent value of 1,06nF.

The equivalent capacitance of 2 capacitors in serie es calculated as:

So what you need is an 2,255 nF capacitor instead of the 1nF that you have placed outside the source.