The key point of the story is:
"The recipe to exploit maximum benefit from a resonant EOM"
 Make a resonant EOM circuit. Measure the impedance Z at the resonance.
 This Z determines the optimum turn ratio n of the stepup transformer.
(n^{2} = Z/Rin where Rin is 50Ohm in our case.)
 This n gives the maximum gain G_{max} (= n/2) that can be obtained with the step up transformer.
And, the impedance matching is also satisfied in this condition.
OK: The larger Z, the better. The higher Q, the Z larger, thus the better.
(Although the relationship between Z and Q were not described in the original post.)
So, how can we make the Q higher? What is the recipe for the resonant circuit?
=> Choose the components with smaller loss (resistance). The details will be provided by Kiwamu soon???
When I was young (3 months ago), I thought...
 Hey! Let's increase the Q of an EOM! It will increase the modulation!
 Hey! Let's use the stepup transformer with n as high as possible! It will increase the modulation!
 Hey! Take the impedance matching! It will increase the modulation!
I was just too thoughtless. In reality, they are closely related each other.
A high Q resonant circuit has a high residual resistance at the resonant frequency. As far as the impedance is higher than the equivalent output impedance of the driving circuit (i.e. Z>Rin n^{2}), we get the benefit of increasing the turn ratio of the transformer. In other words, "the performance of the resonant EOM is limited by the turn ratio of the transformer." (give us more turns!)
OK. So can we increase the turn ratio infinitely? No. Once Rin n^{2} gets larger than Z, you no longer get the benefit of the impedance transforming. The output impedance of the signal source yields too much voltage drop.
There is an optimum point for n. That is the above recipe.
So, a low Q resonant EOM has a destiny to be useless. But high Q EOM still needs to be optimized. As far as we use a transformer with a low turn ratio, it only shows ordinary performance.
