[Radio Wrangler]

I still think something is being missed. In an efficient filter the bandwidth of the resonators should be dominated by coupling to the real components of the source and load impedances. If resonator bandwidths are principally set by the resonators own losses, then the insertion loss is going to be high.

Just have a look in Zverev. He has the derivations for insertion loss versus the ratio of resonator Q (unloaded) to filter Q (filter centre frequency divided by filter bandwidth.

I agree with those derivations for a tuned circuit in isolation, but they are coupled to things in use, and that makes a fairly large difference.


David