# Why is the loaded Q of a critically coupled two-resonator circuit approximately equal to 0.707 times the loaded Q of one of its resonators?

From RF Circuit Design by Christopher Bowick (page 32 bottom of the page left column):

The loaded Q of a critically coupled two-resonator circuit is approximately equal to 0.707 times the loaded Q of one of its resonators.

This isn't clear to me. How can I prove to myself that this is true?

The circuit that describes a critically coupled two-resonator circuit is basically a signal generator with a source resistance Rs feeding a parallel LC filter which is then coupled to another parallel LC filter through a coupling capacitor. The first LC filter is made up of L1 and C1 and the second LC filter is made up of L2 and C2. There is a coupling capacitor C12 between the two LC filters. The output of this two-resonator circuit feeds a load resistor RL.

This is schematically shown as: • Is this the same material? If so, I think you've missed the point. You can't prove what you're asking, because it's actually the definition of critical coupling -- the balance point between under-coupling and over-coupling. – Dave Tweed Feb 23 '13 at 21:43
• Yes, that looks like it's the same material. So the definition of critical coupling is such that the loaded Q of a critically coupled two-resonator circuit is ~0.707 times the loaded Q of one of its resonators? Would you happen to know why is it defined that way? – Learning About Circuits Feb 23 '13 at 21:56
• You should be able to prove it, and I don't see how a Q ratio of sqrt(2)/2 defines it. Critical coupling is defined as maximum power transfer at resonance, so make Zin=Zo to find coupling cap value. Then calculate Q ratio and the idea is to find the constant sqrt(2)/2. I don't know if it will work, but it is a starting point. – apalopohapa Feb 24 '13 at 9:57
• See cc.ee.ntu.edu.tw/~thc/course_mckt/note/note5.pdf , it may shine some light. – apalopohapa Feb 24 '13 at 10:03
• If critical coupling is the maximum power transfer point, then half the power will be in the source, and half in the load. By definition it's a half-power point, 0.707, sqrt(1/2) – david Jul 9 '13 at 3:52