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:

enter image description here

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    \$\begingroup\$ 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. \$\endgroup\$ – Dave Tweed Feb 23 '13 at 21:43
  • \$\begingroup\$ 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? \$\endgroup\$ – Learning About Circuits Feb 23 '13 at 21:56
  • \$\begingroup\$ 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. \$\endgroup\$ – apalopohapa Feb 24 '13 at 9:57
  • \$\begingroup\$ See cc.ee.ntu.edu.tw/~thc/course_mckt/note/note5.pdf , it may shine some light. \$\endgroup\$ – apalopohapa Feb 24 '13 at 10:03
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    \$\begingroup\$ 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) \$\endgroup\$ – david Jul 9 '13 at 3:52

This is just an approximation. If you look at it in terms of voltage 0.707 is 3dB drop or 1/2 power. Adding a second coupler will increase the bandwidth which will cut the Q by 0.707 as compared to a single resonator. If you want to compute this just compute the 3db points for a single resonator vs. a critically coupled set. You can do this by analyzing the transfer functions for each circuit and solving for the 0.707 points.

You'll find the low frequency side of this network to move at 18db/octave and 6db/octave on the high end due to it's poles and zeros. This lack of symmetry in the system make the change in Q (i.e. bandwidth) an approximation.

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