electromagneticelectromagnetismmicrowaveresonanceRF
I was reading this paper about resonant cavities, and on page 6 it introduces the concept of coupling factor β, which is defined as the ratio between the power loss in the total circuit (including the resonant cavity and the source) and the power loss inside the cavity.
But, which is the physical meaning of the coupling factor? What does it mean, in terms of electromagnetic waves, that the system is undercoupled, overcoupled, or critically coupled?
This question is hanging around so I'm going to give some more help. No need for any upvotes.
As Marla said "For applied input current and voltage to be in phase, the reactive components have to cancel out to zero".
Do you understand the implications of this - do you see that the two mutually coupled inductors have to provide a reactance of +j12 so that numerically it becomes a series tuned circuit i.e. the capacitive -j12 is cancelled by the +j12 of the two series coils?
This is the important first step and if you don't understand why this produces a current in phase with the applied voltage then you need to do some back ground work.
If you follow this then the problem boils down to how you make the two coupled coils have a net impedance of +j12. Consider this: -
Do you see that ONLY when the inductor dots are "aiding" do you get an inductance increase over the basic L1 + L2? If you don't understand mutually coupled coils then you will have to do some research.
So, this tells you where to place the dot on coil PQ.
Somewhere between zero coupling and 100% coupling is where you want to be. You are aiming for +j12 but you only have +j8 and +j2 - this means the extra +j2 comes from the mutual coupling. Using the formula above clearly M is equivalent to +j1.
If you did some research on the web you would find this formula: -
Because reactance is proportional to inductance you can substitute the reactances in that formula and derive K thus: -
\$K = \dfrac{M}{\sqrt{L1L2}} = \dfrac{1}{\sqrt{16}}\$ = 0.25
The trick to getting good microwave resonant-cavity Q is to have a good conductor, a smooth finish, precise alignment, light coupling of the input signal, and limited microphonic pickup.
The design in the picture looks like it might have been limited by microphonics, and then reworked to eliminate them. For example, it uses a large heatsink instead of a fan. It also looks like alignment would be a real chore!
The loaded Q specification for the Keysight Split Cylinder Resonator is >20,000 at 10 GHz. If you look into one of the resonator halves, you will see yourself in the mirror surface finish. The resonator is gold plated and precision diamond turned. The parts look so good that they used clear plastic for the instrument covers! Very unusual for Keysight gear.
Here is more background information about the Split Cylinder Resonator, in case anyone is interested:
The alignment is done with a kinematic mount, similar to how a telescope mirror is adjusted. The resonator halves can then be adjusted back and forth, while maintaining the alignment. A measurement sample is placed in the gap. The sample changes the Q and resonant frequency of the resonator. This, along with a Network Analyzer, enables measurement of the sample dielectric constant and loss. The accuracy of the dielectric measurement relies on having a high-Q resonator.
Here are the specifics on the surface finish from the datasheet: "Cylinders are precision diamond turned Al 6061-T6 plated with 0.5 μm Cu, 0.25 μm PdNi, and 2.0 μm Au."
Full disclosure: I am speaking for myself, not Keysight, even though I work there.
Best Answer
Resonant Coupling Factor relates the mutual coupling two resonators (e.g. LC type) with tunable "crosstalk capacitance" or "mutual inductance" such that the overall bandwidth is increased by spreading the high Q centre frequencies apart to obtain steep skirts yet low ripple in the passband.
Below shown with a Capacitive Transformer (or C ratio) instead of inductive transformer as in question. But same effects. Here the C ratio =1.
Note: The crosstalk C = 0.1pF is really tiny and must be well-shielded.
For further reading
https://en.wikipedia.org/wiki/Double-tuned_amplifier
Old radios with 11 MHz FM IF and 455 kHz AM IF used a dual ferrite slug double tuned IF BPF.
You might also call a critically-tuned filter of any order , with maximally flat response AND 0 dB ripple what kind of filter??___ Bonus question.