# Formula for Resonant Frequency of 2 Metal Coaxial Cylinders?

• LarryS
In summary, a resonant cavity can have multiple resonant frequencies due to the geometry of the structure.

#### LarryS

Gold Member
TL;DR Summary
Can resonant frequency always be calculated from solely the inductance and capacitance regardless of the configuration of the resonator circuit/cavity?
Consider an LC circuit consisting of a parallel plate capacitor and a solenoid inductor in series. The formula for the resonant frequency of this circuit is 1/√(LC) where “L” is the inductance of the solenoid and “C” is the capacitance of the capacitor.

Now consider a high-frequency cavity resonator consisting of two concentric metal cylinders with a vacuum between them. Can the same formula as above be used to calculate the resonant frequency of this configuration if both the inductance and capacitance are known?

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LarryS said:
Now consider a high-frequency cavity resonator consisting of two concentric metal cylinders with a vacuum between them. Can the same formula as above be used to calculate the resonant frequency of this configuration if both the inductance and capacitance are know?
You will get more detailed replies from others, but keep in mind that such a resonant structure can have a number of resonant frequencies. There can be a lower frequency resonance due to the lumped L and C of the structure, and there can be multiple physical resonances due to the geometry (reflections).

berkeman said:
You will get more detailed replies from others, but keep in mind that such a resonant structure can have a number of resonant frequencies. There can be a lower frequency resonance due to the lumped L and C of the structure, and there can be multiple physical resonances due to the geometry (reflections).

Below is the link you requested. It's dimensions are specified by variables, not constants. Let me know if you need something more specific.
Coaxial Cylinders

Also, when you say "lumped L and C" do you mean lumped in physical space as in an LC circuit or lumped as in "total L or C" for the structure?

LarryS said:
Also, when you say "lumped L and C" do you mean lumped in physical space as in an LC circuit or lumped as in "total L or C" for the structure?
I mean what you would measure with a typical LC meter in the 1kHz to 10kHz range. Not what you would measure with a wide-range vector impedance analyzer.

LarryS said:
Can the same formula as above be used to calculate the resonant frequency of this configuration if both the inductance and capacitance are know?
No.
It is the capacitance per meter and the inductance per meter that decide the velocity factor of the transmission line. One end of a resonator cavity will usually be shorted, so the capacitance cannot be measured, while the other will be open circuit, so the inductance cannot be measured.

The resonant frequency will depend on how the ends of the inner conductor are terminated, either open circuit, or shorted to the outer conductor.
The insulation that separates and spaces the coaxial conductors will slow the wave and therefore lower the frequency of the line.
It is usual to terminate one end of the line as an open circuit, with a screw to trim the end capacitance and the resonant frequency of the line.

How are the ends terminated?
What insulation is present inside the cavity?

LarryS, DaveE and berkeman
DaveE said:
This sounds just like a coaxial transmission line.
That is correct. The Smith chart is appropriate to both.
The difference is that; the energy in a resonator circulates, while a transmission line is designed to be terminated in the characteristic impedance, such that terminal reflections are minimised.

Baluncore said:
How are the ends terminated?
The paper I was reading avoids the issue by assuming the line is infinitely long. It focuses on capacitance and inductance per unit length.

LarryS said:
It focuses on capacitance and inductance per unit length.
That is the basic derivation of the parameters for a transmission line. From L/m and C/m, the characteristic impedance and the velocity factor can be computed.

To make a resonator from a transmission line requires the ends be either left open, or be shorted. Only then can a reflection occur efficiently at the ends of the resonator, without significant energy losses in the end mirrors. Depending on whether the ends are a short-circuit or an open-circuit, they are being operated with either zero voltage, or zero current. That gives the highest reflection coefficient and least energy losses.

To be of some use, energy must be coupled between the enclosed line resonator and an external circuit. That coupling must be very light, so little of the resonant energy is flowing in the coupling, so the Q of the resonator can remain high.
The coupler usually takes the form of a small loop or a capacitive patch inside the line cavity. Some resonators have two couplers, one for the input and the other for the output.

The resonant wavelength of a transmission line resonator is a simple function of length, corrected for the velocity factor. It is possible to excite a transmission line resonator at harmonics of the resonant frequency, which is NOT possible with a simple lumped LC circuit.
Harmonics can be selected or rejected in a transmission line resonator, by selecting the position of the couplers in the line, relative to the nodes of the standing wave trapped within the line resonator.

LarryS, hutchphd, DaveE and 1 other person
LarryS said:
The paper I was reading avoids the issue by assuming the line is infinitely long. It focuses on capacitance and inductance per unit length.
Which is equivalent to perfect impedance matching at the ends for a finite length line. Infinite or matched lines won't resonate.

LarryS
LarryS