I Off resonance driving of a MW/RF cavity

[Malamala]

TL;DR Summary

How are the E and B fields at 10 GHz distributed relative to the ones at 100 MHz inside a cavity resonant at 10 GHz?

Hello! Let's say I have a cavity resonant at 10 GHz with a Q factor of 1000. Given the Lorentzian shape of the cavity, I can also drive the cavity at, say 100 MHz. Of course the response will be very very weak, but non-zero given that the Loretzian shape never really reaches zero. I am trying to understand how are the magnetic and electric field distributions of the field at 100 MHz relative to the ones at 10 GHz? In particular, if inside the cavity I have some structure, such as 2 plates very close to each other. If these plates are not too close, they will allow the 10 GHz field to enter and build and electric and magnetic field inside (which depends on several things such as the coupling of the MW power inside the cavity). What if on the other hand they are much close than the wavelength of the 100 MHz field, such that certain TE and TM modes are suppressed (if we consider these plates roughly as a waveguide)? Will the electric and magnetic fields inside such plate, at 100 MHz, be suppressed relative to the 10 GHz ones, just by the corresponding Lorentzian drop in power, or will I also have on top of that an exponential decay given that the fields at 100 MHz inside the plates are evanescent fields (the fields in both cases are injected through the same port which is outside the 2 plates).

 

[tech99]

The Lorenzian shape arises because you have placed matter within the cavity. It appears that the material has losses at 100 MHz, which happens to be well below the cavity resonance in this case. So we have the case of a lossy waveguide operating below cut-off. I suspect, therefore, that in addition to the exponential fall-off of the fields, there will also be a linear fall off in decibels per metre.

 

[sophiecentaur]

TL;DR Summary: How are the E and B fields at 10 GHz distributed relative to the ones at 100 MHz inside a cavity resonant at 10 GHz?

Given the Lorentzian shape of the cavity

I think it would be a huge assumption that the response of the cavity would still be Lorentian at 1% of the resonant frequency. A cavity does not consist of 'lumped components' (i.e. L and C) so the ratio of the reactive fields won't be what you might think.

What actually is the context of your query?

 

[Malamala]

I think it would be a huge assumption that the response of the cavity would still be Lorentian at 1% of the resonant frequency. A cavity does not consist of 'lumped components' (i.e. L and C) so the ratio of the reactive fields won't be what you might think.

What actually is the context of your query?

@sophiecentaur @tech99 Thank you for your answers! Basically my question was a long way of asking whether in this case I can still assume I have Lorentzian behavior at frequencies below the cutoff, or a faster decay than that. And if that is the case, how much faster is that decay of the field?

I don't have an actual setup yet, but I want to design an experiment where I want to suppress certain modes at low frequencies (~ 100 MHz), while driving the system at 10 GHz. The modes are well below the cutoff, given the expected dimension of the MW cavity, but in common simulation software (e.g. HFSS, COMSOL), I can only get the cavity response frequency, in a reasonable amount of time, close to the peak, where it looks Lorentzian and all the modes in that neighbourhood can be excited above the cutoff. I was wondering at which point is this assumption failing.

For example, naively, from a Lorentz lineshape, the B or E field at 100 MHz would be ~2000 times smaller than at 10 GHz. Is this value correct, or do I get extra suppression from the fact that the wavelength of these fields is much larger than the size of the cavity?

 

[renormalize]

@sophiecentaur @tech99 Thank you for your answers! Basically my question was a long way of asking whether in this case I can still assume I have Lorentzian behavior at frequencies below the cutoff, or a faster decay than that. And if that is the case, how much faster is that decay of the field?

Let's look at a real-world example consisting of a resonant printed circuit board (PCB) structure driven by coax connectors: https://blog.teledynelecroy.com/2020/05/reading-s-parameters-sharp-dips.html. The swept-frequency insertion-loss through the PCB cavity is found to be:

As you note, the observed resonances are (inverse) lorentzian curves, but below the lowest resonance the loss looks nothing like a lorentzian. The loss simply increases slowly but steadily with increasing frequency. So I don't think your "naive" estimates are likely to be valid.

 

[Malamala]

Let's look at a real-world example consisting of a resonant printed circuit board (PCB) structure driven by coax connectors: https://blog.teledynelecroy.com/2020/05/reading-s-parameters-sharp-dips.html. The swept-frequency insertion-loss through the PCB cavity is found to be:
View attachment 365490
As you note, the observed resonances are (inverse) lorentzian curves, but below the lowest resonance the loss looks nothing like a lorentzian. The loss simply increases slowly but steadily with increasing frequency. So I don't think your "naive" estimates are likely to be valid.

I am actually a bit confused. I thought that at low frequencies, the loss would actually be higher, not smaller. For example, say I use a waveguide to send my radiation inside and I include that region in my Q factor calculation. If the waveguide size is such that its cutoff frequency is 1 GHz, the amount of radiation inside the cavity at 100 MHz should be exponentially (and not just Lorentzian-like) smaller than at 10 GHz, no?

 

[Baluncore]

I am trying to understand how are the magnetic and electric field distributions of the field at 100 MHz relative to the ones at 10 GHz?

At 100 MHz, the 10 GHz cavity would be well below the cut-off frequency. Depending on the cavity coupling technique, it would appear as a high open, or a low short, termination impedance.

Any resonances you see will be due to the length of the transmission lines used to propagate the signal to and from the cavity.

 

[renormalize]

I am actually a bit confused. I thought that at low frequencies, the loss would actually be higher, not smaller.

The PCB cavity looks like a through at DC so when driven by coax it transmits 100% of the DC signal. But a hollow cavity coupled with loop- or probe-antennas looks like a short and, as you say, it should transmit 0% at DC. But as the frequency increases from DC, will it necessarily simply follow the tail of the first resonance? Based on this cavity simulation from https://www.researchgate.net/figure/Simulated-S21-response-of-resonant-cavity_fig102_336278497, the low frequency response of a cavity could possibly just "flat-line"/plateau at a non-zero constant value at low frequency:

 

[tech99]

A waveguide does not follow the shape of the ordinary resonant curve, because it has sharp cut-off on the low frequency side. The PCB example seems to be a TEM line and is not therefore a waveguide.