just a wild guess: mode coupling, non-uniformity of field distribution in a wave-guide? did you try building a larger test waveguide? And I think, the impedance is the most important factor. I don't know what you are doing exactly; I googled something interesting, which may, or may not give you some ideas: R.D. Seager, J.C. Vardaxoglou, "Close coupled resonant aperture inserts for waveguide filtering applications" 2001

Hi Baraa. The reason is that in a rectangular waveguide simulator, the incident wave is the fundamental transverse electric mode (TE10). This mode can be decomposed into two plane waves with the electric field polarised in the direction parallel to the short dimension of the guide. Each of these plane waves propagates at +/- an angle of incidence, theta, with respect to the FSS normal unit vector. The angle, theta, is frequency dependent and given by the formula

cos(theta) = sqrt(1-(lambda/2a)^2) where lambda is the free-space wavelength and a is the broad dimension of the rectangular waveguide. This situation also corresponds (fairly obviously) to the transverse electric plane of polarisation (TE incidence). So, when one sweeps over the frequency, passing through the FSS resonance, the waveguide simulator results will be different for a given frequency compared to free-space since the angle of incidence of radiation in the waveguide is also changing as the frequency is swept. Whether you get a broadening of the bandwidth or not in the waveguide simulator may depend on the type of FSS element you have. What kind of FSS elements have you measured by the way?

Ray, sorry for the previous out of the point comment; I misread the sentence.

Here you are basically speaking about H-vector rotation in H-plane. Please, allow me to argue. The thing you are telling should be right for a cavity. But in a waveguide, the vector field takes all the orientations in H-plane during the period. So there should be no significant difference if the long-side dimension is comparably large.Of course only in case of normal incidence, and symmetry in both topology, and whole composition; when operating point is sufficiently distanced from the wave-guide cut off (where impedance changes the most)

here is an image for WR112 at 8 GHz.

http://www.ecs.umass.edu/mie/labs/mda/fea/fealib/moser/Image12.gif

you can see that the most non-uniformity is near the walls, and it is reduced at higher frequency.

So, do you speak basing on experience, or is it a logical conclusion?

Btw, talking about "significant", how much is the change?

Hi Andrey,

I can't agree with your description. The explanation for the difference between a waveguide simulator FSS measurement and a free-space one when the waveguide is supporting a TE10 mode is as I've stated. The non-uniformity of the field that you refer to (there's a cosine E-field distribution across the width of the guide that is uniform in the orthogonal direction) is a direct result of the plane-wave decomposition that I referred to (two intersecting plane waves propagating at oblique angles plus-theta and minus -heta in the x-z plane). This is a well-known formulation for a TE10 rectangular waveguide mode. In a waveguide simulator FSS measurement, one cannot simulate normal incidence in practice. The incidence angle varies with frequency according to the formula I gave, such that normal incidence would only occur as the waveguide width approached infinity or the wavelength became vanishingly small. This behaviour of waveguide FSS simulators is well known and well established. By the way, I am speaking from a good deal of experience with FSS technology. I even worked closely and co-authored some publications on FSS with Prof. Vardaxoglou of Loughborough University, UK, in years gone by.

Dear Andrey and Ray,

Thank you very much for your valuable discussion.

Let me elaborate more on my design. I’ve simulated and experimentally measured the s-parameters of the unit cell shown in the attached file. This sample has 0.8 FR4 substrate and was tested experimentally inside S-band waveguide (WR-284). There was very good agreement between simulated and measured results. The problem is that their -10 db bandwidth is about 4 times larger than the bandwidth of similar periodic structure. I’m working on a mathematical model and I need to understand why the response is wider in the waveguide and then trying to find a proper way of modelling that phenomena.

Ray’s explanation about wave decomposition sounds right especially I noticed that by increasing the port impedance the bandwidth gets bigger, and if we treat the case similar to a TE oblique incidence that would lead to a higher characteristic impedance as we have to divide it by cos(theta).

I will consider that and see how it goes.

Kind Regards,

Ray, OK, I can agree, because I have not very long experience of "simulating" in a wave-guide, and treated such differences as insignificant, when dimension is 3-4 times wider than the object. that's why I am mentioning symmetry and the magnitude of discrepancy.... But isn't "4 times" too much for non-uniformity?

Baraa... variation of band-edges always happen when changing matching. That depends on steepness of impedance real part plot bending shunt, or open-ended condition. But normally you see some widening + shift, and insertion loss degradation. It is very easy to calculate in matlab, applying multi-cell ABCD apparatus to your FSS cell model.

Dear Baraa,

Thank you for sharing the details of your FSS design with us. I assume that the parameters of your periodic FSS structure are those of a square or rectangular lattice? From what I remember about FSS measurements in rectangular waveguide simulators, the lattice of your perodic structure must be at least rectangular in form and you need to place two unit cells side-by-side across the broad dimension of the waveguide. This is to emulate the correct symmetry and periodic boundary conditions of the full lattice. Did you use side-by-side elements in your waveguide simulations and measurements, or just one element? You can think of this two-element arrangement as generating an infinite set of image elements with respect to the side walls of the waveguide. You would be not able to simulate a triangular lattice in a rectangular waveguide, for example. Whilst I would fully expect your computed and measured results to agree well for whatever FSS element you place in the waveguide, the trick in an FSS waveguide measurement is to mimic the actual periodic boundary conditions that occur in a full lattice. If you've only modelled and measured a single FSS element in the waveguide, the results would not pertain to the same lattice parameters of a fully periodic (free-space) structure involving that element. My previous comments about the plane-wave decomposition and frequency-dependent angle of incidence in the waveguide measurement would still apply to the case of two side-by-side elements placed in the waveguide.

Regards,

Ray

Ray, here, you recommend TE20 mode?

Hi Andrey,

No, not the TE20 mode. I'm still talking about an incident TE10 mode in the rectangualr waveguide (the fundamental mode). For the FSS, one should use a pair of identical elements side by side (two unit cells).

I believe, your unit cell is comparable to the operating wavelength. OR, (May be I can be wrong), Since the K vector in TE10 mode is in two coordinates the incidence would be not a normal incidence inside the wave-guide. Please see the below reference for more details.

https://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=4200922

Dear Ray,

Thanks for your detailed explanation. In fact I only measured one unit cell inside the waveguide. I tried to simulate one unit cell and then two unit cells inside waveguide and the results were exactly the same.However, I noticed that for the periodic structure to have the closest response to the waveguide, it should have a unit cell width half of the waveguide width which proves your point about the symmetry and boundary conditions. There is still a resonance shift and a slight bandwidth difference and this could be a result of oblique incidence phenomena as highlighted by you and Jayaprakash as well.

Thanks everybody for the helpful discussion.

Kind Regards,

Microwave absorbers based on resistively loaded high impedance surfaces are an attractive option for providing radar cloaking where the main design drivers are weight and thickness. However the reflectivity bandwidth of this class of absorber is proportional to the structure thickness, which is determined by the gap between the FSS and ground plane. a -10 dB reflectivity bandwidth of 109% was obtained from a 5 mm thick metal backed resistively loaded single square loop FSS array. This was designed to resonate at the two frequencies where the imaginary component of the FSS impedance is cancelled by the inductance and capacitance .