Microwave Engineering - Science topic

How do we get the dielectric constant from the unloaded Q-factor of a cavity resonator?

Hello,

In the Particle Tracking solver, in the emission model settings (Edit Particle Area Source – Tracking emission model – Emission Settings) for Space Charge Limited Emission, Thermionic Emission, there is a Kinetic Settings tab in CST.

One file is attached for the help.

Thanks,

Miguel Ferrando-Rocher

Hello dear friend, thank you

According to the article:

Do you think it is easy to use the thesis of trapezoidal dielectric shape in reality?

No, a quadrature coupler can't support wide bandwidth.

Sir Szuhsien:

You are welcome, and i wish you good health and happiness....

In this case case I would go for a numerical evaluation which is rather easy these days..even if there is a formula somewhere it will be an approximation and not straightforward to solve

@Baraa F.Al- Azzawi, hei this designer is very useful and related Ansys.

Could you share how to do FSS analysis using Keysight ADs or EMPro?

Amjad Abdulsatar Al-Rahmah You can export your Z-parameters (real and imaginary) from HFSS as a CSV file, then open the attached .mdf file and just replace your .CSV results with the attached one.

Good luck

I am also looking for these data. I found the same article for heat capacity. I also found a dissertation by Liyan Qiu from 2000.

It gives 0.86 J/K/g for NaX. Water content is likely a significant variable.

Good question

Causes

For a fire to start it needs a source of ignition, a source of fuel and a source of oxygen. For example, if a smoker falls asleep with a cigarette still lit, and sets fire to the sofa, the cigarette is the source of ignition, the material on the sofa is the source of fuel and the air is the source of oxygen.

Thank you Mr. Marco Lauro for you suggestion. I published the attached article in this magazine. They lost fast (15 days) and are very professional.

The Journal of Microwaves, Optoelectronics and Electromagnetic Applications (JMOe), published by the Brazilian Microwave and Optoelectronics Society (SBMO) and Brazilian Society of Electromagnetism (SBMag), is a professional, refereed publication devoted to disseminating technical information in the areas of Microwaves, Optoelectronics, Photonics, and Electromagnetic Applications. The review was done within 12 weeks. Authors are not charged for the article Processing and article Submission costs. 

The feeding lines in your design seem to have the characteristic impedance of 50, so there would be no problem for exciting the structure through the 50-ohm lumped ports. Please follow the instructions regarding the lumped port excitation in the attached manual.

@anuraag Misra

Sir , how to design in HFSS? is there any tutorial regarding it...

How to calculate the length of the substrate integrated waveguide antenna

I think you will find it difficult to find a solderable, lumped component that fulfills this requirement. Once you reach these capacitance values, even deviations in soldering and mounting will likely exceed the capacitance value. A small blob of solder will exhibit 100-200fF capacitance. Once you get into X band, varactor tuning is best done as part of an integrated MMIC (like an integrated VCO) where varactor package parasitics are avoided and interconnect performance can be tightly controlled.

If you are using a varactor as a discrete component for tuning a dielectric resonator filter or oscillator (DRO) , your best bet is to use transmission line elements to transform impedances so that you can use a device like the MAVR-011005-12790T. Keep in mind that parasitic inductance must also be accounted for (the MAVR-011005-12790T data sheet specifies 0.6nH for this device).

I have used the Skyworks SMV-2019 for tuning X-band phase-locked DROs with a minimum of fuss.

You are using the so called cold noise measurement technique. This type of measurement using measured s parameter data and noise power values obtained from source pulling about a reflection coefficient with a Gamma of 0.5 or so. Be certain that the device is stable and check to see if the s data returned during measurement makes sense. Is the device matched? As a sanity check that your cal is good, measure a simple item prior to your LNA DUT. A 3 db pad should return ~ a 3 dB NF.

Maybe "transient assisted Harmonic balance" in "initial guess" will help.

Nowadays, It is rarely used of Java applets, an alternative for them is Java Servlet.

Servlets are placed into JAVAEE stack as an standard.

@Abdelhalim Zekry, thanks for your comment. I have now submitted my book proposal to IGI Global and received a positive response within two weeks.

I think this a great question. I am highly surprised that 6 years later there is no clear answer, and several wrong and obtuse answers among the top ones here currently.

Here I try to give a very clear answer, which makes it a bit long... but if you are strong in optics you can easily skim ahead to the main points.

The first part of the correct answer is that the arithmetic posted by Peng Zhang is quite simply and obviously correct.

Next, realize that we are strictly imagining here (the E in the posted question) standing waves in either infinity or an ideal cavity. An ideal standing wave in a cavity do not transport any energy out of the cavity.

But, there is more behind this question than this. (Which is what makes it such a good question with an answer that goes a bit deep, certainly is not trivial.) We do use plane waves as a mathematical basis from which one can (mathematically) construct either wave packets that do model light pulses, or beams of light (or photons), that do transport energy over space in a given direction. The intensity of the energy flux (Poynting vector) is correctly described by the arithmetic above. A hypothetical wave packet with E-field E’ equal to the sum of the two components (E’ = E + E) has twice the energy flux than two separate wave packets with the individual components. In this sense it is perhaps correct to think of "radiation energy density" to be not a linear, but a nonlinear function of electric field intensity. That is quite simply what the math and physics of light shows us. Peng Zhang (original post) is right.

Moreover, this reality is no violation of the conservation of energy - it just appears to be a contradiction, and we will now explore this apparent contradiction a bit deeper. All energy that radiates out from a source is supplied by, and lost by the source. So the more intense beam corresponding to the superimposed scenario would have a source (say star) losing energy faster (proportional to 2E^2) than the total loss (proportional to 0.5E^2 + 0.5E^2 = E^2) by the two other smaller individual sources (separate stars) separately radiating out the two weaker non-superimposed beams.

But, you may say, can't you take the two weaker beams from the two weaker stars, then use some ideal non-lossy prisms and mirrors to guide the two source beams onto a collinear (collimated) path, and create the super-position beam? Then, won't you see the violation of energy conservation suggested by the original question?

In part the issue seems to be thinking of the Poynting vector, and Poynting theorem, as the "energy of the wave" (electric or potential energy). Instead the Poynting vector of a beam (with magnitude E^2) quantifies the energy flux in a given direction. (It's the "flow of energy", not the "amount of energy contained within".) The precise definition of the Poynting vector, as energy flux (or energy transport) that in general gives us an idea of "light intensity" is in “Principles of Optics” by Born & Wolf section 1.1 p. 10.

By utilizing optical superposition that flux can be redirected. (That is what mirrors, AR coatings etc. do.) Consider first this related key example: If you superimpose two linearly polarized standing waves, with the same wavelength, but exactly out-of-phase, then the electric fields sum to zero at all points, the Poynting vector becomes zero, yet the energy is not destroyed or 'disappeared', it is just redirected.

Another perspective is to realize that in the new perfectly interfering plane waves scenario we basically just wrote "nothing" in very fancy way, like 1 - 1 = 0 ... the LHS of this expression, namely "1-1", is just a fancy way to write zero. With perfect interference we have no electric field, no energy flux, no light.

What, you may ask, if you have two rays of light, both ideal plane waves, and you guide them into a collinear path where they are precisely out-of-phase as we imagine in my “key example”. Did we make all the energy in the two beams disappear? Did we “destroy” energy?

The answer is, you described the situation of a perfect mirror. If your two rays are perfectly out of phase they will, in some sense, simply not exist in your hypothetical cavity. They will simply and precisely not enter your cavity, which will indeed be filled with zero net field (indeed zero or nothing). Instead, these EM waves will perfectly reflect back. Their energy will be conserved; you have just described the one scenario (the one precise phase difference choice) where you will not get any fraction of the EM radiation from the two parent beams into your cavity.

A nice similar example, showing how an apparent contradiction of the conservation of energy is just a case of the energy flux being redirected (with total energy conserved globally) is worked out and written up in "Introduction to Modern Optics" by Grant Fowles (section 2.9, p. 55).

I believe the answer to the posted question is inherently the same as the complete interference (perfect mirror) scenario I discussed first. (That is why I discussed it first, it's a bit clearer but in principle the answer is exactly the same.) The complete interference (E-field components perfectly 180 degrees out of phase) scenario is just another variation of the perfectly in phase scenario in the posted question here. (Perfectly in phase vs perfectly out of phase – two special cases of the same general problem.)

With the destructive interference, or out-of-phase case, we realized we were just describing what in the lab would always be the creation of a mirror, both rays will just be reflected and energy perfectly conserved (not destroyed). I believe that the answer for the in phase superposition case is that we will likewise find that for the in phase superimposed beam on our collinear path we will always require two incoming beams whose energy fluxes will sum to the new energy flux (2E^2). I.e. we can mathematically imagine we pull one sine wave (plane wave) from 'somewhere', and another from 'somewhere', and then superimpose them perfectly out-of-phase, then mathematically (and correctly) the field sums to zero, the Poynting vector too becomes zero. We summed two components and got zero. Mathematically simple, precise, kosher. But this is mathematics in our heads. If we do this in the lab with real optical components by collimating two real beams, then the real out-of-phase superposition will also always produce a concurrent real reflection. We made a mirror for our two beams (cleverly made by exploiting destructive interference) and we are just (in our mathematical imagination, 'in our heads') looking at the side of the mirror with no net E-field, with no light… we are looking at the backside. In the lab we will never say, just because we are looking at the backside of our mirror and see no light there, that we made the light, or the energy, disappear. That will be patently silly! Yet, from the pure mathematical super-position point of view that is kinda what it looks like and feels like when we put it like the posted question… we did “take two beams with field E, combined them, and we made energy disappear, we found a violation to energy conservation”. Mathematically we are just looking behind the mirror. We get the field and energy flux exactly right with our math, but, we did not actually find any violation of the conservation of energy.

The answer to the posted question (in phase variation of this scenario) is exactly the same. It is a variation of the same problem. The answer is a variation of the same answer.

All the EM equations that we reliably use to model mirrors, prisms, AR coatings etc. all do conserve energy (talking here only of non-lossy and non-source optical components of course – no lamps, lasers or absorbers of any kind). Hence we find, when in the lab or on paper we construct the out-of-phase superposition scenario by configuring actual optical components to create the out-of-phase superposition then we always find (in experiment and correct optical calculation) that we reflect out beams of equal energy flux than the beams we did not transmit into our cavity.

Likewise, I conjecture that we will find that if we create a steady state version of the superimposed in-phase beam implied by the posted question, we will always need as input to our cavity two (or more) beams with combined energy flux equal to the energy flux (irradiance = 2E^2) of our in-phase superimposed beam. (We assume everything steady state.)

Why? Because the equations (underlying physical laws) for every optical element modeled with the relevant Maxwell description do conserve energy. They all do. Hence any scenario constructed with these optical components will conserve energy. For a worked example, see the Fowles reference above.

In summary. The arithmetic posted by the original question is 100% correct. The conclusion is correct; the superimposed beam will have twice the intensity (irradiance) of the original beams. To construct the 2E^2 intensity beam in steady state by collimating two "source beams" experimentally you will always require more input than the two components in the mathematical E-field decomposition. We can mathematically correctly describe the electric field of the more intense beam as the sum of these two components (E’= E + E), but we cannot physically experimentally construct the more intense beam simply by collimating two beams with the component fields in the lab. This is what "feels a bit wrong" or "a bit strange" about this question and its correct answer. We can mathematically “look behind the mirror” this way, and that is correct. But we cannot really truly physically construct the collimated beam from these two separate component beams. This is the universe our current optical physics suggest we live in. To make an actual super-position beam and do a global energy balance will also require “looking in front of the mirror” (the “mirror” can be a coupler or some steering mechanism for the in phase collimated beam scenario). “In front of the mirror” we will find that to make the superimposed beam with irradiance 2E^2 by combining incoming beams we will always have to use incoming beams with combined irradiance at least 2E^2 (exactly 2E^2 if there is no reflection or other losses in our optical components, and > 2E^2 if there are losses).

Conclusion, no violation of conservation of energy, yet the math in the question is 100% correct. It shows that energy flux in a light beam scales in some sense non-linearly with intensity.

I figured out just now

Use XAxis function solves the problem. As you can see I have f plotted.

Hello Priya Natarajan,

maybe the following post helps you:

There are also more posts linked within the one I have written above, which can help also.

Kindly regards,

Giuseppe

whatever you select , make sure that you are really familiar with the subject and that you have enough background of what you are talking about..and that you like the subject. the audience will and can feel this...there is nothing worse than someone applying for a job with an "inflated"presentation..a single question from a reviewer can kill you ..and never use abbreviations where you dont know the meaning, orginin and context

Hi Smrity,

Thank you very much for your answer and the paper. The SIW paper is very interesting! I could use it for a fanout design in one of our boards.

Best regards,

Wei

Dear Mr Macana, In RF terminology I think you are referring to radomes. Which allow unrestricted RF energy while physically protecting antennas, especially radar and avionics equipment antennas. The most common type is made by application of special resins on E or S glass fabrics. The resins are combination of some of the materials mentioned by Mr Mulla. But they do have a frequency response, at higher frequencies attenuation tends to increase for basic fabric/resin radomes. For higher frequencies usually glass based materials are used. Besides attenuation and heating deviation in the path of energy flow is also a measure of transparency of the material.

Hello Macana..."The microwave radiation as such has no detrimental

effect on the protein quality". Proteins treated with ionizing radiation results in cleavage of large protein molecules into smaller ones. Some studies demonstrated both fragmentation and aggregation occurs.

In general, sulfur containing and aromatic acids are the most sensitive to irradiation (Oxidation of the sulfur of thiol (S–H) and disulfide (S–S) groups of amino acids). In aromatic and heterocyclic amino acids, hydroxylation of aromatic ring is the principal reaction.

I think maybe we are just talking about different protocol layers? It's a bit like asking why IEEE 802.3, the Ethernet standard, doesn't address security. Security is built on top of Ethernet. At least, that's the typical way of implementing security.

The other point is that a telecom does need to implement security measures, but this is primarily to protect itself. For example, their system management messaging, routing tables and routing protocols, have to be secure, to prevent a hacker from disrupting the system. But it doesn't do much good for the telecom to secure user traffic, if ultimately, at the edges, the traffic has to be sent in the clear. If a user of the telecom wants security, that user must install a solution that protects traffic end to end, from inside his PC or smartphone, all the way to inside the PC or smartphone at the other end. Or in the case of an enterprise network, security could be implemented until the link is inside that enterprise's "secure enclave," perhaps not all the way to the end system. So for example, the enterprise may deploy secure tunnels, through a telecom network, where the end points of those secure tunnels are inside secure enclaves.

Otherwise, anyone along that path would be able to monitor the communications. And there is no reason for a user to trust the telecom itself, ultimately, if that user needs security.

Security can and is implemented at any protocol layer. From the standpoint of a user, though, any security measure that is not end to end would be of questionable value. That's why people use Transport Layer Security so often, when dealing with communications across telecom or other network services. It's security built over the telecom link, carried transparently by the telecom or by the digital network.

There are two famous methods for getting phase. The waveguide method (WGM) (its name from waveguide surrounded walls) and Floquet method (famously known as periodic structure). The WGM have some limitations, i.e. this method is valid only for normal incident, however there are many oblique incident angles for ordinary feeding reflectarray system. Secondly, the waveguide is simpler in measurement , just put the unit cell in a rectangular waveguide with cutoff frequency lower than the interested band. The method is based on surrounding the cell by two sided E-walls(bounded the port polarized E-field) while the other walls is H-walls. The top and bottom are free space with termination of waveguide port. The Floquet used periodic structure in all surrounded walls while the top and bottom walls are terminated by ports where it can be easily change its oblique incident angle. The floquet is experimentally difficult .For normal incident, both methods should be matched. WGM faster than Floquet but floquet more general oblique incident.

Thank you Ma'am for your guidance. I think, from dispersion curve, I will get resonant frequency of different mode. I want to know field patterns of different modes. How to identify TM_23 / TE_14 / HEM_10 etc modes...rectangular and cylindrical waveguide are available in Standard books on EM theory such as Harrington, Balanis, Schekunoff etc. I'm waiting for your kind guidance and reply.

Hi,

Take a look at the attached file. it covers both the issue of correct port definition as well as differential ports.

Can you upload your HFSS file so that we can try to solve this problem....

@Paiboon Yoiyod I am doing the same but problem is time axis, I am not getting proper results on required time, second for unknown distance how to know the time delay.

If you do not have these two papers, please inform us. We have. We can share it.

The waveguide won't let low frequencies through.  In waveguide, the impedance changes with frequency.  For the mode you are propagating, the impedance increases as the frequency drops, until it is infinite at the cut-off frequency.  Read about waveguide and cut-off frequency.

still couldn't find the manual.

If the objective is to obtain dual band only such that f2 = 3.5*f1, then it is better to select antenna parameters in such a way that one mode occur at f1 and other band occur at 3.5*f1. Some adjustment may be required for optimal excitation.

Hi,

A very interesting question, even it's a question for all who are working in metamaterials. Under CST microwave studio, in material assign section, we can define permeability and permittivity as well as loss tangent for new material.

Cookbook about metamaterials, I will have to look forward to.

Thanks,

I never used the program itself but, if there is someone who has some experience with it, it would be helpfull. I think in terms of bandwidth, which one is more accurate ?

See Fig. attached.

Please refer following material.

band width of micro-strip antenna (BW)  is proportional to  h/sqrt(epsilon_r)

where 'h' is the height of the substrate and epsilon_r is the dielectric constant.

But for efficient radiation, ratio W/h >> 1 , where 'W' is the width of micro-strip patch.

By increasing the substrate height, BW  can be increased, but  radiation efficiency and gain will decrease.

gain = radiation efficiency * Directivity

 Basically, BI-RME use FEM technique. This paper may help you...

Shape analysis with the ‘Boundary Integral–Resonant Mode Expansion’ method by P. Gambaa , L. Lombardib, Image and Vision Computing 17 (1999) 357–364

.

If you want to use analytical formula, you can use the solution of waveguide. You can consult this book. Advanced Engineering Electromagnetics_C A Balanis_1989

.

As you are working on SIW, please consult Prof. K. Wu's paper or Microstrip Lines and Slotlines_3e_Ramesh Garg_2013

I never seen BI-RME using analytical technique. Every body do prefer to use FEM. Please go on this particular topic.

As per my knowledge goes, SIW antenna is mainly reported based on 3D EM simulation and measurement. Some empirical formula on SIW can be found from Prof. K. Wu's papers or 3rd edition of Microstrip Lines and Slotlines by Ramesh Garg. But limited theory on SIW Antenna is available in literature. You can use CAVITY model bounded by PEC to investigate SIW antenna.

I would suggest you to follow the paper by K. Wu at IEEE (only)

Try the following link:

Move your boundary away from the port and then excite, it should work

The aperture efficiency of the reflectarray antenna is the ratio of the effective aperture to the physical aperture. The formula usually looks like:

ϵap=Ae/Ap

The aperture efficiency is a dimensionless number usually reported as a percentage and Ap is the physical aperture and Ae is the effective aperture (which is related to the antenna gain and operating wavelength) 

Ae=G*lamda*lamda/(4*pi)

the calculated gain using cst at specific wavelength

A patch antenna is a resonant antenna. Therefore the calculated resonance frequency is very sensitive to the mesh. Further, if you use CST Transient Solver (FIT or FDTD method) you have an hexahedral mesh; instead, if you use HFSS, you have a tetrahedral mesh (FEM). I suggest refining both meshes in order to verify convergence in the calculation of the resonance frequency.

Thank you Azita, I'll check them.

Answer me following.

1. How (i,j) considered?

2. where theta is used in converted matrix?

3. theta = tan inv ( (c)* Mkl / Mmn) = 27.22 degree is it correct? 

4. Mail me full paper ( details ) so that i can give complete solution. 

To Mohammad Arinal

thanks for information sharing

but Could not download the paper

plz send the correct link

I am very sorry for the late reply dear Joerg Fricke,

 firstly I admit, I was mistaken after knowing the fact that u want an inverted H field, in such case the retaining field strength of the material will be not so strong so better to use a high reflective mirror of dodecagonal ferrite oxide, or higher atomic nickel or manganese oxide. Reason is the return loss can be adjusted according to the engineered material permittivity which is possible in such artificial materials and u can bring down the S11 upto -5dB in such cases. 

There used to be simple graphs in most antenna design books showing sidelobe level as a function of edge illumination.  This is not only due to space attenuation, but also due to the beam-shape of the feed.  See Volakis Antenna Engineering Handbook 4th Ed page 15-22. for example.

A rule of thumb from the graph there is (for a circular aperture)  no taper = 17 dB sidelobes, 10 dB taper = 24 dB sidelobes, 15 dB taper = 30 dB sidelobes, 25 dB taper = 40 dB sidelobes.  The beam gets wider for more edge taper.

Thank you Peter and Sergei. It seems like Peter understood better this  question and I appreciate your advice. I also kind of sort out where did I do the mistake in calculation.

In my case frequency domain solver yields sharp resonances than time domain on that frequency range.

If the plates are less than a half wavelength apart, and the polarization is parallel to the plates, then it will not propagate, but the fields will taper off exponentially and enter and withdraw from the cavity twice each cycle.

The other polarization (perpendicular to the plates) will propagate as described in the previous answer.  If it is excited at a cylindrical surface then a cylindrical wave will propagate between the plates, with power falling off as 1/radius.

Hello Nuri gulmez.

in my opinion The best way to do that is to measure the S11 of the diode alone and from the results obtained you design the matching circuit to ensure a maximum transfer of power to the input of diode.

Best regards

Hi,

     To assign wave port, there is a simple way to create the same. first, select the face of the feed where you want the wave port. Go to MACROS>>SOLVER>>PORTS>>CALCULATE PORT EXTENSION COEFFICIENT>>CREATE PORT.

With this the software itself will create the waveguide port.

If you are using CST, you can easily match it using Design Studio. Adding the matching network, parameterizing components, tuning it, and even importing the library of components from components manufacturers. The ultimate goal is to match your antenna (load) to a complex conjugate of your source impedance. Assuming that your source reference is 50 Ohms, your antenna (load) should also have 50 Ohms impedance. That's done transforming the antenna impedance through reactances, not resistances.

thank you Marselo

Hi,

Go through this paper, it may help you.

Thanks,

Circular and ring antennas can be fed by other means, such as microstrip over a substrate, or by a via from under a substrate.  A slot ring could be fed by coplanar waveguide.  You just need to get the right voltage and current to the feed point at the same time.  Coax can be quick and easy to use so turns up quite often in papers. 

Hi,

Follow the paper.

Thanks,

During my work, I faced that problem (efficiency > 100%). To overcome that problem, design your antenna structure as per the fabricated design i.e. include thickness of ground plane, use proper coaxial probe to excite the antenna, give at least 4-5 times larger (than max dim of antenna) ground plane, try to avoid hypothetical inputs (these are designed for simulation purpose only such as wave-port, lumped port etc) etc. After incorporating all those corrections, we got excellent results in HFSS. Experimentally, we got similar results. After performing several experiments for the same sample, we got the correct procedure for designing antenna geometry in HFSS (as given above).

HFSS (or other simulators) is a tool only. We have to use it properly in research. All the best for your work.

It is better to design at least one published geometry in HFSS (or CST) and verify those results. Then go for your work using similar design procedure.

Hello dear Sajeed,

I wish that the attached paper can help you.

Best regards

Is it simulate reflection coefficient or generating reflection coefficient graph.

If it is generating reflection coefficient graph then your structure must be simulated without any errors......

Then after

Goto Results->right click->Create modal solution data report->rectangular plot-> select S(1,1)) in dB.......

Hello Mohan,

Check the attached presentation (p18). I wish that can help you

B.regards

Azita

Your waveguide port should encompass the whole cross section of the CPW - both transmission lines and also the ground plane, in case it exists.

Another point to be stressed is the multipin port - when setting the WG port you should tell which mode is to be supported by the structure. Typically, you assign a (+) to the signal line, whereas the other ground conductors should be assigned as (-), the differential mode. Do it and run the solver with "port modes only" checked - once its done check if the computed mode pattern and its impedance is the one youre expecting.

attached are the simulations results S11, 3D radiation pattern, after removing the ground plane of substrate,.

Regards

Hello Aju,

you find attached a useful document on the equivalent circuit. you can export your design from CST as dxf file.

B.regards

The matching of an array with source and line is perfect (Resistive part of load and source should be same and reactive part should be zero) will create non reflective scenario. For best match, use short circuited stub, which can bring load and source impedance same as well as reactive part zero. 

If it is acceptable, keep spacing between adjacent elements of array such a distance where coupling gets minimized. This will help in avoiding reactive impedance. As Mr Marek Klemes suggested overlay of dielectric may help in coupling reduction. However, this may shift resonance frequency and need to design element length at higher frequency