It looks like you have put a thin aluminium washer in the middle of a coaxial transmission line. Aluminium is a good conductor. In CST, if it is touching the inner and outer conductors the contact is perfect and it will reflect all the power, and none will get through, unless it is less than a micron thick (skin depth at a few Gigahertz). The contact between the aluminium and the inner and outer conductors will be good in CST unless you deliberately put in a small gap or resistive layer.

In your experiment the contact between the aluminium and the inner and/or outer conductor may not be good, so you may have a section of low impedance coaxial line where the aluminium is, which looks like a shunt capacitor and will let some power through, although I would expect most of it to be reflected, unless the length of the aluminium was a multiple of a half-wavelength, which might let a lot through by resonance.

Anodised aluminium has an insulating layer all over it.

Thanks a lot, Malcolm.

I edited my simulation and kept 0.01 mm air gap between the outer conductor and aluminium substrate and also changed the aluminium thickness.

I have got these results, but I am sure are these correct, I am not able to find any literature for SE(dB) of only Aluminium metal.

Can you please review my results, are these correct or nor?

This cannot be correct. Shielding effectiveness should only have one sign, positive or negative - yours has both. Usually it is defined so that it should always be positive. How can you have a bigger field on the other side of the washer than on the side the signal arrives? That is what most of your graph indicates.

I also don't know how thick your washer is. Unless it is thinner than a micron any leakage is probably going through the gap rather than leaking through the aluminium.

I used 0.5mm and 0.035mm washers, both are giving the same curve. The magnitude is a bit higher in the thicker washer, rest the curves are identical. I aim to simulate the actual experimental coaxial transmission line method using CST. I have constructed a 50-ohm impedance coaxial aircell and put the metal washers in centre of the aircell and then by adding waveguide ports, I got the S-parameters.

Then using S21 or S12, I plotted SE graph using

SE=10*log(1/(S21^2))

The result I got was uploaded above, and its not correct.

What did I do wrong?

Is it even possible to compute SE of a metal using this method, as this technique is solely experimental and performed using Network Analysers?

Do you know what skin depth is? At 1 GHz in aluminium the skin depth in aluminium is about 1 micron. Your washer is 35 microns thick. There should be about 300 dB of attenuation through the washer. If cst has calculated anything correctly it must be leakage round the edge, and if you have measured anything it must be leakage round the washer, not through it.

S21 must be less than 1 for anything that doesn't have power gain (an amplifier). If you square it and invert it it must be greater than 1. How can the log of this be negative? You should be able to see what is going wrong.

I have kept the thickness of 0.5 microns, the graph looks the same.

The skin depth of Al at 300 MHz is around 4.7 micron, but the SE at 300 MHz is coming out around 32dB. While the skin depth of Al at 2 GHz is 1.8 microns and the SE is only 1.757dB.

How is it possible?

how do I get the SE vs Frequecy plot?? Ravi Kumar

I am not able to get this.

Your SE plot does not make sense next to your S21 plot.

At 1 GHz S21 is about -0.22 dB. This means that 95% of the power gets through, so the shielding effectiveness is poor. At 2 GHz S21 is -0.8dB so 83% gets through, so the shielding is more effective. What is the formula for SE? Why does your SE get worse at higher frequencies when less power is getting through?

I have exported the S21 parameters in text (.txt) file, then imported in MATLAB and used the formula: SE(dB) = 10*log(1/(S21^2)) to plot the SE(dB) v/s frequency graph.

I am attaching the S21 text file.

Dear Ajay,

I have used the MATLAB to plot the SE v/s freq.

I first imported the S21 data into excel, then used the SE(dB) formula " SE(dB) = 10*log(1/(S21^2)) " and then imported the SE and frequency values in MATLAB and plotted them.

I have attached the S21 text file above.

Ravi

The S21 numbers you have sent us are in dB. The formula expects you to use the magnitude which is 10^(S21dB/20).

All you have to do with the S21 in dB is to change the sign to get SE in dB.

Thanks a lot Malcolm White

thanks for the response Ravi Kumar Malcolm White

I have one more doubt is S21 alone calculates total shielding effectiveness or it calculate SE due to reflection only??

Also through CST I got S11 parameter.

I have attached a file for the shielding effectiveness formulae. these are the correct formulae or not??

from the attached file SE (total) = SE(R) + SE(A) + SE(T) , is this correct??

S21 tells you the signal that gets to the second port from the first port. S11 will tell you how much is reflected back to port 1. Do they add up? CST allows you to see where the energy ends up - look under energy in the results tree.

means S21 gives total SE?? which is directly calculated on CST studio. Malcolm White

Your three formulae mean that

the transmission loss (SET) is the same both ways round and is S21 or S12

SEA is the ratio of [power that doesn't get back to the input port] to [transmitted power], so is the inverse of the shielding NOT due to reflection, e.g. ohmic heating, and doesn't include reflection loss.

SEX is the attenuation due to reflection, so doesn't include ohmic heating, for instance.

This assumes that none of the energy is going anywhere else - like radiation for example.

The power and energy budget data in CST lets you know if anything isn't being included. If the losses and the reflection and the transmission don't add up to the incident power (0.5 W) then either CST has gone wrong (it can happen) or there is power going somewhere you haven't thought of.

The formulae should all have modulus signs round the S-parameters, not just in the first equation.

thanks Malcolm White I'll look into it.

I've done simulation in CST MWS to calculate the SE of material.

I got this curve which shows S11 > S21.

Is it correct?? Malcolm White Ravi Kumar

can anyone give me some idea how to do a convergence study in CST.

I don't know if this is correct. Actually it shows S11

I think the equations are wrong because they have confused complex amplitude and power, and in other ways. The 20 at the front means the brackets should be amplitude, however the definition of R makes it power, if the brackets mean modulus. Then R is the reflected power and 1-R is the power that gets into the material. That means the equations should be 10log10, not 20 log10, unless the path goes through the material (and is reflected) twice!

Also, it should not be T/1-R, it should be T/(1-R). T needs to be |S21|^2, which is not defined on the page you give.

The shielding is made up of losses because power is reflected and doesn't get through - as with metal - added to losses by absorption of any that isn't reflected, as in radar absorbing material, added to any losses from resonant materials designed to have a stop-band. Multiple internal reflections is not the only way of doing that, metamaterials can also do that.

Your material has a good match, so there is almost no shielding because of reflection. 1-R is about 1-1/100000000. It has about 40 dB loss, so the shielding is about -40 dB.

Using the absolute value of S11^2 (should be |S11|^2 not [S11]^2, perhaps) gives the power ratio, which is what dB expresses. You are already displaying 20log10(|T|). You need to convert 20log10(|R|) back to power to calculate 1-R.

As defined, all the dB values should be negative.