EMI shielding (SEt) comes from two factors a) Reflection b) Absorption. For example metals shield em waves mostly due to reflection, while lossy dielectric and magnetic materials dominantly shield em waves by absorbing. That's fundamental.

Now come to shielding terminologies.

i) SEt = 10 log (Pt/Pi)=10log(T)=10log (S21^2) is the total power transmitted through the material in dB scale. This should be in negative value because transmitted power cannot exceed the incident power, until there is an another source inside the material (as in case of amplifiers). For example -10dB SEt value means 10% wave passed through the material or other way round 90% wave blockage (either due to reflection or absorption) (mostly researcher use this value).

ii) SEr = 10 log (1-R) which don't have any physical sense, only introduced to satisfy equation SEt=SEr + SEa

For Example if R=0.99 (metal like high reflection) so SEr = -20 dB (which represents the power that is not reflected, its either absorbed or transmitted).

iii) SEa= 10 log(T/(1-R)) is also meaningless in real sense, only introduced, like SEr, to satisfy the above equation.

Let me explain through an example:

Suppose we have a material (a reflection dominant, like metals) with

T=0.01 (1% transmission)

R=0.98 (98% reflection)

A= 0.01 (1% absorption)

and R + T + A = 1 (law of conservation of energy)

With above values

SEt = 10 log(T) = -20 dB = (1%)

SEa = 10 log (T/(1-R)) = -3 dB = (50%)

SEr = 10 log(1-R)= -16.98 dB = (2%)

Now Only SEt is truly showing the transmitted wave, while both SEa (-3dB) and SEr (-16.98 dB) don't show the absorption and reflection value accurately.

iv) RL=reflection loss = 10 log (R)=20 log (S11 or S22) is simply the measure of reflected power. For a given sample RL value without SEt is meaningless, for example paraffin wax has RL value of

Thank you for your valuable answer. That means that we actually require both SET and RL to check the actual performance of our materials. Ubaid Ur Rehman

Ubaid Ur Rehman can you please provide me some references for the Test case you have drawn. I cant found it from authentic source.

I'll be thankful to you.

Ibrar Ahmed I just mentioned guess values, you can check them by converting them into linear scale by employing corresponding formulae.

For example in a)

SEt= -20dB = 0.01=1% (power transmitted, hence shielding is 99%)

RL=-0.5dB= 0.891=89.1% (power reflected)

So hence this material is reflection dominant shielding material.

Ubaid Ur Rehman Got it Sir. Thank you so much for your valuable feedback. This helped me a lot.

Ubaid Ur Rehman has given a good answer, but I do not agree that SEr and SEa are meaningless terms. SEr says how much of the shielding is due to reflection, and SEa how much is due to absorption. These are two very different mechanisms that both result in shielding. If most of the shielding is due to SEa then the shielding will get hotter with high powers than if the shielding is mostly due to reflection. On the other hand, high absorption can be useful for suppressing box modes, which will not be suppressed if most of the shielding is by reflection. If most of the shielding is from SEa, the material may be good as an absorber if it has a metal backing. Where SEa and SEr are both significant, the thickness of the material with a metal backing can sometimes be tuned to create a resonant absorber near chosen frequencies and angles where the two reflections nearly cancel each other.

See these lecture slides (chapter about shielding) for a detailed derivation of the theory: https://cloud.ovgu.de/s/HMKeP8mAw6WwnBi

Malcolm White Thank you for your valuable answer. If we are utilizing VNA for testing instead of closed box/room facility, then how one can determine the hotness of sample during the test. and 2nd question is that yes in practical situation we are using metal back sample to check the shielding effectiveness but how that will be possible in lab facility where again in the case of using VNA. hope you understand my question sir.

Mathias Magdowski Thank You so much for the presentation, such a great source of information.

If you use a very thick sample then SEa may be large enough that you can measure the reflection coefficient and get SEr. You can then measure thinner samples with metal backing and observe the variation with frequency due to the interaction of the surface reflection and the reflection from the back reflector. This will give a periodic variation with frequency which you can use to find the size of the two interfering signals, and from that SEr and SEa. Alternatively, if your sample is thick enough, you can use time gating in the VNA to separate the reflections from the front and back faces and measure them. If SEa is not large you may need to use the full multiple reflection formulae such as used in designing a Fabry Perrot etalon, and in any case remember that the signal from the back face has 2SEA + 2SEr.

Malcolm White Thank you so much for your valuable information.