I have a little bit of problem understanding complex attenuation calculations when there is a shield involved between the transmitter and the receiver.

Let's take a simple example:

We have a theoretical isotropic transmitter with 0 dBi gain radiating 30 dBm (1 W) of power in all directions at 1 GHz frequency. Then we have a shield at 5 cm distance. So before the shield we reduce the power to 23.58 dBm, due to free space path loss. Furthermore the shield provides us 50 dB electric shielding, so after the shield we have −26.42 dBm power. And we have another 50 cm distance between the outer shield and the isotropic receiver antenna with 0 dBi gain. That reduces the power to -52.84 dBm. So given that both the receiver and the transmitter works at 100% efficiency, in a theoretical way, that would leave us with -52.84 dBm power at the receiving end. So the total setup would provide us with 82.84 dB attenuation. And therefore our receiver antenna would pickup only 5.2 Nano Watts from the total 1 Watt radiated.

Is my theory/thoughtprocess valid? Can we add up decibel values like this. I am still learning about dealing with decibels in a complex calculation.

Calculations done with:




  • Indeed sorry for typo and errors @Mike, also as @Andy aka pointed out yes its 0 dBi. So I recalculated everything now using the calculator links provided above (for some reason the other website doesnt load). And yes the calculator apparently uses the free-space-path-loss equation, whereas tomnexus has pointed out that I may have to use antenna path loss between antennas.
  • @Dan Mills pointed out that shielding could be different in the near field, but I specifically referred to the electrical shielding, since the magnetic shielding will be different. So let's just consider 50 dB overall electric attenuation of the shield itself from the incident wave to the transmitted wave, if that is possible.
  • 3
    \$\begingroup\$ 1 Watt is 30 dBm \$\endgroup\$
    – Mike
    Commented May 4, 2017 at 3:59
  • \$\begingroup\$ This would largely depend on the validity of your model of the effect of the shield. That is highly suspect, but then as Mike pointed out, your basic assumptions are wrong too. \$\endgroup\$ Commented May 4, 2017 at 4:15
  • \$\begingroup\$ "So before the shield we reduce the power to 35.57 dBm, due to free space path loss" No, there's no loss in free space! Power just spreads out. The path loss equation lets you calculate the fraction of the transmitted power that is intercepted by a receiving antenna. It's merely the formula for effective area of an antenna, divided by the formula for the area of a sphere. What you need to do is calculate the path loss between the antennas, in dB, and add 50 dB for the shield. \$\endgroup\$
    – tomnexus
    Commented May 4, 2017 at 4:31
  • 1
    \$\begingroup\$ "a theoretical isotropic transmitter" has 0 dBi gain as per the definition of an isotropic antenna. It can't have a gain >1 because that implies free energy. 1 watt is 30 dBm and not 40 dBm. \$\endgroup\$
    – Andy aka
    Commented May 4, 2017 at 7:36
  • 1
    \$\begingroup\$ You have some fundamental issues with your numbers as indicated already but for your underlying question the answer is yes, you can just add up the dB gains and losses like that. \$\endgroup\$
    – Andrew
    Commented May 4, 2017 at 8:18

1 Answer 1


All gains and losses are ratios. In dB this is done by addition, it's easier for mental arithmetic and the numbers are easier to read, but you can also work in raw ratios.

You should use the FSPL formula to find the fraction of a transmitted signal that's received by another antenna.
You can then consider the shielding to be an additional loss (ratio), and multiply the unshielded case by \$10^{-5}\$, or in dB, subtract 50 dB from the unshielded case.

The correct calculation is to use the free space path loss formula: \$ P_r = P_t \frac{G_1G_2\lambda^2}{(4\pi r)^2}\$
(my rule of thumb: \$3\lambda \approx 30 dB\$ )

In your example, 1 GHz, 0 dBi, 0 dbi, 55 cm, expressed in dB:, \$P_r = P_t - 27.2 dB\$

Now add the 50 dB shield, for a total \$P_r = P_t - 77.2 dB\$

If you're transmitting +30 dBm, then \$P_r = -47.2 dBm\$

On shielding:
The faulty logic still in your question is to calculate loss over distance, as though the signal got attenuated as it went along. It doesn't - it just spreads out in a lossless way.

I don't know if there's such a thing as an exact 50 dB shield. In practice the shielding effectiveness of a box is an average, changing fast with frequency, and the radiation that does leak out of the box will come out only in some directions - effectively a highly directional antenna. The figure you get in practice could vary between (say) 45 dB and 60 dB.

  • \$\begingroup\$ "The faulty logic still in your question is to calculate loss over distance, as though the signal got attenuated as it went along." > But the shield does absorb the signal via the skin effect, so I thought calculating the attenuation before and after the shield would be valid. You are basically saying that I should just calculate the overalll FSPL without the shield, and then subtract the shield from it. \$\endgroup\$
    – David K.
    Commented May 6, 2017 at 19:08
  • \$\begingroup\$ Yes obviously the shielding effectiveness changes with frequency, in this example I used a 1 GHz example of electric shielding, at near field condtions for the electric field, as an example. \$\endgroup\$
    – David K.
    Commented May 6, 2017 at 19:09
  • \$\begingroup\$ If you want to use FSPL then that's the only way. The problem with using the FSPL formula twice is that it calculates the effect of spherical spreading of power twice. If the shield is defined as a "50 dB shield" then you should only consider one round of emission, spreading, capture-by-rx-antenna. If the shield is, for example, a metal wall with a small hole in it, you do use FSPL twice. But then it's not a fixed dB shield, the overall shielding achieved depends on the whole geometry. \$\endgroup\$
    – tomnexus
    Commented May 7, 2017 at 4:26
  • \$\begingroup\$ I see. But is it accurate to use the FSPL the way you suggested? And are there other alternatives? \$\endgroup\$
    – David K.
    Commented May 12, 2017 at 7:49
  • \$\begingroup\$ I guess if the shield has a hole in it, then there is no shielding effectiveness for wavelengths smaller than the hole, some people say λ/20 should be the max hole diameter, so there is shielding effectiveness for larger wavelenghts, and FSPL for thin wavelenghts. Its a dual problem then. \$\endgroup\$
    – David K.
    Commented May 12, 2017 at 7:52

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