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I am trying to calculate the power loss in a 2-port system, via the s-parameters. The actual system is a cavity with 2 waveguide ports and a dielectric mass. I know the power that is being pumped into the cavity, and I want to calculate the amount of power that was absorbed by the mass within the cavity. A simulation of the system returned the 4 s-parameters, in [dB].

My first question is regarding the conversion of the [dB] units: Since I am dealing with power, not voltage, can I use S[dB]=10log(S) to calculate the linear s-parameter values? Or do I need to use S[dB]=20log(S), because the S=Vout/Vin and not S=Pout/Pin?

My second question is about calculation of the power loss (power absorbed by the mass): Can I use {Pout}={S}{Pin}, using the linear s-parameters in {S} to calculate Pout, and then subtract Pout from Pin to find the lost (absorbed) power? Or am I going about this all wrong?

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My first question is regarding the conversion of the [dB] units: Since I am dealing with power, not voltage, can I use S[dB]=10log(S) to calculate the linear s-parameter values? Or do I need to use S[dB]=20log(S), because the S=Vout/Vin and not S=Pout/Pin?

Use 20

Can I use {Pout}={S}{Pin},

No, because that's not a correct formula. As you said before, the s-parameter gives the ratio of the wave variables (voltage or current; or E or H fields if you're working in waveguide) in the outgoing vs incoming waves, not the ratio of powers.

Use the S-parameters to find the voltages (or E-fields) of all the outgoing waves. From there calculate the power in the outgoing waves. Then calculate the ratio of the outgoing power to the incoming power. Finally you can calculate the power lost.

You will probably also want to carefully consider whether the absorption of the test sample is truly the dominant loss in the system or if there are other loss mechanisms you need to subtract off of the incoming power before using it to calculate the sample's properties.

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  • \$\begingroup\$ Thank you. But unfortunately, I am unable to solve the problem correctly. I've been given a simulation report which presents the following s-parameters: s-param, and reports an absorbed power of 94155 W. The input power is 50 kW from each of the 2 waveguide ports. I've attempted to calculate the E fields by using the formula for energy density (or intensity): S_av = P/A = (ceps0*E0^2) giving me E0_in = (2*P_in/(Ac*eps0))^0.5 \$\endgroup\$
    – randomname
    Mar 21, 2023 at 18:48
  • \$\begingroup\$ I then calculated E0_1_out = E0_in*(s11 + s12) and E0_2_out = E0_in*(s22 + s21) I then calculated P_1_out and P_2_out, using the energy density formula. Finally, I subtracted P_1_out and P_2_out from 2*P_in to get the lost (absorbed) power. If I use S_dB = 20log(S_linear), I get an absorbed power of 75.2 kW. If I use S_dB = 10log(S_linear), assuming that the reported s-parameters are ratios of the wave powers, not field voltages, I get an absorbed power of 97.9 kW. This is closer to the reported 94.1 kW, but still off. What am I doing wrong? \$\endgroup\$
    – randomname
    Mar 21, 2023 at 18:49
  • \$\begingroup\$ @randomname, your calculations seem to imply you are providing input signals at both ports, with 0 phase difference between the inputs. Is that really how you will drive the ports in your experiment? \$\endgroup\$
    – The Photon
    Mar 22, 2023 at 1:21
  • \$\begingroup\$ Yes. There are 2 identical waveguides that both pump energy into the cavity. I don't know if I can control phase shift between the ports. Is there a way to calculate how a phase shift will affect the fields within the cavity? My understanding is that the waves will bounce around the walls of the cavity, creating different regions of constructive/destructive interference, but the energy being pumped in with either be absorbed by the dielectric sample, lost in the walls, or reflected back to the source. I'm interested in getting the most amount of energy into the sample. \$\endgroup\$
    – randomname
    Mar 22, 2023 at 21:53
  • \$\begingroup\$ There was also a simulation of a single port cavity. Obviously, only S11 was provided along with the absorbed energy. I used S11[dB] = 10*log(S11) to calculate the linear S11. Then I calculated the absorbed energy by subtracting the reflected energy from the forward energy: P_loss = P_forward -P_forward*10^(S11[dB]/10). My results were the same as the reported absorbed energy so I thought I knew what I was doing, but clearly I don't because I should be using S11[dB] = 20*log(S11). I need help. P.S. Did I mention that I'm not an electrical engineer? \$\endgroup\$
    – randomname
    Mar 22, 2023 at 22:05

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