TAGGED: absorptioncoefficient, coplanar, fdtd, MODE, photonics, transmission, waveguides


March 27, 2023 at 1:54 amgpsarSubscriber
I want to match transmission (meaning the abs^2 value of the Sparameters of port 2) from FDTD with MODE. I use this equation (1), α = α_dB*ln(10)/20 to measure the absoprtion coeffient of the MODE, and then use this equation (2), T=T_0*e^(αL), where L is the distance between the 2 ports, to measure transmission of MODE, so as to match the transmission from FDTD. But, the graphs only match when I take into account a factor of 2 either at the exponent of equation 2 or divide equation 1 by 10 instead of 20. Can someone explain to me why that is? How can I match the transmissions? What is the relationship of the Sparameters with the electric field?

March 27, 2023 at 11:05 pmGuilin SunAnsys Employee
You do not need to do such conversion, as in FDE of MODE, it has neff=real+i*imag. so it is
E=E0*exp(j*k0*neff*r)=E0*exp(k0*imag(neff)*r)*exp(j*k0*r)
eg, alfa=2*imag(neff)*k0
Please check this: Working with lossy modes and dB/m to \(\kappa\) conversion – Ansys Optics
I tested using the given formula and the results mtaches each other. Please check your script.
please check your expression α = α_dB*ln(10)/20.

March 27, 2023 at 11:46 pmgpsarSubscriber
To understand this better, you mean that it's not necessary to convert loss in dB/mm to 1/mm, I can just use neff and use this formula alfa=2*imag(neff)*k0 to find alpha?
And what about α = α_dB*ln(10)/20, is this wrong?

March 28, 2023 at 6:03 pmGuilin SunAnsys Employee
I believe you can use alfa=2*imag(neff)*k0 and I got the eaxctly the same value as the loss in dB/cm.

April 3, 2023 at 7:17 pmgpsarSubscriber
In your formula, E=E0*exp(j*k0*neff*r)=E0*exp(k0*imag(neff)*r)*exp(j*k0*r), the formula is expressed in terms of electric field but what we are measuring through the transmission monitor or port, is not electric field, but energy. I agree that neff is related to alpha, but how can I relate it to Sparameters from the FDTD simulation to what I get from the MODE? From the alfa=2*imag(neff)*k0 equation, where does the 2 come from?

April 4, 2023 at 8:48 pmGuilin SunAnsys Employee
Sorry usually we do not use energy which is a timedomain quantity and depends on the time signal. I guess you mean the power transmission. then it is abs(S)^2.
S parameter is similar to Fresnel coefficients. if sum(abs(S))^2 is less than one the system has loss. However,associating S with the loss parameter might not be that direct, as the loss is a propagation property but S is endtoend property. You may need to develop your own method to do this.
"2" is for intensity or power quantities.
DIfferent solvers are for different purposes and have different sources of errors.
This forum is mainly for simulation.

May 8, 2023 at 11:24 pmgpsarSubscriber
I have been trying what you suggested but can't seem to get the transmissions to match, so I just have a follow up question to clarify something in case I'm making a simple mistake. In alfa=2*imag(neff)*k0, is imag(neff) the same as kappa from the link you provided and k0 = 2*pi(n+i*kappa)/λ_0?

May 9, 2023 at 12:12 amGuilin SunAnsys Employee
yes, you are right. You can prove it in theory, as the lossy field is expressed as E0*exp(jk0*z) and k0 is complex. Please note that for such loss calculation MODE FDE is the most accurate method, which does not have any source injection error, numerical dispersion error, and monitor interpoation error ad other errors in FDTD due to discretization and simulation.

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