Awesome!! This particular lecture is on my radar to revise. I have already revised the notes a bit, but have not recorded a new video. You can get links to the latest versions of everything from the course website: empossible.net/academics/emp6303/
@EMPossible The latest version is pretty nice, thanks. I was wondering if 17:03 once we have 't' in the complex format, e.g. R+j*Im, can we just say that the module of t (sqrt(R^2+Im^2)) is equal to exp(-k0*n"*d) and arctan(Im/R) is equal to k0*n'*d ? Because we can get the S-parameters in the complex format and determine t and r from there. I'm trying to apply all these things for an analogous situation with planar transmission lines to get parameter retrieval using an R&S VNA.
@@TheGreatHammys It is completely incorrect to say arctan is k0*n*d because you can add integer multiples of 2pi to k0*n*d and get the same answers. If you have a thick sample, it may be one of these that is the correct answer. This is the branching problem.
Hi Prof Rumpf, There is always so much to take away from your lectures and thanks so much for this resources. Now to my question: 1. you kept mentioning "knowing the refractive index at a previous frequency point " on slide 19. What exactly do you mean by this statement? Is this the refractive index we are working with? 2. Do we always have to take the natural log of the eigenvalue? Couldnt we have just expanded exp(i*k*z) = cos(k*z)+isin(k*z) and just equate cos(k*z) = Re[eigenvalue] like in Yeh's original work P Yeh, A Yariv, CS Hong - JOSA, 1977?
Thank you! 1) The problem is branching. This is all based on waves that repeat. This leads to multiple correct answers during parameter retrieval. Away from strong resonances, the parameters should be smooth. If you know the properties at a previous frequency point, then it becomes easier to determine the properties at the next frequency point because the correct answer will be the one that makes the line most continuous. 2) I suspect you could also use Euler's identity. The problem is not really the natural log, but the branches. When you invert the exponential (or cosine in your case), there is an infinite number of solutions. These are the branches. Hope this helped!
@@empossible1577 Thank you! Prof Rumpf At time stamp 1:05:15 slide 61, what numerical boundary condition did you use for the finite difference estimation? Periodic or Neumann?
Hey Sir, First of all very good lecture. Very interesting topics. One thing still remain unclear for me. Regarding your last technique about differentiating the phase with respect to wavelength/frequency. I did the same procedure as you with TMM algorithm from your lecture, obtain pretty the same effective refractive index, and putting this n_effective back into the TMM algorithm simulating a single layer of thickness (period*(d1+d2)). I was hoping getting the same transmission coefficients as the ones from the periodic thicknesses, but transmission coefficients don't look the same... Is that normal? Best
If you replace whatever structure you simulated with a homogeneous medium, I do expect transmission and reflection to change. They should be nearly the same at the frequency where the parameters were retrieved. Many structures do not fit the effective medium model well, especially things like photonic crystals. Metamaterials and other highly subwavelength structures fit effective medium models the best.
Purushothaman N I actually do not know. The choice of the sign is a convention and I do not think all solvers or all test equipment do it the same. I think you will have to learn a bit more about your solver to figure out which convention it is using.
I have a question on parameter retrieval, particularly, based on K-K relation. I am following the retrieval process based on Zsolt Szabó's paper. The S-parameters that we simulate using a commercial solver are in terms of magnitude and phase (deg). During the retrieval process, I am converting S-parameters into rectangular coordinates. I get correct result only when I use x-jy instead of x+jy. Is this because of time-harmonic dependence mentioned in the paper (exp(-jwt))? Is the same convention used by the commercial solver as well? Kindly clarify.
Purushothaman N That I can tell, the two conventions are used about equally. I think you have the right approach to figure it out. Try both and use the one that you know is correct.
CEM Lectures Thanks for your replies. As I understand, one needs to know about the time-harmonic term used by the commercial solver or the equipment. Is that right? Because, for an arbitrary unit cell, results obtained from using either of the conventions could be right. Maybe, in that case, we need to first verify it with a standard material or a unit cell whose EM properties are well known.
Thanks for this course!! Exactly what I was looking for.
Awesome!! This particular lecture is on my radar to revise. I have already revised the notes a bit, but have not recorded a new video. You can get links to the latest versions of everything from the course website:
empossible.net/academics/emp6303/
@EMPossible The latest version is pretty nice, thanks. I was wondering if 17:03 once we have 't' in the complex format, e.g. R+j*Im, can we just say that the module of t (sqrt(R^2+Im^2)) is equal to exp(-k0*n"*d) and arctan(Im/R) is equal to k0*n'*d ? Because we can get the S-parameters in the complex format and determine t and r from there. I'm trying to apply all these things for an analogous situation with planar transmission lines to get parameter retrieval using an R&S VNA.
@@TheGreatHammys It is completely incorrect to say arctan is k0*n*d because you can add integer multiples of 2pi to k0*n*d and get the same answers. If you have a thick sample, it may be one of these that is the correct answer. This is the branching problem.
@@empossible1577 Did you mean "it is completely INcorrect"? Thanks!
@@TheGreatHammys Woops. Yes...let me edit the above comment so others are not misled. Thanks!
Your lectures are really interesting. Thank you!!!
Purushothaman N You are welcome!
Hi Prof Rumpf,
There is always so much to take away from your lectures and thanks so much for this resources.
Now to my question:
1. you kept mentioning "knowing the refractive index at a previous frequency point " on slide 19. What exactly do you mean by this statement? Is this the refractive index we are working with?
2. Do we always have to take the natural log of the eigenvalue? Couldnt we have just expanded exp(i*k*z) = cos(k*z)+isin(k*z) and just equate cos(k*z) = Re[eigenvalue] like in Yeh's original work P Yeh, A Yariv, CS Hong - JOSA, 1977?
Thank you!
1) The problem is branching. This is all based on waves that repeat. This leads to multiple correct answers during parameter retrieval. Away from strong resonances, the parameters should be smooth. If you know the properties at a previous frequency point, then it becomes easier to determine the properties at the next frequency point because the correct answer will be the one that makes the line most continuous.
2) I suspect you could also use Euler's identity. The problem is not really the natural log, but the branches. When you invert the exponential (or cosine in your case), there is an infinite number of solutions. These are the branches.
Hope this helped!
@@empossible1577 Thank you! Prof Rumpf
At time stamp 1:05:15 slide 61, what numerical boundary condition did you use for the finite difference estimation? Periodic or Neumann?
@@levionelevis5883 I actually do not remember! I either used Neuman or just ignored the end points.
@@empossible1577 Thank you so much prof Rumpf. The prompt response was faster than the speed of light ' :-)) thanks.
Hey Sir,
First of all very good lecture. Very interesting topics. One thing still remain unclear for me.
Regarding your last technique about differentiating the phase with respect to wavelength/frequency. I did the same procedure as you with TMM algorithm from your lecture, obtain pretty the same effective refractive index, and putting this n_effective back into the TMM algorithm simulating a single layer of thickness (period*(d1+d2)). I was hoping getting the same transmission coefficients as the ones from the periodic thicknesses, but transmission coefficients don't look the same... Is that normal?
Best
If you replace whatever structure you simulated with a homogeneous medium, I do expect transmission and reflection to change. They should be nearly the same at the frequency where the parameters were retrieved. Many structures do not fit the effective medium model well, especially things like photonic crystals. Metamaterials and other highly subwavelength structures fit effective medium models the best.
Should I use x-jy or x+jy, if I use experimental S-parameter data instead of the one from commercial solver?
Purushothaman N I actually do not know. The choice of the sign is a convention and I do not think all solvers or all test equipment do it the same. I think you will have to learn a bit more about your solver to figure out which convention it is using.
I have a question on parameter retrieval, particularly, based on K-K relation. I am following the retrieval process based on Zsolt Szabó's paper. The S-parameters that we simulate using a commercial solver are in terms of magnitude and phase (deg). During the retrieval process, I am converting S-parameters into rectangular coordinates. I get correct result only when I use x-jy instead of x+jy. Is this because of time-harmonic dependence mentioned in the paper (exp(-jwt))? Is the same convention used by the commercial solver as well? Kindly clarify.
Purushothaman N That I can tell, the two conventions are used about equally. I think you have the right approach to figure it out. Try both and use the one that you know is correct.
CEM Lectures Thanks for your replies. As I understand, one needs to know about the time-harmonic term used by the commercial solver or the equipment. Is that right? Because, for an arbitrary unit cell, results obtained from using either of the conventions could be right. Maybe, in that case, we need to first verify it with a standard material or a unit cell whose EM properties are well known.
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