Power in path
Hi
I’m simulating a Ti:LiNbO3 waveguide. The attachment shows the profile. The starting field is Mode. I have some questions about the results.
1.Why doesn’t the intensity in the “Power in path “ graph start from the value of 1.00 (it seems only 0.32 in picture 1)? Does it mean that most light propagate out of the waveguide?
2.I couldn’t understand what the “Power overlap integral” graph means even though I have read the user’s reference. Why it goes down at first and then goes up? Are there more references to explain it more clearly?
Thank you for any help!
Attachments:
Responses (10):

June 27, 2014 at 8:15 am #12177
Hi
I’m simulating a Ti:LiNbO3 waveguide. The attachment shows the profile. The starting field is Mode. I have some questions about the results.1.Why doesn’t the intensity in the “Power in path “ graph start from the value of 1.00 (it seems only 0.32 in picture 1)? Does it mean that most light propagate out of the waveguide?
2.I couldn’t understand what the “Power overlap integral” graph means even though I have read the user’s reference. Why it goes down at first and then goes up? Are there more references to explain it more clearly?
Thank you for any help!
Attachments:

June 27, 2014 at 8:47 am #12180
the other attachment
Attachments:

June 30, 2014 at 10:20 am #12203
I think this Knowledge post answers your first question:
Let me know if it helps.

June 30, 2014 at 11:46 pm #12212
Hi,Damian Marek
Thanks a lot! I have read that post and now I can understand the definations of Relative Power and Path Monitor.I also understand why it is less than 1 at the input.In my case,the starting field is Mode so I think the “Power in path “ graph should start from the value closing to 1.00 (maybe 0.9 or 0.8).But actually it’s only 0.32 . I just think it’s too little. Is that reasonable? 
July 2, 2014 at 8:58 am #12213
Hi,
The power in path method gives the power confined to the core of the waveguide. If the waveguide (or the mode you injected) is weakly guiding and most of the power is in the cladding then it is very possible to get a value much smaller than 1.
Regards

July 6, 2014 at 8:56 am #12263
So I conclude as following:(please point out if I’m wrong )
At each z point the Power in path= (power in the waveguide)/(power in the all transverse mesh)
At each z point the Relative Power= (power in the all transverse mesh at z )/(power of input field )I’m still confused about Power overlap integral.
1.What dose the numerator and denominator mean in the formula of picture “POI.png”?
2.If the reference field is the input field, what’s the difference between Power overlap integral and Relative Power?
3.What does people usually use for with the results of Power overlap integral?Attachments:


June 27, 2014 at 8:55 am #12185

July 2, 2014 at 4:56 pm #12231
This issue is just one of precision of simulations. In waveguide.bpd, the Power overlap integral varies between 1 and 0.974. In the same project with finer mesh, larger window, and mode solver tolerance made tighter (waveguide3.bpd), the variation is from 1 to 0.996. This simulation is showing the accuracy of the simulation under the given mesh and mode solver tolerances.
Attachments:

July 6, 2014 at 9:13 am #12265
HI,Steve Dods
I’m afraid you misunderstand my question.I don’t care such little difference in this issue. What make me confused is how to understand the meaning of Power overlap integral.
Thanks anyway!

July 7, 2014 at 4:34 pm #12279
There are two normalizations for power, Global and Local. If we write the power in the transverse mesh at point z as Tm(z), and put the Input Plane at z = 0, then the local Power in Path is
Power In Path(z) = (power in waveguide)(z) / Tm(z),
and the global one is
Power In Path(z) = (power in waveguide)(z) / Tm(0)The Power Overlap Integral is an overlap integral. In the mathematical sense, it is an inner product in an L^2 Hilbert space. The Wikipedia entry on Hilbert space can tell you a lot about the background behind that notion. The reason Hilbert space is relevant in optical waveguide theory is because for waveguides, the Maxwell equations can be reduced to an eigenvalue problem, LE = (lambda) E, where lambda is actually the square of the propagation constant. L is a differential operator, and for lossless materials at least, it turns out to be selfadjoint. That means the eigenvectors (the modes) are mutually orthogonal in the sense of the L^2 inner product.
The power overlap integral is usually used in the case where you have a waveguide in the space Z > 0, and some random field in the space Z < 0. You want to know how much optical power in the space Z < 0 gets into the waveguide mode in the Z > 0 space. Let’s suppose the waveguide mode field pattern is E_2. Then imagine E_1 to be a linear combination of the orthogonal modes to be found in the Z > 0 space (an eigenvalue expansion). Note that E_2 is both an eigen mode in that expansion and the fundamental mode of the waveguide. Since the modes in the expansion are orthogonal, forming the inner product with one of the eigenmodes of that space (E_2) will yield the coefficient of the field that is applied to the fundamental mode of the waveguide. The power in the waveguide is the magnitude of that coefficient.
That explains the numerator. The denominator has the normalizations. The E_2^2 factor is required to normalize the E_2 function to be an eigenfunction of unit size, i.e. the inner product with itself is 1. If the normalization with E_1 is not used, you’ll get the total power transferred to the waveguide in Z > 0. If you divide by the E_1 factor, you’ll get a transmission coefficient instead. The power overlap integral gives you the power transmission, the amplitude of the transmission coefficient. (Insertion loss).
+1
July 13, 2014 at 8:38 am #12363
I think I finally get it. Thank you for teaching me so much and thank you for your patience! You’re really a good teacher making such complicated things easy to be understood!

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