Convert cm^-1 to dB: Confirm Formula

[cyrylabidi]

Hi,

i have found a formula (in Waveguide Loss Measurement Using the Reflection Spectrum, PHOTONICS TECHNOLOGY LETTERS, VOL. 20, NO. 16, 2008) to calculate losses in semiconductor structures such as waveguide. I want to compare the result with another method (described in Photonic crystal waveguides with propagation losses in the 1 dB/mm range, J. Vac. Sci. Technol. B 22(6), Nov/Dec 2004) but from 1st method i get cm^-1 while from 2nd dB. I also have found (i can't remember where) that:
10 cm^-1 = 10 dB,
100 cm^-1 = 20 dB
so

[tex]\alpha [dB] = 10*log \alpha [cm^{-1}] [/tex].

It is correct? Can anybody confirm this, or say what is wrong.

I woluld appreciate yours help.


cyrylabidi

 

[Redbelly98]

I don't think this is right. Isn't dB normally used to represent an actual loss, as opposed to the loss per unit length? That's why they use "dB per mm" or "dB per cm" to express loss-per-length.

I also have found (i can't remember where) that:
10 cm^-1 = 10 dB,
100 cm^-1 = 20 dB

Problem: what if alpha is 0.1 cm^-1, for example? That is still a loss, but your formula gives a negative loss of -10 dB, which really means there is a gain(!)

From what I remember:

S = So e^{- alpha*L}​

where S is the signal strength (in energy or power units), So is initial signal, and L is the length the signal has propagated through a medium with loss rate alpha.

To express the loss in dB,

dB_loss = 10*log(So/S)​

Combine these equations to get

dB (loss) = 10*log(e^alpha*L)

= 10*log(e)*alpha*L

= 4.34*alpha*L​

If you want dB per cm, then

dB per cm = 4.34*alpha, if alpha units are cm^-1​

 

[cyrylabidi]

marcusl, Redbelly98 thanks for your reply.

Redbelly98 i think your equations are good, but i obtain different results from 1st and 2nd method e.g.

1st method 0,449 dB per cm
2nd method 0,223 cm^-1 = 0,99 dB per cm
another sample
1st method 10,21 db per cm
2nd method 1,4048 cm^-1 = 6,223 dB per cm

 

[Redbelly98]

Redbelly98 i think your equations are good, but i obtain different results from 1st and 2nd method e.g.

1st method 0,449 dB per cm
2nd method 0,223 cm^-1 = 0,99 dB per cm
another sample
1st method 10,21 db per cm
2nd method 1,4048 cm^-1 = 6,223 dB per cm

The 2nd method calculations are in close agreement with the formula I posted yesterday.
The 1st method would appear to be wrong. Without seeing the actual formula for Method 1, or details of your calculation, I can't comment further.

 

[cyrylabidi]

From my measurements i obtain Fabry-Perot fringes (Graph1) the formula for 1st method is in graph 1, alpha unit is [dB/cm], R is the facet reflectivity. For calculations with 2nd method i did FFT of F-P fringes (FFT), formula is in pdf file.

 

[Redbelly98]

I must admit I'm not familiar with using this method to measure loss. However, in the 1st method (in graph1.pdf), the formula has no length dependence in it, so I don't see how a dB/cm value can be calculated from this. A dB value, okay, but not dB/cm.

Also:
What is the value of R?
What is the lowercase k along the x-axis in graph1.pdf?

 

[cyrylabidi]

Hi, at the beginning thanks for help.

However, in the 1st method (in graph1.pdf), the formula has no length dependence in it, so I don't see how a dB/cm value can be calculated from this. A dB value, okay, but not dB/cm.

my fault the formula should be: alpha = (10 / L) log ...

The reflectivity is R=0,3 and k along the x-axis is a wavevector (units are reciprocal micrometers).

I must admit I'm not familiar with using this method to measure loss.

so do you know another method for loss calculation from this kind of measurments?

 

[Redbelly98]

Sorry, I'm not familiar with loss calculations using this kind of measurements. My initial reply was just to give the proper relation between dB and loss rate.

The first method appears quite sensitive to what you use for Imax and Imin. If I use 0.0060 and 0.0020, then I get K=3 and dB=0.49 dB (divide by L to get dB/cm). However, if I use 0.0060 and 0.0018, then K=3.33 and dB=0.12, a factor of 4 smaller. That's all with R=0.3; if that changes a little to, say, 0.32 then we get 0.77dB and 0.40dB for those K values. How accurately do you know the value of R?

In the second method, is one supposed to subtract the broad "background" spectrum to calculate the peak values, or just use the peak values read directly from the FFT graph?

Sorry I can't be of more help.

EDIT:
To compare the two methods, you may want to create a simulated signal, one without any noise so that Imin and Imax are consistently the same throughout the entire scan. Then the FFT should be a pretty clean signal too, without the broad background in the spectrum. Hopefully that will answer whether the problem has to do with your measurement technique, or there is a real inconsistency between the two calculation methods.

For a test signal, I would generate a pure sine wave, with a DC offset of course.

 

[cyrylabidi]

How accurately do you know the value of R?

I use a AlGaAs / GaAs waveguide and i calculated value of R for my spectrum of wavelength (for given Al fraction). In my posts i used R=0,3 as a example.

In the second method, is one supposed to subtract the broad "background" spectrum to calculate the peak values, or just use the peak values read directly from the FFT graph?

I used the same spectrum that for 1st method, i have measured broader spectrum about 10 nm. Of course final calculations will be done on broder spectrum.

Thanks for suggestion with simulation of signal and then calculation of losses.

 

[cyrylabidi]

Redbelly98 i have generated 4 sin signal with different amplitudes (different K in 1st method) and calculated losses with 1st and 2nd method. The resuslts are comparable:
e.g. result from 1st method:
6,95E+00 dB/cm
4,52E+00 dB/cm
1,40E+00 dB/cm
1,24E-01 dB/cm

results obtained with 2nd method (FFT with Hamming window):
7,204906231 dB/cm
5,356638983 dB/cm
2,362340179 dB/cm
0,377940168 dB/cm.

Then i compared results from 2nd method with different windows (all for the same generated signal, using 1st method i get 1,40E+00 dB/cm for this signal):
Rectangular 3,321784722 dB/cm
Welch 2,533779803 dB/cm
Hanning 2,146725115 dB/cm
Hamming 2,363943097 dB/cm
Blackman 1,993401144 dB/cm

According the above i have 2 questions:
1) Which one window should i use to calculations, the most comparable results i obtain with Blackman window.
2) How can i calculate an error bar for 2nd method?

 

[Redbelly98]

According the above i have 2 questions:
1) Which one window should i use to calculations, the most comparable results i obtain with Blackman window.

Can you make the period of your test sine-wave match up with the overall length of the signal? I.e. the signal length is exactly an integer multiple of the period. Then using no window should give the "correct" answer, which you can then compare to the different window methods.

It might be good to list the actual I0/I1 ratios for the different windows (including no window), in order to get a good sense of how the different windows are altering the result.

Also, I would think you need to subtract the broad background part of the FFT in order to calculate the peak height. I mentioned this earlier (Post #8), but am not sure if you understood what I meant.

2) How can i calculate an error bar for 2nd method?

You would have to first estimate the errors in the peak heights, then carry those errors through the calculation of alpha.

 

[friebele]

Here's the answer:

Optical Density (OD) = log10(Io/I)
1 OD = 10 dB
alpha = Absorbance (cm-1) = ln(Io/I)/x = 2.303 OD/x = 23.03 dB/cm
or, 1 dB/cm = 0.0434 alpha.

1 dB/km = 10^5 dB/cm

 

[Redbelly98]

Welcome to Physics Forums

1 OD = 10 dB

Actually, it's

1 dB = 10 OD​

That will give the result shown in Post #2.