How can I convert loss units, dB/mm, to mm^-1

In summary: Does that help?I am not good at code, but I don't see these refractive index values being used in your code. You need to include them in the calculations. Also is this a plane wave with linear loss? or is there an exponential factor? (as Baluncore suggested with E vs P)
  • #1
gps57
4
0
I am trying to simulate mode loss for a waveguide and the units I get for loss are dB/mm, but I want to convert the units to just mm^-1 so I can the absorption (α) value. I am guessing there is a specific value that I need to either multiply or divide my values with, but I'm just not sure what that value is.
 
Engineering news on Phys.org
  • #2
Welcome to PF.

The dB is just one way to express a ratio as a logarithm, so addition can replace multiplication, and multiplication can replace exponentiation.

I believe you are correct, you may just need to convert the distance the signal takes to fall by 1 dB, into the distance the signal takes for to fall to 1/e.

Can you please give us the equations you are using that employ the absorption coefficient, α.
 
  • Like
Likes BvU
  • #3
Baluncore said:
Welcome to PF.

The dB is just one way to express a ratio as a logarithm, so addition can replace multiplication, and multiplication can replace exponentiation.

I believe you are correct, you may just need to convert the distance the signal takes to fall by 1 dB, into the distance the signal takes for to fall to 1/e.

Can you please give us the equations you are using that employ the absorption coefficient,
Thank you, and thank you for your reply.

The equation I have is T = I_0 * e^{-x*α}, where T is the transmission and α the absorption coefficient. Essentially, I need graphs that show α vs frequency, but I'm just having trouble with converting the units, as I mentioned in the question.
 
  • #4
Are you considering the power, or the electric or magnetic field?
What is the form of your equation given dB?
Do you use; dB =10⋅Log(power ratio); or, dB = 20⋅Log(amplitude ratio) .
 
  • #5
Baluncore said:
Are you considering the power, or the electric or magnetic field?
What is the form of your equation given dB?
Do you use; dB =10⋅Log(power ratio); or, dB = 20⋅Log(amplitude ratio) .
Hi, I have these 2 equations: V=V_0*e^(-αL) and V=V_0*10^(-α_dB*L/20) where α_dB is loss in dB/mm and α is attenuation which has 1/mm units. When I solve for α, I get α = α_dB*ln(10)/20, but when I use
T = T_0 * e^{-L*α} to convert it to transmission, the values are not correct. The image shows a plot of what I'm trying to achieve. I want to match the MODE with the FDTD transmission. When I divide by 10 rather than 20, I get the correct values, but I can't seem to find these equations anywhere to check if they are correct. Am I using the correct equations? Where do the first 2 equations come from, it's still not clear for me?
 

Attachments

  • T vs f 100um (1).jpg
    T vs f 100um (1).jpg
    10.2 KB · Views: 70
Last edited:
  • #7
TonyStewart said:

I have tried this before, here is the code
Code:
tra1filedata = genfromtxt("T vs f (1mm L=100).txt") #transmission data from FDTD
lossfiledata = genfromtxt("loss vs f xspan 200.txt") #loss data from MODE
loss_dB = lossfiledata[:,1]/1000

lambda_0 = 0.25*2.27 #wavelength
kappa = (loss_dB*lambda_0)/(20*2*np.pi*0.43) 
k0 = (2*np.pi*2.27 + kappa*1j)/lambda_0

alpha1 = 2*kappa*k0

alpha2 = 10**(loss_dB/10)

alpha = loss_dB*((math.log(10))/20) 

T = np.exp(alpha1*-1*0.1) #transmission converted from loss MODE

plt.plot(lossfiledata[:,0]*1e-12, T, label = "T @ 100um (MODE)");

plt.plot(tra1filedata[:, 0]*10e-13, tra1filedata[:, 1], label="T @ 100um (FDTD)");
For the GaAs I'm simulating, the refractive index is 3.6 and the imaginary refractive index is 0.01.
But then I'm not exactly sure how to plot it or if the code I wrote so far is correct. As you can see, I tried 3 different equations for alpha, and none of them have yielded the result I want. Would I still need to use
T = T_0 * e^{-L*α}? Because I'm essentially trying to convert the loss data from MODE to match with the transmission data from FDTD. I know the simulations are correct because the loss from both methods matches, so the error must be in the equations I'm using, I'm just not sure what it is.
 
Last edited by a moderator:
  • #8
I am not good at code, but I don't see these refractive index values being used in your code. You need to include them in the calculations. Also is this a plane wave with linear loss? or is there an exponential factor? (as Baluncore suggested with E vs P)
 

1. How do I convert dB/mm to mm^-1?

To convert from dB/mm to mm^-1, you can use the following formula: mm^-1 = 10^(dB/mm/10). This will give you the conversion in units of mm^-1.

2. What is the relationship between dB/mm and mm^-1?

dB/mm and mm^-1 are both units of measurement for attenuation, or the decrease in intensity of a signal as it travels through a medium. dB/mm is a logarithmic unit, while mm^-1 is a linear unit. They are related by the formula: dB/mm = 10*log(mm^-1).

3. Why is it important to convert from dB/mm to mm^-1?

Converting from dB/mm to mm^-1 allows for easier comparison and analysis of different materials or signals. It also allows for more accurate measurements and calculations in scientific research and engineering applications.

4. Can I convert from dB/cm to mm^-1 using the same formula?

No, the formula for converting from dB/mm to mm^-1 is specific to those units. To convert from dB/cm to mm^-1, you would use the formula: mm^-1 = 10^(dB/cm/10)/10. This takes into account the difference in units (cm vs mm) in the conversion.

5. What other units of measurement are commonly used for attenuation besides dB/mm and mm^-1?

Other units of measurement for attenuation include decibels per meter (dB/m), decibels per kilometer (dB/km), and nepers per meter (Np/m). These units are also related to each other through logarithmic or linear formulas.

Similar threads

Replies
4
Views
648
Replies
6
Views
1K
Replies
55
Views
3K
  • Electrical Engineering
Replies
12
Views
38K
  • Electrical Engineering
Replies
11
Views
1K
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
785
Replies
1
Views
963
  • Other Physics Topics
Replies
2
Views
690
Replies
10
Views
1K
Back
Top