3
\$\begingroup\$

please explain

How is 0.22 dB/km equal to 0.0507 km^(-1)

I tried doing 10 log x = its value in dB but it doesn't work here

\$\endgroup\$
6
\$\begingroup\$

Depending on the units of the loss coefficient \$\alpha\$, there are two ways to calculate optical loss in a fiber, or any other uniform medium.

For \$\alpha\$ in units of [1/length],

$$ \frac{P}{P_0} = e^\left({-\alpha_{1/km} L}\right) $$

For \$\alpha\$ in units of [dB/length]

$$ {P \over P_0} = 10^{-\alpha_{dB/km} L/10} $$

You can set these two expressions for \$P/P_0\$ equal to eachother, and solve algebraically,

$$ \ln(e^{-α_{1/km} L}) = \ln(10^{-\alpha_{dB/km} L / 10} ) $$

$$ -\alpha_{1/km} L = -\alpha_{dB/km} L / 10 \cdot \ln{10} $$

Then, assuming that your units for \$L\$ are the same, (in this case, they are both km), cancel \$L\$.

$$ \boxed{\alpha_{1/km} = \alpha_{dB/km} \cdot \frac{\ln (10) }{ 10}} $$

Putting your numbers in,

$$ 0.0507/\text{km} = 0.23 \cdot 0.22 \text{dB/km} $$

As long as you have the same unit on both sides (km, m, lightyears, angstroms) this conversion ratio will always be \$\ln(10)/10\$. If you need to convert units, you will want to step through the algebra a little bit more carefully, particularly the penultimate step where you cancel out \$L\$.

\$\endgroup\$
1
  • \$\begingroup\$ If this answer was useful, mark it as the accepted answer to help others find it. \$\endgroup\$ Jun 7 '17 at 19:23
3
\$\begingroup\$

As the question is tagged with "optical-fiber", I assume you mean the damping of the fiber.

In this context, both numbers cannot be equal because a damping greater than 0 dB would be equivalent to a damping factor greater than 1.

A damping of 0.22 dB would mean that the input power is \$10^{0.022} = 1.05\$ times higher then the output power.

\$\endgroup\$
1
  • \$\begingroup\$ Let me read out the full line. It is from Fiber optics and optoelectronics by RP Khare. Thus if we take alpha(symbol) to be 0.22 dB km^-1( which is equivalent to a coefficient of 0.0507 km^(-1) at 1.55 um, Leff = 20 km. Please try it now \$\endgroup\$
    – SaJ
    Dec 4 '15 at 16:45
2
\$\begingroup\$

Converting dB's to a power ratio is done by dividing by 10 then raising 10 to that number (antilog). On my calculator it gave me an answer of 1.05196.

However, this is not as per your number of 0.0507.

Maybe the 0.0507 number you have is somehow in error or maybe the 0.22 dB is an approximation. For instance 10log(1+0.0507) = 0.21479 dB and if this is rounded up to 2 decimal places you get 0.22 dB

\$\endgroup\$
0

Not the answer you're looking for? Browse other questions tagged or ask your own question.