0
$\begingroup$

I have trouble trying to interpret the following formula:

$$G_{SCI} = \mu N^{span} G^3 \mathrm{arcsinh}\,(\rho \Delta f^2)$$

$$G_{SCI} = \frac{P_{SCI}}{B},$$ where $P_{SCI}$ is the self-channel interference noise ($W$), $B$ is the bandwidth (Hz), so the unit of $G_{SCI}$ is $W \cdot Hz^{-1}$, $\Delta f$ is also in Hz

$$\mu = \frac{3 \gamma^2}{2\pi \alpha |\beta_2|},\qquad \rho = \frac{\pi^2 |\beta_2|}{2 \alpha}$$

where $\gamma = 1.3 W^{-1} \cdot km^{-1}$, $\beta_2 = -21.3\ ps^2 \cdot km^{-1}$, $\alpha = 0.22 \ dB \cdot km^{-1}$, $N^{span}$ is the number of spans, which is linear, $G$ is the power spectral density ($W/Hz$).

The unit of $\mu$ will be $W^{-2} \cdot Hz^2 \cdot dB^{-1}$, the unit of $\rho$ will be $Hz^{-2} \cdot dB^{-1}$, the overall unit will be :

$$W^{-2} \cdot Hz^2 \cdot dB^{-1} \cdot W^3 \cdot Hz^{-3} \mathrm{arcsinh}\,(dB^{-1}) = W \cdot Hz^{-1} \cdot dB^{-1} \mathrm{arcsinh}\,(dB^{-1})$$

So it seems to me that the $\mathrm{arcsinh}\,(dB^{-1})$ will produce a measure with unit in $dB$, which I don't understand why. Could anybody please explain that to me ?

$\endgroup$
1
  • $\begingroup$ I don't have too much experience with arcsinh, but I have found that dB units tend to behave very differently than others because they represent a logairthmic quantity rather than a linear one. This is actually a major challenge in metrology at this time. $\endgroup$
    – Cort Ammon
    Mar 7, 2020 at 2:02

1 Answer 1

2
$\begingroup$

Functions that can be written as infinite polynomials don't have dimensions. So arcsinh(x) is dimensionless and x has to be dimensionless.

$\endgroup$
3
  • $\begingroup$ But then if you take a look at the term inside arcsinh, the unit is $dB^{-1}$ $\endgroup$ Mar 7, 2020 at 3:36
  • 2
    $\begingroup$ @NguyenQuangAnh dB is dimensionless $\endgroup$ Mar 7, 2020 at 21:34
  • $\begingroup$ Then arcsinh($dB^{-1}$) has unit as $dB^{-1}$ ? If so then for the first formula, the unit at LHS is $W \cdot Hz^{-1}$ while the RHS unit is $W \cdot Hz^{-1} \cdot dB^{-1} \cdot dB^{-1}$, which is very strange $\endgroup$ Mar 8, 2020 at 4:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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