# How do you explain the unit of this formula?

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 ?

• 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. Mar 7, 2020 at 2:02

• But then if you take a look at the term inside arcsinh, the unit is $dB^{-1}$ Mar 7, 2020 at 3:36
• 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 Mar 8, 2020 at 4:55