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I'm doing some experiments with optical power through a glass fiber. The power is measured in dBm. For the graphs that I make (Power in dBm vs time), I normalize the values relative to the initial power in dBm. But if I calculate the power to milliWatts, $P_{mW} = 10^{P_{dB}/10}$, and than normalize it, it gives a different value. This is quite obvious because of the log-scale, but is it wrong to normalize the power output in dBm?

Thanks in advance!

Gregory
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  • By any chance, by "normalize in dBm" do you mean dividing by the initial power in dBm? – AV23 May 30 '15 at 19:32
  • That is true. I normalize it by dividing a measured power with the initial power. – Gregory May 30 '15 at 19:36
  • Putting what @AV23 said in another way - could you please describe your process of normalizing? It might make it easier to give you a sensible answer. – Floris May 30 '15 at 19:36
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    @Gregory - do you divide the dB value of the power, or the power expressed in mW? You can do the latter - but if you want to do the former you have to subtract: $$log{\frac{p_1}{p_2}} = \log{p_1} - \log{p_2}$$ – Floris May 30 '15 at 19:37
  • For example, I have a initial value of -2dB and after some time I have -3dB. I normalize it by $\frac{-2}{-3} = 0,67$. But if I calculate it to milliWatts, I get values of 0,63mW and 0,5mW respectively. This gives than $\frac{0,5}{0,63} = 0,79$. I was wondering if the first way of normalizing it is wrong, because of the different answers. Sorry for the unclear explanation of my problem. – Gregory May 30 '15 at 19:45
  • Yes, it is wrong to normalize by dividing dB values. – The Photon May 30 '15 at 20:55

1 Answers1

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So that we are on the same page, the basic relations are:

\begin{equation} \tag{1} P_\text{dBm} = 10\log\left (\frac{P_{\text{mW}}}{0.001} \right) \end{equation}

\begin{equation} \tag{2} P_{\text{mW}} = \frac{10^{P_\text{dBm}/10}}{1000} \end{equation}

Now imagine you wanted to do a normalisation with miliwatts, you'd simply do: \begin{equation} \tag{3} P_{\text{Normalized}}=\frac{P_\text{mW out}}{P_\text{mW in}} \end{equation}

So that when there is no loss in the experiment you get $P_{\text{Normalised}}=1$, and when all the light is lost you get $P_{\text{Normalised}}=0$ . Now if we sub in equation 2 into equation 3 we get:

\begin{align} P_{\text{Normalized}}&=\frac{10^{P_\text{dBm out}/10}}{10^{P_\text{dBm in}/10}} \\ &=10^{P_\text{dBm out}/10-P_\text{dBm in}/10} \\ &=\left ( 10^{P_\text{dBm out}-P_\text{dBm in}} \right )^{1/10}\\ \tag{4} \end{align}

Now this suggests to us, that if we want something which is related to a normalised value in logarithmic units, we need to subtract instead of divide. This is actually the great thing about logarithmic units, instead of dividing you can subtract, which is much easier. So for example if you input 0dBm into your fibre and collect -10dBm on the other side, you know that you have a 10dB loss (notice that loss is in dB not dBm as it is effectively a normalised value).

Carrying on with the derivation for completeness, if we define the loss $L$ to be:

\begin{equation} L = P_\text{dBm in}-P_\text{dBm out} \tag{5} \end{equation}

Then

\begin{equation} L = -10\log(P_{\text{Normalized}}) \tag{6} \end{equation}

So in conclusion subtract your logarithmic values and divide your linear ones, use the above expression to switch between their respective normalised quantities.

Luka Milic
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  • So, if I take an example with $P_{in}=-2,3$ and $P_{out}=-2,61$. Originally this gave a normalized value of $\frac{2,3}{2,61}=0,88$. I can understand that this is wrong and if I calculate this with your equation 4, I get 0,93 as a normalized value. So I get a higher value, but I was just expecting to get a lower value because of the logarithmic scale right? – Gregory May 31 '15 at 05:16
  • It makes sense to me that the log ratio of 0.88 is smaller than the actual one of 0.93. Because when you convert to the log scale you make small features in your data easier to see, so the ratio between two signals is lowered.

    You might want to plot some examples so you can see visually how this works.

     – Luka Milic May 31 '15 at 09:53