What are valid and invalid operations based on decibel (dB)?

Digital ElectronicsElectronics & ElectricalElectron

Valid and Invalid Operations on dB

Some of the valid operations that can be performed with ‘dB' is discussed in this article. When we add ‘dB’ to a quantity, we are increasing its value and when we are subtracting ‘dB’ from a quantity, we are reducing its value.

We can add ‘dB’ to ‘dBm’ but not ‘dBm’ to ‘dBm’.

37 dBW + 3 dB gives 40 dBW. However, 37 dBm + 37 dBm doesn’t yield 74 dBm. This is an invalid operation.

$$37\:dBm=10log_{10}(\frac{P_{out}}{1mW})$$

$$log_{10}(\frac{P_{out}}{1mW})=3.7;\:P_{out}\sim\:10^{3.7}mW\sim\:5W$$

Also, 74 dBm ≈ 25 MW. When we add 37 dBm to 37 dBm, all we get is around 10 kW, which is nowhere around 25 MW. Thus, it is invalid to add values in dBm. The same rule applies to subtraction.

Some invalid operations are shown below

OperationValid / Invalid
7 dBm + 3 dB = 10 dBmValid
7 dBm + 7 dBm = 14 dBmInvalid
4 dBmV – 2 dBmV = 2 dBmVInvalid
25 dBW x 25 dBW = 625 dBWInvalid
10 dBV + 3 dB = 13 dBVValid

Invalid Operations on dB

1. 4 dBmV – 2 dBmV ≠ 2 dBmV. Let us validate this inequality.

$$4dBmV=20log_{10}(\frac{P_{OUT}}{1mV})$$

$$log_{10}(\frac{P_{OUT}}{1mW})=0.2;\:P_{OUT}\sim\:10^{0.2}mW\sim\:1.585mW$$

Similarly, 2 dBmV maps to 1.259 mW.

4dBmV − 2dBmV ≡ 1.585 mW − 1.259 mW = 0.326 mW

2dBmV=1.259mW ≠ 0.326mW

2. 2 dBµW + 5 dBµW = 7 dBµW

$$2dB\mu\:W=10log_{10}(\frac{P_{OUT}}{1\mu\:W})$$

$$log_{10}(\frac{P_{OUT}}{1\mu\:W})=0.2;\:P_{OUT}\sim\:10^{0.2}\mu\:W\sim\:1.585\mu\:W$$

5 dBμW maps to 3.162 μW.

$$2dB\mu\:W+5dB\mu\:W=1.585\mu\:W+3.162\mu\:W=4.747\mu\:W$$

7 dBμW maps to 5.011μW.

Thus, we observe that 5.011 μ𝐖 ≠ 4.747 μ𝐖

Thus, we have seen how to operate on decibels. The knowledge of such operations is required for applications such as wireless link budget drafting,path loss modeling, and so on.

raja
Published on 23-Jun-2021 14:15:47
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