How to calculate quickly gains and losses (PART 1)
I start with an example:
How to calculate quickly the output if have the next network without using the formula
[input] 3 mW ---> [ + 7 dB (gain)] ---> ? [ output]
Before of all, we know that dB es a logarithmic functions, so is not easy to calculate under that quickly. The formula is dB = 10 log (output/input).
First step.- we know by definition that +10 dB means that the output is 10 times greater than the input and if we want the half of one signal we must reduce in 3 dB.
Second step.- we disarrange the network in a gain of 10 dB and a attenuation of 3 dB, resulting the following equivalent network:
[input] 3 mW ---> [ + 10 dB (gain)] --- [ - 3 dB (attenuation)]---> ? [ output]
Third and final step.- we multiply by 10 the input (30 mW) and the result divide it by 2 (15 mW), and that is the output. So,
[input] 3 mW ---> [ + 7 dB (gain)] ---> 15 mW [ output]
The next lines show the corresponding proportions of a network using dB:
0dB implies that dosn't exist gain or attenuation or loss
10 dB implies that the output is 10 greater than the input
20 dB implies that the output is 100 greater than the input
30 dB implies that the output is 1000 greater than the input
and so on
And for
3 dB implies that the output is the double of the input (a two-times multiplier)
6 dB implies that the output is four times greater than the input (a four-times multiplier)
9 dB implies that the output is eight times greater than the input (a eight-times multiplier)
12 dB implies that the output is sixteen times greater than the input (a sixteen-times multiplier)
and so on
Knowing that, if we want calculate the following output:
[input] 6 mW ---> [ + 24 dB (gain)] ---> ? [ output]
We multiply the input by 1000 and divide by 2 twice or by 4, getting 6000 mw / 4 = 1500 mW or 1.5 W
