## Understanding the deciBel or dB

VHF Marine Band radios, protocol, radio communication theory, practical advice; AIS; DSC; MMSI; EPIRB.
jimh
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Joined: Fri Oct 09, 2015 12:25 pm
Location: Michigan, Lower Peninsula
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### Understanding the deciBel or dB

Many specifications for VHF Marine Band radio performance include expressions using deciBels. The deciBel is a measure of power ratios using logarithms that evolved from early engineering work for the Bell System at the Bell Telephone Laboratory, and was originally the "Bel" in honor of Alexander Graham Bell. The Bel was often too large to be useful, so it was divided by ten and became the deciBel.

If two power levels, P1 and P2, are to be compared with deciBels, the relationship is defined as

(1) dB = 10 × log(P1/P2) where log is to Base-10

Here is an example: how much stronger in deciBels is a power of 25-Watts compared to 7-Watts?

Given:
P1 = 25-Watts
P2 = 7-Watts
dB = 10 × log(P1/P2)
dB = 10 × log(25/7)
dB = 10 × log(3.57)
dB = 10 × 0.553
dB = 5.53

If we reverse the power ratio comparison, and ask how much weaker in deciBels is a power of 7-Watts compared to 25-Watts, the same formula is used:

Given:
P1 = 7-Watts
P2 = 25-Watts
dB = 10 × log(P1/P2)
dB = 10 × log(7/25)
dB = 10 × log(0.28)
dB = 10 × -0.553
dB = -5.53

If a power ratio is stated in deciBels and we want to find the numerical ratio, we can manipulate equation 1 to solve for P1/P2. This gives:

(2) (P1/P2) = 10(dB/10)

For example, if a power ratio is expressed in deciBels to be 5.53 dB, what is the numerical ratio (P1/P2):

Given:
dB = 5.53
(P1/P2) = 10(dB/10)
(P1/P2) = 10(5.53/10)
(P1/P2) = 10(0.533)
(P1/P2) = 3.57

jimh
Posts: 11862
Joined: Fri Oct 09, 2015 12:25 pm
Location: Michigan, Lower Peninsula
Contact:

### Re: Understanding the deciBel or dB

Another form of notation describing two power levels is to calculate the GAIN necessary to increase one power level to the other. For example, if P2 is increase by gain G, the result is P1:

P1 = P2 × G
This is easily restated to be

G = P1/P2

We can say that a dB is related to gain G by

dB =10 × log(G) where log is to Base-10

For example, if an power amplifier has a gain of G=4, what is the gain in deciBels?

dB = 10 × log(4)
dB = 10 x 0.602
dB = 6.02

In a similar manner we calculate a gain of G=2:

dB = 10 × log(2)
dB = 10 x 0.301
dB = 3.01

For G=10
dB = 10 × log(10)
dB = 10 x 1
dB = 10

And, finally, for a gain of 1.25:

dB = 10 × log(1.25)
dB = 10 x 0.97
dB = 0.97

With these simple relationships of 1, 3, 6 and 10-dB, we can calculate many dB rations from gain figures in our heads. Note that Gain multiplies while deciBels add. For example, to roughly figure the gain an amplifier with 30 dB gain, we factor the 30 dB into 10 dB + 10 dB + 10 dB, which is equivalent to a gain of 10 x 10 x 10 = gain of 1,000.

Working from a gain figure to deciBels the process goes the other way. If there is a gain of 25, we factor that into 2 x 2 x 2 x 2 x 1.25 x 1.25 = 25 , and convert to dB as 3 + 3 + 3 + 3 + 1 + 1 = 14 dB. Compared with the exact figure for G=25 of 13.98-dB, our factoring estimate is very close.