can someone explation the formula: 20xlog10(H1/H2) regarding decibels calculations.?

Answer 1

You know 10^1 = 10, 10^2 = 100, 10^3 = 1000, 10^4 = 10000 etc.

Similarly you know log 10 = 1, log 100 = 2, log 1000 = 3, log 10000= 4 etc.

Suppose a quantity is 10 times greater than its initial value, another way of saying this is, it is 1 bel.

Therefore, if we say that the quantity is of one bel we understand that it is ten times its initial value.

Similarly if we say that the quantity is 4 bel, we understand that it is 10^4 times greater than its initial value and so on.

Suppose there are two values of a quantity H1 and H2 and their ratio is (H1/H2).

As already explained, if we say that this quantity is having X bel, we know, H1 is 10 ^ X times greater than the initial value H2.

Or H1/H2 = 10 ^ X.

The above equation can also be written as X = log (H1/ H2).

Suppose a quantity is 10 decibel.

Since deci means (1/10) or 0.1, 10 decibel is 1 bel.

Then (H1/H2) = 10^1

Or 1 bel = log (H1/H2).

Or 10 decibel = 10 log (H1/H2)

In your question some quantity Y = 20 log (H1/ H2).

Therefore Y/ 20 = log (H1/ H2)

Or 10 ^ (Y/20) = (H1/ H2)

Thus the quantity is Y/20 bel. Or H1 is 10^(Y/20) greater than H2.

Y/20 bel can be written as (Y/2) decibel.










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Answer 2

The decibel is a logarithmic unit; this means that each increment represent a multiplication of the value, not simply an added amount; for example, each 6db of voltage decibel represents a factor of two in magnitude. The formula you gave is the formula for voltage db. If you were measuring power (like the output of an amplifier for speakers), the formula would be db= 10log(P1/P2). The logarithm is to the base 10. In this case, every 3db represents a doubling of power level. It is important to know when reading a spec which formula is used. Therefore if H1 = 2*H2, then H1/H2 = 2; log(2) is .30 (get it from a table of logarithms), and 20*.301 = 6; therefore 6db means that H1 is twice H2. If it were the other way around, and H2 = 2H1, then the log becomes log(1/2) = -log(2) = -.30. The result is that H1 is 6db lower than H2.

Answer 3

The general idea represented is that the decibel scale is not linear. The more decibels, the greater the loudness increase when one increases a particular increment.

For example, increasing from 10 decibels to 50 decibels, an increase of 40dB, that is a certan amount louder. But 140 dB is comparably many times louder than 100 dB, though also a 40 dB increase.

Answer 4

holy......... ****, lol. im just now going into algebra 2........ if you really want to, i can try to explain it.... but, it would be the exquivalence of a 2 year old, lol.

Answer 5

no such luck from me, good luck. takare mate