Understanding why we multiply log(P/Po) by 10

  • Thread starter dervast
  • Start date
In summary, log10(P/Po) is multiplied by 10 to convert to dB because log\frac{P}{P_0}, called Bel is too large a unit to be useful.
  • #1
dervast
133
1
Hi i want to ask sth for the db
We have denoted db as log10(P/Po)
I want to ask sth if a^x=theta=>x=log(a)theta
then the db is 10*a^x=10*log(a)theta Is that right?
I want to know why we multiply the log(P/Po) by 10? Why we didnt leave it alone?
 
Engineering news on Phys.org
  • #2
dervast said:
Hi i want to ask sth for the db
We have denoted db as log10(P/Po)
I want to ask sth if a^x=theta=>x=log(a)theta
then the db is 10*a^x=10*log(a)theta Is that right?
I want to know why we multiply the log(P/Po) by 10? Why we didnt leave it alone?
Because [tex]log\frac{P}{P_0}[/tex], called Bel is too large a unit to be useful. So we use the deciBel, that is more manageable.
 
  • #3
If bel is too large to be manageable then why we multiplt by 10?
 
  • #4
Dervast, it is the same thing as microfarads vs farads. Most of the time we deal in signal ratios that make it more suitable to use decibels instead of bels. Same thing with farads and microfarads. However, if you choose NOT to multiply by 10 and work with bels it is up to you. It will work just fine.
 
  • #5
dervast said:
If bel is too large to be manageable then why we multiplt by 10?
I think you are looking for an analytical reason.
Try applying dimensional analysis.
log (Pout/Pin) = y (Bel) equ (i)
how to convert to dB? Since a dB is 1/10 of a Bel, then 10db = 1 Bel
To convert equat (i) to db, multiply both sides by (10db/1Bel)
10 log (Pout/Pin) = y (Bel)(10db/1Bel) <------dimensions Bel/Bel = 1 leaving just db
10 log (Pout/Pin) = 10y (db) (ii)
right side of equation (ii) means we multiply the solution y in Bels by 10 to obtain a value in dB.
 

Related to Understanding why we multiply log(P/Po) by 10

1. Why do we multiply log(P/Po) by 10?

The reason we multiply log(P/Po) by 10 is because log(P/Po) is in the form of a logarithm with base 10. This means that the value of log(P/Po) represents the power to which 10 must be raised to equal the ratio of P to Po. By multiplying log(P/Po) by 10, we are essentially converting the value from a logarithm to a decimal form, making it easier to understand and work with.

2. What does log(P/Po) represent?

Log(P/Po) represents the relative change in a quantity, where P is the current value and Po is the initial or reference value. It is often used in scientific and mathematical calculations to express changes in a scale-invariant manner.

3. Can we use a different base for the logarithm?

Yes, we can use a different base for the logarithm. However, in the context of understanding why we multiply log(P/Po) by 10, using base 10 is most commonly used and makes the most sense. Other common bases used in logarithmic calculations include base e (natural logarithm) and base 2.

4. How does multiplying log(P/Po) by 10 affect the final result?

Multiplying log(P/Po) by 10 does not change the relative change or ratio expressed by the logarithm. It simply converts the value into a decimal form, which may be easier to work with and interpret. The final result will still represent the same relative change in the quantity.

5. What is the significance of using logarithms in scientific calculations?

Logarithms are useful in scientific calculations because they allow us to express large or small numbers in a more manageable form. They also help to simplify complex equations and make them easier to solve. Additionally, using logarithms can help to compare changes in quantities on a scale-invariant basis, making it easier to interpret and understand the data.

Similar threads

Replies
2
Views
1K
  • Electrical Engineering
Replies
7
Views
2K
Replies
19
Views
2K
Replies
10
Views
1K
  • Electrical Engineering
Replies
2
Views
8K
  • Calculus and Beyond Homework Help
Replies
5
Views
378
  • Electrical Engineering
Replies
1
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
684
  • Advanced Physics Homework Help
Replies
2
Views
929
Back
Top