1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dumb question about decibels

  1. May 29, 2014 #1
    1. The problem statement, all variables and given/known data
    Let E1 be the electric field 1cm into a person's skin and E0 be the electric field on the surface of that person's skin. If the field loses -486db/m and E0 = 64.1, find the value of E1.


    2. Relevant equations

    E1/E0 = k, with k < 1


    3. The attempt at a solution

    Well I'm mostly confused about the use of 20log vs 10log. My solution is:

    We have -486dB/m, or -4.86dB/cm.

    10log(E1) = 10log(k) + 10log(E0), where 10log(k) must be negative since k < 1.

    We know that 10log(k) = -4.68, thus k = 10^(-4.68/10) = 0.3404

    However, my teacher wrote the following equation:
    20log(E1) = 20log(k) + 20log(E0)

    Which gives a different value of k, obviously. I thought decibels were always 10log(ratio), and the only reason why 20log(ratio) sometimes appear is when you have a ratio of squared values, such as 10log(P1/P2) = 10log([V1^2/R]/[V2^2/r]) = 10log(V1^2/V2^2) = 20log(V1/V2) in electronics.

    I'd like to know if 10log or 20log should be used in this situation. Thanks!
     
  2. jcsd
  3. May 29, 2014 #2
    One more thing. Would I be right to assume that using the 20log equations give the ratio k of power density loss that corresponds to the electric field loss? I ask this because the teacher uses his ratio k that he found with the 20log equations, then argues that since P1/P0 = E1^2/E0^2, the ratio of P1/P2 = k^2... I thought that using 20log would already give the "correct k" for the P1/P0 ratio without needing to square it.
     
  4. May 29, 2014 #3

    tms

    User Avatar

    You've basically said it yourself:
    $$10\ \log\, \left( \frac{P_1}{P_2} \right) = 10\ \log\, \left( \frac{E_1^2}{E_2^2} \right) = 20\ \log\, \left( \frac{E_1}{E_2} \right). $$
    You are using field strength in this problem, so ##\ldots##
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Dumb question about decibels
  1. Dumb question (Replies: 15)

  2. Quick Decibel question (Replies: 2)

  3. Decibel question (Replies: 4)

Loading...