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Problem with formula - please help.

  1. Dec 14, 2007 #1
    Hello all, this is a formula that calculates DDM (depth of modulation) at certain outputs of the ILS's Glideslope. I posted this question some time ago but I think I left out some important details. Any help is really appreciated.


    This is basic info on the ILS
    http://en.wikipedia.org/wiki/Instrument_Landing_System

    db = 20log(E1/E2)

    E1 is the input voltage before attenuation
    E2 is the output voltage after attenuation

    E1 and db seems like is always given. E2 is what needs to be solved.

    My course book rearranges the formula to this: E2 = E1log-1(db/20)

    -that makes NO sense to me. Log to negative power - or the reciprical?

    this is one of the examples they have:

    db = -3
    E1 = .155

    E2 = .155log-1(-3/20) = 0.109

    more examples:

    2.5log-1(-3/20) = 1.77

    1.0log-1(-3/20) = 0.707

    .494log-1(-3/20) = 0.342

    Is there another way to rearrange the formula that makes sense?

    Thank you.
     
  2. jcsd
  3. Dec 14, 2007 #2

    Dick

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    They mean by log^(-1) the inverse function of log. I'll admit, that's a funny notation. But if you are dealing with base 10 logs, then log^(-1)(x)=10^x. Since log(10^x)=x.
     
  4. Dec 14, 2007 #3
    Dick, I'm not real big on math, Im not sure what you mean.

    Basically

    db = 20log(E1/E2)

    db = -3
    E1 = .155

    How would you solve for E2
     
  5. Dec 14, 2007 #4
    by the way yes, the way its written in the book its

    log^-1

    thanks
     
  6. Dec 14, 2007 #5

    Dick

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    I would write db/20=log(E1/E2), then take 10 to the power of both sides (this is log^(-1)). So 10^(db/20)=E1/E2.
     
    Last edited: Dec 14, 2007
  7. Dec 14, 2007 #6
    Yes, it is Antilog. If the base is 10, then you will find Antilogs for them in logarithmic tables. Base e can be converted into 10 if necessary (by multiplying roughly with 2.303).

    Regards,
    Sleek.
     
  8. Dec 14, 2007 #7

    Dick

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    If it's talking about decibels, then the base of the log is 10.
     
  9. Dec 15, 2007 #8
    So take this example thats in the book:

    E2 = .155log^-1(-3/20) = 0.109

    The answer is already given, but how would I enter this in the calculator? Every way I tried has yielded a syntex or math error.

    thanks in advance
     
  10. Dec 16, 2007 #9

    Hootenanny

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    The argument of a logarithm must be positive, so in this case E2 is undefined.
     
  11. Dec 16, 2007 #10

    Dick

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    It's NOT a log, it's an antilog. On the calculator I'm looking at above the 'log' key is a second function labelled [tex]10^x[/tex]. That's the 'antilog'.
     
  12. Dec 16, 2007 #11
    okay using the antilog key finally worked. I get that 0.109 answer above with no errors.

    Thanks Dick, Sleek, and Hootenanny for responding. Much appreciated!
     
  13. Dec 17, 2007 #12

    Hootenanny

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    Indeed 10^x gives the correct result, I've never seen that notation before.
     
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