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Math Help - Logarithm to understand amplification, get dB value

  1. #1
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    Logarithm to understand amplification, get dB value

    Hey,

    I'm at Uni studying Multimedia Technology, I've managed to avoid complex maths so far but now I'm doing a Networking module I can't hide any longer! It seems Logarithm will crop up a few times in this module.

    To give you an idea of my ability - I don't know trig, soh cah toa, I'd struggle even to work out speed, distance + time functions. In other words, I need a simple explanation!

    I'm reading about the frequency response and amplification of a signal processing system. Here's what it says:

    _____

    The input and output signal amplitudes are Si(f) and So(f), repectively. The ratio of the output signal amplitude to the input signal amplitude So(f)/Si(f) is called the system's amplification a(f) at frequency (f). The frequency response of a system is a plot of the systems amplification a(f) against the frequency f...

    Amplification, or the ratio of the output to the input amplitude, can be described as a logarithm function also. Logarithm1 to the base 10 of the ratio So(f)/Si(f), multiplied by 20 is called the decibel (dB) value of the amplification. This is expressed in the following formula:

    Amplification A(f) in dB = 20 log10 (So(f)/Si(f))

    By expressing the amplification as a logarithm function, the range of values that it takes becomes much more managable, as shown in this table:

    So(f)/Si(f) as a Ratio | A(f) in dB

    1000 | +60
    100 | +40
    10 | +20
    1 | 0
    1/10 | -20
    1/100 | -40
    1/1000 | -60

    _____

    Above, the 10 after 'log10' should be subscript but I'm not sure how to do that here. The 1 after Logarithm1 should have been superscript.

    I understand how to work out the systems frequency response but not how to work out the dB value of the amplification.

    I am really not an expert so please try to explain as simply as possible!

    Cheers,

    Ian
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  2. #2
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    The first thing is to understand what the log function is doing. Like most other operators/functions, log is the inverse of another, like square and square root.

    10^x = N, and log N = x
    10^2 = 100, and log 100 = 2

    Next, consider why we use dB. First it's as you mentioned, to make the values smaller and more manageable. Another reason to use dB is graphing the frequency response. Notice in your table of gain ratios versus dB, the gain ratio values are increasing exponentially, 1, 10, 100, 1000, ... Meanwhile, dB values are increasing linearly, 0, 20, 40, 60. This makes it easier to interpret and compare frequency plots. Distortion shows up as a curve. Easy to see if the graph should be straight in the dB plot. Hard to see in an already curving exponential plot.

    As far as working out the dB value of the amplification, you've already written the formula:

    A(f) dB = 20 log (SO(f)/SI(f))

    Just plug in each measured/calculated amplification ratio into the formula, and create a table or graph of the results, similar to your example table. Relating the formula to log...

    N = (SO(f)/SI(f))
     x = log N
     x = log (SO(f)/SI(f))
    A(f) dB = 20x = 20log (SO(f)/SI(f))

    NOTE: When plotting/graphing log/dB data you need to use semi-log graph paper (log scale only on horizontal axis).

    -Scott
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  3. #3
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    Thanks mate, I get some of what you are saying. What throws me is the (f). I don't know where that comes into the equation. Also, I don't know what properties 'log' has. E.g. if someone says what's 2+2 I know to add 2 on to 2, if someone says 2 log 2 I don't know what to do.

    There's always the possibility I'm just out of my depth here. Like I said before I don't know any functions like trig etc. All I can do is mental arithmetic like multiplications, divisions, percentages etc.

    If you can't be bothered then fair enough, but any chance of 'spelling out' how one of the table results comes to fruition. E.g. '1000 | +60'.

    How can a ratio be defined as '1000'? All I know is 2:3 style of ratios.
    What does 'log' do to that '1000' number?

    Cheers
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  4. #4
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    Quote Originally Posted by 5318008! (sorry) View Post
    Thanks mate, I get some of what you are saying. What throws me is the (f). I don't know where that comes into the equation. Also, I don't know what properties 'log' has. E.g. if someone says what's 2+2 I know to add 2 on to 2, if someone says 2 log 2 I don't know what to do.

    There's always the possibility I'm just out of my depth here. Like I said before I don't know any functions like trig etc. All I can do is mental arithmetic like multiplications, divisions, percentages etc.

    If you can't be bothered then fair enough, but any chance of 'spelling out' how one of the table results comes to fruition. E.g. '1000 | +60'.

    How can a ratio be defined as '1000'? All I know is 2:3 style of ratios.
    What does 'log' do to that '1000' number?

    Cheers
    The (f) in A(f) = SO(f)/SI(f) is the same as the (x) in f(x) = y = mx + b. It just means that A (amplification), SO (signal output), and SI (signal input) are functions of frequency, hence the f. Being a function of something, essentially means depending on something. So A, SO, and SI, depend on frequency or f. Just f(x) or y, depends on the value of x.

    So in an audio system you might test a stereo amplifier at various frequencies, say 1Hz to 50,000Hz, and measure the amplitude of SO, keeping SI constant. At each frequency, SO will have a different value, i.e. will depend on frequency. And of course, this causes the amplification, A(f), to have different values at different frequencies.

    As far as ratio being 1000, this just means the style you're used has been converted to a single decimal number. This is done by dividing the first number by the second. Remember, a ratio is just a fraction. So 2:3 is 2/3 is 0.66666. Thus a amplification ratio of 1000 means 1000:1, or 4362:4.362 etc.

    Also, I don't know what properties 'log' has. E.g. if someone says what's 2+2 I know to add 2 on to 2, if someone says 2 log 2 I don't know what to do.
    Hmm. Maybe tough to explain... First, log (logarithm) only applies to the quantity after it, log (stuff in here). So in 5 log 2, you're not doing log to or between 5 and 2, like you are with + or *. You're only doing log of 2. If you were taking the log of something more complex, you would use ()'s to show what log applies to. For example, log (SO/SI), means divide SO by SI, and then find the log of that result.

    Now, regarding log itself. Remember from my last post that log is related to raising a number to a power.

     10^x = N and x = log N

    Logarithms were developed a very long time ago, to make multiplying large numbers by hand easier (look in the back of old math texts for log tables). How? Well, remember that multiplying two numbers raised to a power, you just add the power (exponent). x^2 * x^3 = x^5. Note, the base, x, must match. Say I want to multiply 10,000 * 100,000. This is 10^4 * 10^5 = 10^9 = 1,000,000,000. Or, (log 10,000) + (log 100,000) = 4 + 5 = 9. Then, reversing the log, 10^9 = 1,000,000,000. Of course this is easy, but what about 17,287 * 106,204. It still works. Try (log 17287) + (log 106204) on your calculator. Then plug the result into the 10^x key.

    So that's what log does. How the actual log value is found, I don't know. Google or Wikipedia to find that.

    A table...

    f SI(f) SO(f) SO(f)/SI(f) log (SO/SI) A(f)=20*log(SO/SI)
    1Hz 1 1 1 0 0 dB
    100Hz 1 10 10 1 20 dB
    1000Hz 1 120 120 2.079 41.58 dB
    5000Hz 1 100 100 2 40 dB
    10000Hz 1 95 95 1.977 39.54 dB
    50000Hz 1 5 5 0.689 13.78 dB

    From the table, the amplitude of SO at f = 1000Hz is 120. Expressed as such:

    SO(1000Hz) = 120

    Further...

    A(1000Hz) = 20 * log (SO(1000Hz)/SI(1000Hz))
    A(1000Hz) = 20 * log (120/1)
    A(1000Hz) = 20 * log 120
    A(1000Hz) = 20 * 2.079
    A(1000Hz) = 41.58dB

    Help?

    -Scott
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  5. #5
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    Thanks Scott,

    Well I think I have it understood when I am given a ratio that is a multiple of 10.

    ---So say it's a ratio of 100000

    First work out how many times to multiply 10 (the base) by itself to get 100000 = 5

    Multiply that by 20 = 100dB---

    As I understand now, you can ignore the (f) in terms of including it in any calculations. It's just basically a reminder 'by the way, we're working at 100Hz here' sort of thing.

    I just set up a little test for myself, to start from the beginning and making up a ratio:

    ---Test a system at 500Hz. The SO/SI is 1:2 or .5

    log10 0.5 is - BAM it's gone!!!

    Ah well, I guess I need to learn negative multiplication or something now? At least I have a much better overall understanding now.

    I was nearly giving up but I glad I stuck to it. I'll look into it further and try and get it imprinted in my mind.

    Really appreciate your help mate, thanks!
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  6. #6
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    Quote Originally Posted by 5318008! (sorry) View Post

    I just set up a little test for myself, to start from the beginning and making up a ratio:

    ---Test a system at 500Hz. The SO/SI is 1:2 or .5

    log10 0.5 is - BAM it's gone!!!
    Ah. That's something I didn't include in my example table, negative dB. Positive dB means you get more out than you put in, ratio SO/SI > 1. Zero dB means you get the same amount out, as you put in, ratio SO/SI = 1. And negative dB means you get less out, than you put in ratio SO/SI < 1.

    The log_{10}N function matches this. If your ratio, SO/SI is less than one, log_{10} (SO/SI) will be negative, and so you'll get negative dB.

    A(500Hz) = 20 * log_{10} (SO/SI)
    A(500Hz) = 20 * log_{10} 0.5
    A(500Hz) = 20 * (-0.301)
    A(500Hz) = -6.020dB

    -Scott
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  7. #7
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    Nice one, I've got it nailed.

    Cheers dude.
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  8. #8
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    There is one mor thing I can add to all this... 'The 3dB value'

    If the amplification is equal to \sqrt{2} we can see that the value in dB would be:

    20 \log A  =  20 \log \sqrt{2} \approx 3 dB

    This value is very frequently used by electrical engineers... In 90% cases band of an amplifier, filter or some other system is defined with 3dB criterion (the band boundaries are frequencies where amplification falls by 3dB)

    What is more, since power of signal is square of its amplitude the -3dB means that the output power is two times smaller than input power...
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  9. #9
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    Cheers Albi!

    *Adds 'The 3dB Value' to grey matter*
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