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What does normalization mean?

  1. Apr 11, 2012 #1
    what does normalization mean?

    for example say i have the guassian input as :

    [tex]A(0,T) = \sqrt{Po}*exp(-T^2/2To^2) [/tex]

    then we can normalize it by defining t=T/To and [tex]A(z,T) = \sqrt{Po}U(z,t)[/tex]

    Po= peak power t= normalized to the input pulse width To. if the peak of the pulse is (arbirtarily) set in t=T=0, we have U(z=0,t=0)=1 . with these notations both t and U are now dimensionless and the normalized form the gaussin input can be written as:

    [tex]U(0,t) = exp(-t^2/2)[/tex]

    i am just a bit confused as to what this means. in the normalized form the peak power dissappears and why is the normalized form uselfull is it because it makes calculations easier?
  2. jcsd
  3. Apr 11, 2012 #2


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    Hey zak8000.

    Normalization in this context of a probability density function for a normal distribution means taking a distribution that relates to N(μ,σ2) to N(0,1) while preserving the probabilities for the un-standardized LHS.

    When you introduce t = T/To, you get this standardization but to preserve the behaviour, what happens is that the limits change just like you would have in any integral substition for a definite integral.

    In short: [tex]U(0,t) = exp(-t^2/2)[/tex] reflects the PDF (need to multiply by 1/SQRT(2pi)) of the standardized normal and this also reflects what is known as the error function which is used in many different contexts.

    You should probably take a look at anything that describes the error function and for probability applications look at the topic of 'standardizing normal distributions'.
  4. Apr 11, 2012 #3
    ok thank you will look into it
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