Understanding Normalization in Gaussian Inputs

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Normalization in Gaussian inputs refers to the process of transforming a Gaussian distribution into a standard form, specifically N(0,1), while maintaining the underlying probabilities. By defining a new variable t = T/To, the Gaussian input can be expressed in a dimensionless form, simplifying calculations. The normalized function U(0,t) = exp(-t^2/2) represents the probability density function of the standardized normal distribution. This transformation is useful as it allows for easier integration and application in various statistical contexts, including the error function. Understanding normalization is essential for accurate probability analysis and mathematical modeling.
zak8000
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what does normalization mean?

for example say i have the guassian input as :

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

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

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:

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

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?
 
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zak8000 said:
what does normalization mean?

for example say i have the guassian input as :

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

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

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:

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

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?

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: U(0,t) = exp(-t^2/2) 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'.
 
ok thank you will look into it
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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