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Normalizing function to 16-bit

by Choisai
Tags: 16bit, function, normalizing
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Choisai
#1
Apr10-14, 09:08 AM
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P: 21
Hey there!
I want to normalise my function. It's a superlorentzian function and looks like this:

A / ( B+ [ ( (x-C)/x0 )^n ] + D

A/B is maximal value of the function
C is the horizontal transliteration of the function
x0 refers to the width of the function (multiply this value by 8 and you have its width)
n is the order of the function
D is the vertical transliteration

I want to normalise this function to it can be used for my algorithm that uses 16-bit numbers. But how can I do this? I tried using normal normalisation (A - aminimal / amax - amin) but that didn't work out for me. How can I normalise this function so it's output will use 16 bit numbers?
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Mark44
#2
Apr12-14, 01:12 AM
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P: 21,261
Quote Quote by Choisai View Post
Hey there!
I want to normalise my function. It's a superlorentzian function and looks like this:

A / ( B+ [ ( (x-C)/x0 )^n ] + D

A/B is maximal value of the function
C is the horizontal transliteration of the function
x0 refers to the width of the function (multiply this value by 8 and you have its width)
n is the order of the function
D is the vertical transliteration

I want to normalise this function to it can be used for my algorithm that uses 16-bit numbers. But how can I do this? I tried using normal normalisation (A - aminimal / amax - amin) but that didn't work out for me. How can I normalise this function so it's output will use 16 bit numbers?
C is the horizontal translation of the function. Transliteration occurs when you change the letters from one language to those in another language. For example, the abbreviation CCCP (in Russian) becomes USSR on transliteration into an English abbreviation.


You didn't say whether your 16-bit output values are signed or unsigned, so I arbitrarily chose unsigned. An unsigned 16-bit integer has a range of 0 through 65,535.

Here's your inequality:

$$0 ≤ \frac{A}{B + (\frac{x}{x_0})^n} + D ≤ 65,535$$
Add -D to all three members:
$$-D ≤ \frac{A}{B + (\frac{x}{x_0})^n} ≤ 65,535 - D$$
Invert the fractions, which changes the direction of the inequalities:
$$\frac{-1}{D} ≥ \frac{B + (\frac{x}{x_0})^n}{A} ≥ \frac{1}{65,535 - D} $$
Multiply all three members by A, and continue a step at a time until you get x all by itself in the middle. That will tell you the interval for your x values so that the output is no larger than 65,535.

If the output can be negative, change the first line to
-32,768 ≤ <your expression> ≤ 32,767

What you wrote is slightly different from the image you attached. What you wrote has x - C. It's a simple matter to change the above to include it.
Choisai
#3
Apr15-14, 04:41 AM
Choisai's Avatar
P: 21
I included the C and isolated the x. But it isn't exactly in a formula form:

$$(\frac{-A}{D}-B)^(\frac{1}{n}){x_0}+C≤ x ≤ (\frac{A}{65,535-D})^(\frac{1}{n}){x_0}+C$$

How can I turn this into a formula? If I want to use it for my algorithm I need to be able to plug my x in and get a normalized function out. Or are these the values that I need to use as my 'New Minimum' and 'New Maximum'?

The problem with that however is that my function doesn't really have a minimum value. My superlorentzian function has a maximum of A/B and goes off nearing zero, but it doesn't actually become zero.
So this is the usual normalization formula:

$$x-OldMinimum\frac{NewMaximum - NewMinimum}{OldMaximum - OldMinimum}+NewMinimum$$

I know the new Maximum (65535) and the new minimum (0). I also know the old maximum (A/B) but the new minimum can only be found using a limit. It goes off to 0, but never becomes zero. Is this is a problem?

PS: I'm not that handy with Latex so I couldn't get the 1/n power to work

DrGreg
#4
Apr15-14, 01:28 PM
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Normalizing function to 16-bit

Quote Quote by Choisai View Post
PS: I'm not that handy with Latex so I couldn't get the 1/n power to work
You need some curly braces: X^{\frac{1}{n}}[tex]X^{\frac{1}{n}}[/tex]
Mark44
#5
Apr15-14, 03:04 PM
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P: 21,261
Quote Quote by Choisai View Post
I included the C and isolated the x. But it isn't exactly in a formula form:

$$(\frac{-A}{D}-B)^(\frac{1}{n}){x_0}+C≤ x ≤ (\frac{A}{65,535-D})^(\frac{1}{n}){x_0}+C$$
You have your inequality symbols going the wrong direction. Here's the corrected inequality (also with the exponents fixed).

Presumably the parameters of your function - A, B, C, D, n, and x0 are known. If you substitute for them, it will give you the range of x values for which f(x) is between 0 and 65, 535.
$$(\frac{-A}{D}-B)^{\frac{1}{n}}{x_0}+C ≥ x ≥ (\frac{A}{65,535 - D})^{\frac{1}{n}}{x_0}+C$$

Quote Quote by Choisai View Post

How can I turn this into a formula? If I want to use it for my algorithm I need to be able to plug my x in and get a normalized function out. Or are these the values that I need to use as my 'New Minimum' and 'New Maximum'?

The problem with that however is that my function doesn't really have a minimum value. My superlorentzian function has a maximum of A/B and goes off nearing zero, but it doesn't actually become zero.
So this is the usual normalization formula:

$$x-OldMinimum\frac{NewMaximum - NewMinimum}{OldMaximum - OldMinimum}+NewMinimum$$

I know the new Maximum (65535) and the new minimum (0). I also know the old maximum (A/B) but the new minimum can only be found using a limit. It goes off to 0, but never becomes zero. Is this is a problem?

PS: I'm not that handy with Latex so I couldn't get the 1/n power to work
Choisai
#6
Apr16-14, 03:42 AM
Choisai's Avatar
P: 21
Quote Quote by Mark44 View Post
You have your inequality symbols going the wrong direction. Here's the corrected inequality (also with the exponents fixed).

Presumably the parameters of your function - A, B, C, D, n, and x0 are known. If you substitute for them, it will give you the range of x values for which f(x) is between 0 and 65, 535.
$$(\frac{-A}{D}-B)^{\frac{1}{n}}{x_0}+C ≥ x ≥ (\frac{A}{65,535 - D})^{\frac{1}{n}}{x_0}+C$$

Thank you for your reply. The constants are indeed known. I have to find the actual values though, but for the sake of my algorithm I presume these to be known (they are actually laser beam parameters, so that is something I have yet to find out). But are these the minimum and respective maximum that I have to plug in the formula I mentioned?
You can find more about it on this Wikipedia page:
http://en.wikipedia.org/wiki/Normali...age_processing)

Are the 'new maximum' and 'new minimum' constants I can fill in in that formula? Or is it a bit more complicated?

EDIT: I am working this out in Labview and because it's a function, I guess I have to use a rounding function too? Or is that not neccesary?
By the way, I tried using it but alas it fails. In most cases the vertical translation of a Lorentzian, the D in the Lorentzian formula, is zero. This means that when finding the minimum you have to divide by zero, so that part is not working out.
Mark44
#7
Apr16-14, 09:12 AM
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P: 21,261
Quote Quote by Choisai View Post
Thank you for your reply. The constants are indeed known. I have to find the actual values though, but for the sake of my algorithm I presume these to be known (they are actually laser beam parameters, so that is something I have yet to find out). But are these the minimum and respective maximum that I have to plug in the formula I mentioned?
You can find more about it on this Wikipedia page:
http://en.wikipedia.org/wiki/Normali...age_processing)
The link you provided is broken, because you omitted the final parenthesis. Here's the link again, corrected. http://en.wikipedia.org/wiki/Normali...ge_processing)

Quote Quote by Choisai View Post

Are the 'new maximum' and 'new minimum' constants I can fill in in that formula? Or is it a bit more complicated?
At the moment I don't know. I won't have a chance to look at the wiki article until later today.
Quote Quote by Choisai View Post



EDIT: I am working this out in Labview and because it's a function, I guess I have to use a rounding function too? Or is that not neccesary?
By the way, I tried using it but alas it fails. In most cases the vertical translation of a Lorentzian, the D in the Lorentzian formula, is zero. This means that when finding the minimum you have to divide by zero, so that part is not working out.
The work that I did started with the formula you provided, which was this:

A / ( B+ [ ( (x-C)/x0 )^n ] + D

Since, as you say, the vertical translation turns out to be zero, retrace my steps using this formula:
A / ( B+ [ ( (x-C)/x0 )^n ]


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