Histogram equalization - Contrast (gray value) distribution using transforming function
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I am looking for some guidance and explanation to be able to solve this exercise.
I understand that I have to come up with a function T(x) such that h(x) becomes H(x) and then with a U(x) which transforms back H(x) to h(x), but I am not sure on how to tackle this problem.
Exercise
calculus image-processing
asked Dec 3 '18 at 21:22
TristanTristan
1
$endgroup$
add a comment |
$begingroup$
I am looking for some guidance and explanation to be able to solve this exercise.
I understand that I have to come up with a function T(x) such that h(x) becomes H(x) and then with a U(x) which transforms back H(x) to h(x), but I am not sure on how to tackle this problem.
Exercise
calculus image-processing
asked Dec 3 '18 at 21:22
TristanTristan
1
$endgroup$
$begingroup$
Can you elaborate, especially on T, h, H, U etc. ? At the moment it's hard to help.
$endgroup$
– Imago
Dec 3 '18 at 21:25
2
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@Imago Please check the image added under Exercise hyperlink.
$endgroup$
– Tristan
Dec 3 '18 at 21:27
$begingroup$
Still Looking for an answer.
$endgroup$
– Tristan
Dec 3 '18 at 23:01
add a comment |
$begingroup$
I am looking for some guidance and explanation to be able to solve this exercise.
I understand that I have to come up with a function T(x) such that h(x) becomes H(x) and then with a U(x) which transforms back H(x) to h(x), but I am not sure on how to tackle this problem.
Exercise
calculus image-processing
asked Dec 3 '18 at 21:22
TristanTristan
1
$endgroup$
I am looking for some guidance and explanation to be able to solve this exercise.
I understand that I have to come up with a function T(x) such that h(x) becomes H(x) and then with a U(x) which transforms back H(x) to h(x), but I am not sure on how to tackle this problem.
Exercise
calculus image-processing
calculus image-processing
asked Dec 3 '18 at 21:22
TristanTristan
1
asked Dec 3 '18 at 21:22
TristanTristan
1
asked Dec 3 '18 at 21:22
TristanTristan
1
asked Dec 3 '18 at 21:22
TristanTristan
1
asked Dec 3 '18 at 21:22
TristanTristan
1
1
$begingroup$
Can you elaborate, especially on T, h, H, U etc. ? At the moment it's hard to help.
$endgroup$
– Imago
Dec 3 '18 at 21:25
2
$begingroup$
@Imago Please check the image added under Exercise hyperlink.
$endgroup$
– Tristan
Dec 3 '18 at 21:27
$begingroup$
Still Looking for an answer.
$endgroup$
– Tristan
Dec 3 '18 at 23:01
add a comment |
$begingroup$
Can you elaborate, especially on T, h, H, U etc. ? At the moment it's hard to help.
$endgroup$
– Imago
Dec 3 '18 at 21:25
2
$begingroup$
@Imago Please check the image added under Exercise hyperlink.
$endgroup$
– Tristan
Dec 3 '18 at 21:27
$begingroup$
Still Looking for an answer.
$endgroup$
– Tristan
Dec 3 '18 at 23:01
$begingroup$
Can you elaborate, especially on T, h, H, U etc. ? At the moment it's hard to help.
$endgroup$
– Imago
Dec 3 '18 at 21:25
$begingroup$
Can you elaborate, especially on T, h, H, U etc. ? At the moment it's hard to help.
$endgroup$
– Imago
Dec 3 '18 at 21:25
2
2
$begingroup$
@Imago Please check the image added under Exercise hyperlink.
$endgroup$
– Tristan
Dec 3 '18 at 21:27
$begingroup$
@Imago Please check the image added under Exercise hyperlink.
$endgroup$
– Tristan
Dec 3 '18 at 21:27
$begingroup$
Still Looking for an answer.
$endgroup$
– Tristan
Dec 3 '18 at 23:01
$begingroup$
Still Looking for an answer.
$endgroup$
– Tristan
Dec 3 '18 at 23:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think this is more probability theory question. The trick is, if $x sim F(x) $ ($F(x)$ is the cumulative distribution function or CDF in abbreviation form) then the transformation $ u = F(x)$ produces the uniform random variable $u$ with uniform distribution on interval $[0,1]$. you can see link below for proof:
Show Y has a uniform distribution if Y=F(X) where F(x)=P[X $le$ x] is continuous in x.
Also to construct an arbitrary distribution $x sim F(x)$ from a uniform random variable $u sim U([0,1])$, it's enough to form the random variable $x = F^{-1}(u)$ (again $F(x)$ is the CDF of $x$). This one is the result of previous one if you change u with x. Therefore hre you need to apply the transformations to get the desired function:
So we have $h(x) = 6x^5 Rightarrow F_x(x) = x^6 , xin [0,1]$.
Then $y = x^6$ has the uniform distribution according to above lemmas.
To get a random variable with distribution $H(X) = 1.8X+0.1 Rightarrow F_X(X) = 0.9X^2 + 0.1X$ you need to calculate the inverse of $F(X)$ which is $frac{-0.1+sqrt{0.81+4X}}{1.8}$ and then introduce the random variable $z = frac{-0.1+sqrt{0.81+4X}}{1.8}$ on the uniform random variable obtained from previous step. combining these two gives the transformation $t = frac{-0.1+sqrt{0.81+4x^6}}{1.8}$.
Therefore if $x$ is distributed as $6x^5$, then $t$, defined as $t = frac{-0.1+sqrt{0.81+4x^6}}{1.8}$ is distributed as $1.8t+0.1$.
answered Dec 3 '18 at 22:06
K.K.McDonaldK.K.McDonald
954618
$endgroup$
$begingroup$
Thanks for the response @K.K.McDonald, I have found your answer quite confusing though. I am looking for two functions. T(x) = ? which will transform h(x) to become H(x) and U(x) = ? which will transform H(x) to become h(x)
$endgroup$
– Tristan
Dec 3 '18 at 22:58
$begingroup$
To achieve $H(x)$ from $h(x)$, first you apply a transform that gets you from $h(x)$ to a uniform distribution $u(x)$. then you apply the second transform $z$ to get to $H(x)$. The uniform transformation is indeed an interface between these two. for further explanationsyou can see the srochastic process book by Papoulis: amazon.com/Probability-Random-Variables-Stochastic-Processes/dp/…. page 140, equation 5-43 and three other equations before this equation.
$endgroup$
– K.K.McDonald
Dec 4 '18 at 11:01
$begingroup$
Thank you for the reply. @K.K.McDonald So what could T(x) and U(x) be in this specific exercise? To be able to validate your thought process I will try to check if those answers are correct and let you know. In this online exercise I have to input those two functions, then we can see if what you said is correct.
$endgroup$
– Tristan
Dec 4 '18 at 15:59
$begingroup$
okay, wish you success. if that worked right, don't forget to accept my question, you know, reputation thingee. although I've already ran it in MATLAB before and checked it.
$endgroup$
– K.K.McDonald
Dec 7 '18 at 14:31
add a comment |
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1 Answer
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$begingroup$
I think this is more probability theory question. The trick is, if $x sim F(x) $ ($F(x)$ is the cumulative distribution function or CDF in abbreviation form) then the transformation $ u = F(x)$ produces the uniform random variable $u$ with uniform distribution on interval $[0,1]$. you can see link below for proof:
Show Y has a uniform distribution if Y=F(X) where F(x)=P[X $le$ x] is continuous in x.
Also to construct an arbitrary distribution $x sim F(x)$ from a uniform random variable $u sim U([0,1])$, it's enough to form the random variable $x = F^{-1}(u)$ (again $F(x)$ is the CDF of $x$). This one is the result of previous one if you change u with x. Therefore hre you need to apply the transformations to get the desired function:
So we have $h(x) = 6x^5 Rightarrow F_x(x) = x^6 , xin [0,1]$.
Then $y = x^6$ has the uniform distribution according to above lemmas.
To get a random variable with distribution $H(X) = 1.8X+0.1 Rightarrow F_X(X) = 0.9X^2 + 0.1X$ you need to calculate the inverse of $F(X)$ which is $frac{-0.1+sqrt{0.81+4X}}{1.8}$ and then introduce the random variable $z = frac{-0.1+sqrt{0.81+4X}}{1.8}$ on the uniform random variable obtained from previous step. combining these two gives the transformation $t = frac{-0.1+sqrt{0.81+4x^6}}{1.8}$.
Therefore if $x$ is distributed as $6x^5$, then $t$, defined as $t = frac{-0.1+sqrt{0.81+4x^6}}{1.8}$ is distributed as $1.8t+0.1$.
answered Dec 3 '18 at 22:06
K.K.McDonaldK.K.McDonald
954618
$endgroup$
$begingroup$
Thanks for the response @K.K.McDonald, I have found your answer quite confusing though. I am looking for two functions. T(x) = ? which will transform h(x) to become H(x) and U(x) = ? which will transform H(x) to become h(x)
$endgroup$
– Tristan
Dec 3 '18 at 22:58
$begingroup$
To achieve $H(x)$ from $h(x)$, first you apply a transform that gets you from $h(x)$ to a uniform distribution $u(x)$. then you apply the second transform $z$ to get to $H(x)$. The uniform transformation is indeed an interface between these two. for further explanationsyou can see the srochastic process book by Papoulis: amazon.com/Probability-Random-Variables-Stochastic-Processes/dp/…. page 140, equation 5-43 and three other equations before this equation.
$endgroup$
– K.K.McDonald
Dec 4 '18 at 11:01
$begingroup$
Thank you for the reply. @K.K.McDonald So what could T(x) and U(x) be in this specific exercise? To be able to validate your thought process I will try to check if those answers are correct and let you know. In this online exercise I have to input those two functions, then we can see if what you said is correct.
$endgroup$
– Tristan
Dec 4 '18 at 15:59
$begingroup$
okay, wish you success. if that worked right, don't forget to accept my question, you know, reputation thingee. although I've already ran it in MATLAB before and checked it.
$endgroup$
– K.K.McDonald
Dec 7 '18 at 14:31
add a comment |
$begingroup$
I think this is more probability theory question. The trick is, if $x sim F(x) $ ($F(x)$ is the cumulative distribution function or CDF in abbreviation form) then the transformation $ u = F(x)$ produces the uniform random variable $u$ with uniform distribution on interval $[0,1]$. you can see link below for proof:
Show Y has a uniform distribution if Y=F(X) where F(x)=P[X $le$ x] is continuous in x.
Also to construct an arbitrary distribution $x sim F(x)$ from a uniform random variable $u sim U([0,1])$, it's enough to form the random variable $x = F^{-1}(u)$ (again $F(x)$ is the CDF of $x$). This one is the result of previous one if you change u with x. Therefore hre you need to apply the transformations to get the desired function:
So we have $h(x) = 6x^5 Rightarrow F_x(x) = x^6 , xin [0,1]$.
Then $y = x^6$ has the uniform distribution according to above lemmas.
To get a random variable with distribution $H(X) = 1.8X+0.1 Rightarrow F_X(X) = 0.9X^2 + 0.1X$ you need to calculate the inverse of $F(X)$ which is $frac{-0.1+sqrt{0.81+4X}}{1.8}$ and then introduce the random variable $z = frac{-0.1+sqrt{0.81+4X}}{1.8}$ on the uniform random variable obtained from previous step. combining these two gives the transformation $t = frac{-0.1+sqrt{0.81+4x^6}}{1.8}$.
Therefore if $x$ is distributed as $6x^5$, then $t$, defined as $t = frac{-0.1+sqrt{0.81+4x^6}}{1.8}$ is distributed as $1.8t+0.1$.
answered Dec 3 '18 at 22:06
K.K.McDonaldK.K.McDonald
954618
$endgroup$
$begingroup$
Thanks for the response @K.K.McDonald, I have found your answer quite confusing though. I am looking for two functions. T(x) = ? which will transform h(x) to become H(x) and U(x) = ? which will transform H(x) to become h(x)
$endgroup$
– Tristan
Dec 3 '18 at 22:58
$begingroup$
To achieve $H(x)$ from $h(x)$, first you apply a transform that gets you from $h(x)$ to a uniform distribution $u(x)$. then you apply the second transform $z$ to get to $H(x)$. The uniform transformation is indeed an interface between these two. for further explanationsyou can see the srochastic process book by Papoulis: amazon.com/Probability-Random-Variables-Stochastic-Processes/dp/…. page 140, equation 5-43 and three other equations before this equation.
$endgroup$
– K.K.McDonald
Dec 4 '18 at 11:01
$begingroup$
Thank you for the reply. @K.K.McDonald So what could T(x) and U(x) be in this specific exercise? To be able to validate your thought process I will try to check if those answers are correct and let you know. In this online exercise I have to input those two functions, then we can see if what you said is correct.
$endgroup$
– Tristan
Dec 4 '18 at 15:59
$begingroup$
okay, wish you success. if that worked right, don't forget to accept my question, you know, reputation thingee. although I've already ran it in MATLAB before and checked it.
$endgroup$
– K.K.McDonald
Dec 7 '18 at 14:31
add a comment |
$begingroup$
I think this is more probability theory question. The trick is, if $x sim F(x) $ ($F(x)$ is the cumulative distribution function or CDF in abbreviation form) then the transformation $ u = F(x)$ produces the uniform random variable $u$ with uniform distribution on interval $[0,1]$. you can see link below for proof:
Show Y has a uniform distribution if Y=F(X) where F(x)=P[X $le$ x] is continuous in x.
Also to construct an arbitrary distribution $x sim F(x)$ from a uniform random variable $u sim U([0,1])$, it's enough to form the random variable $x = F^{-1}(u)$ (again $F(x)$ is the CDF of $x$). This one is the result of previous one if you change u with x. Therefore hre you need to apply the transformations to get the desired function:
So we have $h(x) = 6x^5 Rightarrow F_x(x) = x^6 , xin [0,1]$.
Then $y = x^6$ has the uniform distribution according to above lemmas.
To get a random variable with distribution $H(X) = 1.8X+0.1 Rightarrow F_X(X) = 0.9X^2 + 0.1X$ you need to calculate the inverse of $F(X)$ which is $frac{-0.1+sqrt{0.81+4X}}{1.8}$ and then introduce the random variable $z = frac{-0.1+sqrt{0.81+4X}}{1.8}$ on the uniform random variable obtained from previous step. combining these two gives the transformation $t = frac{-0.1+sqrt{0.81+4x^6}}{1.8}$.
Therefore if $x$ is distributed as $6x^5$, then $t$, defined as $t = frac{-0.1+sqrt{0.81+4x^6}}{1.8}$ is distributed as $1.8t+0.1$.
answered Dec 3 '18 at 22:06
K.K.McDonaldK.K.McDonald
954618
$endgroup$
I think this is more probability theory question. The trick is, if $x sim F(x) $ ($F(x)$ is the cumulative distribution function or CDF in abbreviation form) then the transformation $ u = F(x)$ produces the uniform random variable $u$ with uniform distribution on interval $[0,1]$. you can see link below for proof:
Show Y has a uniform distribution if Y=F(X) where F(x)=P[X $le$ x] is continuous in x.
Also to construct an arbitrary distribution $x sim F(x)$ from a uniform random variable $u sim U([0,1])$, it's enough to form the random variable $x = F^{-1}(u)$ (again $F(x)$ is the CDF of $x$). This one is the result of previous one if you change u with x. Therefore hre you need to apply the transformations to get the desired function:
So we have $h(x) = 6x^5 Rightarrow F_x(x) = x^6 , xin [0,1]$.
Then $y = x^6$ has the uniform distribution according to above lemmas.
To get a random variable with distribution $H(X) = 1.8X+0.1 Rightarrow F_X(X) = 0.9X^2 + 0.1X$ you need to calculate the inverse of $F(X)$ which is $frac{-0.1+sqrt{0.81+4X}}{1.8}$ and then introduce the random variable $z = frac{-0.1+sqrt{0.81+4X}}{1.8}$ on the uniform random variable obtained from previous step. combining these two gives the transformation $t = frac{-0.1+sqrt{0.81+4x^6}}{1.8}$.
Therefore if $x$ is distributed as $6x^5$, then $t$, defined as $t = frac{-0.1+sqrt{0.81+4x^6}}{1.8}$ is distributed as $1.8t+0.1$.
answered Dec 3 '18 at 22:06
K.K.McDonaldK.K.McDonald
954618
answered Dec 3 '18 at 22:06
K.K.McDonaldK.K.McDonald
954618
answered Dec 3 '18 at 22:06
K.K.McDonaldK.K.McDonald
954618
answered Dec 3 '18 at 22:06
K.K.McDonaldK.K.McDonald
954618
954618
$begingroup$
Thanks for the response @K.K.McDonald, I have found your answer quite confusing though. I am looking for two functions. T(x) = ? which will transform h(x) to become H(x) and U(x) = ? which will transform H(x) to become h(x)
$endgroup$
– Tristan
Dec 3 '18 at 22:58
$begingroup$
To achieve $H(x)$ from $h(x)$, first you apply a transform that gets you from $h(x)$ to a uniform distribution $u(x)$. then you apply the second transform $z$ to get to $H(x)$. The uniform transformation is indeed an interface between these two. for further explanationsyou can see the srochastic process book by Papoulis: amazon.com/Probability-Random-Variables-Stochastic-Processes/dp/…. page 140, equation 5-43 and three other equations before this equation.
$endgroup$
– K.K.McDonald
Dec 4 '18 at 11:01
$begingroup$
Thank you for the reply. @K.K.McDonald So what could T(x) and U(x) be in this specific exercise? To be able to validate your thought process I will try to check if those answers are correct and let you know. In this online exercise I have to input those two functions, then we can see if what you said is correct.
$endgroup$
– Tristan
Dec 4 '18 at 15:59
$begingroup$
okay, wish you success. if that worked right, don't forget to accept my question, you know, reputation thingee. although I've already ran it in MATLAB before and checked it.
$endgroup$
– K.K.McDonald
Dec 7 '18 at 14:31
add a comment |
$begingroup$
Thanks for the response @K.K.McDonald, I have found your answer quite confusing though. I am looking for two functions. T(x) = ? which will transform h(x) to become H(x) and U(x) = ? which will transform H(x) to become h(x)
$endgroup$
– Tristan
Dec 3 '18 at 22:58
$begingroup$
To achieve $H(x)$ from $h(x)$, first you apply a transform that gets you from $h(x)$ to a uniform distribution $u(x)$. then you apply the second transform $z$ to get to $H(x)$. The uniform transformation is indeed an interface between these two. for further explanationsyou can see the srochastic process book by Papoulis: amazon.com/Probability-Random-Variables-Stochastic-Processes/dp/…. page 140, equation 5-43 and three other equations before this equation.
$endgroup$
– K.K.McDonald
Dec 4 '18 at 11:01
$begingroup$
Thank you for the reply. @K.K.McDonald So what could T(x) and U(x) be in this specific exercise? To be able to validate your thought process I will try to check if those answers are correct and let you know. In this online exercise I have to input those two functions, then we can see if what you said is correct.
$endgroup$
– Tristan
Dec 4 '18 at 15:59
$begingroup$
okay, wish you success. if that worked right, don't forget to accept my question, you know, reputation thingee. although I've already ran it in MATLAB before and checked it.
$endgroup$
– K.K.McDonald
Dec 7 '18 at 14:31
$begingroup$
Thanks for the response @K.K.McDonald, I have found your answer quite confusing though. I am looking for two functions. T(x) = ? which will transform h(x) to become H(x) and U(x) = ? which will transform H(x) to become h(x)
$endgroup$
– Tristan
Dec 3 '18 at 22:58
$begingroup$
Thanks for the response @K.K.McDonald, I have found your answer quite confusing though. I am looking for two functions. T(x) = ? which will transform h(x) to become H(x) and U(x) = ? which will transform H(x) to become h(x)
$endgroup$
– Tristan
Dec 3 '18 at 22:58
$begingroup$
To achieve $H(x)$ from $h(x)$, first you apply a transform that gets you from $h(x)$ to a uniform distribution $u(x)$. then you apply the second transform $z$ to get to $H(x)$. The uniform transformation is indeed an interface between these two. for further explanationsyou can see the srochastic process book by Papoulis: amazon.com/Probability-Random-Variables-Stochastic-Processes/dp/…. page 140, equation 5-43 and three other equations before this equation.
$endgroup$
– K.K.McDonald
Dec 4 '18 at 11:01
$begingroup$
To achieve $H(x)$ from $h(x)$, first you apply a transform that gets you from $h(x)$ to a uniform distribution $u(x)$. then you apply the second transform $z$ to get to $H(x)$. The uniform transformation is indeed an interface between these two. for further explanationsyou can see the srochastic process book by Papoulis: amazon.com/Probability-Random-Variables-Stochastic-Processes/dp/…. page 140, equation 5-43 and three other equations before this equation.
$endgroup$
– K.K.McDonald
Dec 4 '18 at 11:01
$begingroup$
Thank you for the reply. @K.K.McDonald So what could T(x) and U(x) be in this specific exercise? To be able to validate your thought process I will try to check if those answers are correct and let you know. In this online exercise I have to input those two functions, then we can see if what you said is correct.
$endgroup$
– Tristan
Dec 4 '18 at 15:59
$begingroup$
Thank you for the reply. @K.K.McDonald So what could T(x) and U(x) be in this specific exercise? To be able to validate your thought process I will try to check if those answers are correct and let you know. In this online exercise I have to input those two functions, then we can see if what you said is correct.
$endgroup$
– Tristan
Dec 4 '18 at 15:59
$begingroup$
okay, wish you success. if that worked right, don't forget to accept my question, you know, reputation thingee. although I've already ran it in MATLAB before and checked it.
$endgroup$
– K.K.McDonald
Dec 7 '18 at 14:31
$begingroup$
okay, wish you success. if that worked right, don't forget to accept my question, you know, reputation thingee. although I've already ran it in MATLAB before and checked it.
$endgroup$
– K.K.McDonald
Dec 7 '18 at 14:31
add a comment |
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Can you elaborate, especially on T, h, H, U etc. ? At the moment it's hard to help.
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– Imago
Dec 3 '18 at 21:25
2
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@Imago Please check the image added under Exercise hyperlink.
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– Tristan
Dec 3 '18 at 21:27
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Still Looking for an answer.
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– Tristan
Dec 3 '18 at 23:01