A 4x4 4bitspixel image passes through a pointwise intensity

A 4x4, 4bits/pixel image passes through a point-wise intensity transformation given by:

S=T(r)=alog₂(1+r)+b

where a and b are unknown parameters. Only a few pixels are available in the input and the output images, as shown below.

Input Image:

Output Image (after applying T(r):

(a) Find a and b.

(b) What\'s the value of the pixel with the \"?\" mark in the output image?

(c) What\'s the value of the pixel with the \"?\" mark in the input image?

Step by step answer please.

	3
		15
	?
1			3

Solution

Given

S=T(r)=alog₂(1+r)+b        (1)

In the Table Input Image

      the values of \"r\" is given

and in the table Output Image

      the value of S or T(r) is given

for r = 3 we have T(r) = 5

for r = 15 we have T(r) = 11

so subtituting the complete values given in our equation

T(3) = alog₂(1+3)+b

5 = alog₂(4)+b

5 = alog₂2²+b                        //     4 = 2 *2 => 4 = 2²

5 = 2a log₂ 2+ b                  //          log x^m = m*log x

5 = 2a +b       //      log_a a = 1

2a + b = 5

b = 5 - 2a         (2)

now for r = 15 we have T(r) = 11

T(15) = alog₂(1+15)+b

11 = alog₂(16)+b

11 = alog₂2⁴+b

11 = 4a+b            (3)

a)

from (2) and (3)

11 = 4a +5-2a

11 = 2a +5

6 = 2a

or a = 3

from (2)

b = 5 -2*3

b = -1

b)

we need to find the value of t(r) for r = 1

t(1) = 3log₂(1+1) -1

t(1) = 3log₂2 -1

T(1) = 2

c)

we have T(r) = 8

we need to find the value of \"r\"

T(r) = 3log₂(1+r)-1

8 = 3log₂(1+r)-1

9 = 3log₂(1+r)

3log₂(1+r) = 9

log₂(1+r) = 3

1+r = 2³                  // log_b a = c => a = b^c  

1 +r = 8

r = 7