Step 1: Understanding the 10-bit radiometric resolution.
In a 10-bit image, the number of possible Digital Numbers (DNs) ranges from 0 to \(2^{10} - 1\). This is because a 10-bit resolution allows for 1024 distinct values, which range from 0 to 1023.
Step 2: Determining the maximum DN value.
The maximum DN value is equal to \(2^{10} - 1\), which is: \[ 2^{10} - 1 = 1024 - 1 = 1023 \] Therefore, the maximum DN value of the image is \(1023\).
A 4 × 4 digital image has pixel intensities (U) as shown in the figure. The number of pixels with \( U \leq 4 \) is:
A 4 × 4 digital image has pixel intensities (U) as shown in the figure. The number of pixels with \( U \leq 4 \) is:
The error matrix resulting from randomly selected test pixels for a classified image is given below.
The Producer’s accuracy of class 1 is % (rounded off to 1 decimal place).
Reference Data | |||||
---|---|---|---|---|---|
Class 1 | Class 2 | Class 3 | Class 4 | ||
Classified Data | Class 1 | 320 | 8 | 7 | 3 |
Class 2 | 12 | 270 | 6 | 2 | |
Class 3 | 9 | 6 | 410 | 5 | |
Class 4 | 14 | 2 | 3 | 350 |
The brightness values of four pixels in the input image are shown in the table below. The image is rectified using nearest neighbor intensity interpolation, and the pixel at location (5, 4) in the output image is to be filled with the value from coordinate (5.3, 3.7) in the input image. The brightness value of the pixel at location (5, 4) in the rectified output image is 11. (Answer in integer)
Location of pixels in input image (Row, Column) | Brightness Value |
---|---|
(5, 3) | 9 |
(5, 4) | 11 |
(6, 3) | 14 |
(6, 4) | 12 |
A color model is shown in the figure with color codes: Yellow (Y), Magenta (M), Cyan (Cy), Red (R), Blue (Bl), Green (G), and Black (K). Which one of the following options displays the color codes that are consistent with the color model?
While doing Bayesian inference, consider estimating the posterior distribution of the model parameter (m), given data (d). Assume that Prior and Likelihood are proportional to Gaussian functions given by \[ {Prior} \propto \exp(-0.5(m - 1)^2) \] \[ {Likelihood} \propto \exp(-0.5(m - 3)^2) \]
The mean of the posterior distribution is (Answer in integer)