Science, asked by BAINADARR, 6 months ago

What is the output value for the highlighted DN value after performing linear stretching of the given 16-bit data?​

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Answered by jaimatadi33
0

Answer:

The initial problem with the “16-bit” imagery, as acquired with current mapping cameras, is that it did not

begin life in a 16-bit format. Digital sensors, at present, acquire the data at 12 to 14-bits, while the TIFF

format, which is used to present the data, is only available as 8 or 16 bit (1 or 2 byte) files. (This applies to

most other raster formats as well.) In addition, most image viewers can read only 8 and/or 16-bit images. As a

result, the 14-bit image (in the case of the image used for demonstration purposes in this paper) is put into a

16-bit, or 2 byte, “container” and the columns designating 1024s, and 2048s are set to zero.

As mentioned earlier, a byte is comprised of 8 bits, and is the standard data storage size for computers. This is

why a computer’s internal memory (RAM or Random-access memory) and hard drives are listed according to

the number of bytes they will store, such as a megabyte (MB, a million bytes), a gigabyte (GB, a billion bytes),

or a terabyte (TB, a trillion bytes). To improve read/write access of imagery data, standard image file formats

store pixel information in whole blocks of bytes. Each pixel uses its own byte (or bytes if more than 8-bits) to

store the DN value associated with that pixel. Each byte in computer memory has an address assigned to it,

which allows the computer processor to access any byte directly. This greatly increases the speed at which a

computer can access image data to do things like redraw the image on the screen or modify the brightnesspixel values. An analogy of this memory address would be a very large post office with millions of PO boxes.

Each PO box, which contains a unique mailing address, represents the pixel data that the computer can use to

grab or store DN values. Getting mail directly from the box is a lot faster than standing in line at the post

office to get a stack of mixed letters.

This constraint of using whole bytes to store pixel information can cause confusion in users who are working

with radiometrically uncorrected digital imagery. In addition to doubling the storage memory, the image

appears to be “black” on computer monitors when viewed with some software programs. 12-bit files have a

maximum of 4,096 possible tonal values, but two bytes of data could potentially contain up to 65,536 values.

A computer reads the data as a “normal” 16-bit image and has no way of knowing that the file uses only 1/16th

(4,096 / 65,536) of the possible values that a full 16-bit

image would have.

Figure 8 shows the digital number distribution for a

typical 12-bit image, with no tonal adjustments,

plotted on a histogram which is a 16-bit “container.”

The image would appear almost totally black. The

histogram displays the different tonal values of the

image on the far left side of the chart; the lower values

represent the darkest of the dark tonal values. 61,441

possible values (the difference between 65,536 and

4098), in the brighter range of the spectrum, would be

missing.

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