Computer Science, asked by rj32526, 8 months ago

raster video controller has an 18-bit DAC and comes equipped with 768K of video RAM. a) What is the maximum display resolution (at 4:3) you can provide in a 256 color mode? b) In a 640 x 400 x 16 color mode, how many pages can be stored? c) How many bytes are required for a lookup table in an 8-page 320 x 200 mode?

Answers

Answered by dreamrob
1

Given:

Video Ram available = 768KB

For (a)

Rows : Columns = 4:3

Total number of colors = 256

For (b)

Color mode = 640 × 400 × 16

For (c)

Mode = 320 x 200

To find:

(a) What is the maximum display resolution (at 4:3) you can provide in a 256 color mode?

(b) In a 640 x 400 x 16 color mode, how many pages can be stored?

(c) How many bytes are required for a lookup table in an 8-page 320 x 200 mode?

Solution:

(a) Video Ram available = 768KB.

Let no of rows = 4x

No of columns = 3x

Total pixels: 4x × 3x = 12x²

Total number of colors = 256, therefore 8 bits are required to represent colors.

For each pixel, 1 byte of information is required for storing color.

Therefore, total space needed = 12x² × 1 Bytes

Total available space = 768 * 1024 Bytes

12x² = 786432

x² = 65536

x = 256

Resolution = 4x × 3x = 1024 × 768

(b) In 640 × 400 × 16 color mode:

resolution = 640 × 400

Total colors available = 16

Bits required for colors = 4  

Total pixels = 640 × 400

Bits required for each pixel = 4

Total memory required = 640 × 400 × 4 bits = 640 × 200 bytes

= 128000 Bytes

(c) The size of a lookup table is 2ⁿ × 3m

where m is the number of bits required for storing each of red, blue, and green color value  and n is the number of bits available for each pixel in frame buffer.

768 × 1000 = 320 × 200 × n

n = 12

m = 8, 8 bits are used for storing the information about red, blue, and green color at that pixel

Therefore, lookup table size: 2¹² × 8 bits

= 4096 × 8 bits = 4096 Bytes = 4KB

Similar questions