Question

Consider two raster systems with the resolution of 640*480 and 1280*1024, how many pixels could be accessed per second in each of these systems by a display monitor controller that refreshes the screen at the rate of 60 frames per seconds?

Answer 1

Explanation:

Ans. Because eight bits constitute a byte, frame-buffer sizes of the systems are as follows: 640 x 480 x 12 bits / 8 = 450KB; 1280 x 1024 x 12 bits / 8 = 1920KB; 2560 x 2048 x 12 bits / 8 = 7680KB; b) How much storage (in bytes) is required for each system if 24 bits per pixel are to be stored