1.4 (4 pts) [2] Assume a color display that 60 times a second copies the content
ID: 3742065 • Letter: 1
Question
1.4 (4 pts) [2] Assume a color display that 60 times a second copies the contents of the frame buffer to activate each LCD pixel on the display, producing the image you see on the screen. We will use 8 bits for each of the red, green and blue pixels and we will need to control a display of size 1600 x 1200 pixels (which we call a frame). (2 pts) What is the minimum size in bytes of the frame buffer to store a single one of these screen images? a. b. (2 pts) How long would it take, at a minimum, for the frame to be sent over a 100 Mbit/s network? Hint: assume a Mbit is 10° bits.Explanation / Answer
There are a total of 1600*1200 = 1,920,000 pixels on the screen. Now, since each pixel has three color components - red, blue and green, which need to be stored and each component requires 8 bits of storage, the total amount of storage needed to store a frame = 3 * 8 * 1,920,000 bits = 46,080,000 bits. And, 1 byte contains 8 bits, so the storage required to store a single frame in bytes = 46,080,000 / 8 bytes = 5,760,000 bytes.
A 100 Mbit/s network will transfer a maximum of 100 * 10^6 bits in 1 second. Since we've already calculated that a single frame requires a storage of 46,080,000 bits, we need to transfer 46,080,000 bits over the network to transfer a full frame. The amount of time required to transfer some data = number of bits / rate of transfer (in bits/s).
Hence, the minimum amount of time required to transfer a frame over given network = 46,080,000 / (100 * 10^6) seconds = 0.4608 seconds.
Hope this helps. :)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.