Question

[7] How many MB of video memory is required to display an image with a resolution of 1,920 x 1,080 pixels and 65,536 colors (2^16colors)? However, 1 [MB] = 1,000 [kB] and 1 [kB] = 1,000 (B).

[8] What is the amount of video memory required to display an image with a resolution of 1,600 x 1,200 pixels and 65,536 colors (2^16 colors)? However, 1 [MB]= 1,000 [kB], 1 (kB] = 1,000 (B).

[9] Minimum required for video distribution at 30 frames / sec with resolution 1,280 x 720 pixels and 24-bit full color. What is the bandwidth? However, no data compression, 1 [MB] = 1,000 (kB], 1 (kB] = 1,000 [B]. {{bps? (Total bit / sec)}}

[10] Resolution 1,920 x 1,080 pixels, 24 bit full color 60 What is the minimum bandwidth required for frame / second video distribution? However, no data compression, 1 [MB) = 1,000 [kB], 1 [kB) = 1,000 [B].

Answer 1

[7] How many MB of video memory is required to display an image with a resolution of 1,920 x 1,080 pixels and 65,536 colors (2^16colors)? However, 1 [MB] = 1,000 [kB] and 1 [kB] = 1,000 (B).

Number of pixels in the image = Image resolution = 1,920 x 1,080 = 20,73,600

Total bytes required for the image = Number of pixels x bit depth (the number of bits needed to store colours).

That is, 20,73,600 x 2 bytes = 41,47,200 bytes

That is 4.1472 MB

Thus, the vide memory required to display an image with resolution 1,920 x 1,080 = 4.1472 MB

[8] What is the amount of video memory required to display an image with a resolution of 1,600 x 1,200 pixels and 65,536 colors (2^16 colors)? However, 1 [MB]= 1,000 [kB], 1 (kB] = 1,000 (B).

Number of pixels in the image = Image resolution = 1,600 x 1,200 = 19,20,000

Total bytes required for the image = Number of pixels x bit depth (the number of bits needed to store colours).

That is, 19,20,000 x 2 bytes = 38,40,000 bytes

That is 3.84 MB

Thus, the vide memory required to display an image with resolution 1,600 x 1,200 = 3.84 MB

[9] Minimum required for video distribution at 30 frames / sec with resolution 1,280 x 720 pixels and 24-bit full color. What is the bandwidth? However, no data compression, 1 [MB] = 1,000 (kB], 1 (kB] = 1,000 [B]. {{bps? (Total bit / sec)}}

Number of pixels in a frame = frame resolution = 1,280 x 720 = 9,21,600

Total bytes required for a frame = Number of pixels x bit depth (the number of bits needed to store colours).

That is, 9,21,600 x 3 bytes = 27,64,800 bytes

That is 2.7648 MB

Number of frame in a sec = 30

Total size of 30 frames            = 2.7648 x 30 = 82.944 82.944 MB/Sec

= 663.552x10⁶ bits/Sec = 663.552 Mega bits/Sec

Thus, miniumu video size required per second is = 663.552 Mega bits/Sec

[10] Resolution 1,920 x 1,080 pixels, 24 bit full color 60 What is the minimum bandwidth required for frame / second video distribution? However, no data compression, 1 [MB) = 1,000 [kB], 1 [kB) = 1,000 [B].

[farme per scomd is mentoned is not at proper postion in the question]

Number of pixels in a frmae = frame resolution = 1,920 x 1,080 = 20,73,600

Total bytes required for a frame = Number of pixels x bit depth (the number of bits needed to store colours).

That is, 20,73,600 x 3 bytes = 62,20,800 bytes

That is 6.2208 MB

Number of frame in a sec = 60

Total size of 60 frames            = 6.2208 x 60 = 373.248 MB/Sec

= 2,985.984 x 10⁶ bits/Sec = 2,985.984 Mega bits/Sec

Thus, bandwidth required per second is = 2,985.984 Mega bits/Sec