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Question: a a 3840 x 2160 pixels x 125 10368000...

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A

a. 3840 x 2160 pixels x 1.25 = 10,368,000 pixels

(10,368,000 pixels x 3 [colors]) = 31,104,000 bytes/ frame

b. (31,104,000 bytes / frame x 10bits/byte) / 100E6 bits/sec = 3.11 seconds = 3 seconds

31,104,000 x 10) / 100E6 = 3.11s

c. What bandwidth do you need (at a minimum) to transmit this data at 60 frames per second?

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