Answer :
a) Calculate the data rate of the video only.Data rate of the video only = Frame size x Frame rateFrame size= (1200 x 900 pixels) x 24 bits/pixel= 31,104,000 bitsFrame rate = 30 fps
Data rate of the video only = 31,104,000 x 30 bits/sec= 933,120,000 bits/sec1 Byte = 8 bitsData rate of the video only = 933,120,000/8 Bytes/sec= 116,640,000 Bytes/secb)
Calculate the data rate of the audio only. Data rate of the audio only = Sample rate x Number of bits/sample x No. of channels Sample rate = 50 KHz
Number of bits/sample = 8 bits (1 Byte/sample)No. of channels = 2 (Stereo)Data rate of the audio only = 50,000 samples/sec x 8 bits/sample x 2= 800,000 bits/sec1 Byte = 8 bits
Data rate of the audio only = 800,000/8 Bytes/sec= 100,000 Bytes/sec
Determine the total data size of the video+audio, if the duration is 12 minutes 5 seconds.
Total data size of the video+audio = (data rate of video only + data rate of audio only) x durationDuration of the video+audio = 12 minutes 5 seconds= 725 seconds
Total data size of the video+audio= (116,640,000 Bytes/sec + 100,000 Bytes/sec) x 725 sec= 84,563,600,000 Bytes= 84.56 GBd)
If the video+audio is compressed with a compression ratio of 25:1, calculate its new data size and the time taken to transmit the compressed video+audio over a network with a line speed of 100 Mbps.(i) New data size = Total data size of the video+audio/Compression ratio= 84,563,600,000 Bytes/25= 3,382,544,000 Bytes= 3.38 GB
(ii) Time taken to transmit the compressed video+audio over a network with a line speed of 100 Mbps.
Data transmission rate= 100 Mbps= 100,000,000 bits/sec
Total data size of the video+audio = 3.38 GB= 3.38 x 1024 x 1024 x 1024 bytes= 3,629,512,243 bytes
Total time taken= Total data size/Data transmission rate= 3,629,512,243 bytes/100,000,000 bits/sec= 36.30 secondsTherefore, the time taken to transmit the compressed video+audio over a network with a line speed of 100 Mbps is 36.30 seconds.
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