A person is recording a song on their computer, using a sampling rate of 48 kHz...
A person is recording a song on their computer, using a sampling rate of 48 kHz with each sample having a size of 2 bytes.
- i.How many bits are in each sample?
- ii.How many bits per second are processed while the song is being recorded? Write your answer in scientific notation.
- iii.If the song is 3 minutes and 14 seconds long, how many bits are processed in total? Write your answer in scientific notation.
- iv.If the song had been recorded using a sampling rate of 44.1 kHz with a sample size of 2 bytes, how many bits would be processed during the 3 minutes and 14 seconds? Write your answer in scientific notation.
- v.What is the percentage decrease in the number of bits processed as a result of the sample rate decreasing from 48 kHz to 44.1 kHz? Round your answer to the nearest hundredth of a percent.
Homework Answers
i. Each sample has a size of 2 bytes. 1 byte = 8 bits, Hence 2 bytes = 2*8 = 16 bits. Hence each sample has 16 bits.
ii. The sampling rate is given as 48kHz. This means 48*10^3 samples are processed each second. 1 sample equals 16 bits. Hence 48000 samples equal 48000*16 bits. Hence the number of bits processed per second = 48000*16=768000 bits.
iii. The total time taken for recording the song equals 3 minutes and 14 seconds i.e 3*60 + 14 seconds = 180 + 14 seconds = 194 seconds.
No: of bits processed in one second = 768000
No: of bits processed in 194 seconds = 194*768000 = 148992000 bits
This is the total number of bits processed throughout the recording of the song.
iv. If the song has been recorded using the sampling rate of 44.1 kHz instead, it means that 44100 samples of size 2 bytes are processed in each second.
No of bits processed in one second = 44100*16 = 705600 bits
No: of bits processed in 194 seconds = 194 * 705600 = 136886400 bits.
This will be total number of bits processed if the sampling rate was 44.1 kHz.
v. The percentage of decrease after decreasing the sampling rate = Difference in number of bits / Number of bits processed at sampling rate 48 kHz
This equals (148992000 - 136886400)/148992000 = 0.08125 = 8.125 percent.
Rounding of to nearest 100th = 8.13 percent
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