6. In a CD player, the sampling rate is 44.1 kHz, and the samples are quantized using a 16 bit/sample quantizer. Determine the resulting number of bits for a piece of music with a duration of 50 minutes.e
6. In a CD player, the sampling rate is 44.1 kHz, and the samples are quantized using a 16 bit/sample quantizer. Determine the resulting number of bits for a piece of music with a duration of 50 minutes.e
Introductory Circuit Analysis (13th Edition)
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
Transcribed Image Text:6. In a CD player, the sampling rate is 44.1 kHz, and the samples are quantized using
a 16 bit/sample quantizer. Determine the resulting number of bits for a piece of music
with a duration of 50 minutes.
Answer:
The sampling rate is fs = 44100 meaning that we take 44100 samples per second.
Each sample is quantized using 16 bits so the total number of bits per second is 44100
x 16. For a music piece of duration 50 min = 3000 sec the resulting number of bits per
channel (left and right) is
44100 x 16 x 3000 = 2.1168 ×x 10^9
and the overall number of bits ise
2.1168 x 10^9 × 2= 4.2336 × 10^9
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