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Sagot :
Answer:
a) 40 micro sec
b) 13,100,000 bytes
c) 20,000 samples / sec
d) 10.48 Mbytes
Explanation:
a) The minimum time required for each sample conversion
minimum sample conversion time = max performance of ADC ( 25 k samples /sec )
hence : 1 sample conversion time = 1 / 25k = 1/ 25000
hence minimum time required for each sample conversion time = 40 microsec
b) Determine the number of bytes of storage required to record if the ADC is operating at maximum sample rate
In 1 second the ADC will produce 25000 * 2 bytes of data = 50000 bytes
audio length = 4 min 22 secs = 262 seconds
Hence the number of bytes of storage required =
= 262 * 50000 bytes
= 13,100,000 bytes
c) Determine the minimum sampling rate that would need to be used to accurately reconstruct the song
highest frequency = 10 kHz
hence the minimum sampling rate to reconstruct the song
= 2 * 10 kHz = 20,000 samples / sec
d) Determine The number of bytes of storage required at the lower sampling rate
given that the audio is sampled at 20kHz per sec then each sample will be encoded into a 2 bytes
therefore each each second of the audio will be generated by : 20,000 * 2 bytes
∴ the number of bytes of storage required at this lower sampling rate
= 262 secs * 40,000
= 10.48 Mbytes
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