Answered

Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Our platform provides a seamless experience for finding precise answers from a network of experienced professionals. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

You would like to use a 16-bit ADC to sample and digitally store a stack of vinyl music records you found in your parents’ basement. If the ADC is capable of sampling up to 25 kSamples/sec:________.
a) What is the minimum time required for each sample conversion?
b) How many bytes of storage would be required to record "Uptown Funk" by Bruno Mars and Mark Ronson (4 min, 22 sec) if the ADC is operating at the maximum sample rate?
c) If the highest frequency in "Uptown Funk" is 10 kHz, what is the minimum sampling rate that would need to be used to accurately reconstruct the song?
d) How many bytes of storage are required at this lower sampling rate?

Sagot :

batolisis

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

We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.