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Alecsey [184]
3 years ago
9

An analogue sensor has a bandwidth which extends from very low frequencies up to a maximum of 14.5 kHz. Using the Sampling Theor

em, what is the minimum sampling rate (number of samples per second) required to convert the sensor signal into a digital representation?
If each sample is now quantised into 2048 levels, what will be the resulting transmitted bitrate in kbps?

Give your answer in scientific notation to 1 decimal place.

Hint: you firstly need to determine the number of bits per sample that produces 2048 quantisation levels
Computers and Technology
2 answers:
spin [16.1K]3 years ago
6 0

Answer:

Sorry didn't mean to click answer, and now not sure how to delete

Explanation:

shepuryov [24]3 years ago
3 0

Answer:

3.2*10^5

Explanation:

By Nyquist's theorem we must have 2*14.5kHz=29kHz so 29,000 samples per second. 2048=2^11 so we have 11 bits per sample. Finally we have 29000*11 bits per second (bps) =319000=3.2 * 10^5

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romanna [79]
<h2>Hello!</h2>

The answer is:

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<h2>Why?</h2>

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MaRussiya [10]
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I’m not sure.
8 0
3 years ago
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dusya [7]
I found this:
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Fed [463]

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