Answered

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

Sagot :

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

lhmarianateixeira

In this exercise we want to know how many bits will be transmitted between the two intelligences, so we have that:

[tex]3.2*10^5 bits[/tex]

What is the Nyquist's theorem?

According to the theorem, the reconstructed signal will be equivalent to the original signal, respecting the condition that the original signal does not contain frequencies above or above this limit. This condition is called the Nyquist Criterion, or sometimes the Rahab Condition.

in this way we have:

[tex]2*14.5kHz=29kHz \\2048=2^11 =11 bits \\ 29000*11 =319000=3.2 * 10^5[/tex]

See more about bits at brainly.com/question/2545808

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