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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 quantized into $2048$ levels, what will be the resulting transmitted bitrate in kbps?

EDIT

The answer I get is 11 bits per sample, $2048 = 2^{11}$, is this calculation correct?

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    $\begingroup$ Hi! Why don't you share your own calculation with us? And what's wrong with it? $\endgroup$ – Fat32 Feb 14 at 22:09
  • $\begingroup$ yea good idea ..... please read the edit thanks $\endgroup$ – Alan wood Feb 15 at 11:04
  • $\begingroup$ @Alan wood "The answer I get is 11 bits per sample, 2048=211, is this calculation correct?", yes this calculation is correct. Now you need to identify the bandwidth frequency, in order to proceed to the bitrate calculation. $\endgroup$ – kalgoritmi Feb 15 at 23:52
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It is not clear in which part of the problem statement you experience issues.

However, a few hints may help you get the gist of such problems

  1. The sampling frequency, it is intuitive that in order for a system to monitor the state of another system it must poll it at higher rates than the highest rate in which the monitored signal is able to change its state. But what is the minimum sampling frequency that gives our sampling system the ability to correctly approximate the continuous signal in the discrete time domain?

This frequency threshold is at two times the highest frequency of the continuous signal: $$ f_{sampling} \gt 2\cdot f_{bandwidth} $$

The theorem that the problem statement suggests is the Shannon-Nyquist sampling theorem, the theorem that the above inequality comes from.

In real sampling systems, the sampling frequency is never taken exactly on this threshold, usually the threshold is taken a little bit higher than $2\cdot f_{bandwidth}$, for example at $2.2\cdot f_{bandwidth}$.

However, in theory it is safe to assume a sampling frequency at the Nyquist rate (i.e. the aforementioned frequency threshold).

Your analogue signal's highest frequency is $14.5\, Hz$, ranging from very low frequencies so it is safe to assume, this actually the bandwidth frequency.

All the above summarized in the following equation: $$ f_{sampling} = 2\cdot f_{bandwidth}$$

  1. Bitrate, in order to calculate the bitrate of your sampling system it is necessary to know the sampling frequency. Fortunately, this was the first the step of the exercise. The second component for the bitrate calculation is the actual length of each sample in bits. The minimum number of bits that are necessary in order to represent a quantized value in x distinct levels is given by:

$$ 2^{bitlength}\, =\, x $$

or

$$ bitlength = \lceil\log_2{x}\rceil $$

You could actually deduce the bitrate equation by performing a dimensional analysis, because it is stated that it is measured at kbps(kilobits per second). For instance, we know that $Hz$ is measured in $\frac{samples}{sec}$ and kb stands for kilobits meaning $1000\, \frac{bits}{sample}$, so:

$$ bitrate = bitlength\cdot f_{sampling} \left[\dfrac{bits}{sample}\dfrac{samples}{sec} \right] $$

or in kbps

$$ bitrate = bitlength\cdot f_{sampling}\cdot 10^{-3} \left[kbps\right] $$

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So the sensory signal you are going to collect has the maximum component of 14.5 kHz. Let me guide you through my understanding and you can do the calculation yourself, testing your understanding.

As others have also implied, you need a sampling rate twice the bandwidth (you can check Nyquist Sampling Theorem to get it better).

And according to your question, you want to quantize your samples with 2048 levels, which would require 11 bits as you have also calculated. But that number of bits are only required for only 1 sample while you have taken much more samples per second. Now by simply counting 11 bits for each sample you take (simple multiplication) you can find your bit rate.

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