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I was doing a question related to analogue to digital conversion. Here's the question,

Draw the digital signal corresponding to the analogue signal given below. Assume that the system has a bit rate of 64kbits/s and a sampling rate of 8kHz

enter image description here

I know what is sampling, sampling rate, bit rate and converting samples to binary. What puzzling to me is how I should use those two details (sampling and bit rate) to draw the signal? Any guidance is much appreciated.

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2 Answers 2

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Sampling rate of 8khz means that we must measure the analog signal 8000 times per one second. If the bitrate is 64kbits/s, this means we are allowed to store our data using 64,000 bits of storage for every 1 second's worth of measurements.

Since we are making 8000 measurements per second, this means we are allowed 8 bits to store each individual measurement.

Now each bit contains 2 possible values, 0 or 1. So given 8 bits we are only able to store 256 possible levels of value.

After this, it is quite straightforward to translate all the levels into bits. For example, '22' from the above graph will be translated into 00010110, as so forth.

To actually translate the full data completely from just the sparse graph above is actually impossible by inspection. We need the full analog signal data to be measured 8000 times a second and then translate them one by one into 64000 bits of data.

But if the assignment is to translate 22, 15, 4, 9 from the graph, then we can do it as above.

Hope it helps!

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  • $\begingroup$ Yep, it was super clear now thank you:) $\endgroup$ Oct 21, 2021 at 10:20
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The key is in realising the fact that digitisation is converting a continuous function into a discrete one. For example, if I have a continuous function whose range lies between zero and ten and we have only one bit to represent the function value, then we can chose to represent all signals below five to be $0$ and all signals above five to be $1$. Now consider we have two bits to represent the function value then we can divide the signal into four subdivisions. Similarly for $n$-bits we can subdivide the signal into $2^n$ parts.

Coming to your problem, how many subdivisions can we have for the signal? In other words, how many bits can we spare for each signal instance? We are given that we have a bit rate of $64$kbits/s and we’re sampling at $8$kHz, ie $8000$ signal instances within a second. So that leaves us with $64000/8000 = 8$bits/sample.

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  • $\begingroup$ Yep, it was super clear now thank you:) $\endgroup$ Oct 21, 2021 at 10:20

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