For 2-level and 8-level systems, what is the difference between them (bit rate? noise?) for the
same
.

The formula that applies here is:

Vtm = 2B-W ×log2 Lev

where Vtm is the bit rate, B-W is bandwidth, and Lev is the level. And hence, by the above formula, we find that bit rate is dependent on the level. For two-level we have Lev =2 and for 8 Level we have the Lev=8. And hence the bit rate for 8 level systems, is three times more than what it is for the 2 level systems. as Log 2 base 2=1 and Log8 base 2 =3.

Also, in the year 19444, Shannon came up with the formula for determining the theoretical value of the highest data rate for a noisy channel. And that was, C= B x Log2(1 +SNR), where B is the bandwidth, and SNR is the signal to noise ratio, and C is the capacity of the channel. Hence, noise is dependent on the SNR, and this formula. Noiseless channel means that there is more useful data than useless data. And if SNR is 100db then it means that the audio signal level is 100db more than the noise level. And that way, we can come to the conclusion about the noise.

We can write,

C= Vtm/2Log2 Levx log 2(1+SNR), Thus we see SNR is inversely proportional to Vtm, and hence, more Vtm means less SNR, and hence more noise. Thus, an 8 level system is three times noisier than 2 level system.

Explanation:

Please check the answer section.

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