Question

): A cable has a bandwidth of 3000 Hz assigned for data communication. The SNR is 3162. How many signal levels do we need?

Answer 1

Answer:

$L=2^{25830}$ is the correct answer .

Explanation:

bandwidth = 3000 Hz

SNR =3162

We know that

$bitrate = bandwidth * log2(1 + SNR)$

Putting the value of bandwidth and SNR in the given equation .

$bitrate = 3000 * log_{2} ^\ (1 + 3162) \\\\birate = 3000 * log_{2} ^\ {3163} \\\\birate =3000 * 11.\ 6\\\\bitrate=\ 34860 bps$

Now using the formula

$BitRate = 2 * Bandwidth * log_{2} (L)$

Putting the value of bitRate and bandwidth we get

$34860=2\ * 3000\ * \ log_{2} (L)\\34860=6000 * \ log_{2} (L)$

$31860-6000=log_2L\\25830=log_2L\\L=2^{25830}$