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

`L=2^(25830)` is the correct answer .

Step-by-step 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)`