# NIELIT 2017 OCT Scientific Assistant A (IT) - Section B: 32

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The maximum data rate of a channel of $3000$-Hz bandwidth and SNR of $30$ db is

1. $60000$
2. $15000$
3. $30000$
4. $3000$

edited
0
Maximum number of bps = B log (1 + SNR).
=3000*log(1+30)
=3000*4.95
=14850. So approximately, 15000

Logarithm is in "Base 2".
0

Max data rate = H * log2 ( 1 + S/N ) S/N is called signal to noise ratio

SNR value = 10 * log10 S/N dB (units for SNR is decibals)

SNR = 10 * log10 S/N 30 = 10 * log10 S/N

S/N = 1000

Max data rate =3000 * log2 ( 1 + 1000 ) = 30,000 bps
SNR=SNR=10^SNR dB/10 =10^3.0=1000

Max data rate =3000 * log2 ( 1 + 1000 ) = 30,000 bps

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