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

Question

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

• A. $60000$
• B. $15000$
• C. $30000$
• D. $3000$

Comments

SNR is in db so your answer is not correct

Answer 1

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

Answer 2

SNR=SNR=10^SNR dB/10 =10^3.0=1000

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