Question

Show the calculation steps for each of the problems below.

a. Using Nyquist’ theorem, calculate the channel capacity C of a signal that has 18 different levels and a frequency of 50,000 Hz.

b. Using Shannon’s theorem, calculate the data transfer rate given the following information, signal frequency= 25,000 Hz; signal power = 5000 watts; noise power = 320 watts.

c. Using Nyquist’s theorem and given a frequency of 8000 Hz and a data rate of 90,000 bps, how many signal levels (L) will be needed to convey this data?

d. The signal-to-noise ratio is often given in decibels. Assume that SNRdB = 28 and the channel bandwidth is 8 GHz, what is the theoretical channel capacity?

Answer 1

Solution ::

a) Since the nature of channel(whether noisy or noiseless ) is not explicitly stated, it is assumed to be noiseless.

Capacity of channel C = 2*B*log₂L

= 2*50000*log₂18 = 100000* 4.169925 = 416992.5 ~ 416992bits/second

b) In this case, the channel is noisy.

Shannon's theorem:-  

SNR(Signal to Noise Ratio) :- Signal power/ Noise power = 5000/320 = 15.625

capacity:- B log₂ (1+SNR) = 25000 * log₂ (1+15.625) = 25000 * log₂ (16.625) = 25000 * 4.055

= 101382.061 ~ 101382 bps

c) C = 2*B*log₂L

90000 = 2*8000*log₂L

log₂L = 90000/16000 = 5.625

L = 2^5.625 = 49.35 ~ 50 levels (Answer)

Note:- Generally levels are taken to be power of two. In that case, 64 levels are required(power of 2 just greater than 50)

d)

C = B log₂ (1+SNR) ..(1)

SNR_dB = 10log₁₀SNR

28 = 10log₁₀SNR

log₁₀SNR = 2.8

SNR = 10^2.8 = 630.957

Put this in (1)

C = 8*10⁹ * log₂(1+SNR) [Since 1 GHz = 10⁹Hz]

C = 8*10⁹ * log₂(631.957)

C = 7.443 * 10¹⁰ bps

C = 74.43 * 10⁹ bps

C = 74.43 Gbps (Answer)

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