Question: A 2kHz channel has signal to noise ratio of 24 dB
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  1. Calculate max capacity of this channel.
  2. Assuming constant transmitting power, calculate max. capacity when channel b.w is 1 halved 2, reduced to quarter of its original value.

Subject: Computer Engineering

Topic: Electronic Circuits and Communication Fundamentals

Difficulty: Medium / High

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modified 10 months ago  • written 10 months ago by gravatar for Sayali Bagwe Sayali Bagwe2.2k
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B= 2 kHz and (S/N) = 24dB

$∴24=10 \log_10⁡ \big(\frac{S}{N}\big) \\ ∴\frac{S}{N}=251$

Channel capacity:

$C= B \log_2⁡ \big[1+\frac{S}{N}\big] \\ =2 \times 10^3 \log_2⁡[1+251] \\ C = 15.95 \times 10^3 bits/sec$

B is halved:

$∴ b.w, B_2=1kHz, old b.w., B_1=2 kHz \\ ∴N=N_o B,∴N_1=N_o B_1 \\ ∴N_2=N_o B_2 \\ ∴\frac{N_2}{N_1} =\frac{N_o B_2}{N_o B_1 }=\frac{B_2}{B_1} =\frac{1}{2} \\ ∴\frac{N_2}{N_1} =\frac{1}{2}$

As signal power remain constant,

SNR with new b.w. is,

$\frac{S}{N^2} =\frac{S}{N_1/2}=2 \frac{S}{N_1}$

But, $\frac{S}{N_1} =251 \\ ∴\frac{S}{N_2} =2 \times 251=502 \\ ∴C=B_2 \log_2⁡[i+S/N_2] \\ =2 \times 10^3 \log_{10}⁡(503)/ \log_{10} ⁡2 \\ =17.94 \times 1^3 bits/sec$

¼ of original value:

$\frac{N_2}{N_1} =\frac{1}{4} \\ ∴\frac{S}{N_3} =4 \frac{S}{N_1} =4 \times 251=1004 \\ ∴C=B_3 \log_2⁡(1+ \frac{S}{N_3} )=500 \log_2⁡(1004) \\ C=4.99 \times10^3 bits/sec.$

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written 10 months ago by gravatar for Sayali Bagwe Sayali Bagwe2.2k
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