GATE 2014 EC (EC2) – Question 32
AWGN capacity is $C=W\log_2(1+P/(\sigma^2W))$ bps, where W is bandwidth, P average received power and $\sigma^2$ one-sided noise PSD. For fixed $P/\sigma^2=1000$, find capacity as $W\to\infty$ (kbps).
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Correct answer: (A) 1.44
Explanation
Rewrite C=[W/ln2]ln[1+(P/σ²)/W]. As W→∞, ln(1+a/W)≈a/W, giving C∞=(P/σ²)/ln2=1000/ln2=1442.70bps. Thus capacity is approximately **1.44kbps,A**, even though the bandwidth diverges.