Capacity of each channel in FDMA is given by
A. $$C = W.{\log _2}\left( {1 + \frac{S}{N}} \right)$$
B. $$C = M.{\log _2}\left( {1 + \frac{S}{N}} \right)$$
C. $$C = \frac{W}{M}{\log _2}\left( {1 + \frac{S}{N}} \right)$$
D. $$C = {\log _2}\left( {1 + \frac{S}{N}} \right)$$
Answer: Option C
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