Consider a wireless communication link between a transmitter and a receiver located in free space, with finite and strictly positive capacity. If the effective areas of the transmitter and the receiver antennas, and the distance between them are all doubled, and everything else remains unchanged, the maximum capacity of the wireless link

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GATE EC 2017 Official Paper: Shift 1

Option 3 : remains unchanged

CT 1: Ratio and Proportion

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10 Questions
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30 Mins

__Concept__:

The capacity of a wireless link is given by Shanon Hartley theorem as:

\(C = B~lo{g_2}\left( {1 + \frac{S}{N}} \right)\)

Channel capacity depends on \(\left( {\frac{S}{N}} \right)\), and hence on S.

S = Received power at the receiver = P_{r}

\(S = {P_r} = \frac{{{P_T}{G_t}{G_r}}}{{{{\left( {\frac{{4\;{\bf{\pi }}R}}{\lambda }} \right)}^2}}}\)

\(= \frac{{{P_t}\left( {\frac{{{G_\pi }}}{{{\lambda ^2}}}{A_{te}}} \right)\left( {\frac{{4\pi }}{{{\lambda ^2}}}{A_{re}}} \right)}}{{{{\left( {\frac{{4\pi R}}{\lambda }} \right)}^2}}}\)

__Calculation__:

It is given,

A_{te' }= 2A_{te }

A_{re' }= 2 A_{re}

R' = 2 R

\({P_{r'}} = \frac{{{P_t}\left( {\frac{{4\pi \;{A_{te}}}}{{{\lambda ^2}}}} \right)\left( {\frac{{4\pi }}{{{\lambda ^2}}}{A_{rc}}} \right)2 \times 2}}{{{{\left( {\frac{{4\pi R}}{\lambda }} \right)}^2}{{\left( 2 \right)}^2}}}\)

P_{r}' = P_{r}

S' = S

\(\frac{{S'}}{N} = \frac{S}{N}\)

⇒ C' = C, i.e.

The maximum capacity of the channel remains unchanged.