Past GATE papers of EC 2017 Electromagnetics

GATE EC 2017 Electromagnetics Solution

The voltage of an electromagnetic wave propagating in a coaxial cable with uniform characteristic impedance is V\left(l\right)=e^{-\gamma l+j\omega t} Volts, where l is the distance along the length of the cable in meters, \gamma=\left(0.1+j40\right)m^{-1} is the complex propagation constant, and \omega=2\pi\times10^9 rad/s is the angular frequency. The absolute value of the attenuation in the cable in dB/m is_____.

Ans :- 0.868

Explanation

The expression for V at any l is

V\left(l\right)=V_oe^{-\alpha l}e^{-j\beta l}e^{j\omega t} Attenuation=\frac{\left|Input\right|}{\left|Output\right|}=\frac{\left|V_o\left(0\right)\right|}{\left|V_o\left(l\right)\right|}

Attenuation/meter = \frac{\left|V_o\right|}{\left|V_o\left(1m\right)\right|}=e^\alpha

Attenuation in dB/m = \left(20\log e^\alpha\right)dB/m

= 20(0.1)log e = 0.868 dB/m

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