GATE 2018 EC Q.33: solution and answer

GATE 2018 · ECQ.33 · 2 marks · MCQ

The distance (in meters) a wave has to propagate in a medium having a skin depth of 0.1 m so that the amplitude of the wave attenuates by 20 dB, is

(A)
0.12
(B)
0.23
(C)
0.46
(D)
2.3

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Answer and solution

ANSWER

The wave must travel about 0.23 m, which is 2.303 skin depths, for its amplitude to fall by 20 dB. Answer: (B).

Official answer: BSource: Official answer key, GATE 2018 EC (IIT Guwahati)

Skin depth gives the attenuation constant. In a lossy medium the field amplitude falls as $e^{- α z}$ . The skin depth is the distance over which it falls by a factor of $1/ e$ , so $δ = \frac{1}{α}$ . That gives $α = \frac{1}{0.1} = 10 Np/m$ .
Convert 20 dB to an amplitude ratio. For amplitude (field) ratios the decibel measure is $20 lo g_{10}$ : $20 lo g_{10} \frac{E _{0}}{E ( z )} = 20 \Rightarrow \frac{E _{0}}{E ( z )} = 10$ .
Set up the decay. $e^{- α z} = \frac{1}{10} \Rightarrow α z = ln 10 = 2.303 Np$ . The same number comes out of the unit conversion: $1 Np = 8.686 dB$ , so $20 dB = 20/8.686 = 2.303 Np$ .
Solve for the distance. $z = \frac{ln 10}{α} = δ ln 10 = 0.1 \times 2.303 = 0.2303 m$
Match the options. $z \approx 0.23 m$ , which is option (B).

KEY IDEA

Each skin depth multiplies the amplitude by $1/ e$ , which is an attenuation of 1 Np, or 8.686 dB. A 20 dB amplitude drop means the amplitude falls by a factor of 10. That takes $ln 10 \approx 2.303$ Np, so the wave must travel $z = δ ln 10 = 2.303 δ$ .

WATCH OUT FOR

Treating 20 dB as a power ratio ( $10 lo g_{10}$ , a factor of 100) gives $z = δ ln 100 = 0.46 m$ , which is option (C). The skin depth is defined on the field amplitude, so the $20 lo g_{10}$ form applies. A second error is using $lo g_{10} 10 = 1$ in place of $ln 10$ , which gives $0.1 m$ . Multiplying $δ$ by 23 in place of 2.3 gives option (D).

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GATE 2018 EC Q.33: solution and answer

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