Travelling wave

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A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is –5 dB per km and cable length is 20 km. The power received at receiver is $10^{- x} W .$ The value of ‘x’ is …… $[Given in dB = 10 \log_{10} (\frac{P_{0}}{P_{i}})]$

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Solution

$\begin{array}{l} If sound level decreases by 5 d B every km, then sound level decreased in 20 km = 100 dB \\ β_{o} - β_{i} = 10 \log_{10} (\frac{P_{0}}{P_{i}}) \\ \Rightarrow - 100 = 10 \log_{10} (\frac{P_{0}}{P_{i}}) \end{array}$
$\begin{array}{l} \Rightarrow 100 = 10 \log_{10} (\frac{P_{i}}{P_{0}}) \\ \Rightarrow \frac{P_{i}}{P_{0}} = 10^{10} \\ \Rightarrow P_{0} = \frac{P_{i}}{10^{10}} = \frac{0.1 \times 10^{3}}{10^{10}} = 10^{- 8} W \\ So, x = 8 \end{array}$

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Travelling wave

Remember concepts with our Masterclasses.

A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is –5 dB per km and cable length is 20 km. The power received at receiver is 10−xW. The value of ‘x’ is …… Given in dB=10log10⁡P0Pi

If sound level decreases by 5dB every km, then sound level decreased in 20km=100dBβo−βi=10log10⁡P0Pi⇒−100=10log10⁡P0Pi⇒100=10log10⁡PiP0⇒PiP0=1010⇒P0=Pi1010=0.1×1031010=10−8W So, x=8

Ready to Test Your Skills?

free study material

A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is –5 dB per km and cable length is 20 km. The power received at receiver is $10^{- x} W .$ The value of ‘x’ is …… $[Given in dB = 10 \log_{10} (\frac{P_{0}}{P_{i}})]$