Question 20.WE.5: The input signal to a cable has power 1.2 × 10^−3 W. The sig...
The input signal to a cable has power 1.2 × 10^{−3} W. The signal attenuation per unit length in the cable is 14 dB km^{−1} and the average noise level along the cable is constant at 1.0 × 10^{−10} W. An acceptable signal-to-noise ratio is at least 30 dB.
Calculate the minimum acceptable power for the signal and the maximum length of the cable that can be used without a repeater.
Step 1 The signal-to-noise ratio must be at least 30 dB. Hence, using:
signal-to-noise ratio = 10 lg\left\lgroup\frac{signal power}{noise power}\right\rgroup
we have:
30 = 10 lg \left\lgroup\frac{P}{1 \times 10^{-10}}\right\rgroupwhere P is the minimum acceptable power. Solving for P gives:
= 1.0 \times 10^{-7} WStep 2 A repeater is needed to regenerate the signal when the signal-to-noise ratio falls to 30 dB, i.e. its power is 10^{3} times the noise level, and this is 1.0 \times 10^{-7} W We can calculate the attenuation needed to reduce the signal to this level:
attenuation = 10 lg \left\lgroup\frac{1.2 \times 10^{-3}}{1.0 \times 10^{-7}}\right\rgroup
= 41 dB
Hence the length of cable is \frac{41}{14} = 2.9 km
If the cable is 10 km in length, the total attenuation is:
14 dB km^{-1} \times 10 km = 140 dB
The signal of power 1.2 \times 10^{-3} W is attenuated to a power P where:
140 = 10 lg \left\lgroup\frac{0.0012}{P}\right\rgroupP = 12 \times 10^{-17} W
You can see that the power in the signal is much smaller than the minimum acceptable power – it is even smaller than the noise level. The signal-to-noise ratio is now 10 lg(12 \times 10^{-17} / 1.0 \times 10^{-8}) = -79 dB smaller than the acceptable +30 dB. A repeater is needed well before the end of the 10 km of cable