Page 331 - Physics Coursebook 2015 (A level)

Chapter 20: Communications systems
WORKED EXAMPLE
5 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.
Step1 Thesignal-to-noiseratiomustbeatleast
QUESTIONS
13 A signal has an input power 5.0 mW and an output power of 0.000 2 mW. What is the attenuation in dB?
14 The attenuation of a 6.0 mW signal is 30 dB. What is the final power?
15 What is the signal-to-noise ratio when the signal and the noise have equal power?
16 A signal of 1.0 mW passes through an amplifier of gain 30 dB and then along a cable where the attenuation is 18 dB.
a What is the overall gain of the signal in dB?
b What is the output power at the end of the
cable?
Comparison of different channels
Each type of signal channel has its good points and its disadvantages, which we will now consider.
Wire-pairs and coaxial cables
The earliest telephones used a pair of wires strung
on either side of a pole (Figure 20.13). As the use of electricity became more common, the amount of electrical interference increased, causing crackle and hiss on the line. The potential difference between the two wires is
the signal. Each wire acts as an aerial, picks up unwanted electromagnetic waves and distorts the signal.
30 dB. Hence, using: signal-to-noise ratio = 10 lg
we have:
30 = 10 lg P
1 × 10−10
signal power noise power
where P is the minimum acceptable power. Solving for P gives:
=1.0×10−7W
Step2 Arepeaterisneededtoregeneratethesignal when the signal-to-noise ratio falls to 30 dB, i.e. its power is 103 times the noise level, and this is
1.0 × 10−7 W. We can calculate the attenuation needed to reduce the signal to this level:
attenuation = 10 lg 1.2 × 10−3 1.0 × 10−7
= 41 dB
Hence the length of cable is 14 = 2.9km.
If the cable is 10 km in length, the total attenuation is: 14 dB km−1 × 10 km = 140 dB.
The signal of power 1.2 × 10−3 W is attenuated to a power P where:
140=10lg 0.0012
P P=12×10−17W
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 ratioisnow10lg(12×10−17/1.0×10−8)=−79dB, smaller than the acceptable +30 dB. A repeater is needed well before the end of the 10 km of cable.
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Figure 20.13 An early telegraph pole.
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