The initial section of the transmission line for a signal from a telephone exchange is illustrated in Fig. 5.1.

Question 8

(a) For a signal transmitted along an optic fibre, state what is meant by:

(i) attenuation [1]

(ii) noise. [2]

(b) The initial section of the transmission line for a signal from a telephone exchange is illustrated in Fig. 5.1.

Fig. 5.1

At the exchange, the input signal to the transmission line has a power of 2.5 × 10^-3 W.

After the signal has travelled a distance of 52 km along the transmission line, the power of the signal is 7.8 × 10^-16 W. The signal is then amplified.

(i) Calculate the attenuation per unit length, in dB km^-1, in the transmission line. [3]

(ii) The gain of the amplifier is 115 dB.

Calculate the power of the signal at the output of the amplifier. [2]

 [Total: 8]

Reference: Past Exam Paper – June 2019 Paper 42 Q5

Solution:

(a)

(i) Attenuation is the loss of power from a signal.

(ii) Noise is unwanted power that is added to the signal being transmitted.

(b)

(i)

{As the input signal travels along the transmission line, it gets attenuated – its power decreases.}

attenuation = 10 lg(P2 / P1)

{where P2 is the input signal and P1 is the power of the signal after travelling a distance of 52 km.}

attenuation per unit length = (1 / L) × 10 lg(P2 / P1)

attenuation per unit length = (1 / 52) × 10 lg [(2.5×10^-3) / (7.8×10^-16)]

attenuation per unit length = 2.4 dB km^-1         

(ii)

gain / dB = 10 lg(P2 / P1)

{where P2 is the signal output of the amplifier and P1 is the signal coming into the amplifier.

Note that the numerator in the logarithmic function should always be greater than the denominator so as to avoid negative values. The same applies for the values in the attenuation formula.

Since the power of the signal will increase when output from the amplifier, its power is greater than that going in the amplifier and so, the output signal is used as the numerator.}

115 = 10 lg [P / (7.8×10^-16)]

{ lg [P / (7.8×10^-16)] = 115 / 10

lg [P / (7.8×10^-16)] = 11.5

P / (7.8×10^-16) = e^11.5

P = 7.8×10^-16 × e^11.5 }

P = 2.5 × 10^-4 W