The digital transmission of speech may be illustrated using the block diagram of Fig. 3.1.

Question 5

The digital transmission of speech may be illustrated using the block diagram of Fig. 3.1.

Fig. 3.1

(a) (i) State what is meant by a digital signal. [1]

(ii) State the names of the components labelled X and Y on Fig. 3.1.

X:

Y: [2]

(iii) Describe the function of the ADC. [2]

(b) The optic fibre has length 84 km and the attenuation per unit length in the fibre is 0.19 dB km^-1.

The input power to the optic fibre is 9.7 mW. At the output from the optic fibre, the signal-to-noise ratio is 28 dB.

Calculate

(i) in dB, the ratio

input power to optic fibre

noise power at output of optic fibre,

[2]

(ii) the noise power at the output of the optic fibre. [3]

[Total: 10]

Reference: Past Exam Paper – June 2017 Paper 41 & 43 Q3

Solution:

(a) (i) A digital signal consists of (a series of) 1s and 0s        or offs and ons           or highs and lows

(ii)

component X: parallel-to-serial converter

component Y: DAC/digital-to-analogue converter

(iii) The purpose of the ADC is to sample the (analogue) signal at regular intervals and converts the analogue number to a digital number.

(b)

(i)

{Attenuation per unit length = 0.19 dB km^-1

1 km → 0.19 dB

84 km → 84 × 0.19 = 16 dB}

attenuation in fibre = 84 × 0.19 (= 16 dB)

{If the signal-to-noise ratio at the output is 28 dB then it (the signal-to-noise ratio at the input – it is this quantity that we want to find) must be greater at the input because the noise is constant along the cable.

Along the optical fibre, when the signal has been attenuated (from the input to the output) by a total amount of 16 dB (as calculated above).

Indeed, the signal-to-noise ratio must be greater by the amount that the signal has been attenuated by travelling along the cable, which is 16 dB.}

ratio = 16 + 28 = 44 dB

(ii)

{From the above part, we have found that

(input) signal-to-noise ratio = 44 dB

44 = 10 lg (input signal / noise)

The noise power is constant along the cable. So, we can find the noise power at the output (this is the same as the noise power at the input.).

When taking the number of decibels to be positive, the numerator in the logarithmic function should be greater than the denominator.

The signal power at the input is greater – so, it is the numerator.}

ratio (in dB) = 10 lg (P2 / P1)

44 = 10 lg ({9.7 × 10^-3} / P)

or

{If the number of decibels is taken to be negative, then the numerator should be smaller.}

–44 = 10 lg (P / {9.7 × 10^-3})

{Taking the number of decibels as positive,

44 = 10 lg ({9.7 × 10^-3} / P)

lg ({9.7 × 10^-3} / P) = 4.4

{9.7 × 10^-3} / P = 10^4.4

Power P = {9.7 × 10^-3} / 10^4.4 }

Power = 3.9 × 10^-7 W