The analogue signal from a microphone is to be transmitted in digital form.

Question 10

The analogue signal from a microphone is to be transmitted in digital form.

The variation with time t of part of the signal from the microphone is shown in Fig. 5.1.

Fig. 5.1

 

The microphone output is sampled at a frequency of 5.0 kHz by an analogue-to-digital converter (ADC).

 

The output from the ADC is a series of 4-bit numbers. The smallest bit represents 1.0 mV.

The first sample is taken at time t = 0.

 

(a) Use Fig. 5.1 to complete Fig. 5.2.

 

Fig. 5.2                                                            [2]

 

 

(b) After transmission of the digital signal, it is converted back to an analogue signal using a digital-to-analogue converter (DAC).

 

Using data from Fig. 5.1, draw, on the axes of Fig. 5.3, the output level from the DAC for the transmitted signal from time t = 0 to time t = 1.2 ms.

 

Fig. 5.3

[4]

 

 

(c) It is usual in modern telecommunication systems for the ADC and the DAC to have more than four bits in each sample.

State and explain the effect on the transmitted analogue signal of such an increase. [2]

 

[Total: 8]

 

 

Reference: Past Exam Paper – November 2017 Paper 42 Q5

 

 

Solution:

(a)

(0.2 ms)           8.0 (mV)          1000                B1

(0.8 ms)           5.8 (mV)          0101                B1

 

{4-bit:               2³   2²   2¹   2⁰ }

                        8     4    2     1

8.0 mV            1     0    0     0              = 1(2³) + 0(2²) + 0(2¹) + 0(2⁰) = 8

5.8 mV (consider 5.0 mV, NOT 6.0 mV – NOT the nearest whole number)

5.8 mV            0     1   0     1              = 0(2³) + 1(2²) + 0(2¹) + 1(2⁰) = 5

 

 

(b)

level                 0              8                  10                    15                    5                   8

time / ms         0–0.2     0.2–0.4          0.4–0.6            0.6–0.8            0.8–1.0        1.0–1.2

 

{Consider the initial time for the intervals.

e.g.      for time: 0–0.2ms, consider voltage at t = 0 ms (= 0 mV = level 0)

for time: 0.2–0.4ms, consider voltage at t = 0.2 ms (= 8 mV = level 8)

…}

 

 

(c)

The heights of the step is smaller.

It allows more accurate reproduction (of the input signal). {That is, the output signal represents the input signal more closely (smaller changes between the output and the input).}

The sinusoidal carrier wave has a frequency of 750 kHz and an amplitude of 5.0 V. The carrier wave is frequency modulated by a sinusoidal signal of frequency 7.5 kHz and amplitude 1.5 V.

 Question 9

A carrier wave is frequency modulated.

(a) Describe what is meant by frequency modulation. [2]

 

 

(b) The sinusoidal carrier wave has a frequency of 750 kHz and an amplitude of 5.0 V.

The carrier wave is frequency modulated by a sinusoidal signal of frequency 7.5 kHz and amplitude 1.5 V.

 

The frequency deviation of the carrier wave is 20 kHz V^-1.

 

Determine, for the frequency-modulated carrier wave,

(i) the amplitude, [1]

(ii) the minimum frequency, [1]

(iii) the maximum frequency, [1]

(iv) the number of times per second that the frequency changes from its minimum value to its maximum value and then back to the minimum value. [1]

 

 

 

Reference: Past Exam Paper – November 2015 Paper 41 & 42 Q11

 

 

 

Solution:

(a)

Frequency modulation means that the frequency of the carrier wave varies in synchrony with the displacement of the signal (information) wave.

 

 

(b)

(i) 5.0 V          

{In FM, the amplitude does not change but the frequency changes.}

 

(ii) the minimum frequency, [1]

720 kHz

 

{The frequency of the unmodulated carrier wave is 750 kHz.

 

The frequency deviation of the carrier wave is 20 kHz V^-1. This means that the (above) frequency can deviate (from its value of 750 kHz) by 20 kHz per volt of the signal wave.

 

1 V - - - > deviation of 20 kHz

The amplitude of the signal wave is 1.5 V.

1 V - - - > deviation of 20 kHz

1.5 V - - - > 1.5 × 20 = 30 kHz

 

So, the frequency of the carrier wave can deviate (can increase or decrease) by up to 30 kHz.

Minimum frequency = 750 – 30 = 720 kHz}

 

(iii) 780 kHz

{Maximum frequency = 750 + 30 = 780 kHz}

 

(iv) 7500

{This is the frequency of the signal wave (= 7.5 kHz = 7500 Hz)}

 

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