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).}