A transmission system for speech may be represented by the block diagram of Fig. 12.1.

Question 4

A transmission system for speech may be represented by the block diagram of Fig. 12.1.

Fig. 12.1

(a) Name the component labelled

(i) block X, [1]

(ii) block Y, [1]

(iii) Z. [1]

(b) The variation with time of part of the signal at the input P to the analogue-to-digital converter (ADC) is shown in Fig. 12.2.

Fig. 12.2

Each number of the output from the ADC is a digital number where the smallest bit represents 1 mV.

State

(i) the minimum number of bits in each digital number so that the signal in Fig. 12.2 can be sampled fully, [1]

(ii) the digital number produced by the ADC at time 0.50 ms. [1]

(c) The ADC samples the signal in Fig. 12.2 at a frequency of 4.0 kHz. The first sample is taken at time zero.

Using data from Fig. 12.2, draw, on the axes of Fig. 12.3, the variation with time of the output at point Q for time zero to time 1.5 ms.

Fig. 12.3

[4]

Reference: Past Exam Paper – November 2015 Paper 43 Q12

Solution:

(a)

(i) serial-to-parallel converter

(ii) digital-to-analogue converter or DAC

(iii) (audio) amplifier or AF amplifier

{since this is a transmission system for speech}

(b)

(i) 4

{The signal does not go above 15, so 4 bits would be sufficient to encode the voltage levels. For 4 bits, number of voltage levels = 2⁴ = 16.}

(ii) 1011

{At 0.50 ms, the signal voltage is 11 mV. This is an analogue reading.

In terms of binary number,

2³         2²         2¹         2⁰

Equivalent                   8          4          2          1

Binary                          1          0          1          1         

      = 1(2³) + 0(2²) + 1(2¹) + 1(2⁰)

      = 8 + 0 + 2 + 1 = 11 mV

So, 11 mV = 1011}

(c)

correct levels at 0.25 ms intervals

0, 8, 11, 10, 15                                               

and 7, 4                                                          

series of steps, each of depth 0.25 ms          

voltage levels shown in correct intervals       

{Sampling frequency = 4.0 kHz

Period T = 1 / f = 1 / (4×10³) = 0.25 ms

So, we need to identify the levels after 0.25 ms intervals.

At t = 0 ms, level = 0

At t = 0.25 ms, level = 8 mV

At t = 0.50 ms, level = 11 mV

At t = 0.75 ms, level = 10 mV

At t = 1.00 ms, level = 15 mV

At t = 1.25 ms, level = 7.6 mV = 7 mV

At t = 1.50 ms, level = 4.4 mV = 4 mV}

{Values should be rounded down, and not up. E.g. 7.6 mV becomes 7 mV, not 8 mV}