• Physics 9702 Doubts | Help Page 134

Question 667: [Transmission of Signal > ADC and DAC]

Digital transmission of speech may be represented by the block diagram of Fig.1.

(a) State purpose of the parallel-to-serial converter

(b) Part of the signal from the microphone is shown in Fig.2.

The ADC (analogue-to-digital converter) samples analogue signal at a frequency of 5.0 kHz.

Each sample from ADC is a four-bit digital number where smallest bit represents 1.0 mV.

First sample is taken at time zero.

Use Fig.2 to determine the four-bit digital number produced by the ADC at times

(i) 0.4 ms

(ii) 0.8 ms

(c) The digital signal is transmitted and then converted to analogue form by the DAC (digital-to-analogue converter).

Using data from Fig.2, draw, on the axes of Fig.3, the output level of the transmitted analogue signal for time zero to time 1.2 ms.

(d) State and explain effect on the transmitted analogue waveform of increasing, for the ADC and the DAC, both the sampling frequency and the number of bits in each sample

Reference: Past Exam Paper – June 2013 Paper 42 Q12

Solution 667:

(a) The parallel-to-serial converter takes all the simultaneous digits for one number and ‘sends’ them one after another (along the transmission line)

(b)

{Numbers subjected to 2³       2²         2¹         2⁰         which equal to 8   4   2   1}

{For an explanation on how to convert the voltages into binary numbers, see solution 555 at Physics 9702 Doubts | Help Page 108 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-108.html}

(i)

0.4ms: corresponds to 7mV

0111

(ii)

0.8ms: corresponds to 6.4mV (6 since smallest bit represents 1.0mV)

0110

(c)

t                                   0          0.2       0.4       0.6       0.8       1.0       1.2

microphone output      0          8          7          15        6.4       5.4       8

output level                 0          8          7          15        6          5          8

Correct basic shape of graph, i.e. series of shape

With levels staying constant during correct time intervals

(d) Increasing the number of bits reduces the step height while increasing the sampling frequency reduces the step depth/width. Thus, reproduction of the signal is more exact.

ADC and DAC: Reproduction of an input signal

This is a quick explanation about the reproduction of an input signal and the effects of the number of bits and sampling frequency on the recovered signal. Some of the explanation has been taken from the application booklet provided by Cambridge (credits go to them) and further details have been added to me.

The electrical signals derived from speech are analogue audio-frequency signals (a voltage–time graph would be a smooth curve). To convert an analogue signal into a digital signal (the voltage– time graph would be a series of ‘steps’) involves taking samples of the analogue waveform (i.e. measuring its instantaneous voltage) at regular intervals of time (the sampling frequency affects this time). To instantaneous or sample voltage is converted into a binary number that represents its value.

1. Increasing the number of bits reduces the step height

The number of bits limits the number of possible voltage levels.

With 4 bits, there are 2⁴ = 16 levels.

With 8 bits, there are 2⁸ = 256 levels.

The number of bits has either of the following effects on the reproduction:

(1) It allows more information to be carried.

(2) It allows smaller changes to be detected. This reduces the step height.

Let’s say that the original analogue signal varies between 0V and 10V.

If we use 4 bits (we have 16 levels), the range of 10V will divided by 16. That is, the smallest voltage that can be sampled is 10/16 V (this is 0.625V).

If we use 8 bits (we have 256 levels), the range of 10V will divided by 256. That is, the smallest voltage that can be sampled is 10/256 V (this is about 0.039V).

From this, we can see that a larger number of bits allow smaller voltages to be sampled. That is, smaller changes can be detected. This reduces the ‘step height’ in a voltage–time graph of the recovered signal.

Now, how does a larger number of bits allow more information to be carried?

Let’s say that the smallest voltage that can be sampled for a signal is 0.625V.

With 4 bits, we have 16 levels. The maximum voltage would be 16 (0.625) = 10V.

With 8 bits, we have 256 levels. The maximum voltage would be 256 (0.625) = 160V.

So, here if the original analogue signal varied from, say, 0V to 20V the information from voltages higher than 10V cannot be carried with a 4 bit system, but there would be no problem with the 8 bit system.

2. Increasing the sampling frequency reduces the step depth

As said above, samples of the analogue waveform are taken at regular intervals of time. A regular interval of time means that there is a period involved in measuring the instantaneous voltages.

Period = 1 / Frequency

Here, the frequency is actually the ‘sampling frequency’.

To obtain a better digital signal, the samples need to be taken as quick as possible. That is, the period needs to be as small as possible. In order words, the sampling frequency should be high.

For example,

A sampling frequency of 4 kHz means that the signal is sampled every (period = 1/f) 250 μs. A sampling frequency of 8 kHz means that the signal is sampled every (period = 1/f) 125 μs. Thus, more details of the original signal can be obtained with a higher sampling frequency.

A lower sampling frequency means that less information can be gathered from the analogue signal while a higher sampling frequency enables more detail of the analogue signal to be recovered.

In the signal–time graph for the recovered signal, the step depth is represented by the horizontal lines. At a higher sampling frequency, the period is less and hence, the horizontal lines are shorter.

Applying the 2 changes explained allow the output of the DAC to be smooth, not ‘grainy’.

Question 668: [Communication Systems]

(a) In modern communications systems, majority of data is transmitted in digital form rather than analogue form.

Suggest three advantages of transmission of data in digital form.

(b) A recording is made of some music. For this recording, music is sampled at a rate of 44.1 kHz and each sample consists of a 16-bit word.

(i) Suggest effect on the quality of the recording of

1. sampling at a high frequency rather than a lower frequency,

2. using a long word length rather than a shorter word length

(ii) The recording lasts for total time of 5 minutes 40 seconds.

Calculate number of bits generated during the recording.

Reference: Past Exam Paper – November 2012 Paper 41 & 42 Q11

Solution 668:

(a) Choose any 3:

Noise can be eliminated / filtered / signal can be regenerated

Extra bits can be added to check for errors

Multiplexing possible

Digital circuits are more reliable / cheaper

Data can be encrypted for security

(b)

(i)

1. Higher frequencies can be reproduced

2. Smaller changes in loudness / amplitude can be detected 

{An explanation on the reproduction of signals is available at the end of solution 667 above.}

(ii)

Bit rate = Sample rate x number of bits in each sample (no. of bits per second)

Bit rate = 44.1×10³ × 16 = 7.06×10⁶s^-1

Number of bits generated during recording = Bit rate x Amount of time

5 min 40 sec = 340sec

Number of bits = 7.06×10⁶ × 340 = 2.4×10⁸

Question 669: [Centre of Gravity]

(a) Explain what is meant by centre of gravity of a body.

(b) An irregularly-shaped piece of cardboard is hung freely from one point near its edge, as shown in Fig.1.

Explain why cardboard will come to rest with its centre of gravity vertically below the pivot. You may draw on Fig.1 if you wish.

Reference: Past Exam Paper – November 2005 Paper 2 Q2

Solution 669:

(a) The centre of gravity of a body is the point where the whole weight of the body (allow mass) may be considered to act

(b) When the centre of gravity is below the pivot the weight acts through the pivot. So, the weight has no turning effect about the pivot.

{Moment = Force × perpendicular distance from line of action of force to the pivot

The definition of moment (turning effect of force) is such that it is the force and the distance of the line of action of the force from the pivot should be perpendicular.

When the piece of cardboard is released, its centre of gravity is vertically below the pivot (the plumbline experiment) and the cardboard will be move or rotate (that it is at rest – in equilibrium). This is because the weight, which acts vertically downwards (this is the direction of the force), is vertically below the pivot. So, the ‘perpendicular distance of the force from the pivot’ is zero. Thus there is no turning effect – no moment.

Additionally, from Newton’s 3^rd law, there will be an equal and opposite force (to the weight) at the pivot. This causes the resultant force to be zero. So, the cardboard is in equilibrium.}

Question 670: [Kinematics]

A motor drags a log of mass 452 kg up a slope by means of a cable, as shown in Fig.1.

Slope is inclined at 14.0° to the horizontal.

(a) Show that component of the weight of the log acting down the slope is 1070 N.

(b) Log starts from rest. A constant frictional force of 525 N acts on the log. The log accelerates up the slope at 0.130 m s^–2.

(i) Calculate tension in the cable.

(ii) Log is initially at rest at point S. It is pulled through a distance of 10.0 m to point P.

Calculate, for the log,

1. time taken to move from S to P,

2. magnitude of the velocity at P.         

(c) The cable breaks when log reaches point P. On Fig.2, sketch variation with time t of the velocity v of the log. Graph should show v from the start at S until the log returns to S.

Reference: Past Exam Paper – June 2012 Paper 23 Q2

Solution 670:

(a)

Weight of log (= mg) = 452 × 9.81

Component of weight down the slope = 452 × 9.81 × sin 14° = 1072.7 = 1070 N

(b)

(i)

Resultant force F = ma

{Let the tension in the cable be T. Tension T pulls the log up the slope. The component of the weight causes a force down the slope. The frictional force, which oppose motion, also acts down the slope.}

T – (1070 + 525) = 452 × 0.13

Tension T = 1650 (1653.76) N

(ii)

1.

Equation of uniformly accelerated motion: s = ut + ½ at²

10 = 0 + ½ × (0.130) t²

Time t = [(2 × 10) / 0.13]^1/2 = 12.4 or 12 s

2.

Equation of uniformly accelerated motion: v² = u² + 2as

Speed v = [0 + (2×0.13×10)]^1/2 = 1.61 or 1.6 m s^–1

(c) The graph is a straight line from the origin {constant acceleration}. The line goes down to zero velocity in short time compared to the first stage {as the cable breaks, only the component of the weight and the force of friction are present and both acts downwards. Even if the velocity decreases, the motion is still up the slope until the velocity becomes zero.}. Then, the line is less steep with a negative gradient {the log accelerates in the opposite direction – the component of weight acts down the slope while the frictional force which opposes motion, acts up the slope. The motion of the log is down the slope} and the final velocity is larger than the final velocity in the first part by at least 2 times. {The acceleration is now given by a = (1070 – 525) / 452 = 1.21 ms^-2.}