Thursday, May 7, 2015

Physics 9702 Doubts | Help Page 134

  • 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)

{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 -}
0.4ms: corresponds to 7mV

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

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

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.}

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 3rd 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:
Weight of log (= mg) = 452 × 9.81
Component of weight down the slope = 452 × 9.81 × sin 14° = 1072.7 = 1070 N

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

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

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.}


  1. Here are the remaining questions for now:

    21/O/N/12 Q.6(c)

    22/O/N/12 Q.3(b)

    21/M/J/13 Q.6(c)(d)

    22/O/N/13 Q.3(c)(ii)

    1. For 21/O/N/12 Q.6(c), check solution 671 at

      For 22/O/N/13 Q.3(c)(ii), see

  2. Hi,
    In question 670 the last part you mentioned the graph should shown, 'the line is less steep with a negative gradient', but if acceleration is 1.21m/s^-2(in opposite direction) it is greater than 0.130 m/s^2, so shouldn't it be more steeper because the magnitude is greater?Also why should the final velocity be twice greater than initial?

    Thanks a lot,

    1. Its less step than the deceleration part, not the upward acceleration.

      v^2 = u^2 + 2as. For both the initial acceleration upwards from rest and the acceleration downwards from the zero velocity, the initial velocity is zero. The log should descent the same distance s that it has moved up the slope. Since the acceleration while falling is about 4 times larger than that while moving up, the final velocity which is given by v = √(0 + 2as) should be about two times larger for the falling log (the square root of 4 is 2)

    2. Thank you!!!:)))

    3. Hello there but I still dont get it why the negative acceleration is less steep

    4. The explanation seems enough. Try to read the text in red again and let me know what exactly you are not understanding.

    5. acceleration while falling down is 4 times that the one of moving up but 1.21 divide by 0.13 is about 9 times? and the acceleration in opposite direction is less steep is because only the component of weight is acting down and frictional force is opposing it but since both component of weight and frictional force act down when it is decelerating so that's why it is less steep,is that correct?

    6. the final line is less steep than the middle part, not the first part.

      Also, during the downward motion, the component of the weight acts down the slope and friction, which opposes motion, now acts upwards.

    7. The time taken to reach to point P from S should be greater than the time taken from P to reach S, isn't it?

    8. Yes, because in one case the acceleration due to gravity is in direction of motion and in the other case, it opposes motion.

  3. Hi. I'am not sure where to ask a new question, so I just selected this page. I would appreciate it if you could explain further 9702/41/MJ/12 Q1 (c) . I've read the answers in marking scheme, but what does the 'if r is at h, g is constant' sentence have to do with the explanation? Can't I just simply state Ep=Fxdistance=mgh? Thank you!

  4. I asked 9702/41/MJ/12 Q 1(c) earlier. I would also like to know why (Change in Ep)=(mass)(Change in gravitational potential) ?

    1. See solution 693 at

  5. Can you please upload 2015 all variants of a part 1,2 and 3.I know it's too much to ask for but if it's too much then please the variants of paper 2.I have exam this coming week.Please please ....

    1. You need to specify which question, and in which variant. I won't be solving all the questions in a single paper in one go. It will be one by one.

      + which state about what exactly you are having problems in those questions.

    2. 2015 paper 2 question number 2,4 and 5 and from variant 21 question 2 and 5.AND can you plz give me the links of 21 and 23 variant of paper2(M/J AND O/N) of the following years 2011,2010 AND 2009.As searching for each question on each doubth help page is too time consuming and often i cant find the question

    3. For Q5, see solution 1004 at

      well, here is how you could search for the questions:
      1. Go to Google search
      2. Type '9702 physics ref' (without the quotes), followed by the first line(s) of the question.

      If it has already been solved, you'll obtain the result. If not, it probably means the question has not been solved yet. If it's the latter, then you may ask through the comment box here

  6. Can you please explain how we can obtain the minimum voltage of 0.625V in the binary system?

    1. It's already explained in ADC and DAC: Reproducing an input signal under the number 1. Reducing ....


If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | Physics 9702 Doubts | Help Page 134