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9702 Applications Booklet WEB
Teknik Mesin (2019 , MPD)
Institut Teknologi Sepuluh Nopember
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A Level Science Applications Support Booklet: Physics
Updated October 2009
Contents List
Introduction ....................................................................................................................................................... 1 Gathering and Communicating Information ...................................................................................................... 2 28. Direct Sensing...................................................................................................................................... 2 29. Remote Sensing................................................................................................................................. 13 30. Communicating Information ............................................................................................................... 27
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A LEVEL SCIENCE APPLICATIONS SUPPORT BOOKLET:
PHYSICS
Introduction
Too often the study of Physics at A level can seem theoretical and abstract. The connections between Physics and real life can seem remote.
In reality, Physics is not a purely abstract subject. Like other science subjects, Physics has a “pure” theoretical side and an applied side. The principles of Physics are applied in a vast range of contexts, from the building of bridges to the design of integrated circuits. Much of the technological revolution has its foundations in applied Physics.
To ensure that the syllabus retains a balance between pure and applied Physics, there is a whole section at the end of the syllabus on Applications of Physics. It is at the end of the syllabus because the theoretical principles have to be learned and understood first if the applications are to be understood.
This booklet has been written to support teachers and students as they follow the Applications of Physics part of the syllabus.
In the booklet, each learning outcome is printed in it alics and is followed by a detailed explanation. These explanations have been written by examiners and it is hoped that they will help to illustrate the level of detail that students are expected to master.
It should be stressed that this booklet is not a replacement to the syllabus. While it is hoped that the booklet will help to make the syllabus content clearer to students and teachers, it should not be read as an authoritative guide as to what is and is not included in the syllabus. The examination papers will assess the syllabus, not this booklet.
The sections of the booklet are numbered in the same way as the sections of the syllabus, so that the first part of this booklet is (perhaps rather unusually) section 28. The learning outcomes are covered in syllabus order.
The overarching theme of the Applications of Physics part of the syllabus is “Gathering and Communicating Information”. This is in three sections, as follows.
Section 28, Direct Sensing, covers the electronics necessary to measure temperature, light intensity or strain; to detect sound signals; to amplify signals; and to connect sensors to circuits.
Section 29, Remote Sensing, covers some of the ways in which medical physicists obtain information about the inside of the body without surgery, by using X-rays, ultrasound and NMR; and how the information can be converted into images of the inside of the body.
Section 30, Communicating Information, covers some of the ways in which information is communicated using radio waves, optic fibres, satellites and mobile phones.
These three sections are interconnected. The information that is communicated from one place to another can come from sensors, microphones or scanners. Communications systems contain amplifier circuits. Ultrasound scanners and microphones both use piezoelectric transducers. The equations for the attenuation of X-rays in matter and the attenuation of a signal in a wire are equivalent. Because of the many links, examination papers will often contain questions assessing more than one section.
The Applications of Physics section of the syllabus forms approximately 12% of the full Advanced Level course. It follows that teachers should dedicate about one-ei ghth of their teaching time to the topics outlined in the following pages. Experience shows that students who are left to work through this booklet on their own, without support, supervision or tuition from their teacher, usually do not perform well in the examination.
It is hoped that this booklet will be used in conjunction with a variety of other sources of information, perhaps including visits to hospitals and communications centres, guest speakers, practical work, the internet, textbooks and videos.
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Negative temperature coefficient thermistors have a resistance that becomes less as the temperature of the thermistor rises. The change in resistance R with temperature θ for a typical thermistor is illustrated in Fig. 1.
Fig. 1. It can be seen that there is a comparatively large change in resistance with temperature but this change is non-linear.
(d) Candidates should be able to show an understanding of the action of a piezo-electric transducer and its application in a simple microphone. A transducer is any device that converts energy from one form to another. Piezo-electric crystals such as quartz have a complex ionic structure. When the crystal is unstressed, the centres of charge of the positive and the negative ions bound in the lattice of the piezo-electric crystal coincide. If, however, pressure is applied to th e crystal, the crystal will distort and the centres of charge for the positive and negative ions will no long er coincide. A voltage will be generated across the crystal. The effect is known as the piezo-electric effect (see also the section on 29(h)). Electrical connections can be made to the crystal if opposite sides of the crystal are coated with a metal. The magnitude of the voltage generated depends on the magnitude of the pressure applied to the crystal. The polarity of the voltage depends on whether the crystal is compressed or expanded (increase or decrease in the applied pressure). A sound wave consists of a series of compressions and rarefactions. If the wave is incident on a piezo- electric crystal, a varying voltage across the crystal will be produced. This voltage can be amplified. The crystal and its amplifier act as a simple microphone.
4000
3000
2000
1000
0 0 20 60 40 80 100
θ /°C
R /Ω
4
(e) Candidates should be able to describe the structure of a metal-wire strain gauge.
(f) Candidates should be able to relate extension of a strain gauge to change in resistance of the gauge. A strain gauge is made by sealing a length of very fine wire in a small rectangle of thin plastic, as shown in Fig. 1.
Fig. 1. When the plastic is stretched (the plastic experiences a strain), the wire will also be stretched. This causes the wire’s length to increase and its cross-sectional area to decrease slightly. Both these changes cause the resistance of the wire to increase. Strain gauges are usually glued very securely to the material that is under test. The resistance R of a wire of length L and of uniform cross-sectional area A is given by the expression R = ρL /A,
where ρ is the resistivity of the material of the wire. Assuming that, when the wire extends by a small amount ∆L, the change in the cross-sectional area is negligible, the new resistance will be given by
(R + ∆R) = ρ(L +∆L) /A, where ∆R is the change in the resistance. Subtracting these two expressions,
∆R = ρ∆L /A or, ∆R ∝ ∆L.
Thus the strain which is proportional to the extension ∆L is also proportional to the change in resistance ∆R. Note that the cross-sectional area A is assumed to be constant.
wire
plastic
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When connected to appropriate power supplies, an op-amp produces an output voltage Vout that is proportional to the difference between the voltage V + at the non-inverting input and the voltage V – at the inverting input. Vout = A 0 (V + – V –), where A 0 is the open-loop gain of the op-amp. The ideal operational amplifier (op-amp) has the following properties:
- infinite input impedance (i. no current enters or leaves either of the inputs);
- infinite open-loop gain (i. if there is only a very slight difference between the input voltages, the output will be saturated - the output will have the same value as the supply voltage);
- zero output impedance (i. the whole of the output voltage is provided across the output load);
- infinite bandwidth (i. all frequencies are amplified by the same factor);
- infinite slew rate (i. there is no delay between changes in the input and consequent changes in the output). Real operational amplifiers do deviate from the ideal. In practice, the input impedance is usually between 10 6 Ω and 10 12 Ω and the output impedance is about 10 2 Ω. The open-loop gain is usually about 10 5 for constant voltages. The slew rate (about 10 V μs–1) and bandwidth are not infinite.
(i) Candidates should be able to deduce, from the properties of an ideal operational amplifier, the use of an operational amplifier as a comparator. When an operational amplifier is used in a circuit, it is usually connected to a dual, or split, power supply. Such a supply can be thought to be made up of two sets of batteries, as shown in Fig. 1.
Fig. 1. The common link between the two sets of batteries is termed the zero-volt, or earth, line. This forms the reference line from which all input and output voltages are measured. Connecting the supplies in this way enables the output voltage to be either positive or negative. Fig. 1 shows an input V – connected to the inverting input and an input V + connected to the non- inverting input. The output voltage Vout of the op-amp is given by Vout = A 0 (V + – V –), where A 0 is the open-loop gain (typically 10 5 for d. voltages).
zero volt line (earth)
V – V +
–ve line
+ve line
Vout
7
Consider the examples below. Example 1: +ve supply line = +9 V –ve supply line = –9 V V + = 1 V V – = 1 V
Substituting into the above equation,
Vout = 10 5 × (1 – 1) = 10 000 V Obviously, this answer is not possible because, from energy considerations, the output voltage can never exceed its power supply voltage. The output voltage will be 9 V. The amplifier is said to be saturated.
Example 2: +ve supply line = +6 V –ve supply line = –6 V V + = 3 V V – = 3 V Substituting, Vout = 10 5 × (3 – 3) = –200 V Again, the amplifier will be saturated and the output will be –6 V. The examples show that, unless the two inputs are almost identical, the amplifier is saturated. Furthermore, the polarity of the output depends on which input is the larger. If V – > V +, the output is negative. The circuit incorporating the op-amp compares the two inputs and is known as a comparator. A comparator for use with an LDR is shown in Fig. 1.
Fig. 1. It is usual to connect a potential divider to each of the two inputs. One potential divider provides a fixed voltage at one input while the other potential divider provides a voltage dependent on light intensity. In Fig. 1, the resistors of resistance R will give rise to a constant voltage of ½VS at the inverting input. The LDR, of resistance RLDR is connected in series with a fixed resistor of resistance F. If RLDR > F (that is, the LDR is in darkness), then V + > V – and the output is positive. If RLDR < F (that is, the LDR is in daylight), then V + < V – and the output is negative. It can be seen that by suitable choice of the resistance F, the comparator gives an output, either positive or negative, that is dependent on light intensity. The light intensity at which the circuit switches polarity can be varied if the resistor of resistance F is replaced with a variable resistor. The LDR could be replaced by other sensors to provide alternative sensing devices. For example, use of a thermistor could provide a frost-warning device.
F
RLDR R
R
output
VS
VS
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(k) Candidates should be able to recall the circuit diagrams for both the inverting and the non-inverting amplifier for single signal input.
(l) Candidates should be able to show an understanding of the virtual earth approximation and derive an expression for the gain of inverting amplifiers.
(m) Candidates should be able to recall and use expressions for the voltage gain of inverting and of non-inverting amplifiers. In order to simplify the analysis of the circuits, the power supplies to the op-amps have not been shown. It is assumed that the op-amps are not saturated.
The inverting amplifier
The circuit for an inverting amplifier is shown in Fig. 1.
Fig. 1. An input signal Vin is applied to the input resistor Rin. Negative feedback is applied by means of the resistor Rf. The resistors Rin and Rf act as a potential divider between the input and the output of the op- amp. In order that the amplifier is not saturated, the two input voltages must be almost the same. The non- inverting input (+) is connected directly to the zero-volt line (the earth) and so it is at exactly 0 V. Thus, the inverting input (–) must be virtually at zero volts (or earth) and for this reason, the point P is known as a virtual earth. The input impedance of the op-amp itself is very large and so there is no current in either the non- inverting or the inverting inputs. This means that the current from, or to, the signal source must go to, or from, the output, as shown in Fig. 1.
P P
Fig. 1. Because the inverting input is at zero volts, a positive input gives rise to a negative output and vice versa. This is why the arrangement is given the name inverting amplifier.
+
Vout
Rf
Rin
Vin
P
- –ve output
Rf
Rin
+ve input
- +ve output
Rf
Rin
–ve input
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Referring to Fig. 1, since the input resistance of the op-amp is infinite, current in Rin = current in Rf and
f
f in
in R across p.d across p. RR
=
where Rin and Rf are the resistances of Rin and Rf respectively. The potential at P is zero (virtual earth) and so
f
out in
in 00 R
V R
V − = −
The overall voltage gain of the amplifier circuit is given by
in
f in
gain voltage out R
R V
V
nullThe non-inverting amplifier
The circuit for a non-inverting amplifier incorporating an op-amp is shown in Fig. 1.
Fig. 1. The input signal Vin is applied directly to the non-inverting input. Negative feedback is provided by means of the potential divider consisting of resistors R 1 and Rf. The voltage gain of the amplifier circuit is given by
1
f in
out 1 gain voltage R
R V
V
#######
where R 1 and Rf are the resistances of R 1 and Rf respectively. The non-inverting amplifier produces an output voltage that is in phase with the input voltage.
Vout Vin
Rf
R 1
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A resistor is frequently connected in series with an LED so that, when the LED is forward biased (the diode is conducting), the current is not so large as to damage the LED. A typical maximum forward current for an LED is 20 mA. Furthermore, the LED will be damaged if the reverse bias voltage exceeds about 5 V. Fig. 1 is a circuit using two diodes to indicate whether the output from an op-amp is positive or negative with respect to earth.
D 1 D 2
op-amp output
Fig. 1. When the output is positive with respect to earth, diode D 1 will conduct and emit light. Diode D 2 will not conduct because it is reverse biased. If the polarity of the output changes, then D 2 will conduct and emit light and D 1 will not emit light. The state of the output can be seen by which diode is emitting light. The diodes can be chosen so that they emit light of different colours.
(p) Candidates should be able to show an understanding of the need for calibration where digital or analogue meters are used as output devices. An LED may be used to indicate whether an output is positive or negative. If the output is from a comparator, then LEDs can give information as to, for example, whether a temperature is above or below a set value. However, the LED does not give a value of the temperature reading. Many sensors, for example, a thermistor or an LDR, are non-linear. It was seen in 28(g) that the sensor could be connected into a potential divider circuit so that the output of the potential divider varied with some property, for example temperature or light intensity. This variable voltage could be measured using an analogue or a digital voltmeter. The reading on the voltmeter would vary with the property being monitored. However, the reading on the voltmeter would not vary linearly with change in the property. In order that the property can be measured, a calibration curve is required. The reading on the voltmeter is recorded for known values of the property X. A graph is then plotted showing the variation with the property X of the voltmeter reading. The value of the property X can then be read from the graph for any particular reading on the voltmeter.
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- Remote Sensing
(a) Candidates should be able to explain in simple terms the need for remote sensing (non-invasive techniques of diagnosis) in medicine. Historically, diagnosis consisted of two techniques – observing the patient outwardly for signs of fever, vomiting, changed breathing rate etc, and observing the patient inwardly by surgery. The first technique depended greatly on experience but was still blind to detailed internal conditions. The second quite often led to trauma and sometimes death of the patient. In earlier times there was also the significant risk of post-operative infection. Modern diagnostic techniques have concentrated on using externally placed devices to obtain information from underneath the skin. X-rays have been used for a century. More recently, ultrasound has been used, especially in cases of pregnancy. Magnetic resonance imaging (MRI) is now becoming a frequently-used technique. Other techniques involve lasers that can shine through a finger or can be used in a very narrow tube that can be inserted into the body through various orifices. In all these situations, the aim is to obtain detailed information concerning internal structures. This may be concerned, for example, with the functioning of an organ or the search for abnormalities. This is achieved without the need of investigative surgery and is described as a non-invasive technique. Non- invasive techniques are designed to present a much smaller risk than surgery and are, in general, far less traumatic for the patient.
(b) Candidates should be able to explain the principles of the production of X-rays by electron bombardment of a metal target. X-rays are produced by bombarding metal targets with high-speed electrons. A typical spectrum of the X-rays produced is shown in Fig. 2.
Fig. 2. The spectrum consists of two components. There is a continuous distribution of wavelengths with a sharp cut-off at short wavelength and also a series of high-intensity spikes that are characteristic of the target material. Whenever a charged particle is accelerated, electromagnetic radiation is emitted. The greater the acceleration, the shorter is the wavelength of the emitted radiation. This radiation is known as Bremmstrahlung radiation. When high-speed electrons strike a metal target, large accelerations occur and the radiation produced is in the X-ray region of the electromagnetic spectrum. Since the electrons have a continuous distribution of accelerations, a continuous distribution of wavelengths of X-rays is produced. There is a minimum wavelength (a cut-off wavelength) where the whole of the energy of the electron is converted into the energy of one photon. That is, kinetic energy of electron = eV = hc / λ, where e is the charge on the electron that has moved through a potential difference V, h is the Planck constant, c is the speed of light and λ is the wavelength of the emitted X-ray photon.
wavelength
intensity
0
15
The quality of the shadow picture (the image) produced on the photographic plate depends on its sharpness and contrast. Sharpness is concerned with the ease with which the edges of structures can be determined. A sharp image implies that the edges of organs are clearly defined. An image may be sharp but, unless there is a marked difference in the degree of blackening of the image between one organ and another (or between different parts of the same organ), the information that can be gained is limited. An X-ray plate with a wide range of exposures, having areas showing little or no blackening as well as areas of heavy blackening, is said to have good contrast. In order to achieve as sharp an image as possible, the X-ray tube is designed to generate a beam of X- rays with minimum width. Factors in the design of the X-ray apparatus that may affect sharpness include
- the area of the target anode, as illustrated in Fig. 2,
Fig. 2.
- the size of the aperture, produced by overlapping metal plates, through which the X-ray beam passes after leaving the tube (see Fig. 2),
Fig. 2.
- the use of a lead grid in front of the photographic film to absorb scattered X-ray photons, as illustrated in Fig. 2.
Fig. 2. In order to improve contrast, a ‘contrast medium’ may be used. For example, the stomach may be examined by giving the patient a drink containing barium sulphate. Similarly, to outline blood vessels, a contrast medium that absorbs strongly the X-radiation would be injected into the bloodstream. The contrast of the image produced on the photographic film is affected by exposure time, X-ray penetration and scattering of the X-ray beam within the patient’s body. Contrast may be improved by backing the photographic film with a fluorescent material.
full shadow
partial shadow
partial shadow
anode object
electrons
full shadow
partial shadow
partial shadow
anode object
electrons
film
patient
X-ray beam
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(e) Candidates should be able to show an understanding of the purpose of computed tomography or CT scanning. The image produced on an X-ray plate as outlined in the section on 29(d) is a ‘flat image’ and does not give any impression of depth. That is, whether an organ is near to the skin or deep within the body is not apparent. Tomography is a technique by which an image of a slice, or plane, of the object may be obtained. In this technique, a series of X-ray images are obtained from different angles through one section, or slice, of the object to be examined. The images are all in the plane of the slice, as illustrated in Fig. 2.
Fig. 2. Computer techniques make it possible to combine these images to give an image of the slice. The technique is called computed (axial) tomography or CT scanning. Images of successive slices can be combined to give a three-dimensional image. The three-dimensional image can be rotated and viewed from any angle.
(f) Candidates should be able to show an understanding of the principles of CT scanning.
(g) Candidates should be able to show an understandi ng of how the image of an 8-voxel cube can be developed using CT scanning. The aim of CT scanning is to make an image of a section (or slice) through the body from measurements made about its axis, as illustrated in Fig. 2. The section (or slice) through the body is divided up into a series of small units called voxels. The image of each voxel would have a particular intensity, known as a pixel. The pixels are built up from measurements of X-ray intensity made along a series of different directions around the section of the body. Suppose a section consists of four voxels with intensities as shown in Fig. 2.
Fig. 2. The number on each voxel is the pixel intensity that is to be reproduced.
X-ray detectors
collimator
X-ray tube
object
4
2 7
1
The final images are taken after rotating the X-ray tube and the detectors through a further 45°. The
readings added to voxels
third set of detector readings - - Fig. 2. - Fig. 2. result is shown in Fig. 2. - - - - - - - - - - fourth set of detector readings - - - - - - + - - - detector - - - - - - - readings added to voxels
19
The final pattern of pixels is shown in Fig. 2.
####### Fig. 2.
In order to obtain the original pattern of pixels, two operations must be performed.
- The ‘background’ intensity must be removed. The ‘background’ intensity is the total of each set of detector readings. In this case, 14 is deducted from each pixel.
- After deduction of the ‘background’, the result must be divided by three to allow for the duplication of the views of the section. These processes are illustrated in Fig. 2.
Fig. 2. The pattern of pixels for the section now emerges. In practice, the image of each section is built up from many small pixels, each viewed from many different angles. The collection of the data and its construction into a display on a screen requires a powerful computer and complicated programmes. In fact, the reconstruction of each pixel intensity value requires more than one million computations. The contrast and brightness of the image of the section as viewed on the TV screen can be varied to achieve optimum results. In order to build up an image of the whole body, the procedure would be repeated for further sections (or slices) through the body. All the data for all the sections can be stored in the computer memory to create a three-dimensional image. Views of the body from different angles may be constructed.
(h) Candidates should be able to explain the principles of the generation and detection of ultrasonic waves using piezo-electric transducers. Ultrasonic waves may be produced using a piezo-electric transducer. The basis of this is a piezo- electric crystal such as quartz. Two opposite sides of the crystal are coated with thin layers of silver to act as electrical contacts, as illustrated in Fig. 2.
Fig. 2.
thin layers of silver
two-dimensional representation of a quartz crystal
4
2 7
12 1
6 21
26 3
20 35
17 deduct 14 divide by 3
26
20 35
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9702 Applications Booklet WEB
Course: Teknik Mesin (2019 , MPD)
University: Institut Teknologi Sepuluh Nopember
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