EEEE3089 Sensing system and signal processing 2020-2021 Exam Questions and solutions
Q1(a) Describe a sensing application where both high accuracy and high precision is important, contrast this with an additional example where high accuracy and precision is less important. How do these requirements impact the choice of sensors used?
High accuracy and precision – absolute position sensors where the exact position is required – such as scanning microscope stages where the stage position needs to be known so that image artefacts are not introduced.
Low accuracy – temperature sensors for air conditions / heating systems. +/-1 degree is sufficient for comfort.
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High accuracy / precision usually means higher cost devices as they need to be reliable and robust as the consequences of malfunction / errors could be higher. Low risk / low cost applications have greater flexibility of choice and may be dominated by price point, size or power efficiency.
[2 marks each for discussion of applications. 1 mark for discussion on choice.]
Q1(b) A square variable area capacitor has a relative permittivity of 2, a length of 10mm and a gap of 10 microns. If the object moves laterally by 300 microns how much does the capacitance change? Does this device have a linear or non-linear response?
We are told the device is square, so the device has an area of 10mx10mm. The width is therefore also 10mm. the gap, d, is 10×10-6
We can work out C for both cases and compare to get the change in capacitance
The value for 𝜀0𝜀𝑟 is 2*8.85×10-12
Using the Equation above we get C0=0.177nF (x=0) and with it displacing (x=300
microns) we get C = 0.1717nF
The change in capacitance is therefore: -5.31pF
[3 marks for calculating values, 1 for units]
The change in capacitance is linear with changes in area. [1 mark]
Q1(c) With the aid of a diagram explain how a photomultiplier tube works. What applications can this device be used for?
Photocathode – converts an incident photon into a single electron (photoelectric effect) Focusing electrodes -guide the electron into the electron multiplying dynodes electrostatically accelerates the electron
Dynodes – on impact with the dynode, each electron generates several secondary electrons
Anode – collects the electrons and produces an electrical (current) output PMT characteristics (vs. APD)
• Bulky and expensive
• Low quantum efficiency
• Low dark current
• Can be operated in linear mode, or used with a pulse discriminator to
count single photons
Application: photon counting in fluorescent imaging.
2 marks for simple diagram, 2 marks for explanation of how it works, 1 mark for suitable application.
Q1(d) How many bits does the ADC need if the SNR limit due to quantisation error is to be 90dB and the input range of the device is +/-2V. You may assume we are trying to digitise a single frequency sinusoidal signal.
𝑀𝑎𝑥𝑆𝑁𝑅 = 20log( max𝑟𝑚𝑠 ) = 20log( √2 ) = 90𝑑𝐵
𝑉∆ 𝑛𝑜𝑖𝑠𝑒 𝑟𝑚𝑠
The maximum input signal (V max rms) has an RMS amplitude of half the input the range / sqrt(2). Here we are assuming it is sinusoidal.
𝑉 = ∆ = √2 =44.72μV 𝑛𝑜𝑖𝑠𝑒 𝑟𝑚𝑠 √12 10(90)
∆= 44.72𝜇𝑉 ∙ √12 = 154.9𝜇𝑉
∆=𝑟𝑎𝑛𝑔𝑒=2−−2= 4 𝑙𝑒𝑣𝑒𝑙𝑠 2𝑁 2𝑁
𝑁 = 154.9𝜇𝑉 = 14.65 𝑏𝑖𝑡𝑠 log(2)
We need at least 15 bits to make this specification. 15 bit ADC are not very common so a 16 bit device would typically be used in this situation. [5 marks]
Q1(e) Briefly describe how signals and signal processing are used in biomedical applications. Explain the impact this has to the wider world giving examples of specific applications.
Sentence on impact this has had (e.g. disease diagnosis, treatment monitoring, drug discovery etc) [2 marks]
Specific applications mentioned (bio markers, brain imaging etc) [1 mark]
Q1(f) MRI produces signals whose characteristics can be analysed to produce images with varying contrast. Describe the key steps in the process of generating MRI signals in a subject. In addition to intensity at a particular frequency, name two parameters that can be extracted from MRI signals to allow construction of images.
MRI produces signals whose characteristics can be analysed to produce images with varying contrast.
Describe the key steps in the process of generating MRI signals in a subject (3 marks).
Correct answers scoring full marks will mention:
(i) alignment of proton spins to a high standing magnetic field (superconducting magnet)
(ii) introduction of RF pulse to induce precession of nuclei
(iii) detection of RF emission from relaxing nuclei. #
Strong answers may mention imposition of gradient fields to encode position along XYZ axes, which may substitute for one of these key steps if well described.
In addition to intensity at a particular frequency, name two parameters that can be extracted from MRI signals to allow construction of images (2 marks).
Correct answers could include:
(ii) Frequency
(iii) Time constant (T1/T2 etc)
Q1(g) Describe a situation where buffers are needed when working with sensors. What are the ideal properties of a buffer that make them useful?
Description of suitable sensors (Ultrasonics, fiber sensors, MRI etc.) [2 marks]
Buffers are needed where a high impedance source needs to interface with a device with a low input impedance. For example a high impedance microphone and a loud speaker driver. [2 marks]
A buffer should have:
a high input impedance – so the voltage get drops across it’s inputs
a gain of 1 – stable operation and it doesn’t change the signal scale
a low output impedance – so it can interface to other circuits with moderate input
impedance (the driver in this case) [3marks]
Q1(h) Two unknown, but identical, strain gauges are used in a half bridge configuration. The bridge is initially balanced when zero strain is applied. When 50 micro-strain is applied to the bridge, the output voltage is measured as 0.764mV. The supply voltage is 5v during this test. What is the Gauge factor for these unknown strain gauges? Comments should include any assumptions made.
Assumption is that the gauges are in the correct positions for the bridge and the direction of the applied strain. This must be true as there is an output voltage measured when strain is applied, it would be zero if this was not the case. [2marks]
Balanced Half bridge equation is: Vo = 1⁄2 VsGεr G=2Vo /(Vs εr)
G = (2*0.764mV) / (5*50e-6)
Total [5 marks]
(i) Describe how a resistive temperature device (RTD) works, discussing the type of response and typical working ranges.
Resistance Temperature Detectors (RTDs)
These are made out of metals, such as Platinum, Nickle and Copper.
They provide almost linear change in resistance with changes in temperature and usually have a positive temperature co-efficient.
Standards exist for particular materials, for example platinum. This material is easy to obtain in pure form and therefore have reliable material properties.
A common standard for the temperature coefficient (𝛼) is 3850 ppm/K, a common nominal resistance is 100 Ω (pt100 family of devices). This nominal resistance is the resistance of the sensor at 0 degrees Celsius. There are also standards relating to tolerances of devices.
Wide temperature range -200 °C +650 °C 𝑅𝑡 =𝑅0(1+𝛼(𝑡−𝑡0))
𝑅𝑡 =1+𝛼𝑡 when𝑡0 is0oC 𝑅0
𝑅𝑡 resistance at temperature t
𝑅0 resistance at standard temperature 𝑡0
𝛼 temperature coefficient of resistance (0oC-1)
1 mark for sketch/equation of response, 3 marks for explanation of how they work and 1 mark to cover the working range for total 5 marks.
(ii) Given the specs for the RTD in Table 1, calculate the temperature if the resistance measured is 228Ω.
𝑅𝑡 =𝑅0(1+𝛼(𝑡−𝑡0))
(𝑅𝑡)−1 (228)−1
𝑡=[𝑅0 ]−𝑡0=[ 100 ]−0 𝛼 3850 × 10−6
𝑡 = 332.5oC
1 mark answer, 1 mark units,1 mark working
(iii) Describe the limitations of the simple circuit a shown in Figure 1, Show how a bridge connected three wire configuration is superior to the two wire version.
2 wire version is the simplest – but also the most susceptible to errors.
Recall Initial / null operating condition 𝑅1 = 𝑅4 or 𝑅1𝑅3=𝑅2𝑅4
𝑅1𝑅3 = 𝑅4(𝐿1 + 𝑅𝑡 + 𝐿2)
𝑅3 = 𝐿1 + 𝑅𝑡 + 𝐿2
Ideally 𝑅𝑡 ≫ 𝐿1, 𝐿2
So measured change in R also contains the change the lead resistance which is not what we want as we only want the change of R of the sensor.
The 3 Wire version allows the cancelation of some wire resistance.
[ 4 marks for diagram and explanation with equations]
𝑅1(𝐿2 +𝑅3 +𝐿3)=𝑅4(𝐿1 +𝑅𝑡 +𝐿2) 𝑅1 = 𝑅4
𝐿2 + 𝑅3 + 𝐿3 = 𝐿1 + 𝑅𝑡 + 𝐿2
𝑅3 = 𝑅𝑡 + (𝐿1 − 𝐿3)
Ideally 𝐿1 = 𝐿3cancelling lead resistance
Even if using same length and gauge of wire, the lead resistance will have a tolerance and so wont be exactly the same. But now we have an output that has an error due to the difference (not the sum) of the lead resistances so if they change in a similar manner then most of the change cancels.
[4 marks for diagram and explanation with equations]
(iv) Explain how Negative temperature coefficient (NTC) thermistor devices differ from metallic RTDs including a sketch of a typical response curve. What impact can this have on bridge circuit configurations?
Thermally sensitive resistor or thermistor is usually made from a bulk semiconductor with a large change in resistance with temperature. [1 mark]
Temperature coefficients can be very large ~-5% per degree change. Typically have negative temperature
coefficients (NTC thermistors)
Due to high sensitivity they are usually used over a narrower range than RTDs (typically -90+120°C)
Over a very small range the response can be considered linear, but over wider range the exponential response is used with the given 𝛽 parameter for the device.
𝛽(1−1) 𝑅(𝑇)=𝑅0𝑒 𝑇 𝑇0
𝑅0 = 10kΩ at 25 °C
[2 marks + 1mark for graph sketch]
Advantages
Good accuracy 0.05-1.5 °C
Ok stability
Generally good immunity to electrical noise
Protected tips work well in harsh environments
Vibration resistance
Disadvantages
Narrower range (-100 °C +300 °C)
Non-Linear response
Self heating
[1 mark for some general specs]
Large change in R with temperature can unbalance the bridge significantly, effecting operation. [1 mark]
(v) What is the temperature if the measured resistance for the NTC device in table 1 is 5kohms? Assuming a two wire connection with negligible lead resistance what is the power dissipated by the device if the excitation voltage is 5V?
𝛽(1−1) 𝑅(𝑇)=𝑅0𝑒 𝑇 𝑇0
𝛽 =3435 in K s need to convert temperatures to 0 is 25+273K = 298K.
ln[𝑅𝑡] ln[ 5000 ]
1 = 𝑅0 + 1 = 10000 + 1 = 3.15 × 10−3
𝑇 𝛽 𝑇 3435 298 0
T=317.1K or 44 oC
Resistance is given as 5kΩ at 44 oC excitation voltage is 5V so power dissipated is V2/R = 25/5000 = 5mW.
4 marks for Temperature calculation with units in Celsius, 1 mark for power calculation with units correct.
(vi) Give two example applications where NTCs may be used.
Two good example applications: any temperature control systems that require accurate control without needed a wide operating range.
e.g. oil temperature in automotive, air conditioners, power supply temperature monitoring etc.
[2 marks for 2 examples]
Figure 1 – RTD circuit, where L1, L2 are the lead resistances and RTD is the temperature sensor. Table 1 Temperature sensor specifications
PT100 (RTD) NTC
Reference Resistance (R0) 100 Ω
Sensitivity
3850 (ppm/K) 3435 (K)
Reference Temperature (T0) 0 o Celsius
25 o Celsius
a) What is the difference between incremental and absolute position encoders? Give an example where each type could be used.
Incremental encoders produce a signal that changes when the device moves, they do not know the absolute position only the amount by which it has moved. [1 mark]
These are found in cheap rotation stages or cheap linear stages using limit switches. [1 mark]
Absolute encoders use unique patterns to identify the current absolute location of the positioner. [1 mark]
These are used in high end scientific instruments where precision and accuracy is most important. [1 mark]
b) How many tracks (bits) would be needed for a linear stage that used an absolute encoder if the travel range is 175mm and the positional accuracy needs to be better than 5 μm? Include the final resolution your choice obtains.
The travel is 175mm and the positional accuracy should be better than 5 microns. This means we need to use at least 175e-3/5e-6 patterns to encode the positions. [1 mark]
We therefore require 35000 codes. This corresponds to 15.1 bits requiring 16 bits to satisfy the minimum resolution criteria. [2 marks]
[1 mark working]
175 × 10−3 log10( 5×10−6 )
𝐵𝑖𝑡𝑠 = log10 2 = 15.1
The obtained resolution of the encoder is therefore 175e-3/(2^16) = 2.67e-6m or 2.6μm [2 marks answer and units]
c) Whatisthepotentialworstcaseerrorifbinaryencodingisusedtoencodethe positions for this device? Assume that only a single bit changes before the other bits during a transition between positions. Give your answer as a percentage and as a distance.
Suggest an alternative encoding scheme to minimise the error. Explain why this encoding scheme is better and state the new possible error.
[10 marks]
In binary encoding the number of bits that change from position to position varies. The worst case is when a transition from 2^(N-1)-1 2^N takes place. For example for the 16 bit device here:
Current Position (LSBMSB)
Next Position Momentary position
1111 1111 1111 1110
0000 0000 0000 0001 1111 1111 1111 1111
65536 error 32768
If the MSB changes before the other bits then momentarily the position will appear to be the highest value as all bits will be ‘1’.
This leads to a 100% error in the true value and corresponds to an error of 50% of the travel range.
This will be 32768*2.67um = 87.5mm error. [6 marks]
Using a gray code the difference between successive positions will only differ by 1 bit and so momentary errors are reduced to the positions next to the true position.
Decimal Binary Gray 0 00 00
This gives and error equivalent to +/- 1 positions which corresponds to the resolution of the device. Here it is +/-2.67 μm
d) What is the maximum speed the stage can be moved if the data acquisition system has a maximum sampling frequency of 100 kHz. You may assume that the acquisition system can read all of the parallel tracks simultaneously.
The maximum sampling frequency is 100kHz, this means we can record at most 100 000 patterns per second.
Our stage has a full range of 175mm with 2^16 patterns across this range.
100kHz/ 2^16 = 1.525 full stage motions per second Giving the max speed to be 175mmx1.525=266.9mm/s
Similarly, we could multiply the minimum step size of 2.67um by 100k to get the same answer.
[5 marks for explanations, working and answer with correct units]
e) What measurand could be exploited for this encoder system? Explain how the 1’s and 0’s could be encoded on the tracks?
The optical properties could be used for the measurand. The easiest would be to use the amplitude of light reflected or transmitted to a photodetector.
For example, we could use a track consisting of variations in reflectivity with high reflectivity being a 1 and low reflectivity being a zero.
This could be achieved in a number of ways, for example using black paint for the zeros and white paint for the ones or using a polished surface for high reflectivity and etched surface for low reflectivity.
Other suggestions will be judged on merit.
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