Answer one:
Example Exam Paper solutions:
Describe how signals and signal processing are used in ultrasonic applications. Explain the impact this has to the wider world giving examples of specific applications. Discuss the types of techniques and approaches used. [5]
Medical ultrasound uses an array of piezoelectric transducers to produce images of soft tissues according to mechanical contrast, particularly at discontinuities in acoustic properties. Although various modes of operation are available, typically B-scans are used to produce images along the direction of propogation of ultrasound energy – where back-reflections are presented as varying contrast along each penetration vector. Applications include scanning the abdomen of pregnant women to track fetal development and morphology, or to detect injuries or malformations of soft tissues. Impacts include the diagnosis of disease, abnormalities, ultrasound-guided surgery and injection.
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Alternative answer:
Ultrasound waves can be used in biosensing to detect the deposition of mass on substrates composed of resonant features, whose resonant frequency to surface waves (SAW devices) or standing waves (QCM – quartz crystal microbalance) are detected by lock-in or other frequency-domain methods (such as FFT). Applications include the sensing of low-availability bioanalytes from medical samples by adsorption to surface- functionalised sensors, the quantification of applied coatings or precipitated analytes from chemical reactions, and the sensitive detection of toxins or pathogens from environmental samples.
Alternative answer:
Ultrasound use in NDE and importance of NDE
(b) Using a device of your choice define sensitivity in the context of sensing. [5]
Sensitivity is the amount by which the output of a device responds to unit change of the stimulus.
Sensible example sensor, describing the change in response with stimulus
Mention response type, linear, non-linear etc.
𝑅𝑡 =𝑅0൫1+𝛼ሺ𝑡−𝑡0ሻ൯
𝑅𝑡 = 1 + 𝛼𝑡 when 𝑡0 is 0oC 𝑅0
Mention cross talk e.g. temperature as well as thing you want to measure.
For example: An RTD has an output that is:
Here the sensitivity is 𝛼 as this is what changes with temperature
The response is linear see graph.
Different materials have different 𝛼 for example 3850 ppm/K for platinum devices Resistive devices can be effected by strain as well as temperature, and also the environment can cause some materials to degrade which effects long term performance.
(c) Describe the different ways in which signals can be classified. [5]
Periodic signals repeat, such as a fixed frequency sine wave which is continuous over all time.
Aperiodic signals have no observable periods but may not be random.
Deterministic signals have a structure that when analysed allows one to predict or determine the future course of the signal.
Random signals have no structure, but they do have properties. They are indeterministic in that the data could have occurred in any order and the signal would not appear to be different.
Stochastic signals combine both deterministic and random components. Most real engineering signals fall into this class because we have the ‘useful’ components of the signal with unwanted electronic noise superimposed.
A signal can belong to more than one class as it can have multiple features.
Marks awarded for descriptions and/ or diagrams that clearly show different classes
(d) With reference to Fourier theory, Explain how to obtain the form of the Fourier transform of a triangle function from the transform of a rectangular window (top hat) function. [5]
The triangle function is obtained by the time convolution of the top hat / rectangular function with itself.
The Fourier transform of the top hap function is the sinc(x) function, see working below:
𝐹ሺ𝑠ሻ = ∫ 𝑟𝑒𝑐𝑡ሺ𝑠ሻ𝑒−𝑗2𝜋𝑠𝑡𝑑𝑡 −∞
𝐹ሺ𝑠ሻ = ∫ 𝑒−𝑗2𝜋𝑠𝑡𝑑𝑡 −0.5
1 1/2 𝐹ሺ𝑠ሻ = [−𝑗2𝜋𝑠 𝑒−𝑗2𝜋𝑠𝑡]
𝐹ሺ𝑠ሻ = 1 [𝑒−𝑗2𝜋𝑠𝑡 − 𝑒+𝑗2𝜋𝑠𝑡] −𝑗2𝜋𝑠
𝐹ሺ𝑠ሻ= 1 1[𝑒𝑗2𝜋𝑠𝑡−𝑒−𝑗2𝜋𝑠𝑡] 𝜋𝑠 2𝑗
𝐹ሺ𝑠ሻ = 𝑠𝑖𝑛ሺ𝜋𝑠ሻ = 𝑠𝑖𝑛𝑐ሺ𝑥ሻ 𝜋𝑠
To obtain the transform of the triangle function we square the spectrum from the top hat function. This is because a convolution in the time domain is a multiply in the Fourier domain. This gives the triangle function spectrum as the F(s) = sinc2(x).
(e) Using a diagram and equations, briefly describe the operation and limitations of a variable gap capacitive displacement sensor. [5]
The capacitance of a capacitor depends on the plate separation (d), the permittivity of the dielectric filling (𝜀) and the plate area (A).
All of these parameters can be used for displacement sensing depending on what the moving object is attached to.
Using variable gap can sense movement from sub-micron to a few millimetres of displacement
One limitation is that they can be very temperature / pressure sensitive depending on what dielectric is used.
For the variable gap design the capacitance changes due to the change in d, this creates a response that is non linear (proportional to 1/d) see equations and diagram below
Required info: Constantan, GF 2.1 , Resistivity 49e-8, poisons ratio 0.33
(a) If a 120 strain gauge is 2mm long what is the thickness of the gauge (assuming cylindrical)
Thickness is therefore 1.6mm
𝐴 = 𝜌 𝑅 = 49𝑒−8 ∗ 120 = 2.04𝑒−6
𝜋𝑟2 = 2.04𝑒−6 𝑟 = 8.05𝑒−4
(b) What is the relative change in resistivity per unit strain 𝜌 𝜀𝑙
𝑑𝜌 𝑆≜ 𝑅 =ሺ1+2𝜈ሻ+ 𝜌
for constantan?
𝜌 =2.1−ሺ1+2∗0.33ሻ=0.44 𝜀𝑙
(c) Give that typical engineering strains are in the range 10-610-3 calculate the range for the change in resistance of the 120ohm constantan strain gauge.
R*S*𝜀𝑙=120*2.1*10e-6 = 0.252mOhms
R*S*𝜀𝑙=120*2.1*10e-3 = 0.252Ohms
(d) Describe the operation of the Wheatstone bridge and why it is used. Show that the output voltage Vo is:
𝑉 = 𝑉 [ 𝑅3𝑅1 − 𝑅4𝑅2 ] 𝑜 𝑠 ሺ𝑅3 +𝑅2ሻሺ𝑅1 +𝑅4ሻ
Strain gauges are usually used in a bridge circuit because the change in R is small it is hard to measure directly. So a circuit is used to convert the small change into a voltage which is easier to measure as it can be more easily amplified.
(e) Describe the conditions for balanced operation and show how this leads to an expression for the output voltage Vo of 1 𝑉 𝑆𝜀. Where Vs is the input voltage, S is
the gauge factor and 𝜀 is the longitudinal strain.
This is ‘Balanced’ when 𝑉 = 0. For this we have two conditions Either : 𝑜
𝑉 = 𝑉 [ 𝑅3𝑅1 − 𝑅4𝑅2 ] = 0 𝑜 𝑠 ሺ𝑅3 +𝑅2ሻሺ𝑅1 +𝑅4ሻ
If𝑅 =𝑅 =𝑅 =𝑅 then𝑉=0 3142𝑜
OrIf𝑅1 =𝑅4 then𝑉=0 𝑅2𝑅3 𝑜
Use Potentiometer at R2 to allow the circuit to be balanced for initial conditions. Assume Balanced Initial conditions, so:
𝑅1=𝑅2=𝑅4=𝑅
And Strain is applied
𝑉=𝑉[ 𝑅3 − 𝑅4 ] 𝑜 𝑠 𝑅3 + 𝑅2 𝑅1 + 𝑅4
𝑉=𝑉[ 𝑅+∆𝑅 − 𝑅 ] 𝑜 𝑠𝑅+∆𝑅+𝑅𝑅+𝑅
𝑉 = 𝑉 [ 𝑅+∆𝑅 − 𝑅 ] 𝑜 𝑠2𝑅+∆𝑅2𝑅
𝑉 = 𝑉 [ሺ𝑅+∆𝑅ሻ2𝑅 − 𝑅ሺ2𝑅+∆𝑅ሻ]
𝑉 =𝑉[2𝑅2+∆𝑅2𝑅−2𝑅2−∆𝑅𝑅]
2𝑅ሺ2𝑅+∆𝑅ሻ 4𝑅2 + ∆𝑅2𝑅
𝑉 =𝑉1[ 𝑅∆𝑅
𝑜 𝑠4 𝑅2+2𝑅∆𝑅 4
𝑉 =𝑉1[ 𝑆𝜀 ]≈1𝑉𝑆𝜀forsmallstrains
𝑠4 1+1𝑆𝜀 4 𝑠 2
If 𝑉 = 20 V, S = 2.1, the maximum strain to be measured is 1×10-3 𝑠
𝑉=𝑉1[ ∆𝑅 ]
𝑜 𝑠4 𝑅+2∆𝑅 4
𝑉=𝑉1[ ∆𝑅/𝑅 ]
𝑜 𝑠4 1+1∆𝑅/𝑅 2
calculate the gain of an amplifier (A) required to give an output signal of 1 V at the maximum strain.
𝑉𝑆𝜀 20×2.1×1×10−3
𝜕𝑉 ≈ 𝑠 = = 0.0105𝑉
Gain A is 1V/0.0105 = 95.2 or 20 log(95.2) =39.6dB
Q2 Additional example: Component choice question
Using the data given in Table 1 for a range of photodiodes and the following specifications for the application, choose the most appropriate photodiode to use.
Wavelength = 1200nm
Required Bandwidth = 20kHz
Light power incident at photodiode = 0.01mW and is 5x5mm Load resistance is 50 Ohms
Desired voltage range for digitisation circuit is ~5V;
your choice by discussing:
the bandwidth of the photodiode
the responsivity
the dark shot noise, the Johnson noise and NEP of the device
the photocurrent and shot noise when illuminated for the input power given in the specifications
Discuss ways to convert the photocurrent to a voltage and calculate the required voltage gain to match the specifications
wavelengt h range (nm)
Diode capacitance (F)
Parallel Shunt resistance (Ohms)
Dark current
ma max x (mA
Activ e Area mm2
200 – 1100 340 – 1100
400 – 1100 800 – 1700 800 – 1700 900 – 2600 800 – 1800 800 – 1800
6.00E-12 3.80E-10
6.50E-13 2.00E-12 1.00E-10 5.00E-10 6.00E-09 8.00E-08
5.00E+08 5.00E+08
5.00E+08 1.00E+07 1.00E+07 1.00E+07 2.50E+04 2.00E+03
3.00E-10 2.00E-10
3.00E-11 5.00E-11 6.00E-09 3.00E-06 4.00E-06 5.00E-05
0.9 0.8 0.9 100
0.01 0.9 8
0.8 0.01 0.8 3.1 0.8 0.79 0.7 7.1 0.7 100
20 1 20 4 3 10 1.8 10 3 10 1 10
Choose diode type for wavelength range, diode 1-3 are out, 4-8 are still possible
Look at area for this case, as BW is low and the area of the beam is quite big 5×5 mm, number 8 looks like a good choice. So work out some specs to see if this is ok
Work out BW from C and R load (stated), C depends on Area so high speed will be small as first go to reduce. Assume Tj can be ignored as C is ‘big’.
𝑅𝑡𝑜𝑡𝑎𝑙 =𝑅𝑗\\𝑅𝑠 +𝑅𝐿 ≈𝑅𝑠 +𝑅𝐿 ≈𝑅𝐿
𝐵𝑊 = 1 2𝜋൫𝜏𝑗 +𝜏𝑅𝐶൯
So BW is ~40kHz as 1/(2*pi*8e-8*50) this is ok! Work out R(λ)
R(λ) = QE.λ / 1.24 R(λ) is in AW-1
λ is in microns
= 0.7*1.2/1.24=0.68 – this is pretty good! We will get good conversion from our input light
Work out shot noise for ideal device, assume B = 1 to compare to other devices IS = (2.e.iD.B)0.5
Work out Johnson noise
IS is the shot noise current (Arms)
e = 1.6 x 10-19 C is the electronic charge iD = dark leakage current (A)
B = system bandwidth (Hz)
This is dark current first = Is=Id =sqrt(2*1.6e-19*5e-5*1) = 4e-12
IJ = (4.k.T.B / R)0.5
IJ = Johnson noise current (Arms)
k = Boltzmann’s constant (1.33 x 10-23 JK-1)
T = absolute temperature (K) – assume room temp 293K
R = shunt resistance (Ω) – of the device to compare noise not with load
= sqrt(4*1.33e-23*293*1/2e3) = 2.7917e-12
Work out NEP – comment on which dominates and why IN = (IS2 + IJ2)0.5
NEP = IN / R(λ) = 4.88e-12
They are both close for this diode as the shunt resistance is low so the thermal noise is high for this diode
For given incident power work out photocurrent
Ip = QE.*e.*P./(h.*v);
Ip = R.*Pin
Ip = 0.68*0.01mW = 6.77 uA
For this current can work out shot noise in actual use with our known max bandwidth. (2.e.iD.B)0.5
Sqrt(2*e*6.77e-6*20e3) = 2e-10
Thermal noise in actual use, same bw and load of 50 ohms as this is smaller to shunt resistance and as in parallel it dominates the impedance. Use: (4.k.T.B / R)0.5 Sqrt(4*k*293*20e3/50) = 2.5e-9
So here thermal noise dominates due to the low resistance and low input light level. Generated signal is : v = ip*Rl = 6.77uA*50 ohms = 0.3385 mV
To get to a 5 voltage range we need a lot of gain 5/0.34m = 14700, 20log(14700) is 83 dB,
Use op amp circuit and chose gain appropriately.
Vout=-Rf.Iin
Use inverting amp configuration and large rf to get the gain required. This is ok as bandwidth is not high so gain bandwidth product for opamp will be ok. Use buffer to match to subsequent circuit to match 50 Ohm requirement and reinvert the signal so we get +5v output
So best diode is 8 because it is the right size and for the right wavelength, the BW is ok.
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