Coverage: through entire course material (updated July 18, 2017) Terminology/miscellaneous. for a linear instrument:

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1 Boston University ME 310: Summer 017 Course Notes Prof. C. Farny DISCLAIMER: This document is meant to serve as a reference and an overview, and not an exclusive study guide for the course. Coverage: through entire course material (updated July 18, 017) Terminology/miscellaneous Precision Accuracy Resolution Full Scale Input, Output Significant figures Static sensitivity for a linear instrument: K = f(x) x K = output input where f(x) is the output (the functional response of the instrument) and x is the input parameter. Probability density function p(x) p(x) = 1 σ π exp [ ] (x x) σ The absolute probability percentage P is defined over a defined interval of deviation ±δx about the average value x: P ( x δx x x + δx) = x+δx x δx (1) () (3) p(x)dx (4) Convert to a simplified scale by defining β, the normal variate for all samples x, and z 1, the normal variate for a particular sample value x 1 : β = x x σ ; z 1 = x 1 x ; σ dx = σdβ (5) and substituting into Eqn 4, P ( z 1 β z 1 ) = 1 π z1 z 1 exp( β /) dβ (6) or the more common version to reflect the symmetrical distribution assumption (note change in the integration bounds and constant term): P ( z 1 β z 1 ) = π z1 1 0 exp( β /) dβ (7)

2 Analog-Digital Conversion (ADC) The ADC process discretizes the analog signal at discrete time points (sampling frequency) and discrete voltage amplitudes. Amplitude discretization: The resulting amplitude discretization corresponds to n = V F SR V res (8) where n is the number of bits (an integer value, commonly even). The full scale range must be greater than the amplitude of the analog signal in order to properly resolve the signal. Temporal discretization: The ADC can only sample incoming data at a certain rate. This rate is generally adjustable, and should be set according to the Nyquist frequency. The Nyquist sampling criterium states that at least two data points are required to in order to resolve the highest frequency present. This results in: f s f max (9) (10) f Nyq = f s = 1 (11) t s Aliasing: Effect occurs when f Nyquist is lower than one or more frequencies in the analog signal. The energy from these higher analog frequencies appears as energy at a frequency lower than f Nyquist. ADC Uncertainty: Due to its discrete nature, there are two types of uncertainty related to ADC, both of which act like resolution-based bias sources: Temporal: The digitizer is sampling the signal at discrete time points, so the amplitude at any given time has an inherent uncertainty: U B,t = ± t s Magnitude resolution: The vertical resolution due to the number of bits and full scale range introduces a resolution error: (1) U B,V = ± V res (13)

3 Uncertainty analysis Precision Uncertainty: Inherent scatter in a measurable x about a mean x due to uncontrolled fluctuations. 1. Infinite Statistics: Large sample sets (sample size N > 63) σ x U P,x = ±z 1 ; (14) N z 1 = x 1 x ; (15) σ (xi x) σ x =. Finite Statistics: Small sample sets. Degrees of freedom ν = N-1 N (16) S x U P,x = ±t ν,95% ; (17) N (xi x) S x = (18) ν Note: The subscript P stands for precision (not pressure!). If the measurement were for a pressure, it would be denoted U P,p and if it were for a voltage, it would be denoted U P,V. Bias Uncertainty: Systematic uncertainties from instrumentation. Affects every data point in a data set of measurable x. Common bias sources: Resolution: Arises from rounding error. U B,resolution,x = ±resolution/ (19) Hysteresis: Arises from discrepancy in output due to direction-dependent loading effects. Commonly found in elastic systems. U B,hysteresis,x = ±%FS (0) Linearity (also referred to as Nonlinearity): Arises from instrument response, when output deviates from a linear response over the full scale range. U B,linearity,x = ±%FS (1) Sensitivity (also referred to as a Calibration Error): Arises from drift in sensitivity over time. Total bias uncertainty: U B,sensitivity,x = ±%FS () U B,x,total = (U B,x,i ) (3) 3

4 Total Uncertainty: Combine bias and precision sources for each measurable x using root-sum-square (RSS) method U total,x = (U P,x,i ) + (U B,x,i ) = U P,x,total + U B,x,total (4) Class 5 Error propagation: Combining uncertainty from multiple measurables x, y, z, where f(...) represents the generic analytical expression that relates x, y, z to the resultant parameter R: U total,r = U total,r = f(u x, U y, U z ) (5) ( f ) ( ) ( ) f f x U x + y U y + z U z (6) where each individual measurable uncertainty (U x, etc) is the total uncertainty for that parameter. Error propagation can be used in a single-instrument setting for the purpose of converting its input and output units. Measurement analysis Input/output impedance: Recall the voltage divider scenario that the function generator and oscilloscope describe: [ ] R L V L = V S (7) R S + R L where V is the generic instrument voltage, R is the generic instrument impedance, L represents the load (the downstream measurement instrument, or the oscilloscope), and S represents the source (the signal source instrument, or function generator). Linear system modeling Given 3 linear instruments 1,, and 3 in line with each other, an instrument-specific static sensitivity K i and static offset O i, and an input parameter x, the instrument I output is y, instrument II output is z and instrument III output is W. In generic functional response terms: y = f 1 (x); z = f (y); W = f 3 (z) = g(x); (8) Plugging in for the instrument-specific sensitivities: W = f 3 (f (f 1 (x))) = g(x); (9) y = K 1 x + O 1 ; z = K y + O ; W = K 3 z + O 3 ; (30) W = K sys x + O sys ; (31) where K sys = K 1 K K 3, O sys = O 3 + K 3 O + K K 3 O 1 (3) 4

5 Resistance measurement methods RTDs Strain gauge/wheatstone bridge Bridge constant Compensation techniques Linear System Calibration Linear Regression Analysis: The method of least squares error gives the following relationship for a linear system (with additional terms for an appropriately-nonlinear relationship): y fit = f(x) = a 0 + a 1 x (33) x a 0 = i ym,i x i xi y m,i ; a 1 = N x i y m,i x i ym,i (34) B B B = N x i ( xi ) (35) where x i is each control point (input) value, y m,i is each measured response (output) point value, y fit is the calibration value for a given input x, and N is the number of unique control points involved in the calibration. Note that it is helpful at times to invert the functional relationship to instead solve for x in terms of a measurement of y: x = f 1 (y) = y a 0 a 1 (36) For an exercise involving multiple output measurements per control point, a weighted scheme is used: wi x i wi y m,i w i x i wi x i y m,i a 0 = (37) B w wi wi x i y m,i w i x i wi y m,i a 1 = (38) B w B w = w i wi x i ( wi x i ) (39) w i = 1 U i where U i is the precision uncertainty value per control point. Calibration uncertainty: Potentially the most confusing of the uncertainty calculations. The goal is to understand the relationship between an output y (likely a voltage) and a physical input parameter x for an uncalibrated instrument A. Remember that A is being used to measure x, so x is unknown, and y is simply a voltage (and thus not helpful on its own). The functional response of A may only be known via use of a calibrated 5 (40)

6 instrument C that is capable of knowing what the input x actually is. C tells us what x is, and a regression analysis can then reveal how to get the constants in Eq 34 for instrument A. It gets confusing because both instruments can have bias uncertainties, and both may have their own precision uncertainties. We re dealing with multiple units (for x and for y), so a final, total uncertainty value can only be obtained via propagation of the multiple uncertainties to a final uncertainty in units of x or of y. Try to clearly differentiate which uncertainties belong with each instrument; keeping careful track of units helps a good deal. Here U meas refers to the uncertainties from the uncalibrated instrument A and U cal refers to those from the calibrated instrument C. Obtaining a total uncertainty from C is familiar and straightforward: U cal,x = ± ( U B,x,cal,i) + U p,x,cal,max (41) Remember that these will all be in units of x. Here the precision uncertainty would come from the scenario where you ve taken multiple measurements of x i at each control point value, for the purpose of accounting for scatter in the measurement. This would give rise to N values of U p,x,cal ; since only one such value would be considered for the uncertainty, you should take the maximum such value over the N that you ve calculated. For the uncalibrated instrument A, the various uncertainty parameters will likely be in units of y, or voltage. The new component of the analysis here involves the precision uncertainty of instrument A. The precision uncertainty is now a scatter of the data about a linear trend (the calibration curve), known as the precision uncertainty of the fit U p,fit that relies on a standard error of the fit S yx : U p,y,fit = ±t ν,95% Syx (4) N (yfit,i y m,i ) S yx = (43) ν The number of statistical degrees of freedom in this case (specifically, calibration uncertainty analysis) ν is N (I sometimes notate ν here as ν fit as a reminder of the calibration aspect), and N is the number of control points on the x axis. The parameter y fit,i represents the known/calibrated output value at each control point. The uncertainty from instrument A may be then found per the usual method: U meas,y = ± ( U B,y,i) + U p,y,fit (44) Since the values of the calibration for instrument A depend on the inherent uncertainties of instrument C, the two uncertainties U meas and U cal must be combined, to obtain a total uncertainty. Error propagation must be used to properly combine the different units of x and y: ( f ) U total,y = ± x U cal,x + Umeas,y (45) 6

7 ( f 1 or: U total,x = ± y U meas,y ) + U cal,x (46) where f represents the calibration equation y = a 1 x + a 0 and f 1 is the inverse of that equation, or x = y a 0 a 1. Signal conditioning Signal-to-Noise Ratio: SNR = 0 log 10 ( Vsig V noise ) Noise sources: Electrical (broadband), frequency-specific (60 Hz power/ground loops, environmental sources) Filtering techniques Hardware: Passive (RC); Active (op amp, powered) Software: Sampling issues Amplification Measurement Systems: The response of a measurement device to a dynamically-changing physical parameter (pressure, temperature, etc) that it is seeking to measure. All measurement systems must interact with the physical parameter in order to measure it, so the inherent response of the measurement device will be involved with the measurement signal output. The output signal will be affected by how the measurement is performed (ie, the physical characteristics of the measurement instrument). The input parameter may be thought of as a driving, or stimulus energy, since it is causing the instrument to respond to it. We are now incorporating the effect of time in the measurement response dynamic signals. We have studied types of measurement systems (1st and nd-order) and their response to different types of driving parameters: Step input: where H(t) is the Heaviside function. Sinusoidal input: where x(t) = AH(t) (47) x(t) = A sin(ωt + Φ 0 ) + A 0 (48) T = 1 ; ω = πf (49) f 7

8 First Order Systems 1st-Order systems only have a storage component. One example is a thermocouple or thermometer. They have no inertia, so there is no oscillatory reaction to a driving system. The storage aspect means that the system will have some inherent response time to a change in input parameter. The system (regardless of the input parameter x(t)) may be modeled as: τẏ(t) + y(t) = Kx(t) (50) where K is the static sensitivity and τ is the time constant. Response to input parameter x(t): Step input: x(t) = AH(t) The solution to the model is: y(t) = KA + (y 0 KA)e t/τ + O (51) Sinusoidal input: x(t) = A sin(ωt + Φ 0 ) + A 0 The steady-state solution to the model is: y(t) = KAM(ω) sin(ωt + Φ(ω) + Φ 0 ) + KA 0 + O (5) where M(ω) is the Magnitude Ratio, Φ(ω) is the phase lag, and O is the instrument-specific DC offset. The frequency-dependent terms may be modeled as: M(ω) = (ωτ) (53) (54) Φ(ω) = tan 1 (ωτ) (55) Quantifying the response Dynamic error Γ(t): Frequency-based dynamic error δ(ω): Γ(t) = y(t) y y 0 y = e t/τ (56) δ(ω) = M(ω) 1 (57) Rise time, t rise : Time that it takes to reach 90% of the steady state value: y(t rise ) = 0.9(y y 0 ). This occurs when t =.3τ. For 1st-order, step input functions, t rise = t response. The system reaches 99.3% of the steady-state value when t = 5τ At t = τ, the system has reached 63% of the steady-state value 8

9 Second Order Systems nd-order systems inherently have a mass, spring, and damping, which are all involved in the response of the system to the driving signal. The system (regardless of the input parameter x(t)) may be modeled as: mÿ(t) + cẏ(t) + ky(t) = Kx(t) (58) where m is the mass, c is the damping coefficient, k is the spring constant, and K is the static sensitivity. Since the system has a second-order dependence on time, its response will depend on highly on the values of these three parameters (m, c, k). We can simplify the equation slightly, by dividing through by m, and restating as: ÿ(t) + c mẏ(t) + ω ny(t) = Kx(t) (59) where ω n = k/m. The response is based on the value of the generic solution to the homogeneous equation (x(t) = 0), Be λt : λ 1, = c m ± c 4km m (60) The solution is imaginary for a critical value of c: c cr = km. This defines the damping ratio: ζ = c c cr = c. km Simplifying the solution: λ 1, = ζω n ± ω n ζ 1 (61) It s now easy to see the boundary for the importance of ζ, since it dictates the solution to the model. 3 solutions arise: ζ < 1: Underdamped solution y(t) = Ce ζωnt sin(ω d t + Φ) (6) where ω d = ω n 1 ζ : the damped natural frequency. ζ = 1: Critically damped solution y(t) = C 1 e λ 1t + C te λ t (63) ζ > 1: Overdamped solution y(t) = C 1 e λ 1t + C e λ t (64) Not much happens in the critically and overdamped cases: most of our attention has been focused on the underdamped scenarios. 9

10 Response to input parameter x(t): We ve mostly focused on a sinusoidal input, for underdamped systems. Sinusoidal input: x(t) = A sin(ωt + Φ 0 ) + A 0. The steady-state solution is the same as 1st-order systems: y(t) = KAM(ω) sin(ωt + Φ(ω) + Φ 0 ) + KA 0 + O (65) The difference between systems is the magnitude ratio and phase lag form: M(ω) = 1 ( ( ) ) ω 1 ω n + ( ζ ω ) Φ(ω) = tan 1 ω n 1 (ω/ω n ) ( ζ ω ω n ) (66) Resonance: When output amplitude exceeds input amplitude. Resonance occurs at the maximum amplitude, M(ω = ω r ) = M max, where ω r is the resonance frequency: (67) (68) ω r = ω n 1 ζ (69) Q = ω nτ relax = ω r ω ω 1 = The Quality Factor Q defines the sharpness of resonance. (70) 1 ζ 1 ζ = M max (71) Step input: x(t) = AH(t) + A 0 The solution to the model for an underdamped system, where A 0 is the previous amplitude of the input and A is the new change in input amplitude: y(t) = KA + O + (KA 0 KA) exp( ζω n t) [ ζ 1 ζ sin(ω dt) + cos(ω d t) Rise time, t rise : Time that it takes to first reach 90% of the steady state value Response time, t response, or t settle : Time that it takes to stay within 10% of the steady-state value Think of as the free decay: No driving signal is applied. The system s natural response takes over, and its damping governs the decay. The system responds at its damped natural frequency ω d : ] (7) ω d = ω n 1 ζ (73) Special case: IF the scenario involves an initial input amplitude A 0 and a final amplitude of A = 0, its envelope has an exponential decay: y(t) = KA 0 exp( ζω n t) (74) 10

11 τ relax is the time that it takes the undriven signal to decay from its previous excitation state to y(t = τ relax ) = 0.37y 0. Quantifying the response Time-based dynamic error Γ(t): Miscellaneous Frequency-based dynamic error δ(ω): Optimal value: ζ = Γ(t) = y(t) y y 0 y (75) δ(ω) = M(ω) 1 (76) First order systems asymptote to Φ = π/ for large frequencies Second order systems asymptote to Φ = π for large frequencies First order systems pass through Φ = π/4 when ω = 1/τ Second order systems pass through Φ = π/ when ω = ω n The measurement always lags (ie, is behind) the driving signal, so Φ(ω) 0. Because the system has a physical inertial response, it cannot mechanically respond instantaneously! Quality Factor, Q A unitless parameter that describes the sharpness of the instrument s resonance response to an oscillatory input: where and Q = M max = 1 ζ 1 ζ = ω r ω BW (77) M max = M(ω r ), (78) ω BW = ω ω 1, (79) and ω 1, ω correspond to the frequencies at which M(ω) = 0.707M max, or when the response is at its half-power relative to its maximum amplitude (at the resonant frequency). The Quality Factor may also be visualized during a step input response, or free decay scenario. In this scenario, the instrument s natural response takes over, and its damping governs the decay. The system responds at its damped natural frequency ω d : ω d = ω n 1 ζ (80) 11

12 and its envelope has an exponential decay: y(t) = A 0 exp( ζω n t) (81) The time constant τ relax is the time that it takes the instrument output to decay from its previous excitation state to y(t = τ relax ) = 0.37A 0. The Quality Factor is related to this time constant: Q = ω nτ relax Frequency analysis Note: See the course text (Chapter: The Analog Measurand: Time-Dependent Characteristics) for the full derivation of the equations. Premise: If a measured signal y(t) contains energy at multiple frequencies, it will likely be difficult to visually determine the amplitude of each frequency. To better understand the processes that are going on in the signal, we can use Fourier Analysis. Fourier analysis allows us to decompose the measured time-domain signal into all of its frequency components. This is ultimately accomplished by multiplying the time-domain signal by its sin and cos components and integrating the signal over some time window, in an operation called the Fourier Transform: Y (ω) = F(y(t)) = (8) y(t) exp iωt dt (83) Y (ω) = A(ω) ib(ω) = Y (ω) exp iφ(ω) (84) 1 Im(Y (ω)) Φ(ω) = tan Re(Y (ω)) (85) where Y (ω) is the complex signal at each frequency ω in the signal spectrum, A represents the real (amplitude) component and B represents the imaginary (phase) component. Note that these equations can all be expressed in terms of f instead of ω. Fourier analysis and instrumentation: To apply the Fourier Transform to digitized data, you will want to use a Fast Fourier Transform (FFT) in Matlab. There are a few important things to remember for this process: The FFT returns a complex vector that has the same number of points as the data y(t). The absolute value represents the amplitude and the imaginary values represent the phase angle. In the context of ME310, we only care about the amplitude. The maximum frequency represented by Y (f) is equal to the Nyquist frequency, or f s /, and the frequency resolution f res is equal to f s /N, where N is the number of data points. The FFT returns the negative and positive frequency components, of which we only need to use the positive components. Practically speaking, this means we only need half of the points contained in Y (f). In Matlab, the first index value in Y (f) is the DC offset, or A 0. The second index point corresponds to the lowest frequency f min, the third point corresponds to f min + f res, and so on. This is important to remember when plotting the frequency amplitude or accessing individual index values. 1

13 The amplitude Y (f) does not have the same units as y(t)! The amplitude scaling in Matlab involves a complicated relationship between the y(t) units, frequency, and the number of data points, so it is best to simply think of the amplitude as a relative scale. The time-domain data can be recovered from Y (f) by applying the inverse FFT, or IFFT ( ifft in Matlab). This involves both the complex value Y (f), not just the amplitude. 13

14 Boston University ME 310: nd exam review questions Prof. C. Farny Review problems; solutions on next page: 1. A pressure transducer behaves as a second-order system. If the natural frequency is 4 khz and the damping ratio is 0.75, determine the resonance frequency, as well as the frequency range(s) over which the measurement error is not greater than 5%. Repeat this exercise with a damping ratio of A mercury glass thermometer that initially reads 5 is suddenly immersed into a liquid that is maintained at 100. After a time interval of.0 s, the thermometer reads 76. Assuming the thermometer is a first-order system, estimate its time constant. 3. The output from a temperature sensor is expected to vary from.5 mv to 3.5 mv. Its output is sampled by a 1-bit AD converter with a ±5 V range. a) Estimate the resolution of the AD system. How does this compare to the temperature magnitude from the sensor? b) You implement a preamplifier to improve the measurement accuracy/reduce uncertainty. What standard gain setting (state in both linear and in db scale) is appropriate for such a preamplifier? Most preamplifiers offer steps of either 6 or 0 db. 14

15 Solutions: 1. For ζ = 0.75, the system has no resonance, so there is no concept of a resonance frequency. Solve for the frequency range by plugging into Eqn 76 and solving for M. M will be flat up until some frequency, and this is the frequency you re interested in finding. Plug into Eqn 66 and solve for f. For ζ = 0.75, f = 1900 Hz. For ζ = 0.5, the system will be resonant, so there will be a resonant frequency (see Eqn 69), where ω r = krad/s, or f r =.88 khz. There will exist two frequency regions that satisfy the criteria for the error: 0 < f < 1300 Hz; 3785 < f < 4195 Hz.. Solve by plugging into Eqn 51 and solving for τ: 76 = (5 100) exp ( /τ) (86) τ = 1.76 s (87) 3. a) Use Eqn 8, which returns a resolution of.44 mv on the same magnitude as the temperature output you re looking to measure. b) Comparing the V F S of the AD converter with the maximum output from the sensor, the ratio is 149. Choosing a standard amplification value, you should choose a gain of 1000, or 60 db. 15