soluciones a tu vida

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LULEA UNIVERSITY OF TECHNOLOGY Course E7021E

Dept. of Computer Science and Electrical Engineering Date 2009-10-20

Time 15:00-19:00

Exam in: Measurement Technology & Uncertainty Analysis

Attending teacher: Johan Carlson (070-580 82 52)

Problems: 5 (5 points per problem)

Tools allowed: BETA (Mathematics Handbook), Physics handbook,

Language dictionary, calculator

Text books: Principles of Measurement Systems, by John Bentley

Introduction to Empirical Model Building and Parameter Estimation,

by Johan E. Carlson

1. A displacement sensor has an input range of 0.0 to 3.0 cm and a supply voltageVs=0.5V. Results from a calibration experiment are given in the table below

Displacement x (cm) 0.0 0.5 1.0 1.5 2.0 2.5 3.0

Output voltage (mV) 0.0 16.5 32.0 44.0 51.5 55.5 58.0

(a) Calculate the maximum non-linearity as a percentage of the full-scale deflection

(f.s.d.), assuming the steady state sensitivity is calculated as in the book, i.e. (2p)

K=OMAXOMINIMAXIMIN

Solution:The slope of the straight line is

K=58.00.0

3.00.0 mV/cm 19.3 mV/cm

The intersection is (page 10, below Eq. 2.2)

a=OMINKIMIN= 0.0

58.0

3.00.0=0.0

So, the ideal straight line has the equation

OIDEAL=58

3 I

The non-linearities, given by the outputs in the table are thus

N(I) =O (I)KI,

which becomes

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Displacement x (cm) 0.0 0.5 1.0 1.5 2.0 2.5 3.0

Output voltage (mV) 0.0 16.5 32.0 44.0 51.5 55.5 58.0

Ideal straight line 0 9.67 19.33 29.0 38.67 48.33 58.0Non-linearity 0.0 6.83 12.67 15.0 12.83 7 .17 0.0

We see that the maximum non-linearity isN=15.0, which in terms of the full-scaledeflection (O=58.0) is

15

580.0100 25.9%

(b) The performance of the system can easily be improved by instead fitting theoptimal

straight line, using the principle of least-squares. Give the necessary equations for

finding the slope Kand the intersection a of the straight line using the data in the

table above.

Note:You do not need to solve for the actual numerical values ofKanda. (2p)

Solution:Fitting the straight line by the method of least squares means solving the

following over-determined system of equations

O = Ia,

where

O =

0

16.532

44

51.555.558

,I =

1 0

1 0.51 1

1 1.51 2

1 2.51 3

,a = aK .

This gives the least-squares estimate ofa andKas

a = aK

=

ITI1

ITO

7.69619.393

(c) Explain why the method in (b) gives a better average result.

Solution: Since the principle of least-squares minimizes the sum of squares of the

errors over the entire range, the result is better than for the straight line fit in (a),

where only two of the calibration points are used. (1p)

2. A temperature measurement system consists of the following elements

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where is the true temperature and M is the measured temperature (in Kelvin). Themodel equations and the corresponding uncertainties are given in the table below.

Thermistor Bridge Recorder

Model equations R= K1 exp

VO= VS

1

1+ 3.3R

a1

M= K2VO+a2

Mean values K1=5104 k VS= 3.00 V K2=50.0 K/V

=3103 K a1=0.77 a2=300 KStandard deviations K1= 0.510

4 VS= 0.03 K2= 0=0 a1= 0.01 a2= 3.0

(a) Calculate the mean outputMand the mean error E= Mfor an input tempera-ture of 320 K. (2p)

Solution:Mean output (at 320 K):

R= K1 exp

=5104 exp

3103

320

5.895 k

VO= VS

1

1 + 3.3R

a1

= 3

1

1 + 3.35.8950.77

V 0.3867 V

M= K2VO+a2=50.0 0.3867 + 300 K 319.34 K

The mean error is then

E=M= 320319.34 K 0.66 K

(b) Calculate the standard deviation of the output errorEfor an input of 320 K. (3p)

Solution: The standard deviation of the error is the same as the standard deviation

of the output, since the standard deviation only depends on the uncertainty of the

elements, not the mean values of them. Doing the error propagation calculations we

get that

2R=RK122K1= exp22K1= exp3103320 2 0.51042 0.3475

2VO=

VO

22R+

VOVs

22Vs+

VOa1

22a1=

=

330Vs

(10R+ 33)2

22R+

1

1 + 3.3R

a1

22Vs+ (Vs)

22a1 0.00257

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2M=

Ma2

22a2+

MVO

22VO 15.4265

This gives the standard deviation of the error as

M=2M 3.93 K

3. A force measurement system consisting of a piezoelectric crystal, charge amplifier and

recorder is shown in the figure below.

Calculate the system output and the corresponding dynamic error when the force input

signal is (5p)

F(t) =50

sin10t+

1

3sin30t+

1

5sin50t

.

Solution:The overall system transfer function is (the product of the three transfer func-

tions in the figure)

G(s) =201031 + s 5012ns2 + 2n s+ 1

= Inserting the numbers==20

103 0.1

1 + 0.1s

50

1502

s2 + 20.250 s+ 1

=

= 0.1

(1.0 + 0.1s) (.0004s2 + .008s+ 1.0)

Now, replacings with jwhere is the angular frequency of the input, we can computethe output by the principle, sinusoid in gives sinusoid out, i.e. the output signal is given

by

|G(j)|sin (t+G(j)) ,where |G(j)| is the absolute value of the transfer function and G(j)is the phase ofthe transfer function. Doing this for each of the sinusoids in the input signal we obtain

|G(j)| =

0.1(1.0 + 0.1j).0004 (j)2 + .008j+ 1.0=

= 0.1

(1.0 + 0.1j)

.0004 (j)2 + .008j+ 1.0

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G(j) =tan1 0.1 tan1

Re

(1.0 + 0.1j).0004 (j)2 + .008j+ 1.0

Im(1.0 + 0.1j).0004 (j)2 + .008j+ 1.0

= tan1

Re

(1.0 + 0.1j).0004 (j)2 + .008j+ 1.0

Im

(1.0 + 0.1j).0004 (j)2 + .008j+ 1.0

4. An ultrasonic transit-time flowmeter is used to measure the average flow velocityv (m/s)

through a pipe.

average velocity, v (m/s)

D

Transducer A

Transducer B

A short ultrasonic pulse is first transmitted from Transducer B to Transducer A and then

from Transducer A to Transducer B. The cross-correlation functionR()between the twopulses is shown in the figure below.

30 20 10 0 10 20 30 406

4

2

0

2

4

6

8

lag, [samples]

R()

The flow velocity is given by

v= T c2

2Dcot,

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whereis the angle between the transducers and the center axis of the flow, cis the speedof sound through the fluid, andD is the pipe diameter.

Assuming that = /6 rad,c=1480 m/s,D=10 cm, and the sampling frequency of theA/D converter used to measure the pulses is fs=10 MHz:

(a) Determine the average flow velocity,v. (2p)

Solution: We know that the cross-correlation has its maximum at the lag cor-

responding to the time delay between the pulses. From the figure we see that

Tsamp=6 samples. This corresponds to a time delay of

T=Tsamp

fs=

6

107 seconds =6107 s=0.6s.

Using this together with the numbers given in the problem, we obtain the average

flow velocity as

v= T c2

2Dcot=

6107 14802

2 0.1 cot(/6)m/s 3.79 m/s

(b) In reality, the speed of sound, c, is temperature dependent. Assuming it can be

modeled as a second-order polynomial function of temperature, make the necessary

modifications to the equation for the flow velocity. (1p)

Solution:Assuming that the speed of sound can be modeled as

c (T) =a0+a1T+ a2T2,

the expression for the average flow velocity becomes

v=T

a0+a1T+ a2T

22

2Dcot

(c) Assuming there is some uncertainty regarding the mounting of the transducer, mod-

eled as= 0.01 rad, what is the effect on the overall uncertainty of the measuredflow velocity?

Note:You do not need to estimate the total uncertainty, only the contribution of thetransducer angle uncertainty. (2p)

Solution:The sensitivity of the velocity measurement to the angle of the transducer

is given by v

22=

T c2

2Dcot

22=

=

Tc2

1 + cot2

2Dcot2

22

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For a deviation of 2=0.02 rad, this would mean (for the example in (a)), that theuncertainty is increased by

2v=

6107 14802

1 + cot2 (/6 + 0.02)

2 0.1 cot2 (/6 + 0.02)

2(0.01)2 8.05103,

meaning that the total standard deviation is increased by

v 8.97102 m/s

5. We have a differential pressure flowmeter setup designed for incompressible fluids. The

volume flow rate is given by

Q= A21A2A1

22 (P1P2) ,whereA1 andA2 are the cross-sectional areas of the pipe where the pressures P1 andP2are measured, respectively. The problem now is that we do not know the fluid density.

Assuming we can control the volume flow rate, Q to within some uncertainty, i.e. the

flow rate can be said to be normally distributed as N(Q,Q), derive an expression for theleast-squares estimator of the flow rate, using the Gauss-Newton linearization method,

that based on the calibration measurements also estimates the unknown fluid density. (5p)

Solution:Given that the cross-sectional areas are constant, and only the pressures change,lets assume that the volume flow rate we are actually able to set is

ym=Qm (P1,P2;) +em,

where

em N(0,Q).

We then have the problem of finding an estimate of the flow rate,Q so that the sum ofsquares of the errors is minimized. For repeated calibration measurements, this can be

written on vector form as

y =

y1y2...

yM

= f(X;) + e =

Q1 (X1;)Q2 (X2;)...

QM(XM;)

+

e1e2...

eM

,where

X = P1,m P2,m

T.

The subscript n denotes pressures measured for experiment m, where m= 1,2, . . . ,M.Now, assuming we have an initial guess of the density, lets say

=0. We can then use

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the Gauss-Newton linearization method to find the estimate of that miminizes the sumof squares of the errors

J= yQTyQ .The iteration is given by (see text book)

i+1=i+HT(i) H (i)1 HT(i)yQ (i) ,whereH ()is the gradient of the model with respect to the unknown parameter, i.e.

H () =Q (X;)

=

h1h2

hM

,

where

hm= d

d

A21A2A1

2

2 (P1,mP2,m)

== d

dK

Lm

=

Lm

22K

Lm,

where

K= A21A2A1

2Lm= 2 (P1,mP2,m)

So, for each element in the gradient vector, the only things that vary are the measured

pressures.

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