˚ UNIVERSITY OF TECHNOLOGY LULEA Dept. pt. of Comp omputer ter Scien ciencce and and Electri ctrica call Eng Engineer eering EISLAB
Exam in: Teac eacher: her: Prob Probllems: ems: Tools allowed: allowed:
Course SME101 Date 2005 005-12-20 2-20 Time 9.00–13.00
Measurement Measurement Systems
Joha Johan n Ca Carl rlso son, n, Ext. Ext. (ank (ankn. n.). ). 2517 2517 5 (5 poi points per per prob probllem) em) BETA BETA (Mathematics (Mathematics Handbook), Handbook), Physics Physics handbook, Language dictionary, calculator Text book: Principles of Measurement Systems, by John Bentley
Solutions 1. A thermocouple sensor has an electromotive force (e.m.f.) in µV. E (T (T )) = 38. 38.74T 74T + 3. 3.319
× 10
−
2
T 2 + 2. 2.071
× 10
−
4
T 3
− 2.195 × 10
−
6
T 4 ,
for the range 0 to 400 C. For T = 0 C, E (T ) T ) = 0 µV and for T T = 400 C, E (T ) T ) = 20869 µV. ◦
◦
◦
(a) Calculate Calculate the expression expression for the ideal straight straight line relationship, relationship, E (T (T )) = K T . Solution: According to the text book, the solution should be...
·
E (400) (400) = 20869
× 10
−
6
V = 400 K
·
V/ C ≈ 52. 52.17 µV/ C. ⇒ K = 20869 400
=
◦
◦
This gives the equation for the ideal straight line as: E L (T (T )) = 52. 52.17 T
·
(in µV).
However, the examp example le in the book is wrong, wrong, so a better better solut solutio ion n would would be
based on E (400) (400) = which gives K =
−22131 × 10
−
6
V,
−22131 = −55. 55. 328 µV/ C ◦
400 (b) Determine Determine the sensitivity sensitivity of the sensor. dE (T ) which gives gives Solution: Sensitivity is defined as dT , which
dE (T (T )) d = 38. 38.74T 74T + 3. 3.319 10 2T 2 + 2. 2.071 10 4 T 3 2.195 dT dT 38. 38. 74 + .06638T 06638T + 6. 6. 213 10 4 T 2 8. 78 10 6 T 3 ,
≈
×
×
−
×
which is temperature dependent. 1
−
−
−
×
−
−
× 10
−
6
T 4 =
(c) Determine the maximum non-linearity of the system, as function of the full-scale (400 C). Solution: The non-linearity is given by E (T ) E L (T ), where E L (T ) is the ideal straight line from exercise (1a), ◦
−
4
1
x 10
0.5 0 −0.5
) T ( E
−1 −1.5 True model −2
Linear approximation
−2.5 0
50
100
150 200 250 Temperature, T
i.e.
N (T ) = E (T ) E L (T ) = = 38.74T + 3.319 10 2 T 2 + 2.071
−
×
−
× 10
−
4
T 3
300
350
400
− 2.195 × 10
−
6
T 4 + 55. 328 T
·
The derivative w.r.t. T is then d 38.74T + 3.319 10 2T 2 + 2.071 10 4 T 3 2.195 10 6 T 4 + 52.17 T dT = 90. 91 + .06638T + 6. 213 10 4 T 2 8. 78 10 6T 3 ,
×
−
×
−
× −
−
− ×
×
−
−
·
Setting to zero and solving we get the maximum nonlinearity at T = 256. 98 C, which gives the maximum non-linearity as percentage of the full-scale: ◦
E (256.98) E L (256.98) = E (400)
−
×100% = −91.8%
2. Two strain gauges are bonded onto a cantilever as shown in the figure below. Given that the gauges are placed halfway along the cantilever and the cantilever is subject to a downward force F. Use the tabulated data below to: Cantilever data
Length Width Thickness Young’s modulus
Strain gauge data
l = 25 cm w = 6 cm t = 3 mm E = 70 109 Pa
Gauge factor G = 2.1 Unstrained resistance R0 = 120 Ω
×
2
(a) Calculate the resistance of each strain gauge for F = 0.5 N and F = 10 N . Solution:
∆R = GR 0 e, where (from [8.57]) 6 (l x) F, wt2E m. That is, with all the numbers put in
−
e =
and x = l/2 = 12.5 e =
6
×
× 10
−
2
6(12.5 10 2) 10 2 (3 10 3 )2 70 −
· ×
×
−
· ×
−
So, for F = 0.5 N, we get
109
· F ≈ 1. 9841 × 10
R1 = R 0 (1 + G e) = 120 1 + 2.1 1. 9841 R2 = R 0 (1 G e) = 119. 9975 Ω.
· − ·
·
5
5
−
F.
× 10 · 0.5 = 120. 0025 Ω −
Similarly, for F = 10 N, we get
−
5
R1 = 120 1 + 2.1 1. 9841
−
R2 = 120 1
−
· × 10 · 10 = 120.05 Ω 2.1 · 1. 9841 × 10 · 10 = 119. 95 Ω 5
(b) Design a resistive deflection bridge suitable for force measurements in the interval in (a). Motivate your design choices clearly. Solution: See text book or lab. 3. A resistive sensing element with resistance R p is connected to a deflection bridge according to the figure below. The supply voltage was measured 10 times and stored in the vector Vs =
5.02 4.99 5.0 4.98 5.01 4.99 5.01 5.0 5.01 4.98
T
.
Assume that σR = 0.1 and σR = σ R = σ R = 0.05, and that R1 = R 2 = R 3 = 200 Ω. 1
p
2
3
(a) Estimate the mean V s and the standard deviation σV of the supply voltage. Solution: The mean is given by s
N
10
1 1 V s = V s [n] = N n=1 10 The standard deviation is given by σV = s
−
Note that the book uses especially for small N .
1
N
n=1
N
1
N
V s [n] = 4.990
1
V s [n]
n=1
2
− V ≈ 0.0137 s
even for standard deviation. This is WRONG,
3
2 (b) Derive an expression for the total variance, σE of E , as a function of the resistance R p , where R1 R2 E = V s . R1 + R p R2 + R3
−
Solution: The total variance is given by 2
σE =
∂E ∂V s
2
2
∂E 2 σV + ∂R 1 s
∂E σR2 + ∂R 2 2
2
2
∂E σR2 + ∂R 3 2
∂E σR + ∂R p 2
3
2
σR2
p
The partial derivatives are:
∂E ∂V s
∂E ∂R 1 ∂E ∂R 2 ∂E ∂R 3
∂E ∂R p
2
= 2
= 2
= 2
= 2
=
−
R1 R1 + R p
−
R2 R2 + R3
R p V s (R1 + R p )2
2
2
R3 V s (R2 + R3 )2
R2 V s (R2 + R3 )2
2
R1 V s (R1 + R p )2
−
2
2
.
(c) Looking at a 2σE , how much of the total variation does the supply voltage variation account for, when R p = 150 Ω? Solution: Putting in numerical values, we get 2
σE = +
− −
200 200 + 150
−
200 200 + 200
200 4.99 (200 + 200)2
200 + 4.99 (200 + 150)2 1. 2521 10 6
2
2
150 (0.0137)2 + 4.99 (200 + 150)2
200 0.0025 + 4.99 (200 + 200)2
2
E
2
2
0.0025+
0.0025+
0.01
≈ × √ =⇒ 2σ = 2 1. 2521 × 10 ≈ 2.2380 × 10 −
−
6
−
3
This corresponds to a 95 % confidence interval for the output, E . The contri-
4
bution from V s is
∂E ∂V s
2
2
σV = s
R1 R1 + R p
−
200 = 200 + 150 1. 6294 10
≈
R2 R2 + R3
−
2 σV
s
200 200 + 200
−
×
2
3
2
(0.0137)2
As a percentage of the 2σE interval, this is
2
∂E ∂V s
2
2 σV
s
2σE
· √ 9. 5760 × 10 × 100 % ≈ 87. 45 %. 2.2380 × 10
2
=
−
−
7
3
4. A temperature measurement system is given in the figure below, with transfer functions of the individual elements as in the figure. (a) Calculate the dynamic error E (T ) of the system. Which element in the system is the dominant cause of this error? Solution: We want to know the dynamic error, E (t), when the system input is a unit step, u (t). This was unclear in the problem description, but was clarified during the exam. From [4.45] in the text book we have the system Laplace transform 1 1 1 1 = s 1 + 10s (1 + 10 4 s) (1 + 1/200s)2 1 A B Cs + D = + . s s + 0.1 s + 104 (s + 200)2
∆T M (s) =
−
−
−
The error is thus E (s) = ∆T M (s) =
− ∆T (s) = 1s − s +A0.1 − s +B10 T
− s +A0.1 − s +B10
4
+
4
+
Cs + D (s + 200)2
− 1s =
Cs + D , (s + 200)2
which gives E (t) =
−
Ae
−
0.1t
+ Be
−
104 t
+ Ee
200t
−
(1 + 200t) .
We see that the B and E terms decay rapidly to zero, while the first term takes about 50 s to decay. Thus, the first block is dominant when it comes to the dynamic error . 5
(b) Assume that the input temperature is varying as T T (t) = 5 sin(10 2 t) + 2 sin (50t) . −
What is the output of the system? Explain this behavior. Solution: Sinusoidal input gives sinusoidal output, with output
|G ( jω )| sin(ω t + arg G ( jω )) . 0
0
0
In this case, G ( jω) , is
| | |G ( jω)| =
With ω = 10 2 , this is −
1 = j10
−
1 1 1 1 jω 1 + 10 jω (1 + 10 4 jω) (1 + 1/200 jω)2 −
1 2 1 + 10 j10
−
1
1 2 (1 + 10 4 j10 2 ) (1 + 1/200 j10 2 )2 −
−
−
and with ω = 50 this is 1 1 1 1 i50 1 + 10i50 (1 + 10 4 i50) (1 + 1/200i50)2 −
≈
4
≈
× 10
−
5
80
.
The same principle is used to calculate the argument. The important part here is, however to realize that the 50 Hz part will almost negligible in the output. This is because the system acts as a low-pass filter. Fast temperature fluctuations will not be captured using this system. 5. An ultrasonic Doppler flowmeter is used to measure the flowrate of particles suspended in a liquid. The setup is shown in the figure below. The transmitting transducer is transmitting a continuous sound wave at f = 1 MHz and the velocity of the particles is in the range of 5 to 20 m/s. The angle θ = 30 , and the speed of sound in the fluid is c = 1485 m/s. ◦
(a) Explain the working principle of this flowmeter and give an approximate expression for the flow velocity, v, in terms of the Doppler shift ∆f . You may assume that only the frequency of the Doppler signal is used, not the amplitude. Hint: You may also assume that v/c is small. Solution: The motion will cause a frequency shift of the received signal, which is proportional to the velocity of the fluid. Using the assumptions given in the problem, we can use equation [16.58] in the text book, giving 2f (cos θ) v c c∆f = v = . 2f cos θ
∆f =
⇒
6
(b) What is the minimum sampling frequency required for a analog-to-digital converter (ADC) recording ∆f ? Solution: For a flow of 5 m/s, we have ∆f 5 =
2
6
× 10 cos (30 ) · 5 ≈ 5831. 8 Hz ◦
1485
For a flow of 20 m/s we have 2
6
× 10 cos (30 ) · 20 = 23327 Hz ◦
1485
The sampling theorem states that we need to sample faster than twice the maxmimum frequency. This means we need to sample faster than 2 23327 = 46654 Hz, i.e. faster than 46.7 kHz.
·
(c) Mistakes when installing the flowmeter causes the angle to vary randomly around θ = 30 , so that the true angle is Gaussian distributed with mean θ = 30 and standard deviation σθ = 1 . Determine a 95 % confidence interval (corresponding to 2σ) for the flow velocity, v. What is the maximum error due to this misalignment as percentage of the maximum flow velocity? Solution: From (a) we have ◦
◦
◦
v =
c∆f . 2f cos θ
The variance of the measured velocity is 2
σv =
2
∂v θ
2
σθ =
c ∆f sin θ 2 f cos 2 θ
2
σθ2
For the numerical values given in the problem, this gives 2σv
≈ 0.001∆f.
The maximum error is at the highest velocity, and with an angular error at one extreme end of the interval. Given that the error is Gaussian, it is unlikely to have angular errors of more than 3σθ or 4σθ . For 20 m/s and an angle of 34 , we will receive a ∆f of ◦
2f 2 106 ∆f = (cos θ) v = (cos 34 ) 20 = 22331 Hz, 1485 c
×
◦
·
and given that we believe the angle to be 30 , we will estimate the velocity according to c∆f 1485 22331 v= = 19. 15 m/s, 2f cos θ 2 106 cos30 ◦
×
7
·
◦
≈
thus the error is error =
1485 22331 2 106 cos30
×
·
◦
− 20 ≈ −0.85 m/s
As a percentage of the full scale velocity this is
−0.85 × 100 % = −4. 25 % 20
8
