industrial control and instrumentation notes with …
TRANSCRIPT
INDUSTRIAL CONTROL AND
INSTRUMENTATION
Notes with solved problems
Vlastimil MasekMemorial University of Newfoundland
Ver. 0.3.0
Contents
1 Sensors & Instrumentation 11.1 Introduction to Sensor Based Measurement Systems . . . . . . . 1
1.1.1 Sensor Classification . . . . . . . . . . . . . . . . . . . . . 11.1.2 Static Characteristics . . . . . . . . . . . . . . . . . . . . 11.1.3 Exercises in Error Analysis . . . . . . . . . . . . . . . . . 21.1.4 Dynamic Characteristics . . . . . . . . . . . . . . . . . . . 71.1.5 Problems in System Dynamics . . . . . . . . . . . . . . . 71.1.6 Other Sensor Characteristic . . . . . . . . . . . . . . . . . 191.1.7 Minimizing Internal & External Perturbations . . . . . . 19
1.2 Resistive Sensors . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.3 Sensors in Process Industry . . . . . . . . . . . . . . . . . . . . . 321.4 Sensors in Manufacturing Industry . . . . . . . . . . . . . . . . . 501.5 Capacitive Sensors . . . . . . . . . . . . . . . . . . . . . . . . . . 51
1.5.1 Parallel plates . . . . . . . . . . . . . . . . . . . . . . . . . 511.5.2 Disk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 511.5.3 Concentric cylinders . . . . . . . . . . . . . . . . . . . . . 511.5.4 Parallel cylinders . . . . . . . . . . . . . . . . . . . . . . . 511.5.5 Cylinder and plane . . . . . . . . . . . . . . . . . . . . . . 511.5.6 Two cylinders and plane . . . . . . . . . . . . . . . . . . . 521.5.7 Interfacing the Sensor . . . . . . . . . . . . . . . . . . . . 52
1.6 Inductive Sensors . . . . . . . . . . . . . . . . . . . . . . . . . . . 531.6.1 Variable self-inductance transducers . . . . . . . . . . . . 531.6.2 Variable mutual-inductance transducers . . . . . . . . . . 54
i
1 Sensors & Instrumentation
1.1 Introduction to Sensor Based Measurement Systems
1.1.1 Sensor Classification
• Need for power supply
– modulating (aux power source)
– self generating
• Output signal
– analog (V,A)
– digital (counts, series, parallel bus)
– time data (frequency, PWM, phase)
• Operating mode
– deflection
– null mode
• I/O dynamics (zero, 1st, 2nd order)
• Measurand
– temperature, pressure, flow, level, humidity, pH, chemical composi-tion, density, moisture, ...
– position, velocity, acceleration, force, torque, ...
1.1.2 Static Characteristics
• Accuracy - proximity to the true (ideal) value
• Absolute Error = Result - True Value (discrepancy between measurandand the True Value [units of the measurand, or %FSO, %span]
• Relative Error = Absolute Error / True Value [relative % of reading]
• Uncertainty of measurement = error range
• Precision = agreement between successive readings
• Sensitivity = the slope of calibration curve (it is often desirable to havehigh and constant sensitivity)
• Linearity = closeness between the calibration curve and a specified straightline (Note: Digital instruments today require more precision than linear-ity.)
– Independent linearity (Least Squares)
1
– Zero based linearity (Least Squares plus passing through zero[0,0])
• Resolution = the minimal change of the input necessary to produce adetectable change at the output
• Threshold = Resolution when the increment is from zero.
• Hysteresis = difference between two output values that correspond to thesame input, depending on the direction of input values.
• Systematic Errors
– yield measurement bias
– detectable and correctable by static calibration
• Random Errors = errors that remain after eliminating the systematic er-rors
– mean x
– std deviation σ
– variance σ2
1.1.3 Exercises in Error Analysis
HHHJJJ,,1.1: In order to measure the current in a line, we consider two
alternative methods: (1) Use an ammeter, whose accuracy is ±0.2% ofthe reading. (2) Use a voltmeter to measure the drop in voltage acrossa resistor, in which the voltmeter’s accuracy is ±0.1% of the readingand the resistor has ±0.2% tolerance.
By using a simplified Taylor expansion formula in Eq.1 determinewhich method is more accurate?
uf(x1,x2,...xn) =
∣∣∣∣ ∂f∂x1
∣∣∣∣ux1+
∣∣∣∣ ∂f∂x2
∣∣∣∣ux2+ . . .+
∣∣∣∣ ∂f∂xn∣∣∣∣uxn (1)
u : absolute uncertainty
I ±0.2% [A]
U ±0.1% [V ]
R ±0.2% [Ω]
2
I(1) = I ± 0.2% [A]
I(2) =U
R±?% [A]
uI(2) =
∣∣∣∣∂I(2)
∂U
∣∣∣∣uU +
∣∣∣∣∂I(2)
∂R
∣∣∣∣uRuI(2) =
1
RuU +
U
R2uR
∣∣∣∣∣× 1
I(2)
uI(2)I(2)
=uUU
+uRR
= 0.1 + 0.2 = 0.3
=⇒ I(2) ± 0.3% [A]
The uncertainty when measuring current directly by method (1) is ±0.2%,hence lower.
HHHJJJ,,1.2: A capacitor of 10nF is paralleled with a capacitor of 100nF,
each having a tolerance ±10%. What is the resulting capacitance andthe maximal/probable uncertainty? What would be the capacitanceand uncertainty if the two elements were connected in series.
Using absolute uncertainty uC :
C1 = 10 [nF ] ± 10%
uC1= 1.0 [nF ]
C2 = 100 [nF ] ± 10%
uC2= 10.0 [nF ]
C = ? [nF ] ±?%
Maximum uncertainty1 uCmax :
uCmax =
∣∣∣∣ ∂C∂C1
∣∣∣∣uC1 +
∣∣∣∣ ∂C∂C2
∣∣∣∣uC2 (2)
Probable uncertainty2 uCprob :
uCprob =
√(∂C
∂C1uC1
)2
+
(∂C
∂C2uC2
)2
(3)
Part A
C = C1 + C2 = 10 + 100 = 110 [nF ] (4)
1represents the worst case uncertainty2represents the more realistic uncertainty
3
Maximum uncertainty uCmax :
uCmax = |1|uC1+ |1|uC2
uCmax = 1.0 + 10.0 = 11.0 [nF ]
C = 110 [nF ]± 10%
Probable uncertainty uCprob :
uCprob =
√(1× uC1
)2
+ (1× uC2)2
uCprob =
√(1.0)
2+ (10.0)
2 .= 10.05 [nF ]
C = 110 [nF ]± 9.1%
Part B
C =1
1C1
+ 1C2
=C1C2
C1 + C2=
10× 100
10 + 100= 9.09 [nF ] (5)
Maximum uncertainty uCmax :
uCmax =
∣∣∣∣∣(
C2
C1 + C2
)2∣∣∣∣∣uC1
+
∣∣∣∣∣(
C1
C1 + C2
)2∣∣∣∣∣uC2
uCmax =
(100
10 + 100
)2
× 1.0 +
(10
10 + 100
)2
× 10.0 = 0.90 [nF ]
C.= 9.09 [nF ]± 10%
Probable uncertainty uCprob :
uCprob =
√√√√(( C2
C1 + C2
)2
× uC1
)2
+
((C1
C1 + C2
)2
× uC2
)2
uCprob =
√√√√(( 100
10 + 100
)2
× 1.0
)2
+
((10
10 + 100
)2
× 10
)2
.= 0.83 [nF ]
C.= 9.09 [nF ]± 9.13%
Alt SolutionUsing relative uncertainty uC
C :
4
C1 = 10 [nF ] ± 10%uC1
C1= 0.1 (10%)
C2 = 100 [nF ] ± 10%uC2
C2= 0.1 (10%)
C = ? [nF ] ±?%
Part A
C = C1 + C2 = 10 + 100 = 110 [nF ] (6)
Maximum uncertainty(uCC
)max
:
uCmax = |1|uC1+ |1|uC2
∣∣∣∣∣× 1
C(uCC
)max
=|1|
1 + C2
C1
uC1
C1+
|1|1 + C1
C2
uC2
C2(uCC
)max
=|1|
1 + 10010
× 10% +|1|
1 + 10100
× 10% [%]
C = 110 [nF ]± 10%
Probable uncertainty(uCC
)prob
:
(uCC
)prob
=
√(1
1 + 10010
× 10%
)2
+
(1
1 + 10100
× 10%
)2
[%]
C = 110 [nF ]± 9.1%
HHHJJJ,,1.3: A transit time flowmeter in Figure 1 is based on the variation
in transit time between transmitter and receiver, depending on whetherthe ultrasonic signal propagates in the same direction as the flow (Eq.7)or in the opposite direction (Eq.8). By combining the two equationswe obtain the flow rate q being proportional to t21−t12
t12t21(Eq.9) which
eliminates the dependency of the flow measurment on the speed ofsound.
Determine the relative uncertainty of the flow rate q when the transittime measurement has ±2% uncertainty and the pipe diameter hasbeen measured with ±1% uncertainty.
5
t12 =D/ sinα
c+ v cosα(7)
t21 =D/ sinα
c− v cosα(8)
q =πD2
4v =
πD3
4 sin 2α
t21 − t12
t12t21
[m3
s
](9)
α
E1 R1
E2R2
t21
t12
v
D
Figure 1: Transit time flowmeter
∆q =
∣∣∣∣ ∂q∂D∣∣∣∣∆D +
∣∣∣∣ ∂q∂t12
∣∣∣∣∆t12 +
∣∣∣∣ ∂q∂t21
∣∣∣∣∆t21 +
∣∣∣∣ ∂q∂α∣∣∣∣∆α
∆q =3πD2
4 sin 2α
t21 − t12
t12t21∆D +
πD3
4 sin 2α
1
t212
∆t12 +πD3
4 sin 2α
1
t221
∆t21 +πD32 cos 2α
4 sin2 2α
t21 − t12
t12t21∆α
∣∣∣∣∣× 1q
uq% = 3uD% +t21
t21 − t12
∆t12
t12︸ ︷︷ ︸ut%
+t12
t21 − t12
∆t21
t21︸ ︷︷ ︸ut%
+2cos 2α
sin 2α∆α
∣∣∣∣∣insert Eqs.7 + 8
uq% = 3uD% +c
v cosαut% + 2 cot 2α∆α
cot is reaching zero at π/2 which results in an optimal angle α of 45. Thiswould be in case a protractor is used in the angle measurement. Determine theuncertainty of the flowmeter that involves a longitudinal dimension x betweenthe transducers along the pipe instead of the angle α. Explain which method ofthe angle measurement yields a more practical solution with lower uncertainty.
6
1.1.4 Dynamic Characteristics
• Zero order (e.g. potentiomenter) y(t) = k x(t)
• 1st order Y (s)X(s) = k
τs+1
– corner frequency ωc = 1τ
– dynamic error
∗ step
∗ ramp
∗ harmonic sin(ωt)
• 2nd order Y (s)X(s) =
kω2n
s2+2ζωns+ω2n
1.1.5 Problems in System Dynamics
HHHJJJ,,1.4: Consider a simple closed loop circuit where a capacitor C
is connected with a resistance R and a voltage source E (of smallimpedance) in series.
Calculate the time constant τ for voltage across the capacitor C. Ifan inductance L is substituted for the capacitor C, calculate the timeconstant τ for voltage across the inductance L.
Part A
E = iR+Q
C[V ] (10)
i =dQ
dt[A] (11)
E
R=
dQ
dt+
Q
RC
Step response with zero initial conditions (L):
E0
sR= sQ(s) +
Q(s)
RC
Q(s) =E0
sR× 1
s+ 1RC
=A
s+
B
s+ 1RC
∣∣∣∣∣L−1
Q(t) = A+Be−tRC = A+Be−
tτ
Ecap = Q/C ≈ e− tτ
τ = RC is the time constant.Part B
7
E =di
dtL+ iR (12)
Step response with zero initial conditions (L):
E0
sL= sI(s) + I(s)
R
L
I(s) =E0
sL× 1
s+ RL
=A
s+
B
s+ RL
∣∣∣∣∣L−1
i(t) = A+Be−t
L/R = A+Be−tτ
Eind =di
dtL ≈ e− t
τ
τ = LR is the time constant.
HHHJJJ,,1.5: What time constant τ is associated with the circuit in Figure
2? Draw into the provided chart in Figure 2 approximate response ofVC to the step input
Vi =
0[V ] t < 0Vo[V ] t ≥ 0
Consider VC = 0[V ] at t < 0.
t
ViVo
00
04τ t
Vo
50%
τ 2τ 3τ 5τ
Vc
0
C
R
Vi
Vc
Figure 2: RC circuit
τ = RC
[Ω F =
V
A
C
V=V
A
A s
V= s
]
8
~63%
4τ t
Vo
50%
τ 2τ 3τ 5τ
Vc
00
~100%
HHHJJJ,,1.6: Consider a simple closed loop circuit where a capacitor C is
connected with a resistance R, an inductance L, and a harmonic voltagesource E (of small impedance) in series.
Calculate the resonant frequency (natural frequency) for voltage acrossthe capacitor C.Calculate the critical resistance (critical damping) forvoltage across the capacitor C.
E0 cos Ωt =d2Q
dt2L+
dQ
dtR+
Q
C[V ]
∣∣∣∣∣L (13)
E0
L
s
s2 + Ω2= s2Q(s) + sQ(s)
R
L+Q(s)
1
LC
Q(s) =E0
L× s
s2 + Ω2× 1
s2 + sRL + 1LC
=As+B
s2 + Ω2+
Cs+D
s2 + sRL + 1LC
=
=As+B
s2 + Ω2+
Cs+D
(s+ R2L )2 +
√ 1
LC−(R
2L
)22
︸ ︷︷ ︸ω2nd
ωnd =
√1
LC−(R
2L
)2
=
√1
LC︸ ︷︷ ︸ωn
√√√√√1− R2C
4L︸ ︷︷ ︸ξ2
(14)
ξ =R
Rc=⇒ Rc = 2
√L
C(15)
9
HHHJJJ,,1.7: A pressure transducer uses a diaphragm as a pressure-
summing device. In application the diaphragm and fluid behave asa second-order, single-degree system. The static displacement is pro-portional to the applied force (pressure). If the natural undampedfrequency of the system is 3600 Hz and the total viscous damping is75% of critical, determine the frequency range(s) over which the ra-tio of dynamic amplitude to static amplitude (inherent error) deviatesfrom unity by an amount no greater than 6%.
PdPs
=1√[
1−(
Ωωn
)2]2
+[2ξ Ω
ωn
]2 (16)
ξ = 0.75
ωn = 2πfn = 2π3600 [rad s−1]
PdPs∈ (94%, 106%)
Ω = ±ωn
√±√
4(PdPs
)2
ξ4 − 4(PdPs
)2
ξ2 + 1− 2 PdPsξ2 + Pd
Ps√PdPs
(17)
Ω94%.= ±2π1831.7 [rad s−1] ⇒ f94%
.= ±1831.7 [Hz]
±2π2568.1i [rad s−1] ⇒ no physical meaning
Ω106%.= ±2π3600
√0.307211i− 0.125 [rad s−1] ⇒ no ph. meaning
±2π3600√−0.307211i− 0.125 [rad s−1] ⇒ no ph. meaning
The inherent error deviates from unity by an amount no greater than 6%within the theoretical frequency range f ∈ (−1831.7, 1831.7)[Hz], and practi-cally within f ∈ (0, 1831.7)[Hz]. The deviation above 100% is not applicable forξ = 0.75, however, in case of very small damping, e.g. ξ = 0.05, all four rootsmight be ∈ < (refer to MAXIMA console output C(29)∼D(30) and Figure 3).
10
Numerical evaluation by MAXIMA Symbolic Math Tool:3
Alt.SolutionReading from the Frequency Response in Figure 3:
f94%.= 0.5fn
.= 3600/2
.= 1800 [Hz]
xi=0.75−2
0
2
4
6
8
10
0.10 1.00 10.000
20
40
60
80
100
120
140
160
180
Gai
n (d
B)
Phas
e (d
egre
es)
OMEGA−−−−−−−−−−
omega
Frequency Response of 2nd Order System
106%
94%xi=0.75
xi=0.05
−4
−2
0
2
4
6
8
10
0.10 1.00 10.000
20
40
60
80
100
120
140
160
180
Gai
n (d
B)
Phas
e (d
egre
es)
OMEGA−−−−−−−−−−
omega
Frequency Response of 2nd Order System
106%
94%xi=0.75
xi=0.05
−4
−2
0
2
4
6
8
10
0.10 1.00 10.000
20
40
60
80
100
120
140
160
180
Gai
n (d
B)
Phas
e (d
egre
es)
OMEGA−−−−−−−−−−
omega
Frequency Response of 2nd Order System
106%
94%xi=0.75
xi=0.05
−4
−2
0
2
4
6
8
10
0.10 1.00 10.000
20
40
60
80
100
120
140
160
180
Gai
n (d
B)
Phas
e (d
egre
es)
OMEGA−−−−−−−−−−
omega
Frequency Response of 2nd Order System
106%
94%
xi=0.05
−4
Figure 3: 2nd order system frequency response
HHHJJJ,,1.8: Consider the pressure transducer from the previous problem to be
damaged such that its viscous damping ratio becomes changed to some un-known value. If the transducer is subjected to a harmonic input of 2400 Hz,the phase angle between output and input is measured as 45 degrees. Withthis in mind, determine the inherent error (attenuation) of the transducerwhen used to measure a harmonic pressure signal of 1800 Hz. What will bethe phase angle between the output and input at this frequency?
3Open source: http://maxima.sourceforge.net
11
φ = tan−1
2ξ Ωωn
1−(
Ωωn
)2
(18)
PdPs
=1√[
1−(
Ωωn
)2]2
+[2ξ Ω
ωn
]2 (19)
ωn = 2πfn = 2π3600 [rad s−1]
Ω1 = 2πf1 = 2π2400 [rad s−1]
φ1 = 45π/180 [rad]
ξ =?
Ω2 = 2πf2 = 2π1800 [rad s−1](PdPs
)2
=?
φ2 =?
Numerical evaluation by MAXIMA Symbolic Math Tool:
ξ = 5/12 MAXIMA (D3)
φ2.= 29 MAXIMA (D6)(
PdPs
)2
.= 116.6% MAXIMA (D9)
HHHJJJ,,1.9: An experimental arrangement is assembled to determine the time
constant of a first-order measuring system A . The measured process can becontrolled to produce a sinusoidal output with respect to time. Measuringsystem B is used to monitor the process. At all frequencies used in the test,the response of system B may be assumed as perfect.
The following data is obtained:
Process amplitude measured by system A is 1.2 units.Process amplitude measured by system B is 1.4 units.Process frequency indicated by both systems is 1.25[Hz].The output from the system A lags behind that from system B by 0.07[s].
The data provide for two different time-constant calculations for system A.Determine the two values; in which would you place greater confidence?Why?
Part AA harmonically excited first order system:
12
TF (s) =K
τs+ 1
M(ω) =
∣∣∣∣ K
τjω + 1
∣∣∣∣ =K√
1 + τ2ω2
ϕ = 6
(K
τjω + 1
)= arctan
−τω1
For a measuring instrument, the K=1.Using Phase Lag (ϕ < 0)
|ϕ| = tan−1 ωτ = tan−1 2πfτ (20)
τ =tanϕ
2πf=
tan(time lag × 2πf)
2πf[s]
τ =tan(0.07× 2π1.25)
2π1.25= 0.078 [s]
Using Amplification Ratio
aAaB
=1√
1 + (ωτ)2=
1√1 + (2πfτ)2
(21)
τ =
√(aBaA
)2
− 1
2πf=
√(1.41.2
)2 − 1
2π1.25= 0.0765 [s]
Part BThe greater confidence measurement is one of smaller uncertainty, i.e. of smaller
sensitivity to the inaccuracies.Sensitivity:
S1 =dτ1dϕ
=1
ω
1
cos2 ϕ
S1.=
1.375
ω
limS1ϕ−→0 =1
ω
S2 =dτ1d(aBaA
)=
1
ω
aBaA√(
aBaA
)2
− 1
S2.=
2.86
ωlimS2
(aBaA
)−→1+
= ∞
13
HHHJJJ,,1.10: Voltage outputs from the two measuring systems A and B in
the previous problem are connected to a cathode-ray oscilloscope, one tothe vertical (A) and the other (B) to the horizontal plates. The resultingLissajous diagram is being used to measure the amplification (amplituderatio) and the phase between system A and system B. The X-Y screen isrectangular and all voltage readings from the screen have a constant absoluteuncertainty given as %FSO.
Determine time constant uncertainty based on (1) amplitude measurement,and (2) phase measurement.
Determine time constant at which the advantage of using one method overthe other method alters.
Z
ϕ Y
X
Y
Z
Figure 4: Lissajous Diagram: ϕ = sin−1 xy
aBaA
= yz
Uncertainty from Phase Lag based measurement
14
τ1 =tanϕ
ω
ϕ = sin−1 x
yFig.4
⇓τ1 = τ1(x, y, ω)
uτ1 =
∣∣∣∣∂τ1∂x∣∣∣∣ux +
∣∣∣∣∂τ1∂y∣∣∣∣uy +
not considered︷ ︸︸ ︷∣∣∣∣∂τ1∂ω
∣∣∣∣uωuxyz = ux = uy = uz absolute uncertainty
u∗τ1 =
(∣∣∣∣∂τ1∂x∣∣∣∣+
∣∣∣∣∂τ1∂y∣∣∣∣)uxyz (22)
Using Amplification Ratio
τ2 =
√(zy
)2
− 1
ωFig.4
⇓τ2 = τ2(y, z, ω)
uτ2 =
∣∣∣∣∂τ2∂y∣∣∣∣uy +
∣∣∣∣∂τ2∂z∣∣∣∣uz +
∣∣∣∣∂τ2∂ω
∣∣∣∣uω︸ ︷︷ ︸not considered
u∗τ2 =
(∣∣∣∣∂τ2∂y∣∣∣∣+
∣∣∣∣∂τ2∂z∣∣∣∣)uxyz (23)
15
Evaluation by MAXIMA Symbolic Math Tool:
16
Parametric plot by GNUPlot :4
y
z=
1√1 + (Ωτ)2
φ = tan−1 Ωτx
y= sinφ
Ω =2π
T
τ = 0+ ∼T
4parameter
T = 1 selected (24)
z = 1 selected (25)
y =1√
1 + (Ωτ)2(26)
x = y sin(tan−1 Ωτ) (27)
by phase by amplitude
lower uncertainty
unce
rtai
nty
wei
ght
τ / Τ
0
2.5
3
3.5
4
4.5
5
5.5
0 0.05 0.1 0.15 0.2 0.25
by amplitude
1.5
by phase
1
0.5
2
Figure 5: Uncertainty of time constant measurement by two methods
4Open source: http://sourceforge.net/projects/gnuplot
17
####################################################################
# Parametric plot of uncertainty of 1st order system time constant #
# based on phase and amplitude measurement using Lissajous diagram #
####################################################################
pi=3.1415926
z = 1
OMEGA = 2*pi
set parametric
set dummy t
set trange [0.01:0.25]
plot t,abs(1./(OMEGA*sqrt(1.-((1./sqrt(1.+(2.*pi*t)**2.))* \
sin(atan(2.*pi*t)))**2./(1./sqrt(1.+(2.*pi*t)**2.))**2.)* \
(1./sqrt(1.+(2.*pi*t)**2.)))+((1./sqrt(1.+(2.*pi*t)**2.))* \
sin(atan(2.*pi*t)))**2./(OMEGA*(1.-((1./sqrt(1.+(2.*pi*t)**2.))* \
sin(atan(2.*pi*t)))**2./(1./sqrt(1.+(2.*pi*t)**2.))**2.)**(3./2.)* \
(1./sqrt(1.+(2.*pi*t)**2.))**3.))+abs(((1./sqrt(1.+(2.*pi*t)**2.))*\
sin(atan(2.*pi*t)))/(OMEGA*sqrt(1.-((1./sqrt(1.+(2.*pi*t)**2.))* \
sin(atan(2.*pi*t)))**2./(1./sqrt(1.+(2.*pi*t)**2.))**2.)* \
(1./sqrt(1.+(2.*pi*t)**2.))**2.)+((1./sqrt(1.+(2.*pi*t)**2.))* \
sin(atan(2.*pi*t)))**3./(OMEGA*(1.-((1./sqrt(1.+(2.*pi*t)**2.))* \
sin(atan(2.*pi*t)))**2./(1./sqrt(1.+(2.*pi*t)**2.))**2.)**(3./2.)* \
(1./sqrt(1.+(2.*pi*t)**2.))**4.)) \
title "by phase" , \
t,abs(z)/(abs(OMEGA)*(1./sqrt(1.+(2.*pi*t)**2.))**2.* \
sqrt(z**2./(1./sqrt(1.+(2.*pi*t)**2.))**2.-1.))+ \
z**2./(abs(OMEGA)*abs((1./sqrt(1.+(2.*pi*t)**2.)))**3.* \
sqrt(z**2./(1./sqrt(1.+(2.*pi*t)**2.))**2.-1.)) \
title "by amplitude"
set output "lissajous_uncertainty.fig"
set terminal fig color
replot
HHHJJJ,,1.11: The output voltage of LVDT is described by Eq.28. By rearrang-
ing the characteristic part of Eq.28 into the form of Eq.29 and by usingEq.30, determine the damped natural frequency fd[Hz], the undamped nat-ural frequency fn[Hz], and the damping ratio ξ[1]
18
E0 =s(M1 −M2)E1RL
2L1L2s2 + (R2L1 + 2R1L2)s+R1R2(28)
R1 : resistance in the primary winding [Ω]
R2 : total resistance in the secondary windings [Ω]
RL : load resistance [Ω]
L1 : self inductance in the primary winding [H]
L2 : self inductance in the secondary winding [H]
M1 : mutual inductance between primary and sec.winding 1 [H]
M2 : mutual inductance between primary and sec.winding 2 [H]
E1 : input voltage [V ]
s2 + as+ b = (s+ τ)2 + ω2d (29)
ωd = ωn√
1− ξ2 (30)
1.1.6 Other Sensor Characteristic
• Impedance,
– Z(s) = effortvar(s)flowvar(s)
, e.g. U(s)I(s)
– Note: Admitance Y (s) = 1Z(s)
– e.g., pot with large friction, large thermal mass thermometer, etc.
• Loading error = when power extraction from the measuring system modifies thevalues of the measurand variable.
• Note: Sensor’s input impedance/admitance shall be very large in order to min-imize the loading error when measuring an effort/flow variable.
1.1.7 Minimizing Internal & External Perturbations
• Negative feedback = Null-measurement (Balancing) systems (linearity of thesensor is not critical)
• Opposing inputs = Differential Method, e.g. LVDT, Wheatstone Bridge, diff.capacitor, etc.
• Smart sensors, data fusion
19
1.2 Resistive Sensors
Change in resistance is not by itself a signal. The simplest signal conditioning is donethrough a voltage source and a resistive voltage divider in which one resistor has aconstant value RC and the other one is the actual sensor R = R0(1 + x).
Maximum power dissipation as well as sensitivity occurs when R = RC :
P =U2
(R+RC)
dP
dR=
dP
dR= 0
HHHJJJ,,2.1: In temperature measurements by using resistive sensors we must
minimize self-heating. Otherwise, the sensor temperature T would be higherthan that of the surrounding medium Ta. For a sensor in a given environ-ment, the heat dissipation capability is given by the heat dissipation constantδ [mW/K], which depends on the surrounding fluid and its velocity.
Write a steady-state relation for the heat dissipated by RTD in Fig. 6.For a constant voltage U determine RTD’s temperature T . Consider(R+R0) >> R0αT .
U
δRt = Ro(1 + T)α
R
Figure 6: Voltage divider for temperature measurement
δ(T − Ta) = i2RT =
(U
R+RT
)2
RT
20
δ(T − Ta) =
(U
R+Ro(1 + αT )
)2
Ro(1 + αT )
.=
(U
R+Ro
)2
Ro(1 + αT )
T.=
U2Ro + δ(R+Ro)2Ta
δ(R+Ro)2 − U2Roα
HHHJJJ,,2.2: Write a formula for capacitance of a parallel plate capacitor and
include dimensions for each parameter.Select the appropriate sensor that matches the signal conditioning circuitsin Figure 7 in order the output voltage to be directly proportional to theindicated parameter (area or spacing).
Vi
spacing variationarea variation
C2
C1Vo
spacing variationarea variation
C2
C1Vo
Vi
Figure 7: Low-Z amplifier for capacitive sensors
C =ε0εRA
d
[Fm
1 m2
m= F
]
Vo = −Z2
Z1Vi = −
1jωC2
1jωC1
Vi = −C1
C2Vi
21
=⇒ Area variation measurement circuit Vo = k1A must have sensor at the positionof C1 whereas spacing variation measurement circuit Vo = k2d must have sensor at theposition of C2.
HHHJJJ,,2.3: Figure 8 shows three schemes of Wheatstone bridge in the deflection
mode configuration. The following components R1, R2, R3, R4 listed beloware available in any quantity.
Select the right components (write into the schemes) in order to obtain atemperature compensated measurement of x with bridge constant k=1,2,4.(Note: bridge constantk = output of bridge
output of bridge with one sensor
)
R1 = R1(+x,+T ) = R0(1 + αx+ βT )
R2 = R2(−x,+T ) = R0(1− αx+ βT )
R3 = R3(+T ) = R0(1 + βT )
R4 = const = R0
where x : measured parameter
T : temperature
Note 1: Bridge output is zero (balance) when the ratio of adjacent arm resistanceson one side is equal to the ratio of adjacent arm resistances on the other side.
Note 2: Bridge output is directly proportional to the difference of adjacent arm re-sistances.
Note 3: Bridge output is directly proportional to the sum of opposite arm resistances.
Vo
Vi
Vo
Vi
Vo
Vi
Bridge constant: 1 Bridge constant: 2 Bridge constant: 4
−
+
−
+
−
+
Figure 8: Wheatstone bridge
22
R1
Vo
Vi
Vo
Vi
Vo
Vi
Bridge constant: 1 Bridge constant: 2 Bridge constant: 4
R3
R4
R2
R1
R4
R1
R1 R3
R3 R1
R1
R2
R4
R4
R2−
+
−
+
−
+
Test:
k = 1 Vo = Vi
(1
2− R3
R1 +R3
)= Vi
((R1 +R3)− 2R3
2(R1 +R3)
)=
= ViR1 −R3
2(R1 +R3)
.= Vi
αx
4
k = 2 Vo = Vi
(R1
R1 +R3− R3
R1 +R3
)= Vi
R1 −R3
R1 +R3
.= Vi
αx
2
k = 2 Vo = Vi
(1
2− R2
R1 +R2
)= Vi
(R1 +R2)− 2R2
2(R1 +R2)
)=
= ViR1 −R2
2(R1 +R2)
.= Vi
αx
2
k = 4 Vo = Vi
(R1
R1 +R2− R2
R1 +R2
)= Vi
R1 −R2
R1 +R2
.= Viαx
HHHJJJ,,2.4: A simple Wheatstone bridge shown in Figure 9 is used to accurately
determine the value of an unknown resistance R1. If upon the initial nullbalance R3 is 127.5[Ω], and if when R2 and R4 are interchanged null balanceis achived when R3 is 157.9[Ω], what is the value of the unknown resistanceR1.
Null ballance occures when:
R1
R2=R3
R4(31)
therefore:
23
R4R3
AC eo
R1 R2
B
D
ei
+ −
Figure 9: Wheatstone bridge
R1
R2=
127.5Ω
R4
R1
R4=
157.9Ω
R2
R4 =R2 × 127.5Ω
R1
R1 =R4 × 157.9Ω
R2=
R2×127.5ΩR1
× 157.9Ω
R2=
157.9Ω× 127.5Ω
R1
R1 =√
157.9Ω× 127.5Ω.= 141.9 [Ω]
The unknown resistor’s value is 141.9[Ω].H
HHJJJ,,2.5: A commonly used differential-shunt configuration of the Wheatstone
bridge (Figure 10) employs a multiturn potentiometer R6 for either initialnulling of the bridge or as a read-out means. R1 is often a resistance-typetransducer, and R2, R3, and R4 may or may not be fixed resistances (oftenthey, too, are transducer elements). The variable K is a proportional termvarying from 0 to 1 (or 0 to 100%). R5 is called a scaling resistor. Its valuelargely determines the range of effectiveness of R6. If the nominal resistanceof R1 is 1000[Ω], R2 = R3 = R4 = 1000[Ω](fixed), R5 = 8000[Ω], and R6 =10000[Ω], what null-balance range of ∆R1 can the bridge accommodate?Devise a computer program (using Matlab, Octave, Maple, ...) that can beused to solve this problem.
i5 = i3 − i4i7 = i6 + i5 = i6 + i3 − i4
24
R1 R2
R4R3
AC
ei
eo
D
B
R6
R5k R6
−+
Figure 10: Differential shunt bridge
1
C
ei
eo
D
B
R6
R5i3 i4
i5
E
i1
i=0
i1
i6 i73
4
2
R1 R2
R4R3
A
−+
Figure 11: Currents in differential shunt bridge
25
R1 + ∆R1 +R2 0 0 0
0 R3 0 R4
0 (1− k)R6 R6 (k − 1)R6
0 −R3 −R5 kR6 R5
i1i3i6i4
=
eieiei0
R× I = E
I = R−1 ×E
or
i1 =|R1||R|
i4 =|R4||R|
R1 =
ei 0 0 0ei R3 0 R4
ei (1− k)R6 R6 (k − 1)R6
0 −R3 −R5 kR6 R5
R4 =
R1 + ∆R1 +R2 0 0 ei
0 R3 0 ei0 (1− k)R6 R6 ei0 −R3 −R5 kR6 0
BALANCE:
i = 0 =⇒ eB = eD =⇒ i1R2 = i4R4
R1nominal = R2 = R3 = R4 = R0
R1 = R4
Using MAXIMA Symbolic Math:
∆R1 ∈(− R2
0
R5 +R0,R2
0
R5
)H
HHJJJ,,2.6: Using the program developed in the previous problem with the same
basic resistance values, investigate changes in the measurement range of thebridge as affected by changes in R5, and changes in R6. Investigate thelinearity of the circuit when used in the null-balance mode, using K as thecalibrated read-out.
A. R5 scales the region of ∆R1
B. R6 influences linearity (R5 >> R6, R0)
26
Fun1Fun2Fun3Fun40.510 0
0.
5
0.
5
1 1
-1
00
-1
00
00
10
0100
Fo
r
S
at
N
ov
2
65
9:
46
:5
19
AM
N
ew
fo
un
dl
an
d
St
an
da
rd
T
im
e
20
05
Figure 12: Differential shunt bridge: ∆R1 vers k
27
HHHJJJ,,2.7: A modified Wheatstone bridge in Figure 13 that includes a single
linear sensor R0(1 + x) is used to obtain a voltage proportional to any sizechange in the sensor’s resistance.
Derive the output voltage Vo as a function of x while considering the op ampas ideal.
Vr
R0 R0
R0(1 + x)
Vo
R0
−
+
Figure 13: Linearized Wheatstone bridge
∑inv node
i = 0
Vo − V−Ro(1 + x)
+Vr − V−Ro
= 0
∣∣∣∣∣V−=V+=Vr
2
Vo − Vr2
Ro(1 + x)+Vr − Vr
2
Ro= 0
Vo = −Vr2x [V ]
HHHJJJ,,2.8: RC oscillator using CMOS 555 IC is employed in a capacitance
measurement. Determine the frequency as a function of the time constantτ = RC. (Note: voltage across the capacitor varies between 1
3and 2
3of the
rail voltage. 50% duty cycle.)
First-order system:
28
C
555thres.
trig.out Vo
R
Figure 14: CMOS 555 capacitance meter
T
0time
Vo
Vc
rail voltage Vr
1/3 Vr
2/3 Vr
Figure 15: Timing of CMOS 555 capacitance meter
29
VC − VC∞VCA − VC∞
= exp− tτ
VC∞ = VC
∣∣∣∣∣t−→∞
VCA = VC
∣∣∣∣∣t=0
VC∞ = VR
VCA =VR3
VC =2VR
32VR
3− VR
VR3− VR
= exp− tτ
1
2= exp− t
τ
∣∣∣∣∣ lnt = − ln 0.5τ
.= 0.693τ
50% duty cycle:
T = 2t.= 1.386τ
f =1
T
.= 0.72τ−1 [Hz]
The frequency of the oscillator is 0.72τ−1 [Hz].HHHJJJ,,2.9: A Wheatstone bridge in Figure 16 that includes a single linear sensor
R0(1 + x) is nonlinear because the current through the sensor depends onits resistance. To obtain a voltage proportional to any size change in thesensor’s resistance, we modify the structure of the bridge shown in Figure17 to keep constant the current through the sensor.
Derive the output voltage vo as a function of x while considering the op ampas ideal.
HHHJJJ,,2.10: We need to scale voltage from a linear sensor to get the proper
resolution. Figure 18a shows the output voltage of the sensor e = f(x) andFigure 18b shows the desired scaled voltage e′ = g(e) = h(x). Using theappropriate equations illustrate how would you realize the scaling.
30
Vo
R0 R0
R0R0(1 + x)
Vr
Figure 16: Wheatstone bridge
Vo
R0 R0
R0R0(1 + x)
Vr
Figure 17: Linearized Wheatstone bridge
31
e2’=0V
X1 X2
e[V]
X
(a)
e1
e2
X1 X2
e’[V]
X
(b)
e1’=5V
Figure 18: Scaling a linear sensor voltage.
1.3 Sensors in Process IndustryHHHJJJ,,3.1: The circuit in Figure 19 is used to linearize thermistor RT to the
detriment of lower sensitivity. What is the overall sensitivity SP , i.e. thesensitivity of parallel combination of R and RT , and the sensitivity ST of
the thermistor alone? Compare both sensitivities by evaluating the ratio SPST
.
R
RT
Figure 19: Linearizing a thermistor RT = R0 eβ(
1T − 1
T0
)
Sensitivity of a measurement system (sensor system) y = f(x) is the derivative
S = dydx
.
HHHJJJ,,3.2: For the measurement circuit shown in Figure 20 determine temper-
atures T1 and T2 in order to obtain a voltage reading dependent on T butnot on T1 and T2.
From Figure 21 and Figure 22 we obtain:
V = emfT,0 − 2emfT1,0 + emfT2,0
32
B A Cu
V
A B CuT1 T2
Figure 20: Simulation of a cold junction by two temperature-controlled cham-bers
where V = emfT,0 is the desired measurement. Therefore
−2emfT1,0 + emfT2,0 = 0
emfT2,0 = 2emfT1,0
emfT,0 = C1T + C2T2 + C3T
3 + . . .
J thermocouple: C1 = 5.04× 10−5 [V/C]
C2 = 3.04× 10−8 [V/C2]
C3 = −8.60× 10−11 [V/C3]
...
T thermocouple: C1 = 3.87× 10−5 [V/C]
C2 = 3.32× 10−8 [V/C2]
C3 = 2.07× 10−10 [V/C3]
...
For T1 and T2 small: emfT,0.= C1T
C1T2 = 2C1T1
T2 = 2T1
T2 = 2T1 simulates 0C at the reference point.
33
Figure 21: Application of the Law of intermediate wires.
34
Figure 22: Law of intermediate temperatures: emfH,C = emfH,0 − emfC,0
HHHJJJ,,3.3: The circuit in Figure 23 measures a temperature from 400oC to
600oC by a type J thermocouple and cold junction compensation. TheLM134 is a current source whose output is i[µA] = 227[Ω] × (273 +Ta[oC])/R4.
(A) If the op amp is assumed ideal, derive the equation for its output volt-age.(B) Determine the gain needed to obtain a −10V to +10V output range fora 400oC to 600oC temperature range.(C) Design the resistances in order to obtain a cold junction compensationfor an ambient temperature between 10oC and 40oC.
Part A
V0 − V−R2
+VR − V−R3
= i+V−R1
i =a
R4+
b
R4Ta
a = 227× 273× 10−6 [V ]
b = 227× 10−6 [V/C]
V0 = R2i−R2
R3VR + V−(1 +
R2
R3+R2
R1)
Ideal OP AMP ⇒ V−.= V+
35
-
+
2
31 1
1VI
2GND
3 VO
Vout
R1
R2
R3
R4 +15V-15V
Ta+T
-
LM134
Figure 23: Cold junction compensation by LM134
V+ = eT,Ta
eT,Ta = eT,0 − eTa,0
V0 =R2
R3(a+ bTa)− R2
R3VR + (eT,0 − eTa,0)(1 +
R2
R3+R2
R1)
Part B
10V = V0
∣∣∣∣∣T=600C
=R2
R4(a+ bTa)− R2
R3VR + (e600C,0 − eTa,0)(1 +
R2
R3+R2
R1)(32)
−10V = V0
∣∣∣∣∣T=400C
=R2
R4(a+ bTa)− R2
R3VR + (e400C,0 − eTa,0)(1 +
R2
R3+R2
R1)
20V = V0,T=600C − V0,T=400C = (e600C,0 − e400C,0) (1 +R2
R3+R2
R1)︸ ︷︷ ︸
gain G
G =20 V
e600C,0 − e400C,0=
20 V × 103
33.11− 21.85 [mV ]︸ ︷︷ ︸from table
= 1776.2
Part C
V0 =R2
R4a− R2
R3VR︸ ︷︷ ︸
5
+eT,0G+R2
R4bTa − eTa,0G︸ ︷︷ ︸
cold junction comp. (= 0)
5LM134 bias current compensation & adding offset for scaling (Part B).
36
R2
R4=
eTa,0G
bTa
∣∣∣∣∣Ta=25C
R2
R4=
1.28 [mV ] × 10−3 × 1776.2
227 [µV/C] × 25 [C] × 10−6= 400.6 (33)
1 +R2
R3+R2
R1= 1776.2 (34)
We have only three equations Eq.32 Eq.33 Eq.34 for four unknowns R1, R2, R3, R4.Hence we need to select one of the resistances, e.g. R4 = 100Ω =⇒ R1 R2 R3.
HHHJJJ,,3.4: Figure 24 shows a thermometer based on a type T thermocouple
and the AD590 for cold junction compensation.
(A) Derive the equation for the output voltage as a function of the thermo-electromotive forces and Vc.(B) Derive vc when the op amp is assumed ideal.(C) Design R1 and R2 to achive cold junction compensation when 10oC <Ta < 60oC.(D) If the Zener diode yields 6.9V when its reverse current is from 0.6mAto 15mA, design R,R3, R4, Rp to obtain a null output at 0oC.
-
+
1
2
31 1
1
-15V
R
R4 LM329C
Rp
R3
R2
-15V
R1
Ra
1k
Rb
10k
Vc
Vo VoAD590
BT
A Ta
Figure 24: Cold junction compensation by AD590
T Thermocouple
AD950 Absolute temp. sensor :iAD = 298.2 + (Ta − 25C) [µA]
37
iAD = (273.2︸ ︷︷ ︸a
+Ta)× 10−6 [A]
Part A
V0 = eT,0 − eTa,0 + VC
Part B
V0 − V−Rb
= iAD
V0 = RbiAD + V−
V+.= V−
VC = V0R2
R1 +R2
Part CTa ∈ (10, 60) [C]
VC ∼= eTa,0 . . . compensationFirst approximation: eTa,0 = C1Ta C1 = 38.7× 10−6 [V ]
VC = C1Ta
(RbiAD + V+)R2
R1 +R2= C1Ta
(Rb(a+ Ta)× 10−6 + V+)R2
R1 +R2= C1Ta
R2
R1 +R2RbTa × 10−6 = C1Ta (35)
(Rba× 10−6 + V+)R2
R1 +R2= 0 (36)
Note: By using the first approximation we obtain cold junction compensation for theentire range of Ta.
From Eq.35:
R1
R2=Rb × 10−6
C1
.= 249 (37)
From Eq.36:
Rba× 10−6 = −V+
V+ = VZR3 + kRP
R3 +RP +R4
iZ = 0.6 to 15 mA
We select iZ = 1 mA⇒ R = 8100Ω
Note: The voltage drop across R3 +RP +R4 without the zener diode must be > 6.9V.
38
HHHJJJ,,3.5: Figure 25 shows a cold junction compensation circuit for a thermo-
couple of type J. The cold junction is at ambient temperature and includesAD592CN, which yields 1[µA/K].
Determine the circuit resistors to obtain cold junction compensation withoutput voltage Vo = 0[V ] at 0[C] and sensitivity of S[mV/C]. ConsideriAD592CN = a+ bTa [mA] and eJT,0 = c1T + c2T
2 + . . .︸ ︷︷ ︸≈0
[mV ]
block
+15V
−15VType J
AD592CN
R1 R3
R2
+10V
OP177A
−15V
100ΩT
Ta
Vo
isothermal
Figure 25: Cold junction compensation by AD592CN
∑inv node
i = 0Vo − V−R3
+0− V−R2
+10− V−R1
+ iAD592CN = 0
∣∣∣∣∣V−=V+=eJ
T,Ta=c1(T−Ta)
VoR3
+10
R1− c1(T − Ta)
(1
R1+
1
R2+
1
R3
)− (a+ bTa) = 0
Vo = R3
(c1R1
+c1R2
+c1R3
)︸ ︷︷ ︸
=S sensitivity
T +R3
(b− c1
R1+c1R2
+c1R3
)︸ ︷︷ ︸
=0 cold junction comp.
Ta + R3 (a− 10
R1)︸ ︷︷ ︸
=0 offset at 0C
39
HHHJJJ,,3.6: We wish to measure a temperature between −100oC and +1000C
by using a K thermocouple and cold junction compensation according to thecircuit in Figure 26. RT is a Pt100 with α = 0.385%/K at 0oC.
(A) Derive the equation for the output voltage as a function of the sensitiv-ity SK for the thermocouple and the value for RT .(B) Determine the conditions to be fulfilled by the resistors in order to havea null output when Ta = 0oC.(C) If R2 and R3 are chosen so that the output voltage of the bridge circuitthey constitute with RT and R1 can be considered linear, determine theirvalue in order to compensate the cold junction in the entire range of Ta.(D) Determine the gain for the instrumentation amplifier (op amp configu-ration with feedback) in order to have an output voltage from −1V to +1V .(E) Determine the maximal offset voltage for the instrumentation amplifierin order to limit the error to a maximal 0.1oC.
-
+
2
31 1
1VI
2GND
3 VO
R3
Vo
RT
R1 R2
+15V
Ta V1A VoT
B
Vr=10V
Figure 26: Cold junction compensation by RTD
RT : Pt 100 :
RT = R0(1 + αTa)
R0 = 100 [Ω]
α = 0.385 [%C
]
Part A
40
Figure 27: Problem 4
41
V0 = f(SK , RT )
V0 = G(VBridge + VTh) = G
[(RTa
R2 +RTa− R3
R1 +R3
)VR + (eT,0 − eTa,0)
]
V0 = G
VR ( RTaR2 +RTa
− R3
R1 +R3
)+ SK(T − Ta)︸ ︷︷ ︸
first approximation
Part B
Ta = 0CR1
R3=
R2
R0
Part C Compensation :
VR(RTa
R2 +RTa− R3
R1 +R3) = SKTa
For R2 >> RTa , R1 >> R3
VR(RTaR2− R3
R1) = SKTa
VR(R0(1 + αTa)
R2− R3
R1) = SKTa
VR(R0αTaR2
+R0
R2− R3
R1) = SKTa
R0
R2− R3
R1= 0from Part B
VRR0α
R2= SK
R2 =VRR0α
SK=
10× 100× 0.00385
3× 89× 10−5
.= 100 kΩ
Part D
4 [mV ] = 100C
=⇒ Gain = 250
4 [mV ] = 100C
4 [µV ] = 0.1C
42
HHHJJJ,,3.7: A given sensor with unipolar output has a temperature sensitivity
α[
%C
]. To obtain an output voltage constant with temperature, the cir-
cuit in Figure 28 is proposed. The NTC thermistor has β = 3500[K] andR25 = 10[kΩ]. Both the sensor and the NTC thermistor are at an ambienttemperature of about 20C.
Design the circuit in order to have a gain of 1000 at 20C and temperaturecompensation for the sensor; consider the op amps as ideal. If the thermistorhas δ = 1[mW/K], what restriction does this parameter introduce?
Rt
Vi
Vo
R1
R2R3
R4
R5 R6−
+
+
−
Figure 28: DC amplifier with a gain-temperature characteristic tailored to theinput signal temperature coeficient.
Vo = G× Vi = G× Vs(1 + αT )
G =R6RFB
R5R1− R6
R4(38)
RFB =RTR2
RT +R2+R3
RT = R0eβ( 1T− 1T0
)= 10× 103e
3500( 1T [C]−273.15
− 1273.15+25
)
G
∣∣∣∣∣T=20C
= G0 = 1000 (39)
Vo
∣∣∣∣∣T∼20C
= G0 × Vs
∣∣∣∣∣T∼20C
=⇒ G =G0
1 + αT(40)
Our gain Eq.38 is different from the desired gain in Eq.40 and is nonlinear. There-fore, we require that at 20C the variation of Vo with the temperature is zero. Byusing the Taylor expansion formula we obtain:
43
G(T )Vi(T, x) = G0Vs(x)
∣∣∣∣∣T∼20CG0 +
dG
dTdT +
d2G
dT 2dT 2 + . . .︸ ︷︷ ︸≈0
Vs(x) +
∂Vi∂T
dT +∂2Vi∂T 2
dT 2 + . . .︸ ︷︷ ︸≈0
= G0Vs(x)
∣∣∣∣∣T∼20C
G0Vs(x) +G0∂Vi∂T
dT + Vs(x)dG
dTdT +
∂Vi∂T
dG
dTdT 2︸ ︷︷ ︸
≈0
= G0Vs(x)
⇓
G0∂Vi∂T
dT = −Vs(x)dG
dTdT
dG
dT= − G0
Vs(x)
∂Vi∂T
− R6R22βRT
R5R1(R2 +RT )2T 2= −G0α
∣∣∣∣∣T=20C
(41)
The unknown resistances must fulfill the two equations Eq.39 and Eq.41.We select R1 = 1kΩ, R4 = 1kΩ, R5 = 1kΩ. Then for RT (T = 20C) = we obtain
R6 = f1(R2, R3) from Eq.39 and R6 = f2(R2) from Eq.41.HHHJJJ,,3.8: A given capacitive level sensor consists of two concentric cylinders
with diameter 40mm and 8mm. The storage tank is also cylindrical, 50cmin diameter and 1.2m in height. The stored liquid has εr = 2.1. Calculatethe minimal and maximal capacitance for the sensor, and sensor’s sensitivity[pFL
]when used in the storage tank.
HHHJJJ,,3.9: The circuit in Figure 29 is the signal conditioner for a capaci-
tive level sensor from the previous problem. Design the circuit components(R2, R3, Ve, C, C1) to obtain a voltage that is 0V for the empty tank and 1Vfor the full tank.
Note 1: The circuit is an AC bridge whose output current is converted intoa voltage through the op amp. Because the equivalent output impedance forthe voltage divider that includes the sensor is capacitive, the transimpedanceof the op amp feedback loop must also be capacitive. Hence, R is added onlyto bias the op amp and is of very high value (R >> 1/(ωC)). Do not con-sider R in the design.
Note 2: Applications that do not need to determine the phase of the ACsignal can use one of the three basic methods to obtain a DC voltage froman AC voltage:root mean square (RMS): VRMS = Vpeak/
√2 peak detection mean absolute
value (MAV): VMAV = 2Vpeak/π
44
R2
R3Cx
C1
Ve +
− Vmav
C
R
Vo MAV
Figure 29: Signal conditioner for capacitive level sensor
HHHJJJ,,3.10: Design ultrasonic CW liquid level measurement system shown in
Fig.30. Consider the following parameters:
liquid Castor Oil C11H10O10 ρ = 0.969[ kgL
]
velocity of sound c25C = 1477[ms
] ∆c∆t
= −3.6[ msC ]
liquid level range h ∈< 0, 2.5 > [m]
US transducer f ∈< 50, 200 > [kHz]
emitter - receiver spacing s = 0.05[m]
s
liquid
air
h
RE
Figure 30: US CW level measurement
45
HHHJJJ,,3.11: Using MATLAB, plot the absolute and the relative uncertainty of
the level measurement for h ∈< 0, 2.5 > [m] when the liquid temperaturevaries between 0C and 50C, and when the phase angle measurement bythe circuit in Fig.31 has constant uncertainty of 0.75%π
2[rad] 6. Consider
also the systematic error due spacing s 6= 0.
p
sin(wt)
d/d(wt)
cos(wt)
sin(wt+p) LPF
LPF2x AD 532
(AD 538)
X
Y
arctg(Y/X)
Figure 31: Circuit for phase shift measurement
HHHJJJ,,3.12: Table 1 represents the Temperature-E.M.F. function of J thermo-
couple for temperatures between 0C and 500C by 1C increments. Thetable read-out is referenced to 0C, i.e. “cold junction” at 0C.
(A) Determine the E.M.F.[V] generated by J thermocouple with “hot junc-tion” at 100C and “cold junction” at 36C.(B) Determine approximately the sensitivity SJ [ VC ] of J thermocouplewithin the temperature range of 0C ∼ 100C.
HHHJJJ,,3.13: Cross-correlation flowmeter measures flow rate of a liquid or gas
by determining the time delay τ associated with the peak of the cross-correlation function (Figure 32). Eq.42 represents the sampled data formof the cross-correlation function.
(A) What is the flow rate q = q(τ) [m3/s] ?(B) Describe the difference between point-by-point and evolutionary calcu-lation of the cross-correlation function.
Rxy(0) = x1y1 + x2y2 + x3y3 + . . .+ xnyn
Rxy(1) = x1y2 + x2y3 + x3y4 + . . .+ xnyn+1
Rxy(2) = x1y3 + x2y4 + x3y5 + . . .+ xnyn+2 (42)
...
Rxy(j) = x1yj+1 + x2yj+2 + x3yj+3 + . . .+ xnyn+j
46
C 0 1 2 3 4 5 6 7 8 9 10
0 0.000 0.050 0.101 0.151 0.202 0.253 0.303 0.354 0.405 0.456 0.50710 0.507 0.558 0.609 0.660 0.711 0.762 0.814 0.865 0.916 0.968 1.01920 1.019 1.071 1.122 1.174 1.226 1.277 1.329 1.381 1.433 1.485 1.53730 1.537 1.589 1.641 1.693 1.745 1.797 1.849 1.902 1.954 2.006 2.05940 2.059 2.111 2.164 2.216 2.269 2.322 2.374 2.427 2.480 2.532 2.58550 2.585 2.638 2.691 2.744 2.797 2.850 2.903 2.956 3.009 3.062 3.11660 3.116 3.169 3.222 3.275 3.329 3.382 3.436 3.489 3.543 3.596 3.65070 3.650 3.703 3.757 3.810 3.864 3.918 3.971 4.025 4.079 4.133 4.18780 4.187 4.240 4.294 4.348 4.402 4.456 4.510 4.564 4.618 4.672 4.72690 4.726 4.781 4.835 4.889 4.943 4.997 5.052 5.106 5.160 5.215 5.269100 5.269 5.323 5.378 5.432 5.487 5.541 5.595 5.650 5.705 5.759 5.814110 5.814 5.868 5.923 5.977 6.032 6.087 6.141 6.196 6.251 6.306 6.360120 6.360 6.415 6.470 6.525 6.579 6.634 6.689 6.744 6.799 6.854 6.909130 6.909 6.964 7.019 7.074 7.129 7.184 7.239 7.294 7.349 7.404 7.459140 7.459 7.514 7.569 7.624 7.679 7.734 7.789 7.844 7.900 7.955 8.010150 8.010 8.065 8.120 8.175 8.231 8.286 8.341 8.396 8.452 8.507 8.562160 8.562 8.618 8.673 8.728 8.783 8.839 8.894 8.949 9.005 9.060 9.115170 9.115 9.171 9.226 9.282 9.337 9.392 9.448 9.503 9.559 9.614 9.669180 9.669 9.725 9.780 9.836 9.891 9.947 10.002 10.057 10.113 10.168 10.224190 10.224 10.279 10.335 10.390 10.446 10.501 10.557 10.612 10.668 10.723 10.779200 10.779 10.834 10.890 10.945 11.001 11.056 11.112 11.167 11.223 11.278 11.334210 11.334 11.389 11.445 11.501 11.556 11.612 11.667 11.723 11.778 11.834 11.889220 11.889 11.945 12.000 12.056 12.111 12.167 12.222 12.278 12.334 12.389 12.445230 12.445 12.500 12.556 12.611 12.667 12.722 12.778 12.833 12.889 12.944 13.000240 13.000 13.056 13.111 13.167 13.222 13.278 13.333 13.389 13.444 13.500 13.555250 13.555 13.611 13.666 13.722 13.777 13.833 13.888 13.944 13.999 14.055 14.110260 14.110 14.166 14.221 14.277 14.332 14.388 14.443 14.499 14.554 14.609 14.665270 14.665 14.720 14.776 14.831 14.887 14.942 14.998 15.053 15.109 15.164 15.219280 15.219 15.275 15.330 15.386 15.441 15.496 15.552 15.607 15.663 15.718 15.773290 15.773 15.829 15.884 15.940 15.995 16.050 16.106 16.161 16.216 16.272 16.327300 16.327 16.383 16.438 16.493 16.549 16.604 16.659 16.715 16.770 16.825 16.881310 16.881 16.936 16.991 17.046 17.102 17.157 17.212 17.268 17.323 17.378 17.434320 17.434 17.489 17.544 17.599 17.655 17.710 17.765 17.820 17.876 17.931 17.986330 17.986 18.041 18.097 18.152 18.207 18.262 18.318 18.373 18.428 18.483 18.538340 18.538 18.594 18.649 18.704 18.759 18.814 18.870 18.925 18.980 19.035 19.090350 19.090 19.146 19.201 19.256 19.311 19.366 19.422 19.477 19.532 19.587 19.642360 19.642 19.697 19.753 19.808 19.863 19.918 19.973 20.028 20.083 20.139 20.194370 20.194 20.249 20.304 20.359 20.414 20.469 20.525 20.580 20.635 20.690 20.745380 20.745 20.800 20.855 20.911 20.966 21.021 21.076 21.131 21.186 21.241 21.297390 21.297 21.352 21.407 21.462 21.517 21.572 21.627 21.683 21.738 21.793 21.848400 21.848 21.903 21.958 22.014 22.069 22.124 22.179 22.234 22.289 22.345 22.400410 22.400 22.455 22.510 22.565 22.620 22.676 22.731 22.786 22.841 22.896 22.952420 22.952 23.007 23.062 23.117 23.172 23.228 23.283 23.338 23.393 23.449 23.504430 23.504 23.559 23.614 23.670 23.725 23.780 23.835 23.891 23.946 24.001 24.057440 24.057 24.112 24.167 24.223 24.278 24.333 24.389 24.444 24.499 24.555 24.610450 24.610 24.665 24.721 24.776 24.832 24.887 24.943 24.998 25.053 25.109 25.164460 25.164 25.220 25.275 25.331 25.386 25.442 25.497 25.553 25.608 25.664 25.720470 25.720 25.775 25.831 25.886 25.942 25.998 26.053 26.109 26.165 26.220 26.276480 26.276 26.332 26.387 26.443 26.499 26.555 26.610 26.666 26.722 26.778 26.834490 26.834 26.889 26.945 27.001 27.057 27.113 27.169 27.225 27.281 27.337 27.393C 0 1 2 3 4 5 6 7 8 9 10
Table 1: J thermocouple: Thermoelectric Voltage in [mV]
47
vRxy
τ t[sec]
LX Y
D
Figure 32: Cross-correlation flowmeter
48
A:
q =πD2
4v =
πD2
4
L
τ
[m3
s
]B:
point−by−point
Rxy
τ t[sec]
Rxy
τ t[sec]
evolutionary
49
1.4 Sensors in Manufacturing IndustryHHHJJJ,,4.1: A 1 [kΩ] strain gauge having gauge factor G = 2.0 is attached to
a load cell (E = 100× 103 [MPa]). The cross-sectional area of the load
cell A is 200 [mm2]. Using Eq.43, Eq.44 and Eq.45 calculate the changein resistance ∆R when the load cell supports a 1000 [kg] load. Consider
g.= 10 [ms−2].
∆R
R= G ε (43)
σ = E ε (44)
σ =F
A
[N
m2= Pa
](45)
∆R = RGε = RGσ
E= RG
FA
E= RG
mgA
E=
=RGmg
AE
[Ω 1 kg ms−2
m2Pa=
Ω 1 N
m2Pa=
Ω 1 Pa
Pa= Ω
]R = 1000 [Ω]
G = 2 [1]
m = 1000 [kg]
g.= 10
[ms−2]
E = 100× 109 [Pa]
A = 200× (0.001)2 = 200× 10−6 [m2]∆R =
1000 2 1000 10
200× 10−6 100× 109=
2× 107
2× 107= 1 [Ω]
HHHJJJ,,4.2: A given LVDT has a primary winding with a dc resistance of 67Ω
and two series-opposition connected secondary windings that have 2800Ω intotal. At 1 kHz, the primary has impedance of 290Ω, the series-connectedsecondary windings have 4800Ω, and the sensitivity (normalized to the ex-
citing voltage) is 270[µV/Vµm
] when the load resistance is 500kΩ.
(A) Calculate the respective inductance for the primary and each secondarywinding while assuming the output resistance of the exciting oscillator isnegligible.(B) Calculate the excitation frequency that yields zero phase shift betweenthe primary and secondary voltages when the load resistance is 500kΩ.(C) Calculate the normalizd sensitivity when the LVDT is excited at 60 Hz,first when the load resistance is 500kΩ and then when the load resistance is10kΩ.
Note: Sensitivity: S = |E0/E1|x
50
1.5 Capacitive Sensors
1.5.1 Parallel plates
C =ε0εrS
d[F ] (46)
ε0 = 8.854 × 10−12
εr = relative dielectric constant
S = area
d = spacing
The capacitance with 1[cm2] plate area and 1[mm] spacing is 0.8[pF ].
1.5.2 Disk
The capacitance of a single thin plate with a diameter d[m] to a ground at infinity is
C = 35.4× 10−12εr × d [F ] (47)
A 1[cm] plate in the air has a capacitance of 0.354[pF ]. As conductive objects approachthe plate, this capacitance increases.
1.5.3 Concentric cylinders
The capacitance between two concentric cylinders of length L[m] and radius R1[m]and R2[m] (R2 > R1) is
C =2πε0εrL
ln(R2/R1)[F ] (48)
1.5.4 Parallel cylinders
The capacitance for cylinders of length L[m] and radius R[m] separated by d[m] is
C =πε0εrL
ln
(d+√d2−4R2
2a
) [F ] (49)
1.5.5 Cylinder and plane
The capacitance between a cylinder of length L[m] and radius R[m] located d[m] abovean infinite plane is
C =πε0εrL
ln
(d+√d2−R2
a
) =πε0εrL
acosh(d/R)[F ] (50)
51
1.5.6 Two cylinders and plane
The mutual capacitance between two cylinders of length L[m] and radius R[m] spacedby d[m] is reduced by the proximity of a ground plane dg[m]:
C ≈πε0εrL ln
(1 +
2dgd
)[ln(
2dgR
)]2 [F ] (51)
1.5.7 Interfacing the Sensor
Direct DC
• charging a capacitor Q = CV [C] ≈ constant• very high impedance amplifier
• τ = Ramp inC >> measurement period
• used in electret microphones
RC oscillator using CMOS 555
• grounded capacitor
• 50% duty cycle
• output swings to the power rails (R is large)
• triger points 1/3 and 2/3
• frequency or time counter
• (guarding from stray capacitances can be utilized by using Hi-Z amplifier)
Hi-Z amplifier
• noninverting amplifier as voltage follower
• AC current source
• output is proportional to the impedance of sensor C⇒ linear output with spac-ing variations
• input is usually guarded by a shield connected to the output (floating shield)
Lo-Z amplifier
• noninverted input grounded, AC voltage source
• low input impedance (virtual ground amplifier): Zinput ≈ 1gain×Zfeedback
• input can be guarded by a grounded shield
• the output is proportional to C/Cfeedback or Zfeedback/Z
• can be used to linearize
– area sensors Carea
– spacing sensors Cspacing
52
1.6 Inductive Sensors
1.6.1 Variable self-inductance transducers
• Single coil
• Two-coil (single-coil with center tap) connected for inductance ratio
x
x
Figure 33: Gap variation inductive sensor
L = NΦ
i[H] (52)
N : number of turns [1]
Φ : magneticflux [webers]
i : current [A]
Φ =Fm< [webers] (53)
Fm : magnetomotive force : Fm = iN [At] (54)
< : magnetic reluctance
The reluctance indicates the amount of magnetic flux it links due to an electriccurrent.
53
L =N2
< [H] (55)
For a coil having a cross section A and a length l much longer than its transversedimensions, < is given by:
< =1
µ0µr
l
A+
1
µ0
l0A0
(56)
µr : relative magnetic permeability for the magnetic core
inside the coil
µ0 : magnetic permeability for the air [Hm−1]
l : length of the coil [m]
A : cross− section of the coil [m2]
l0 : length of field lines through the air
(outside the coil) [m]
A0 : the path cross− section [m2]
If the magnetic circuit includes paths through the air and paths through a ferro-magnetic material placed in series:
< =∑ 1
µ0µr
l
A+∑ 1
µ0
l0A0
(57)
1.6.2 Variable mutual-inductance transducers
L2’
L2
L1
eo2
eo1eo
e1 M3
linear range
|eo|
xM2
M1
x
Figure 34: Series-opposition connected LVDT
Linear Variable Differential Transformer (LVDT) Using the equivalentcircuit in Figure 35 the voltage-current relation of the primary winding can be ex-pressed as:
54
M3L1
L2
L2’
x
M1
M2
R1
Rc2
Rc2’i1
i2
E1
E2RL
Figure 35: Equivalent circuit of LVDT
E1 = I1(R1 + sL1) + I2(−sM1 + sM2) (58)
Secondary windings:
0 = I1(−sM1 + sM2) + I2(R2 + sL2 + sL′2 − sM3) (59)
R2 = Rc2 +Rc2′ +RL
(Equation 58) + (Equation 59) =⇒
I2 =s(M2 −M1)E1
s2 [L1(L2 + L′2 − 2M3)− (M2 −M1)2] + s [R2L1 +R1(L2 + L′2 − 2M3)] +R1R2
E0 = I2RL (60)
In the center position M2 = M1 =⇒ e0 = 0[V ]For other core positions, L1, L2, L
′2,M3 and M2 −M1 undergo the following ap-
proximate variations:
dL1
dx,dM3
dx<<
d(M2 −M1)
dx≈ const >>
d(L2 + L′2 − 2M3)
dx
.= 0
L2 + L′2 − 2M3.= 2L2
2L2L1 >> (M2 −M1)2
Then the output voltage is:
55
E0 =s(M2 −M1)E1RL
s22L1L2 + s(R2L1 + 2R1L2) +R1R2=
=s (M2−M1)
2L1L2E1RLs+
R2L1 + 2R1L2
4L1L2︸ ︷︷ ︸τ
2
+[√ R1R2
2L1L2−(R2L1 + 2R1L2
4L1L2
)2]2︸ ︷︷ ︸
ω2nd
ωnd =
√R1R2
2L1L2︸ ︷︷ ︸ωn
√√√√√√1− (R2L1 + 2R1L2)2
8L1L2R1R2︸ ︷︷ ︸ξ2
(61)
If the primary winding is excited at the natural frequency then the output voltagedoes not depend on the excitation frequency and the phase shift between primary andsecondary voltages is zero:
E0 =jω(M2 −M1)E1RL
(jω)22L1L2 + jω(R2L1 + 2R1L2) +R1R2
∣∣∣∣∣ω = ωn
E0 =jωn(M2 −M1)E1RL
(−R1R2 + jωn(R2L1 + 2R1L2) +R1R2
E0 =(M2 −M1)RLR2L1 + 2R1L2
E1 (62)
If there is no load RL →∞:
limRL→∞
E0 =s(M2 −M1)E1
sL1 +R1= s(M2 −M1)I1 (63)
where I1 is primary current.
Resolvers
RESOLVER = rotary transformer with output signal being a function of the rotorposition
Brushless resolvers use a transformer to couple signals from stator to the rotor aremore rugged and reliable than common resolvers with brushes or slip rings.
Vector resolvers = single winding on the rotor acts as a primary and two windingson the stator act as secondaries.
eS13 = KeR13 sinα
eS42 = KeR13 cosα
Electric resolvers = two windings on the rotor at 90 and two windings around thestator, also at 90. Only one of the primary windings (on the rotor or on the stator)is used.
56
α
R1
S2 S4
S3
S1R2
R4
R3
Figure 36: Resolver is a variable transformer with windings at 90 both in statorand in the rotor
eR13 = K(eS13 cosα+ eS24 sinα)
eS42 = K(eS24 cosα− eS13 sinα)
Resolvers can perform calculations:
• coordinate system transformation
– polar to Cartesian: short S2 to S4, in S13(r), out R13(X) and R24(Y)
– Cartesian to polar: feedback from R13(α), out R24(r), in S13(X) andS24(Y)
• Cartesian frame rotation: in S13(X) and S24(Y), out R13(X’) and R24(Y’)
Data transmission resolvers ≈ synchros.
Synchros
SYNCHRO = rotary transformer that converts an angular displacement into an ACvoltage, or an AC voltage into an angular displacement.
eS10 = keR12 cos(α+ 120)
eS20 = keR12 cosα
eS30 = keR12 cos(α− 120)
57
R2
α
S3
S2
S1
R1
Figure 37: Electric circuit for a synchro
eS13 = eS30 − eS10 = KeR12 sinα
eS32 = eS20 − eS30 = KeR12 sin(α+ 120)
eS21 = eS10 − eS20 = KeR12 sin(α+ 240)
Types of synchros:
control transmitter CX, control transformer CT single rotor winding, three star-configuration stator windings 120 apart
control differential CDX three star-configuration windings 120 apart
Applications:
Control synchro angular displacement between two rotating shafts (CX&CT)
adjustment of the angular relationship between two rotating shafts (CX&CDX&CT)
Torque synchro (selsyns) transmission of anglular information (CX&CT)
Inductosyn
Inductosyn = variable transformer for sensing rotary or linear displacement
Inductosyn consists of a precision printed circuit patterns with parallel hairpinturns repeated along a flat bar, or radial hairpin turns on the flat surface of a disk. Asimilar pattern is attached to a ruler sliding on the scale with a small air gap.
Vo = kVe cos 2πx
PP = pitch [mm]
Because of the addition of the flux from each conductor, the output voltage goes:
max aligned conductors
zero P/4 displacement
min P/2 (aligned conductors)
Applications:
• HDD position control
58
Vo
P
Ve
Stator
Ruler
Figure 38: Working principle for the Inductosyn. The ruler slides on the scale,but is shown moved aside for clarity
• NC machine tools
• radar, antenas, radiotelescopes
• scanners
• remote manipulator of the space shuttle
59