TMMS04
Lesson 1 – Sensors and Characteristics
2014 HT-1
1. A linear position sensor based on a potentiometer is shown in Figure 1. This potentiometer has a
total resistance of 100Ω where the wiper (middle connector) can travel in a range of [5, 95]%. The
mechanical measurement range of this sensor is [1.8, 3.0]m. The sensor is connected to the data
acquisition box with four wires. Each wire has a resistance of 1.7Ω/m and is 5m long. The two
connectors of the differential analog input (measures the difference of the voltages on the inputs)
have infinite input resistance thus no current will flow at this connectors. The data acquisition
box is exciting the sensor with 10V DC.
Figure 1: Linear position sensor (potentiometer)
(a) Draw a simple connection diagram showing the potentiometer, the resistors of the wires, and
the data acquisition box ports.
Solution:
(b) Calculate the sensor gain Ks in V/m.
Solution:
Rw = 8.5Ω
0.95 − 0.05
Rp1 = Rp tot 0.05 +
(x − 1.8m)
3.0m − 1.8m
Rp1
ua = Ue
2Rw + Rp tot
∂
Ks =
ua
∂x
1
0.95 − 0.05
= Ue
2Rw + Rp tot 3.0m − 1.8m
= 6.41V/m
The sensor gain is Ks = 6.41V/m
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TMMS04
Lesson 1 – Sensors and Characteristics
2014 HT-1
(c) Will the sensor gain be affected if the supply voltage used for exciting the sensor varies?
Solution: Yes. It will vary linearly with the supply voltage.
(d) What will the output signal be when a position 2.3m is measured?
Solution: 3.63V
(e) The data acquisition box has a input range of [0.0, 10.0]V and a resolution of 1mV. Which
resolution (distance) may this sensor system have in the best case? How many distinct
position can be measured?
Solution:
0.001V
= 0.156mm.
Ks
The resolution is 0.156mm.
10.0V − 0.0V
+ 1 = 10001
0.001V
This system can distinguish 10001 position.
2. A “HEDS 5540E” optical quadrature encoder is used to measure a linear motion. The encoder is
attached to a rack and pinion (a gear wheel that rotates with the linear motion). The diameter
of the wheel is 100mm and the gear ratio is 1:1.
(a) If the linear motion is 200mm long, how many pulses will the encoder give?
Solution:
Data sheet: Pulses per revolution : N = 200
0.200m
Number of revolutions: nr = 2π0.100m/2
Pulses: N · nr = 127.3pulses
Depending on counting and start position: 127 or 128 pulses. There are 127 rising or 127
falling edges.
(b) What is the system’s resolution (distance) of the translational motion? Hint: Can the data
acquisition box know the position within the pulse distance?
Solution:
dpp = Circumference
= 2π·0.100m/2
= 1.57mm
N
200
This sensor has two channels with 90 deg phase shift. So the achievable resolution is 1/4
pulse and thus 0.393mm.
(c) What is the velocity limit for this system? What are the requirement on the data acquisition
box to allow this velocity?
Solution:
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TMMS04
Lesson 1 – Sensors and Characteristics
2014 HT-1
Data sheet: fmax = 100kHz
dpp · fmax = 1.57mm · 100kHz = 157m/s
The maximum velocity is 157m/s.
The data acquisition box has to be able to receive optical quadrature signals which a
frequency of at least 100kHz (pulses).
Note: Think about a systems with high resolution, e.g. 1e − 6m and less.
3. Measurement results from a pressure transducer IDA354-1,5C-10V are shown in Figure 2. The
transducer exhibits a non-linear behavior.
10
9
8
7
Volt
6
5
4
3
2
1
0
0
20
40
60
80
bar
100
120
140
160
Figure 2: The pressure transducer measurement showing the nonlinear characteristic.
(a) By approximating the curve with a straight line an approximate linear voltage/pressure
relationship can be derived. Determine the value.
1 V
V
10V
= 16
Solution: ≈ 160bar
bar = 0.0625 bar
There are various ways how to place this straight line. Here we used approximately the
method of minimizing the highest deviation, L∞ norm, whereby the value at 0bar is fixed
to 0V.
(b) Find the largest deviation (due to non-linearity) from the ideal transducer characteristics. Is
it within the limits promised by the data-sheet?
1.1V
Solution: Highest deviation: 1.1 V.
10V = 11%
According to the data sheet this sensor shall have an accuracy of ±0.5% of the full scale.
Thus this sensor seems to be faulty.
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TMMS04
Lesson 1 – Sensors and Characteristics
2014 HT-1
(c) In your application it is necessary to measure pressure up to 350 bar, select a suitable transducer IDA***-***-*** for that range. What is the maximum error due to nonlinearities or
hysteresis?
Solution: IDA 3**-3,5C-**
emax = 0.005 · 350bar = ±1.75bar
(d) In your application it is necessary to measure pressure up to 30 bar, select a suitable transducer IDA***-***-*** for that range. What is the maximum error due to nonlinearities or
hysteresis?
Solution: IDA 37*-35-**
emax = 0.01 · 35bar ± 0.35bar
4. Sensors and actuator may have a deadband or neutral zone.
Input
Output
1
Value []
0.5
0
-0.5
-1
0
0.1
0.2
0.3
0.4
0.5
Time [s]
Figure 3: Actuator dead-band
(a) What are the dead-band limits in the actuator response shown in figure 3.
Solution: −0.2
+ 0.4
(b) How could this be compensated in the controller generating the input signal to the actuator?
Can this rise any problems?
Solution:
if (Uref > 0)
Uc = Uref + 0.4
else if (Uref < 0)
Uc = Uref - 0.2
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TMMS04
Lesson 1 – Sensors and Characteristics
2014 HT-1
else
Uc = Uref
end
This compensation will generate a step when crossing zero. Even minimal noise around
zero will generate a big change in the input signal to the actuator and thus in the actuator.
This compensation may destabilize the system.
5. Bias (offset) in a sensor signal can cause problems if not handled correctly. In this example,
figure 4, a robot arm is rotating between four stations making a very brief stop at each one.
Measurements from an angular velocity sensor (gyro), figure 5, are integrated to determine the
absolute robot arm angle, however, a bias is affecting the measurement.
500
Integrated Angle [degrees]
400
300
200
100
0
Biased integration
Un−biased integration
−100
0
2
(a)
4
Time [s]
6
8
(b)
Figure 4: (a) A robot arm rotates between four stations. (b) Comparison between biased and unbiased
integrated measured angular velocity, (absolute angle).
(a) Find the seemingly constant bias value
Signal with bias
Angular Velocity [rad/s]
1.5
1
0.5
0
0
Solution:
b = 0.2rad/s
1
2
3
4
Time [s]
5
6
7
8
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TMMS04
Lesson 1 – Sensors and Characteristics
2014 HT-1
Signal with bias
Angular Velocity [rad/s]
1.5
1
0.5
0
0
1
2
3
4
Time [s]
5
6
7
8
Figure 5: Angular velocity signal with an unwanted bias
(b) Find the actual angular velocity motion frequency and determine the signal function. Subtract
the bias “in your head”. Hint! The actual signal amplitude is π4 .
Solution: A = π4 (≈ 0.75, difficult to see in plot)
Function: F (t) = π4 (1 − cos (2π · 0.5 · t))
f = 0.5 Hz
(c) Approximately calculate the integrate angle values (with and without bias) at t = 2s and at
t = 5s. Comment on the deviation between true and biased results. Why are they different
at the two time instances?
Solution: With bias:h
R Ti
0 F (t) + b (t) dt = b t +
t : 2 ⇒ 113◦
π
4
t−
t : 5 ⇒ 282◦
Without bias:
h Rt
π
f
(t)
dt
=
0
4 t−
1
π
sin (πt)
i2;5
0
1
π
sin (πt)
i2;5
0
t : 2 ⇒ 90◦
t : 5 ⇒ 225◦
deviation t : 2 ⇒ 22.9◦
t : 5 ⇒ 57.3◦
◦
This makes 11.5 /s as the deviation is increasing linearly with time.
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TMMS04
Lesson 1 – Sensors and Characteristics
2014 HT-1
6. Comprehension questions
(a) Potentiometer:
Why are four wires used to connect the sensor? What are the advantages to use four wires
compared to use three, two or . . . wires to connect the sensor? What is the minimum amount
of wires to connect the sensor to make a reading possible?
For further information: This thought are relevant when connecting strain gauge, too.
Solution: The wiring may disturb the sensor reading by adding probably unknown resistors to the system. Using more wires enables partial or full cancellation of these
disturbances. Five wire wiring allow full cancellation of the disturbance by wire resistance. With four wires the assumption that the two wires for excitation have the same
resistance has to be made to fully compensate these disturbances. Ue = Ue∗ + 2 uAin−
where Ue∗ = 10V
Two wires are the minimum to measure the resistance and thus determine the position.
(b) Quadrature Encoder: Which problem may arise at low speed?
Solution: At low speed the resolution of the position may become a problem especially
for determining the speed as there may happen no or only a few position increments in
the considered time period.
(c) Quadrature Encoder: Explain two methods of determining the speed / velocity from a quadrature encoder signal. What are the advantages of each method?
Solution:
Method 1:
Measure the position increments (¼ pulses) in one time period.
This is the usual method.
Method 2:
Measure the time for one position increment (¼ pulses).
This method is more accurate at low speed.
(d) Which characteristic does an integrator have so bias may become an infinite problem?
Solution: An integrator is not a stable system.
The pole of the integrators transfer function is at the imaginary axis. Thus it is only
marginal stable and not stable. Thus derivations will not decline.
(e) How does an offset influence the result when differentiating a signal with offset?
Solution:
(f) How does noise influence the result when integrating / differentiating a signal with high
frequency noise?
Solution: Integration is acting like a low-pass filter and thus improving the signal to
noise ratio.
Differentiation is acting like a high-pass filter and thus worsen the signal to noise ratio.
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