11/3/2014
DC Circuits
Transducers
A transducer is any device used to convert a
physical phenomenon into an electrical signal
(e.g., microphone, thermocouple, and voltmeter)
Characteristics of measurement design:
• Sensitivity
• Linearity
• Accuracy
• Precision
• Stability
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DC Circuits
Resistance Temperature Detectors (RTDs)
Make use of changes in their resistance to determine the changes in
temperature.
Example
A resistance temperature detector (RTD) that is not perfectly linear is
used for a temperature measurement. The temperature coefficient is
3.900 × 10–3ºC–1, the reference temperature is 0ºC, and the
reference resistance is 500.0 . The resistance measured when the
actual temperature is 400ºC is 1247 . Determine the error in the
temperature measurement.
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DC Circuits
Resistance Temperature Detectors (RTDs)—Example
Solution
If the RTD was perfectly linear the
resistance would be given by


RT  R0 1 (T T0 )
However, the temperature that the RTD indicates is not the actual
temperature of 400ºC, so rearranging the RTD equation to solve for the
yields
temperature that the RTD indicates
T


RT  R0  R0T0
R0
1247   500.0   (3.900 103 C1)(500C)(0C)
(3.900 103 C1)(500)
 383.1C
The error in the measurement is:
Error =
383.1ºC – 400ºC = –16.9ºC

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DC Circuits
Strain Gages
Metal or semiconductor foils that change resistance linearly with the
strain.
R
GF  R
L
L
Example
A strain gage is measured to determine the gage
factor. A strain gage with an initial resistance of
200.00  and final resistance of 199.79  when

subjected to a strain that causes the gage to
compress to 0.9994 cm. The initial length of the
gage was 1.0000 cm. What is the gage factor?
Solution

 1.75
199.79 200.00
200.00
0.9994 1.0000
1.0000
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(A) 0.15
(B) 0.42
(C) 1.8
(D) 4.0
Answer is (C)
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11/3/2014
DC Circuits
Wheatstone Bridges
Balanced
Quarter-bridge
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DC Circuits
Wheatstone Bridges—Example
There are three high-precision resistors known to be 10.00 k in a
quarter bridge circuit. R1 is a sensor with a small resistance difference
from 10 k. Find the resistance if Vin = 5.00 V and Vo = 0.03 V.
Solution
R 
4RVo
Vin

(4)(10.00)(0.03)
5.00
 0.24 k

R1  10.00 k 0.24   10.24 k
For the strain
 gage
 quarter-bridge circuit, R
can be substituted.

1
Vo  (GF)Vin
4
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DC Circuits
Sampling
Sampling Rate or
Frequency:
Shannon’s Sampling Theorem
Determines the sampling rate to reproduce accurately in the discrete
time system.
(where fI is the
frequency of interest)
Nyquist Rate:
Reproducible Sampling:
fs  fn

[reproducible sampling]
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DC Circuits
Sampling—Example
An analog signal is to be sampled at
0.03 s intervals. What is most
nearly the highest frequency that can
be accurately reproduced?
(A) 4.0 × 106 Hz
(B) 12 × 106 Hz
(C) 16 × 106 Hz
(D) 18 × 106 Hz
Solution
The sampling frequency is
fs 
1
1

t
0.03 106
 33 106 Hz
The sampling frequency must be greater than the Nyquist rate for
accurate reproduction.

 f  2f

s
N
The greatest frequency that can be reproduced at this sampling rate is
f 33 106 Hz
fN  s 
 16.7 106 Hz (16 106 Hz)
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Answer is (C)
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DC Circuits
Analog-to-Digital Conversion
Voltage Resolution of n bit representation
The range from a high voltage, VH, and a low voltage, VL, is divided up
into the 2n ranges.
For example, if all the bits are “1” then the analog value is somewhere
between VH and VH – V. To calculate the analog value from the digital
value use
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DC Circuits
Analog-to-Digital Conversion—Example
A 16 bit analog-to-digital conversion has a resolution of 1.52588 × 10–4 V
and the lowest voltage measured has half the magnitude of the highest
voltage. Both the high and low voltages are positive. Determine the
highest voltage.
Solution
VH VL  2n v
 (2)16 (1.52588 104 V)
 10.0 V
The problem
statement also says that

VH  2VL
Since VH and V
L are positive, this equation becomes

VH = 2VL
Substituting into the first equation
2VL – VL = 10 V
VL = 10 V
VH = 20 V
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11/3/2014
DC Circuits
Measurement Uncertainty
Kline-McClintock Equation
A method for estimating the uncertainty in a function that
depends on more than one measurement.
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DC Circuits
Measurement Uncertainty—Example
A function is given by
R  5x1  3x22 .
Find the measurement uncertainty at (1, 2) if the uncertainty in the
variables is ±0.01 and ±0.03 respectively.
Solution

R
R
5
 6x2
x1
x2
At the point (1, 2) the partial derivatives are

R
5
x1 (1,2)
R
 (6)(2)  12
x2 (1,2)

 R   R 
wR  


w1 x
  
w2 x

1 (1,2 ) 
2 (1,2 ) 


2
2

(0.01)(5)  (0.03)(12)
2
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

2
 0.36
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11/3/2014
DC Circuits
Measurement Uncertainty
If the function R is the sum of the measurements,
R = x1 + x2 + x3 + … + xn, then the Kline-McClintock method reduces
to
wR  w12  w22 
 wn2
This is called the root sum square (RSS) value.

If the function R is the sum of the measurements times constants,
R = a1x1 + a2x2 + a3x3 + … + anxn, then the Kline-McClintock method
reduces to
wR  a12w12  a22w22 
 an2wn2
This is called a weighted RSS value.

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