MEASUREMENT AND INSTRUMENTATION
BMCC 3743
PC BASED
DATA-ACQUISITION SYSTEMS
Mochamad Safarudin
Faculty of Mechanical Engineering, UTeM
2010
 Measurement process
Sensor/transducer
measurand
Signal
conditioning
Recorder/display/
processor
 Analogue signal conditioning - done
 DIGITAL SIGNAL CONDITIONING
2
ANALOG AND DIGITAL
•Most measurands originate in analog form
•Analog signal varies smoothly in time, without discontinuty
•Example: 220 V ac, 60 Hz power line voltage
Example of analog signal
•Digital information is transmitted and processed in form of bits
•Each bit defined by one or other of two predefined “logic level”
•The time interval assigned to it called bit interval
•Most common two logic states is predetermined voltage levels
(say 0 and 5 V dc)
3
Why digital?
1.
Digital electronics easier to design
and fabricate
ex: IC, low cost, mass product
compare to capacitor etc
2. Ease of data recording, storage and
display
ex: digital voltmeter provides a
direct numerical display
of voltage compared with analog
voltage that has to be
visually interpolated if the pointer
is between two scales
3. Inherently noise resistant
4
COMPUTER AS A
MEASUREMENT SYSTEM
5
Contents
 Components of computer
systems
 Representing numbers in computer
systems
 Components of data-acquisition
systems
 Configuration of data-acquisition
systems
6
Components of computer systems
Display
Digital
input-output
(ports or expansion bus)
Printer
CPU and RAM
Mass storage
(disk drives)
Keyboard
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COMPUTERIZED DATA ACQUISITION SYSTEM
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Typical computer components




Central processing unit (CPU)
Program (software)
Random access memory (RAM) - ROM
Mass storage system – magnetic tape recorder,
magnetic disk drive, optical disk drive
 Display/monitor/screen
 User input device (keyboard, mouse,
joystick,etc)
 Printers and plotters
9
Contents
 Components of computer systems
 Representing numbers in
computer systems
 Components of data-acquisition
systems
 Configuration of data-acquisition
systems
10
Representing numbers in computer
systems
 Computers use bistable flip-flops to store
information, which have only 2 possible
states: on (1) or off (0)
E.g. 1001 2
MSB
LSB
 1 byte = 8 bits
4 bit binary number
MSB:Most Significant Bit
LSB: Least Significant Bit
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Examples (binary/decimal)
1. Convert the 8-bit binary number
01011100 to decimal
2. Find the 8-bit binary number with the
same value as that of the decimal
number 92.
12
1. 01011100
N10=0(27)+1(26)+0(25)+1(24)+1(23)+2(22)+0(21)+0(20)
=0+64+0+16+8+4+0+0
=92
2. By a series of divisions by 2
remainder
2
92
LSB
2
46
0
2
23
0
2
11
1
2
5
1
2
2
1
2
1
0
MSB
0
1
Answer:
1011100 but we
are asked for 8 bit:
01011100
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What about negative number?
Most commonly represented using:
2’s complement binary
Procedure:
1. Convert the integer to binary as if it were positive
2. Invert all of the bits – change 0’s to 1’s and 1’s to 0’
3. Add 1 LSB to the final result
e.g. convert –92 to an 8-bit 2’s
complement binary number
answer: from previous, 01011100
invert
10100011
+1 LSB
101000112 + 12 become 10100100

Note that, positive numbers always have 0 as MSB and negative
numbers have 1 as MSB

In a computer often a special code is used : ASCII – American
Standard Code for Information Interchange, e.g. k = 011010112 =
10710
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ASCII Characters
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Contents
 Components of computer systems
 Representing numbers in computer
systems
 Components of dataacquisition systems
 Configuration of data-acquisition
systems
16
Components of data-acquisition
systems
 Multiplexer
 Simultaneous sample-and-hold subsystem
 ADCs
 DACs
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Multiplexer (MUX)
 Works as an electronic switch – computer
will ask MUX to select a particular channel
to be read and processed, sequentially.
 Can have crosstalk errors and transfer
accuracy.
Illustration of a multiplexer
In this figure, channel 1 is connected
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Simultaneous sample-and-hold subsystem
 Need to be used to record data from different channel of MUX, precisely at
the same time.
e.g. Measuring tire forces using 6 component force transducers
simultaneously
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Analogue-to-Digital Converters
 Converts continuous analogue waveform
into discrete digital signals
 Examples: audio amplifiers, TV, output
voltage from transducers, etc
 Output of ADCs has 2N possible values
 If N , no. of possible output states , hence
results more accurate
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Types of ADCs
 Unipolar single-slope integrating converter
(ramp type – quite slow, not very accurate)
 Successive-approximations converter (quite fast
– typical 12-bit completes a conversion in 10 –
25 μs)
 Parallel or flash or half-flash converter (the
fastest – can be 10 ns, using lots of
comparators)
 Dual-slope integrating converter (used in digital
voltmeter)
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Unipolar single-slope integrating converter
1.
2.
3.
4.
5.
6.
A fixed reference voltage is used to charge an integrator at a constant rate
The integrator output voltage then increase linearly with time
A digital clock (counter) is started at the same time that the charging is begun
The integrator output voltage is compared continuously with the analog input
voltage using a comparator
When the integrator voltage exceeds the analog input voltage, digital clock is
stopped
The count of the digital clock is the digital output of the A/D converter
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Example
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Formula to estimate A/D converter digital output
 The output of a 2’s-complement, given the analogue input voltage, is
 Vi  Vrl N  2 N
Do  int
2 
Vru  Vrl
 2
where max. positive output is (2N/2 –1) and max. negative output is
(-2N/2)
 The output of an offset binary or simple binary converter is given by
 V V

Do  int i rl 2 N 
Vru  Vrl

where output will range from 0 to (2N-1) max.
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Example:
From example before, estimate the digital output for
6.115 V analog input to A/D converter
Answer:
Since this is a simple binary devices the second equation
is applicable:
 6.115 0 4 
Do  int
x2   int(9.78)  10
 10  0

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Quantisation error
 Resolution uncertainty (or treated as random
error, analogous to the reading error of a
digital display) due to output of ADC with
discrete steps, given by
Input resolution error =
Vru  Vrl
 0.5
volts
N
2
 The quantisation error is thus ±0.5 LSB
26
Successive-approximations converter (most common type)
1.
2.
3.
4.
A series of known analog voltages are created and compared to the analog input
voltage
In the first trial, a voltage interval of one-half the input span is compared with
the input voltage
If the input voltage is in the upper half of the range, the MSB is set to1; otherwise
it is set to zero
This process is repeated with an interval half the width of the interval used in the
first trail to determine the second MSB and so forth until LSB is determined
Successive aproximation
method for 4 bit A/D
converter
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Example:
28
Example:
A 12-bit A/D converter has an input range of -10 to +10 V.
Find the resolution error of the converter for the analog input.
Answer:
Using above equation
10  (10) 
input resolutionerror  0.5
  0.00244
212

The resolution uncertainty of ±0.00244 is the best that
can be achieved
Comment: if input voltage=0.1 V (low end of input range),
The quantization error would represent 2.5% of the reading,
which is probably not acceptable. The input signal should be
amplified probably before the signal enters the converter
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Digital-to-Analogue Converters
 Converts discrete digital signals into
continuous analogue waveform
 Examples: To operate heaters or valves
under computer control
 Similar specs as ADCs, i.e. depends on
no. of input bits, analogue output range
and conversion speed.
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4 bit D/A converter
1. Rn=2nRf
2. When the switched is closed, in flows to the summing
bus
vR
vR
in 
 n
Rn 2 R f
3. The op-amp converts the currents to voltages
k
vo   R f  in
n 1
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Example:
A digital code 1011 (equivalent to 11) for the circuit above
with Rf= 5 kW and vs=-10 V. then
i1=-1 mA
i2=0
i3=-1/4 mA
i4=-1/8 mA
Summing these currents and multiplying by Rf gives
Vo=6.875 V which is 11/16 of the full scale (ref) voltage
32
Contents
 Components of computer systems
 Representing numbers in computer
systems
 Components of data-acquisition
systems
 Configuration of dataacquisition systems
33
Configuration of data-acquisition
systems
Several configuration of DAS
 Internal single board plug in system (PCI slot)
 External system (USB, IEEE1394, Ethernet,
RS-232, GPIB)
 Virtual Instruments (modular) ex: Labview
from NI
 PC turns into digital storage oscilloscope
 Data loggers (simpler and specialized)
ex: flight data recorder in airplane
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INTERNAL SINGLE BOARD PLUG IN SYSTEM
PCI DAQ
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EXTERNAL SYSTEM
USB DAQ from NI
36
VIRTUAL INSTRUMENTS
LABVIEW from NI
37
DATA LOGGER
38
EXAMPLE:
How many bits are required for a digital device to represent the decimal
number 27541 in simple binary? How many bits for 2’s complement binary?
Solution:
The following table presents the maximum decimal number versus the number of bits for simple
binary:
No. Bits
Max. Dec. No. Simple Binary
12
212 -1 = 4095
13
213 -1 = 8191
14
214 -1 = 16383
15
215 -1 = 32767
16
216 -1 = 65535
Consequently, 15 bits are needed to represent 27541 in simple binary. For a two's complement
binary number, the MSB will be zero so 16 bits will be required.
39
An A/D converter is to operate with a full-scale voltage of 10V. How many
bits should be employed to obtain a resolution of 0.01 %?
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A 12-bit A/D converter has an input range of ± 8 V, and the output code is
offset binary. Find the output (in decimal) if the input is:
a. 4.2 V
b. -5.7 V
c. 10.9 V
d. -8.5
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Solution:
N = 12
Vru = 8V
Vrl = -8V
Vin = input voltage
(a) By second equation
 V  Vrl  N 
D0  int  in
2 

 Vru  Vrl  


 4.2   8 12 

 int 
2 
 8   8



 int3123.2
 3123
(b) By second equation :

 5.7   8 12 

D0  int 
2 


 8   8

 int588.8
 589
(c) Since 10.9V falls outside the input range, Do will have the maximum output:
D0  212  1  4095
(d) Since -8.5V falls outside the input range, D0 will take the minimum value:
D0  0
42
A 12 bit A/D converter has an input range of ± 10V and an amplifier at the
Input with a gain of 10. The output of the A/D converter is in 2’s complement
Format. Find the output of the A/D converter if the input to the amplifier is:
a. 1.5 V
b. 0.8 V
c. -1.5 V
d. -0.8 V
43
We need the first equation to solve this problem
(a) When the 1.5V signal is amplified with a gain of 10, it becomes 15V which exceeds the input range
of the A/D converter (it is saturated). According to the formula, the maximum output is 2N/2-1 = 212/2-1
= 2047
(b) With the gain of 10, the input becomes 8V. The output, in decimal, is then:
Vin  Vrl N  2N
 8  ( 10) 12  212
Do  int 
2 
 int 
2 
 1638
10  ( 10)  2
Vru  Vrl
 2
(c) When amplified, -1.5V results in an input to the A/D converter which is
below the input range (it is saturated). The largest negative output is –2N/2 = 2048
(d)With the amplifier, this voltage results in an input to the A/D of –8V. The
output is then:
V V
 2N
  8  (10) 12  212
Do  int in rl 2 N  
 int
2 
 1638
 10  (10)
 2
Vru  Vrl
 2
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A 3.29 V signal is input to a 12 bit successive approximations converter with
an input range of 0 to 10 V and simple binary output. Simulate the successive
approximation process to determine the simple binary output
45
The reference voltage increment is:
 input span   10 
V  
   12   0.0024414

 2 
2N
Trial digital output (D0)
D0V
Pass/Fail
Actual digital output
100000000000
(2048)
5.0
F
010000000000
(1024)
2.5
P
010000000000
(1024)
011000000000
(1536)
3.75
F
010000000000
(1024)
010100000000
(1280)
3.12
P
010100000000
(1280)
010110000000
(1408)
3.44
F
010100000000
(1280)
010101000000
(1344)
3.28
P
010101000000
(1344)
010101100000
(1376)
3.36
F
010101000000
(1344)
010101010000
(1360)
3.32
F
010101000000
(1344)
010101001000
(1352)
3.30
F
010101000000
(1344)
010101000100
(1348)
3.2901
F
010101000000
(1344)
010101000010
(1346)
3.286
P
010101000010
(1346)
010101000011
(1347)
3.289
P
010101000011
(1347)
The output is 010101000011 or 1347 in decimal.
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Thank You
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