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lesson2et438a.pptx
LESSON 2: PERFORMANCE OF
CONTROL SYSTEMS
1
ET 438a
Automatic Control Systems Technology
LEARNING OBJECTIVES
After this presentation you will be able to:




Explain what constitutes good control system
performance.
Identify controlled, uncontrolled, and unstable
control system response.
Analyze measurement error in measurement
sensors.
Determine sensor response.
Apply significant digits and basic statistics to
analyze measurements.
lesson2et438a.pptx

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CONTROL SYSTEM PERFORMANCE
System control variable changes over time so error
changes with time.
lesson2et438a.pptx
E(t) = R - C(t)
Where E(t) = error as a function of time
R = setpoint (reference) value
C(t) = control variable as a function of time
Determine performance criteria for adequate control
system performance. System should maintain
desired output as closely as possible when subjected
to disturbances and other changes
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CONTROL SYSTEM OBJECTIVES
1.) System error minimized. E(t) = 0 after changes or
disturbances after some finite time.
lesson2et438a.pptx
2.) Control variable, c(t), stable after changes or
disturbances after some finite time interval
Stability Types
Steady-state regulation : E(t) = 0 or within tolerances
Transient regulation - how does system perform under
change in reference (tracking)
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TYPES OF SYSTEM RESPONSE
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250
200
Control Variable
lesson2et438a.pptx
2
150
100
C( t ) 100
50
1
0
50
0
10
20
30
40
50
t
Time (Seconds)
1.)
Uncontrolled process
2.) Process control activated
3.) Unstable system
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TYPES OF SYSTEM RESPONSE
Damped response
Control R esponses
120
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R2
Control Variable
100
80
R1
60
td
40
Setpoint
Change
20
0
0
5
10
15
20
25
Time (Seconds)
30
35
40
45
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Control variable requires time to reach final value
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TYPES OF SYSTEM RESPONSE-TRANSIENT
RESPONSES
Setpoint Change
Response To Setpoint Changes
120
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Control Outputs
100
80
60
40
20
0
5
10
15
20
25
Time (Sec)
30
35
40
45
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Ideal
Typical
TYPES OF SYSTEM RESPONSE-TRANSIENT
RESPONSES
Disturbance Rejection
Response to Disturbance
120
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100
Control Variable
SP
80
60
40
20
0
0
5
10
15
20
25
Time (seconds)
30
35
40
45
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ANALOG MEASUREMENT ERRORS AND
CONTROL SYSTEMS
Amount of error determines system accuracy
Determing accuracy
Accuracy ± 5%
FS =10V
E=10V(± 5%/100)
E= ± 0.5 V
Percent of Span
Span=max-min
Span∙(%/100)
Percent of Reading
Reading∙(%/100)
Accuracy ± 3%
Accuracy ± 2%
min=20, max= 50 psi
E=(50-20)(± 3%/100)
E= ± 0.9 psi
Reading = 2 V
E=(2V)(± 2%/100)
E= ± 0.04 V
lesson2et438a.pptx
Percent of Full
Scale (FS)
FS∙(%/100)
Measured Value
Reading ± Value
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100 psi
± 2 psi
SYSTEM ACCURACY AND CUMULATIVE
ERROR
Subsystem errors accumulate and determine accuracy limits.
Consider a measurement system
K±DK
Sensor
G±DG
V±DV
Sensor amplifier
lesson2et438a.pptx
Cm
K = sensor gain
G = amplifier gain
V = sensor output voltage,
DG, DV, DK uncertainties in measurement
What is magnitude of DV?
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SYSTEM ACCURACY AND CUMULATIVE
ERROR
Output
With no uncertainty:
Input
V = K∙G∙Cm
Multiple out and simplify to get:
Where :
Component
Tolerance/100
lesson2et438a.pptx
With uncertainty V±DV = (K±DK)∙(G±DG)∙Cm
DV DG DK


V
G
K
DV
 normalized uncertaint y of output
V
DG
 normalized uncertaint y of sensor amp
G
DK
 normalized uncertaint y of sensor
K
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COMBINING ERRORS
Use Root-Mean-Square (RMS) or Root Sum Square (RSS)
2
2
This relationship works on all formulas that include only multiplication
and division.
Notes:
lesson2et438a.pptx
DV
 DG   DK 
 
 

V
 G   K 
DV
is fractional RMS uncertaint y. Multiplyby 100 to get %
V
DG DK
,
are fraction uncertaint y. Divide tolerances by 100%
G
K
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CUMULATIVE ERROR EXAMPLE
Example 2-1: Determine the RMS error (uncertainty) of the OP AMP
circuit shown. The resistors Rf and Rin have tolerances of 5%. The input
voltage Vin has a measurement tolerance of 2%.
Rf
Vo
Determine uncertainty from
tolerances
 DR f

 Rf
  Rf
Vo  Vin  
 R in
  DR in   5%
  
 
 0.05
  R in  100
 DVin   2%

 
 0.02
 Vin  100
 DVo 



 Vo  RMS
 DR f

 Rf
2



lesson2et438a.pptx
Vin
Rin
2
  DR in   DVin 
  
  

  R in   Vin 
2
 DVo 


  0.052  0.052  0.022  0.073
 Vo  RMS
 DV 
%U  100%   o 
 7.3%
 Vo  RMS
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ANS
SENSOR CHARACTERISTICS
Sensitivity - Change in output for change in input.
Equals the slope of I/O curve in linear device.
Resolution - Smallest measurement a sensor can
make.
lesson2et438a.pptx
Hysteresis - output different for increasing or
decreasing input.
Linearity - How close is the I/O relationship to a
straight line.
Cm = m∙C + C0
Where C = control variable
m = slope
C0 = offset (y intercept)
Cm = sensor output
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SENSOR SENSITIVITY EXAMPLE
Find a sensors sensitivity using data two points. Use
point-slope equations to find line parameters
y  y1  m  x  x1 
y 2  y1
x 2  x1
Example 2-2: Temperature sensor has a linear resistance change of 100
to 195 ohms as temperature changes from 20 - 120 C. Find the sensor I/O
relationship
Define points :
lesson2et438a.pptx
m
Where:
(x1, y1) = (20 C, 100 W) x = input y = output
(x2, y2) = (120 C, 195 W)
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SENSOR SENSITIVITY EXAMPLE (2)
Compute the slope and the equations:
m
195  100 W 95 W

 0.95 W / C
120  20 C 100 C
y  0.95 W / C  x  0.95 W / C  (20 C )  100 W
y  0.95  x  81 W / C
y  100 W  0.95 W / C  x  20 C 
lesson2et438a.pptx
m
y 2  y1
x 2  x1
ANS
Can plot above equation to check results
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SENSOR RESPONSE
bf
Cf
bi
Ci
lesson2et438a.pptx
Measured Control
Variable
Ideal first-order response-ideal
No time delay
Time
Step change in the measured variable- instantly changes value.
Practical sensors exhibit a time delay before reaching the new
value.
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PRACTICAL SENSOR RESPONSE
Let b(t) = sensor response function with respect to time.
Sensor Responses
80
60
Control Variable
bf
Step
Decrease
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lesson2et438a.pptx
Step
Increase
bi
20
bf
bi
0
0
5
10
15
20
25
Time (Seconds)
30
35
40
45
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MODELING 1ST-ORDER SENSOR RESPONSE
For step increase:
For step decrease:
bf = final sensor value
bi = initial sensor value
t = time
t = time constant of sensor
lesson2et438a.pptx
Where
t


b( t )  bi  b f  bi   1  e t 


 t 
b( t )  bi  b f    e t 
 
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SENSOR RESPONSE EXAMPLE 1
lesson2et438a.pptx
Example 2-3: A control loop sensor detects a step
increase and has an initial voltage output of bi = 2.0 V Its
final output is bf = 4.0 V. It has a time constant of t =
0.0025 /s. Find the time it takes to reach 90% of its final
value.
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EXAMPLE 2-3 CONTINUED (2)
Complete the calculations to find t
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SENSOR RESPONSE EXAMPLE 2
lesson2et438a.pptx
Example 2-4: A sensor with a first order response
characteristic has initial output of 1.0 V. How long does it take
to decrease to 0.2 V if the time constant of the sensor is 0.1/s.
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SIGNIFICANT DIGITS IN
INSTRUMENTATION AND CONTROL
Readable output of instruments
Resolution of sensors and transducers
lesson2et438a.pptx
Significant Digits In
Measurement
Calculations Using
Measurements
Truncate calculator answers to match
significant digits of measurements and
readings,
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SIGNIFICANT DIGIT EXAMPLES
Example 2-5: Compute power based on the following
measured values. Use correct number of significant digits.
3 significant digits
4 significant digits
P = V∙I=(3.25 A)∙(117.8 V) = 382.85 W
lesson2et438a.pptx
3.25 A
117.8 V
Truncate to 3 significant digits P = 383 W
Significant digits not factor in design calculations. Device values
assumed to have no uncertainty.
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SIGNIFICANT DIGIT EXAMPLES
Example 2-6: Compute the current flow through a resistor that has
a measured R of 1.234 kW and a voltage drop of 1.344 Vdc.
4 significant digits
4 significant digits
I = (1.344)/(1.234x 103) = 1.089 mA 4 digits
lesson2et438a.pptx
R = 1.234 kW
V = 1.344 V
Since both measured values have four significant digits the
computation result can have at most four significant digits.
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BASIC STATISTICS
Measurements can be evaluated using statistical measures
such as mean variance and standard deviation.
Arithmetic Mean ( Central Tendency)
x
Where
x
i 1
i
n
lesson2et438a.pptx
n
xi = i-th data measurement
n = total number of measurements taken
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BASIC STATISTICS – VARIANCE AND
STANDARD DEVIATION
Variance ( Measure of data spread from mean)
di  (x i  x)2
n
2 
d
i 1
i
n 1
lesson2et438a.pptx
Deviations of measurement
from mean
2 = variance of data
Standard Deviation
 = standard deviation
n

d
i 1
i
n 1
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STATISTICS EXAMPLE
A 1000 ohm resistor is measured 10 times using the same
instrument yielding the following readings
Reading (W) Test #
Reading (W)
1
1016
6
1011
2
986
7
997
3
981
8
1044
4
990
9
991
5
1001
10
966
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Test #
Find the mean variance and standard deviation of the tests What is
the most likely value for the resistor to have?
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STATISTICS EXAMPLE SOLUTION
lesson2et438a.pptx
Variance Calculations
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lesson2et438a.pptx
END LESSON 2: PERFORMANCE
OF CONTROL SYSTEMS
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ET 438a
Automatic Control Systems Technology
15