Lesson 2: Pe · 2015-08-28 · 8/27/2015 3 TYPES OF SYSTEM RRESPONSE 5 x 0 10 20 30 40 50 50 0 50...
Transcript of Lesson 2: Pe · 2015-08-28 · 8/27/2015 3 TYPES OF SYSTEM RRESPONSE 5 x 0 10 20 30 40 50 50 0 50...
8/27/2015
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LESSON 2: PERFORMANCE OF
CONTROL SYSTEMS ET 438a
Automatic Control Systems Technology
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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.
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CONTROL SYSTEM PERFORMANCE
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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
System control variable changes over time so error
changes with 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
CONTROL SYSTEM OBJECTIVES
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1.) System error minimized. E(t) = 0 after changes or
disturbances after some finite time.
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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0
50
100
150
200
250
Time (Seconds)
Co
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ol
Var
iab
le
100
C t( )
t
1.) Uncontrolled process
2.) Process control activated
3.) Unstable system
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3
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TYPES OF SYSTEM RESPONSE
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Damped response
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20
40
60
80
100
120Control Responses
Time (Seconds)
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Var
iable
R1
R2
Setpoint
Change
td
Control variable requires time to reach final value
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TYPES OF SYSTEM RESPONSE-TRANSIENT
RESPONSES
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0 5 10 15 20 25 30 35 40 4520
40
60
80
100
120
Ideal
Typical
Response To Setpoint Changes
Time (Sec)
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Ou
tpu
ts
Setpoint Change
TYPES OF SYSTEM RESPONSE-TRANSIENT
RESPONSES
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Disturbance Rejection
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20
40
60
80
100
120Response to Disturbance
Time (seconds)
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Var
iab
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SP
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ANALOG MEASUREMENT ERRORS AND
CONTROL SYSTEMS
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Amount of error determines system accuracy
Determing accuracy
Measured Value
Reading ± Value
100 psi
± 2 psi
Percent of Full
Scale (FS)
FS∙(%/100)
Accuracy ± 5%
FS =10V E=10V(± 5%/100)
E= ± 0.5 V
Percent of Span
Span=max-min
Span∙(%/100)
Accuracy ± 3% min=20, max= 50 psi
E=(50-20)(± 3%/100)
E= ± 0.9 psi
Percent of Reading
Reading∙(%/100)
Accuracy ± 2% Reading = 2 V
E=(2V)(± 2%/100)
E= ± 0.04 V
SYSTEM ACCURACY AND CUMULATIVE
ERROR
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Sensor Sensor amplifier
Subsystem errors accumulate and determine accuracy limits.
Consider a measurement system
K±DK Cm
G±DG V±DV
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
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With no uncertainty: V = K∙G∙Cm
With uncertainty V±DV = (K±DK)∙(G±DG)∙Cm
Multiple out and simplify to get:
Input Output
K
K
G
G
V
V D
D
D
sensor ofy uncertaint normalized K
K
ampsensor ofy uncertaint normalized G
G
output ofy uncertaint normalized V
V
:Where
D
D
D
Component
Tolerance/100
COMBINING ERRORS
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Use Root-Mean-Square (RMS) or Root Sum Square (RSS)
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K
K
G
G
V
V
D
D
D
This relationship works on all formulas that include only multiplication
and division.
Notes:
100%by s toleranceDivide y.uncertaintfraction are K
K ,
G
G
%get to100by Multiply .yuncertaint RMS fractional is V
V
DD
D
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CUMULATIVE ERROR EXAMPLE
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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%.
VoRin
Rf
Vin
in
fino
R
RVV
Determine uncertainty from
tolerances
02.0100
%2
V
V
05.0100
%5
R
R
R
R
in
in
in
in
f
f
D
D
D
2
in
in
2
in
in
2
f
f
RMSo
o
V
V
R
R
R
R
V
V
D
D
D
D
073.002.005.005.0V
V 222
RMSo
o
D
%3.7V
V%100U%
RMSo
o
D ANS
SENSOR CHARACTERISTICS
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Sensitivity - Change in output for change in input.
Equals the slope of I/O curve in linear device.
Hysteresis - output different for increasing or
decreasing input.
Resolution - Smallest measurement a sensor can
make.
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
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Find a sensors sensitivity using data two points. Use
point-slope equations to find line parameters
11 xxmyy
12
12
xx
yym
Where:
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 : (x1, y1) = (20 C, 100 W) x = input y = output
(x2, y2) = (120 C, 195 W)
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SENSOR SENSITIVITY EXAMPLE (2)
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12
xx
yym
C/ 95.0
C 100
95
C 20120
100195m W
W
W
C 20xC/ 95.0 100y WW
C/ 81x 95.0y
100)C 20(C/ 95.0xC/ 95.0y
W
WWW
Compute the slope and the equations:
Can plot above equation to check results
ANS
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SENSOR RESPONSE
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Ideal first-order response-ideal
bi
bf
Ci
Cf
No time delay
Time
Measu
red
Con
trol
Vari
able
Step change in the measured variable- instantly changes value.
Practical sensors exhibit a time delay before reaching the new
value.
PRACTICAL SENSOR RESPONSE
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Let b(t) = sensor response function with respect to time.
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40
60
80Sensor Responses
Time (Seconds)
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Var
iable
bi
bf
Step
Decrease
Step
Increase
bi bf
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MODELING 1ST-ORDER SENSOR RESPONSE
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For step increase:
Where bf = final sensor value
bi = initial sensor value
t = time
t = time constant of sensor
t
t
ifi e1bbb)t(b
For step decrease:
t
t
fi ebb)t(b
SENSOR RESPONSE EXAMPLE 1
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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)
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Complete the calculations to find t
SENSOR RESPONSE EXAMPLE 2
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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
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Significant Digits In
Measurement
Readable output of instruments
Resolution of sensors and transducers
Calculations Using
Measurements
Truncate calculator answers to match
significant digits of measurements and
readings,
SIGNIFICANT DIGIT EXAMPLES
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Example 2-5: Compute power based on the following
measured values. Use correct number of significant digits.
Significant digits not factor in design calculations. Device values
assumed to have no uncertainty.
3.25 A 3 significant digits
117.8 V 4 significant digits
P = V∙I=(3.25 A)∙(117.8 V) = 382.85 W
Truncate to 3 significant digits P = 383 W
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SIGNIFICANT DIGIT EXAMPLES
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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.
R = 1.234 kW 4 significant digits
V = 1.344 V 4 significant digits
I = (1.344)/(1.234x 103) = 1.089 mA 4 digits
Since both measured values have four significant digits the
computation result can have at most four significant digits.
BASIC STATISTICS
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Measurements can be evaluated using statistical measures
such as mean variance and standard deviation.
Arithmetic Mean ( Central Tendency)
n
x
x
n
1i
i
Where xi = i-th data measurement
n = total number of measurements taken
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BASIC STATISTICS – VARIANCE AND
STANDARD DEVIATION
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1n
d
)xx(d
n
1i
i2
2
ii
Variance ( Measure of data spread from mean)
Deviations of measurement
from mean
2 = variance of data
Standard Deviation
= standard deviation 1n
dn
1i
i
STATISTICS EXAMPLE
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A 1000 ohm resistor is measured 10 times using the same
instrument yielding the following readings
Test # 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
Find the mean variance and standard deviation of the tests What is
the most likely value for the resistor to have?