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...

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8/27/2015 1 LESSON 2: PERFORMANCE OF CONTROL SYSTEMS ET 438a Automatic Control Systems Technology 1 lesson2et438a.pptx lesson2et438a.pptx 2 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.

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...

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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 10 20 30 40 5050

0

50

100

150

200

250

Time (Seconds)

Co

ntr

ol

Var

iab

le

100

C t( )

t

1.) Uncontrolled process

2.) Process control activated

3.) Unstable system

1

3

2

TYPES OF SYSTEM RESPONSE

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Damped response

0 5 10 15 20 25 30 35 40 450

20

40

60

80

100

120Control Responses

Time (Seconds)

Co

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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)

Co

ntr

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Ou

tpu

ts

Setpoint Change

TYPES OF SYSTEM RESPONSE-TRANSIENT

RESPONSES

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Disturbance Rejection

0 5 10 15 20 25 30 35 40 450

20

40

60

80

100

120Response to Disturbance

Time (seconds)

Co

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Var

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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)

12

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.

0 5 10 15 20 25 30 35 40 450

20

40

60

80Sensor Responses

Time (Seconds)

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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?

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STATISTICS EXAMPLE SOLUTION

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Variance Calculations

END LESSON 2: PERFORMANCE

OF CONTROL SYSTEMS

ET 438a

Automatic Control Systems Technology

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