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GENERALIZED PERFORMANCECHARACTERISTICS OF
INSTRUMENTS
Lecture 8Instructor : Dr Alivelu M Parimi
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Example
2
IQ scores of result are normally distributed to mean of 100 and standard deviation of 13.
Find (i) Fraction of students having an IQ > 133
(ii) Fraction of students having an IQ > 90
(iii) IQ value exceeded by upper quartile (from 75% to 100%)
Solution
(i) IQ >133
z =13
100-133= 2.54, Area corresponding to z = 2.54 is 0.4945, students having IQ 133
are 0.5 + 0.4945 =0.9945, i.e. 99.45% have IQ 133
Fraction of students having IQ > 133 are ( 0.50.4945 = .0055) . i.e. 0.55% will have IQ >
133
(ii) For IQ = 90, z =13
100-90=0.77 Area from 0 to 0.77 = 0.2799 which is same as
area for z = -0.77 to 0.
Total area 0.5 + 0.2794 = 0.7794, i.e. 77.94% will have IQ score greater than 90
(iii) Upper quartile implies area > 0.75
from Table 3.2, find z corresponding to area of 0.25 which is z
xx = 0.6745
So xx = 0.6745 x 13 = 8.76,
x = 108.76, hence upper quartile will be for students having IQ > 108.76
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Gaussian distribution
3
Table 3.2: Values of Area under Gaussuain Distribution Curve
Table list areas under the
curve between zero andvarious values of z
This table will have areas
from 0 to maximum 0.5
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DYNAMIC PERFORMANCE
CHARACTERISTICS Instruments rarely respond instantaneously to changes in the
measured variables.
Instead, they exhibit slowness or sluggishness due to factors
such as mass, thermal capacitance, fluid capacitance or
electrical capacitance. In addition, pure delay in time is often encountered where the
instrument waits for some reaction to take place.
Such industrial instruments are nearly always used for
measuring quantities that fluctuate with time. Therefore the dynamic and transient behavior of the
instrument is as important as the static behavior.
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DYNAMIC PERFORMANCE
CHARACTERISTICS The parameters characterizing the dynamic characteristics of
an instrument are:
Speed of response: It is the rapidity with which an instrument
responds to changes in the measured quantity.
Fidelity: It is the degree to which an instrument indicates, thechanges in the measured variable without dynamic
error(faithful reproduction)
Lag: It is the retardation delay in the response of an
instrument to changes in the measured variable. Dynamic error: It is the difference between the true value of a
quantity changing with time and the value indicated by the
instrument . 6
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DYNAMIC PERFORMANCE
CHARACTERISTICS When measurement problems are concerned with rapidly
varying quantities, the dynamic relations between the input
and output are generally defined by the use of differential
equations. For most measuring instruments, generalized
relation between any particular input x and output y can bewritten as:
7
ann
n
dt
yd + an-1
1-n
1-n
dt
yd . .+ a1
dt
dy+ a0y = bn
n
n
dt
xd + bn-1
1-n
1-n
dt
xd..+ b1
dt
dx+ b0x
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DYNAMIC PERFORMANCE
CHARACTERISTICS Types of Standard Inputs
Once response of an instrument to certain standard inputs is
known, its response to more complicated inputs can be
determined using mathematical techniques.
Step input:System is subjected to an instantaneous and finitechange in measured variable.
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DYNAMIC PERFORMANCE
CHARACTERISTICS Linear/Ramp input: System is subjected to a measured
variable which is changing linearly with time. Like
thermometer is put in a bath whose temperature is rising
linearly.
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DYNAMIC PERFORMANCE
CHARACTERISTICS Pulse/Impulse input:System is subjected to step magnitude
for small time and then is brought back to its initial state.
When this small time is tending to zero, pulse becomes
impulse, which is a theoretical concept not realizable
practically as it implies in no time system is subjected to largechange. If the automobile running on road suddenly hits a pot
hole on the road, then the force experienced by shock
absorbing system may be modeled by impulse function
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DYNAMIC PERFORMANCE
CHARACTERISTICS Sinusoidal input: System is subjected to measured variable,
the magnitude of which changes in accordance with a
sinusoidal function of constant amplitude.
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DYNAMIC PERFORMANCE
CHARACTERISTICS Order of a System
The order of the system relates to the number of energy
storage elements in the measurement system.
Zero order,
First order,
Second order instrument
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DYNAMIC PERFORMANCE
CHARACTERISTICS Order of a System: examples
For example, if you use a mercury-in-glass thermometer to
measure temperature there is a single energy store (the heat
capacity of the mercury), thermometer acts predominantly
like a Firstorder system. RC and RL circuits are examples of firstorder system as only
electrostatic or electromagnetic energy is involved.
RLC circuit is an example of secondorder system as both
electromagnetic and electrostatic energies are involved. Accelerometers and seismographs having mass-spring-damper
arrangement have two energy storesthe spring (elastic
potential energy) and the mass (kinetic energy).
Thus, these devices respond like Secondorder systems.
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DYNAMIC PERFORMANCE
CHARACTERISTICS Zero Order instruments
14
ann
n
dt
yd + an-1
1-n
1-n
dt
yd . .+ a1
dt
dy+ a0y = bn
n
n
dt
xd + bn-1
1-n
1-n
dt
xd..+ b1
dt
dx+ b0x
y =0
0
a
bx
Since y = Kx is algebraic, it is clear that no matter how x
might vary with time, the instrument output (readings )follows it perfectly with no distortion or time lag of any
sort. Thus, the zero order system represents ideal or
perfect dynamic performance and is thus a standard
against which less perfect instruments may be compared.
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DYNAMIC PERFORMANCE
CHARACTERISTICS A practical instrument of a zero order type is the
displacement potentiometer shown in Figure, where a strip of
resistance material is excited with a voltage and provided with
a sliding contact. If the resistance is disturbed linearly along
length L, we may write
15
e0 =L
xiEb= K xi
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DYNAMIC PERFORMANCE
CHARACTERISTICS An ideal potentiometer acts like a zero order instrument.
Output reflects input faithfully without any time delay.
Ideal Operational amplifier used as an inverter is a zero order
system with a gain of (1.).
A wire strain gauge in which the change in the electrical
resistance of the wire is proportional to the strain in the wire
is also an example of zero order system.
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DYNAMIC PERFORMANCE
CHARACTERISTICS Step response of a zero order instrument
In a zero order instrument, y (t) = Kx (t), we have y (t) = 0 for t
< 0, and y = K for t 0. Therefore the response to the unit step
function is a step function with height K
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Dynamic performance characteristics:
First Order instruments
18
The governing equation for the First order system is
a1dt
dy+ a0 y = b0 x
Any instrument which follows this equation is by definition a first order instrument.
The equation can be rewritten as
0
1
a
a
dt
dy+ y =
0
0
a
bx
=0
1
a
a, where is the time constant.
K =0
0
a
b, where K is static sensitivity/gain.
Kxydt
dy
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Dynamic performance characteristics:
First Order instruments
Example
Time constant of measurement system is determined by thephysical characteristics of the system.
A measurement system with a small time constant will
respond quicker to changes than a system with a large timeconstant.
The time constant depends on the size of the energy storageelement as well as how quickly the energy can get in/out ofthat storage element.
For example, mercury in glass thermometer will respondquicker when plunged into water than it does in air (fasterheat transfer with water).
A thermometer with a large bead will respond slower than athermometer with a small bead
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Dynamic performance characteristics:
First Order instruments
Step response of first order instruments
Assuming system is in equilibrium at t = 0 and then a step
input of magnitude A is applied.
20
Kxydtdy
y(t) = AK (1e- t/)
Steady state value ( as t ) of y(t) = AK = yss
Error = em= inputoutput =
/t/t Ae)e1(AAK/)t(y)t(x
Steady state error ( as t ) = em,ss= A
So /
,
t
ssm
m
e
e
e
X(s)Y(s) =
)s(1K
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Dynamic performance characteristics:
First Order instruments
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Nondimensional Response of First Order
System to Step Input
Non Dimensional Error Curve of First Order
System
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Dynamic performance characteristics:
First Order instruments
Ramp response of first order instruments
Measuring instruments may be placed in an environment
whose temperature, pressure is increasing or decreasing in
linear (ramp) fashion.
22
X(s)
Y(s)=
)s(1
K
X(s) =2s
M
KM
y(t)= -+ t + e
-t/
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Dynamic performance characteristics:
First Order instruments
23
K
)t(y= M ( t - + e
-t/)
At larger values of t, the exponential term will tend to zero.
y = KM ( t - ); Input = Mt, implying there is a steady state lag of
Measurement error emis given by:
em = InputOutput = x(t)y(t)/K = MtM ( t - + e-t/
)
em = M- Me-t/
= M( 1- e-t/
)
Steady state error = em,ss( emas t ) = M, Transient error = Me-t/
For em,ssto be small, M and should be small.
ssm,
m
ee = 1- e
-t/
At steady state, measurement error is equal to the steady state error.
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Dynamic performance characteristics:
First Order instruments
Impulse response of the first order system
A unit impulse function is defined as a signal which has zero
values at all times except at t = 0, where its magnitude is
infinite. In real, practical systems, it is not possible to produce
a perfect impulse to serve as input for testing. Therefore, abrief pulse is used as an approximation of an impulse.
Provided that the pulse is short compared to the impulse
response, the result will be near enough to the true,
theoretical, impulse response
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Dynamic performance characteristics:
First Order instruments
25
For 0 < t < T:
( D + 1) y = K x =T
KA
Since, up until time T, this is no different from a step input of sizeT
A. For initial condition y
= 0 at t = 0, the complete solution is
y =T
KA ( 1-e
-t/)
at t = T, y =T
KA ( 1-e
-T/) (A)
For t > T, x = 0
Therefore, (D+1) y = Kx = 0
y(t) = Ce-t/
Therefore, y(t) = Ce-t/
(B)
At t = T, y(t) = Ce-T/
Y(t) = Ce
-t/
= T
KA
/
-T/e-1Te
e
-t/
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Dynamic performance characteristics:
First Order instruments
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Dynamic performance characteristics:
First Order instruments
Response of FOS to a sinusoidal input function
The response of a system to sinusoidal input is known as the
frequency response.
If the input is sinusoidal, the steady state output is also
sinusoidal.
The output frequency equals the input frequency and the
output phase lags behind the input phase.
It is important to study the sinusoidal response as many
natural phenomena are sinusoidal in character like vibration ofguitar string, currents in oscillating circuit. In generation and
transmission of electric power and communication also
sinusoidal signals are used. 27
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Dynamic performance characteristics:
First Order instruments
28
For example: if x = xmsin t then
(t) =)1(
)tsin(Kx22
m
On comparing the input with the output, it is observed that
1. Frequency does not change in first order system.
2. There is attenuation in amplitude by a factor)(1
K
22
.
3. Output lags behind the input by tan-1
(- ).
The frequency response has two elements which can be plotted graphically
(i) The ratio of output amplitude to input amplitude plotted against frequency.
(ii) The phase lag between output and input as a function of frequency.
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Dynamic performance characteristics:
First Order instruments
29
Frequency response is often specified by itsbandwidth: this is the range of frequency over
which variation in magnitude is less than 70%
(3dB).
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Problem
A certain thermometer has time constant of 15 sec and an
initial temperature of 200 C. It is suddenly exposed to
temperature of 1000C. Determine the time it would take to
reach 90% of step size.
30
Solution
= 15 sec. Step size = final valueinitial value = 10020 = 800C
90 % of 800
= 720C
Therefore, total rise = 200C + 72
0C = 92
0C
Y (t) = AK (1-e
-t/
) + 20
(assuming K=1)
A = 800C
92 = 80 (1-e-t/
) + 20 therefore t = 34.54 sec.
The thermometer will take 34.54 seconds to reach 90 % of the step size.
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Problem
A mercury thermometer is kept in a bath of 400C and bath is
heated such that its temperature increases linearly at the rate
of 50C/min. Calculate the thermometer reading after 5
minutes when is (i) 10 sec (ii) 2 min.
31
(i) = 10 sec.
After 5 minute, actual temperature becomes 40 + 5*5 = 650C. But due to error thermometer
indicates different value.
K
)t(Y= M (t+ e
t/)
t
= sec30
min5
= 30 .
Transient term is 10 e30
0 and can be neglected
Steady state error = M= 5min
0C
* min60
10= 0.83
0C
Thermometer reading = 650.83 = 64.170C
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Solution contd
32
(ii) = 2 min,2
5t
= 2.5
et/
= e2.5
= 0.165
Y (t) = M (t+ e2..5) + 40
Y (t) = M ( 52 + e2.5) + 40 = 55.820C
Error at t = 5 min is 6555.82 = 9.180C
Steady state error = M= 5x 2 = 100C
Note: Smaller is the time constant lesser is the steady state error.