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The type of instrument to be used is decided on the characteristics required. A accuracy instrument is allowable for human body (feeling) while it may be useless for an instrument in a control system. So for selection the performance characteristics of measuring instruments must be known.
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CHAPTER 2
GENERAL IZED PERFORMANCE
CHARACTERISTICS OF INSTRUMENTS
1
2.1 INTRODUCTIONThe type of instrument to be used is decided on the
characteristics required. A accuracy instrument is allowable for human body (feeling) while it may be useless for an instrument in a control system. So for selection the performance characteristics of measuring instruments must be known.
Instrument performance characteristics is generally broken down into two, namely
Static characteristics Dynamic characteristics,
C5.0 o
2
2.2 STATIC CHARACTERISTICSThe set of performance for applications that involve
the measurement of quantities which are constant or vary only quite slowly are defined in terms of static characteristics. These characteristics generally show up as non-linear and statistical effects.
Since static characteristics affects the dynamic behavior, the overall performance is then judged by a semi-quantitative superposition of the static and dynamic characteristics.
3
2.2.1 Static CalibrationThis consists of determining or checking the systems scale.
This is done by using known standard quantities as inputs (fixed or variable) and noting the outputs and finally developing input output relationships. This procedure can be used as a periodic check for a known instrument or for scaling a new instrument. The output of the new instrument can be a different measurable item from the one which is of interest. As an example for measuring temperature variation of resistance is registered. Theoretically interfering and modifying inputs are held constant during calibration.
4
Calibration procedures involve a comparison of the particular instrument with either (1) a primary standard (2) a secondary standard (3) a known input source (weighing a tank of water in the given interval of time to determine the flow rate.
2.2.2 Readability This term indicates the closeness with which the scale of the
instrument may be read. An instrument which sweeps through 180o will have a higher readability than another instrument which sweeps through 90o for the same range of measurement.
5
2.2.3 Least CountThe smallest difference between two indications that can be
detected on the instrument scale2.2.4 SensitivityStatic sensitivity is defined as the ratio of the change in output
to the corresponding change in input under static or steady-state conditions ie. the slope of the calibration curve. The sensitivity can be linear or non-linear (fig_chp2\fig2.1.pptx). Sensitivity can be affected by interfering and modifying inputs. The effects are indicated as zero and sensitivity drift respectively as shown in fig_chp2\fig2.2.pptx .
6
Zero drift or bias describes the effect where the zero reading of an instrument is modified by a change in the ambient conditions. This causes a constant error that exists over the full range of measurement of the instrument.
Sensitivity drift or scale factor drift defines the amount by which an instruments sensitivity of measurement varies as ambient conditions change. It is quantified by sensitivity drift coefficient.
Example 2.1 example2.pptx
7
Both effects can be evaluated by running suitable calibration tests.
If elements of a system having static sensitivities of K1, K2, K3,, etc. are connected in series or cascaded as shown in Fig. 2.3 then the overall system sensitivity K is given by K = K1 x K2 x K3 x
provided that there is no alteration in the values of K1, K2, K3,, etc. due to loading.
Sensitivity, gain, and magnification are the terms used to mean the same.
8
Fig. 2.3 Overall system sensitivity9
2.2.5. LinearityIt is normally desirable that the output reading of an
instrument is linearly proportional to the quantity being measured. Based on this convenience linearity is defined as the maximum deviation from a linear relationship between input and output i.e. from a constant sensitivity (least square fitting) and is usually expressed as a percentage of full scale reading.
2.2.6 ThresholdIf the instrument input is increased very gradually from zero,
there will be some minimum value below which no output change can be detected.
10
This minimum value defines the threshold of the instrument. Manufacturers specify it as absolute value or percentage of full scale reading.
2.2.7 ResolutionThis defines the smallest measurable input change that will
permit the detection of change in the output.2.2.8 HysteresisThis is an effect of producing different readings when the
measured quantity is approached from above or below. It may be the result of mechanical friction, magnetic effects, or thermal effects. 2nd law-irreversibility.
11
Fig. 2. 4 Instrument characteristic with hysteresis
Dead Space
Maximum input hysteresis
Maximum output hysteresis
Measured variable
Outputvariable
Curve A-Variable increasing
Curve BVariable decreasing
12
2.2.9 AccuracyThis is the term used to indicate the closeness with which the
indications of an instrument approach the true values of the quantities measured. Inaccuracy is the extent to which reading might be wrong and it is usually expressed as a percentage of full scale reading. A 1% accuracy over a full scale reading pressure of 100 kPa will be accurate within 1 kPa. If this instrument is used to measure 5 kPa, then the possible error will be 20% !
Such plus or minus inaccuracies are also termed as measurement uncertainties.
13
2.2.10 PrecisionIt indicates the ability of an instrument to reproduce a certain
reading with a given accuracy. If there is no reproducibility, then the instrument is said to have a drift.
As an example for a true value of 100 V, measured values are 104, 103, 105, 100, 105.
The accuracy is 5 V.Precision is maximum deviation from mean.Mean = 104 Max. deviation = 1 VPrecision is 1%.
14
It may be noted that the instrument could be calibrated so that it could be used to dependably measure voltages within 1 V. This conveys the message that accuracy can be improved but not beyond the precision of the instrument.
15
Fig.2.5 Comparison of accuracy and precision
Low precisionLow accuracy
High precisionLow accuracy
High precisionHigh accuracy
16
2.2.11 ErrorsThere are two types of errors in measurements, namely
systematic and random errors. Systematic or fixed errors or bias
These types of errors will cause repeated readings to be in error by the same amount. These are related to calibration errors and they can be eliminated by correct calibration.Accuracy is related to such type of errors
Random errorsThese are caused by personal fluctuations, random electronic fluctuations in the instruments, various
17
influences of friction, etc. They usually follow certain statistical distribution.
Such errors are related to precision.2.3 DYNAMIC CHARACTERISTICSThe dynamic characteristics of a measuring instrument
describe its behavior when input varies with time such that the instrument response will have transient and steady state parts.
The most widely used mathematical model for the study of measurement-system dynamic response is the ordinary linear differential equation with constant coefficients of the form
18
(nth order)If we define the differential operator the above
equation can be written as
ioi
11mi
1m
1mmi
m
m
ooo
11no
1n
1nno
n
n
qbdtdqb...
dtqdb
dtqdb
qadt
dqa...dt
qdadt
qda
++++=
++++
dtdD
io11m
1mm
m
oo11n
1nn
n
q)bDb...DbDb(q)aDa...DaDa(++++=
++++
19
The method of undetermined coefficients or Laplace-transform method can be used to get the solution. Here the former will be used.
In this method the general solution is given by qo = qocf + qopi
whereqocf = complementary function part of solutionqopi = particular-integral part of solution
The solution qocf is determined from the algebraic characteristic equation
(Refer to handout for solutions)0aDa...DaDa o1
1n1n
nn =++++
20
2.3.1 Zero-Order InstrumentThe simplest possible special case of the general
dynamic equation is when all the as and bs other than ao and bo are assumed to be zero. The differential equation then degenerates into the simple algebraic equationaoqo = boqi
Any instrument that closely obeys the above equation is defined to be a zero-order instrument. The two constants can be combined to give
iio
oo Kqqa
bq ==21
where K=bo/ao = static sensitivity The above equation shows that no matter how qi might
vary with time, the instrument output follows it perfectly with no distortion or time lag of any sort. Thus the zero-order instrument represents ideal or perfect dynamic performance.
A practical example of a zero-order instrument is the displacement measuring potentiometer. Referring to Fig. 2.6(a) and fig_chp2\fig2.6.pptx linear distribution of resistance along length L, the output voltage eo can be written as
Measurement error em = Kqi - eo = o (ideal)ib
io KqEL
xe ==22
Fig.2. 6 (a) Zero-order instrument (linear response)23
2.3.2 First Order InstrumentsAll as and bs except a1,ao,bo are taken as zero. The
resulting equation is
Division by ao reduces the No. of coefficients by one as
Defining a1/ao
ioooo
1 qbqadtdqa =+
o
oi
o
oo
o
o
1
abKq
abq
dtdq
aa
==+
24
The above equation gives(D + 1) qo = Kqi
is a time constant and always has dimension of time.As an example of a first order instrument let us apply
this to a thermometer (liquid in a glass) fig_chp2\fig2.7.pptx
The input is Ti(t) which varies with time and the output is xo (thermometer liquid level change). If the liquid level is at xo and using the following definitions
Ttf=temperature of liquid in bulb (uniform), Ttf = 0 at xo =0
25
Kex = differential expansion coefficient of thermometer fluid and bulb glass, (m3/m3 oC)
Vb = Volume of bulb, (m3)Ac = cross-sectional area of capillary tube, (m2)Displaced volume and volumetric expansion of liquid
in the bulb are related asxo Ac = Kex Vb Ttf
This will give
To get the differential equation we will use the conservation of energy over an infinitesimal
c
tfbexo A
TVKx =
26
time dt for the thermometer bulb as follows:Heat rate in Heat rate out = energy storage rate
U=overall heat transfer coefficient across bulb wall, (W/m2)
Ab = heat transfer area of bulb wall, (m2) = mass density of thermometer liquid (kg/m3)c = specific heat of thermometer liquid (J/kg m2 ) The above equation can be rewritten as
dtdTcV0)TT(UA tfbtfib =
ibtfbtf
b TUATUAdtdTcV =+
27
Using the equation that relates Ttf and xo which is
And substitution in the differential equation gives
Defining
dtdx
VKA
dtdT o
bex
ctf =
ibobex
cbo
ex
c
ibbex
cob
o
bex
cb
TUAxVKAUA
dtdx
KcA
TUAVK
AxUAdt
dxVK
AcV
=+
=+
bobex
cbo
ex
c1 UAbVK
AUAaKcAa ===
28
This will give
Again defining
results in the following ODE
ioooo
1 Tbxadtdxa =+
)C/m(A
VKAbK)s(
UAcV
aa o
c
bex
o
o
b
b
o
1 ====
io
ioo
KTx)1D(
orKTxdt
dx
=+
=+
29
As can be seen clearly, the solution depends on the type of input. Four types of inputs will be dealt with.
Step InputInitially the system is in equilibrium with qi=qo=0 and at time
t=0+ the input quantity increases instantly by an amount qis. This will make the initial condition qi=qis at t=0+.
The differential equation is(D +1)qo = Kqis
The solutions areqocf=Ce-t/ and qopf = Kqis
30
This gives the complete solution asqo = Ce-t/ + Kqis
Applying the initial condition qo=0 at t=0gives C= -Kqis and the solution becomes
qo = Kqis(1 e-t/)This is shown in fig_chp2\fig2.8.pptxNon-dimensionalizing gives
e-t/ = 0 makes it an ideal instrument (req. small )
t
is
o e1Kqq
=
31
The speed of response depends on the value of only and is faster if is smaller.
Measurement error is given by
Non-dimensional form
The non-dimensional output and error are shown infig_chp2\fig2.9.pptx
/tis
/tisis
oim eq)e1(K
KqqKqqe ===
)timewithdecays(eqe /t
is
m =
32
There are some dynamic characteristics that are useful in characterizing the speed of response of any instrument. These are
Rise Time: time required to achieve a response of 90% of the step input.
A response is usually assumed to be complete after t=5 s since
303.2t1.0e
e19.0q
K/q
/t
/t
is
o
==
==
%3.99993.0e1e1 5/t === 33
Settling time: time to reach and stay within a stated plus-and-minus tolerance band value around its final value. A 5% settling time is equivalent to t=3
Knowing now that fast response requires a small value of , application to our thermometer example shows that may be reduced by
1. Reducing , C, and Vb2. Increasing U and AbFor the thermometer fluid use small C product.
Reducing Vb decreases Ab (unwanted effect). Reduced Vb , moreover reduces the sensitivity K. Results in trade-off between speed and sensitivity.
34
Also as U is dependent on the flow situation (free or forced convection), caution has to be taken with respect to specification of the instrument.
Ramp ResponseInitially the system is in equilibrium, with qi=qo=0.
The differential equation becomes
Solution of complementary and particular function
000
===
ttqqtqq
isi
oi
tqKq)1D( iso =+
)t(qKqandCeq isopf/t
ocf ==
35
The complete solution will be
Applying the initial condition will give
This will give the final solution as
Define measurement error em by (qi-qo/K). Then
transient error(em,t) steady state error(em,ss)
)t(qKCeq is/t
o +=
isis qKCqKC0 ==
)te(qKq /tiso +=
is/t
is
/tisism
qeq)te(qtqe
+=
+=
36
Note that the unsteady state output and error die out with increasing time.
The non-dimensionalized error is given by
The response and the non-dimensionalizedmeasurement error are shown in fig_chp2\fig2.10.pptx
The solution shows that there is a steady state error given by em,ss and there is a fixed time lag of .
/t
is
is/t
is
ss,m
m e1q
qeqee =+=
37
Frequency ResponseHere the input is harmonic (sine or cosine function)For a sinusoidal input
qi=Ai sin t A graphical representation of a possible solution is
shown in fig_chp2\fig2.11.pptxThe response shows both the steady and unsteady
parts. Usually the transient part decays with time. Some of the terms used are shown in the figure.
38
The differential equation will be(D + 1)qo = Kqi =KAi sin t
The complementary function from previous solutions is
(Refer to handout for details or example2.pptx )
)tsin()(1
KAq
andCeq
2i
opf
/tocf
++
=
=
39
The complete solution will be
For steady state ie. large t, the complementary function decays
The phase shift angle is given by
And the amplitude ratio becomes
)tsin()(1
KACeqqq2
i/topfocfo
+
++=+=
)(tan 1 =
2i
o
)(11
KAq
+=
40
It can also be shown that the time lag isThe amplitude ratio and phase lag are shown infig_chp2\fig2.12.pptxThe steady state solution shows that for an ideal
frequency response, the phase angle must approach zero. This will also make the amplitude ratio to be one. This occurs if the product approaches zero. This requires
a. For any , there will be some frequency input below which measurement is accurate
b. For high , the instrument must have a sufficiently small example2.pptx
/t =
41
Impulse ResponseThe impulse (Fig.2.13 a) function of strength (area) A
is defined by the limiting processImpulse function of strength A lim T0 p(t)
Fig. 2.13 (a) Impulse of strength A
42
Its time duration is infinitesimal. Its peak is infinitely high and its area is A.
For pulse 0 < t < T, it is the same as a step input (Fig.2.13b) where the input is qi = A/T = qis and the differential equation to be solved will be(D + 1)qo = Kqis = KA/T
And the complete solution
This solution is valid only up to time T. At this time we have
)e1(T
KAq /to=
)e1(T
KAq /TTt@o
= =43
44
For t > T, our differential equation will be(D + 1)qo = Kqi = 0 (Fig.2.13c)
Which gives the solution as
And imposing the initial condition at t=T which gives
/to Ceq
=
/T
/t/T
o
/T
/T/T/T
Tee)e1(KAq
givingTe
)e1(KACCe)e1(T
KA
=
==
45
46
As T is made shorter, the first part (t < T) of the response becomes of negligible consequence (Fig.2.13d), so that we can get an expression for qoby taking the limit T0
In such a case LHospitals rule will be used. Differentiate the numerator and denominator with respect to T and take the limit. This will give the approximate solution as ( fig_chp2\fig2.13e.pptx)
00e
Te)e1(KA0Tlim /t/T
/T
=
/t
o eKAq =
47
48
2.3.3 Second Order InstrumentsIt follows the equation
The essential parameters can be reduced to three, as follows:
static sensitivity
undamped natural frequency, rad/time
ioooo
12o
2
2 qbqadtdqa
dtqda =++
o
o
abK
2
oaa
n49
Anddamping ratio, dimensionless
With these definitions the equation becomes
A good example of a second order instrument is the force-measuring spring scale shown in fig_chp2\fig2.14.pptx
Assuming frictional effect proportional to velocity we will use the following values.
=2o
1
aa2a
ion
2n
2
Kqq1D2D =
++
50
Ks = spring constantB = damping coefficient (constant)Considering xo = 0 when fi = 0, application of
Newtons second law yields
Division by Ks will give
Comparing with the general 2nd order equation
ios2
2o
2
oso
i fx)KBDMD(dtxdMxK
dtdxBf =++=
s
io
s
2
s Kfx1
KBD
KM
=
++
soo
s1
s2 K
1b,1a,KBa,
KMa ====
51
For the spring system this will give
This will allow the definition of critical damping as
MK2B
K2Bwhere
KB2),s/rad(
MK),N/m(
K1K
ss
n
sn
sn
s
==
===
c
sc
CB
ratiodampingandMK2C
=
=
52
With step input size qis, the differential equation becomes
Initial condition: at t=0 qo=0 and dqo/dt = 0The characteristic equation for the complementary
function will depend on the roots of
Depending on the value of (>1 overdamped =1 critically damped
The complete solutions are given as follows:
( )11)t1sin(1e
Kqq
)1(1e)t1(Kqq
)1(1e12
1
e12
1Kqq
n2
2
t
is
o
tn
is
o
t)1(
2
2
t)1(
2
2
is
o
n
n
n2
n2
+
+
+=
+
54
And
The above solutions are plotted as functions of the product nt in fig_chp2\fig2.15.pptx.
The following can be observed from the solution n is a direct indication of speed of response.
Doubling n will halve the response time. An increase in reduces oscillation, but also slows
the response. Many commercial instruments use =0.6 to =0.7.
21 1sin =
55
Ramp ResponseThe differential equation for this case is
With initial conditionsThe solutions are found to be :Overdamped
tqKq1D2D ison
2n
2
=
++
0tat0
dtdq
q oo
===
)e14
1212
e14
12121(q2qKq
t)1(
2
22
t)1(
2
22
n
isis
o
n2
n2
+
++
+=
56
Critically damped
Underdamped
+= tn
n
isis
o netqtqKq
2112
1212tan
t1sin(12
e1q2tqKq
2
2
n2
2
t
n
isis
on
=
+
=
57
For steady state all the above will give
Steady state time lag can be shown to be 2/n.Measurement error, em can be determined from
Steady state error will be
fig_chp2\fig2.16.pptx and fig_chp2\fig2.17.pptx
n
isis
o q2tqKq
=
n
isois
oiss
q2Kqtq
Kqqe
===
Kqtq
Kqqe oisoim ==
58
respectively show ramp response and non-dimensional ramp response error.
Frequency Response For a cosine input, the equation to be solved will be
The solutions can easily be determined (for example for the underdamped case) as
which decays with time
tcosKFq1D2D oon
2n
2
=
++
)1sin1cos( 222
1 tCtCeq nnt
ocfn +=
59
And the particular integral function for all the cases will be
The above will be the steady state solution.
tcos)(4
KF)(
tsin)(4
KF2q
22n
22n
2o
22n
2n
22n
22n
2o
3n
opf
+
+
+=
60
Trigonometric manipulation will give
where
=0.6 to 0.7 is the practical choice for instruments.fig_chp2\fig2.18.pptx shows the amplitude ratio and
phase lag fig_chp2\fig2.19.pptx .
22
n
2
n
2o
o
14
)tcos(F
K/q
+
=
n
n
2tan
=
61
Impulse ResponseFor impulse strength of A solutions are found to beOverdamped:
Critically damped
)e
e(12
1KA
q
t)1(
t)1(
2n
o
n2
n2
+
=
tn
n
o nteKA
q
=
62
Underdamped:
The results are shown graphically in fig_chp2\fig2.20.pptx
)t1sin(e1
1KA
qn
2t2
n
o n
=
63
2.4 IMPEDANCE MATCHINGThe introduction of any measuring instrument into a
measured medium always results in the extraction of some energy from the medium, thereby changing the value of the measured quantity from its undisturbed state thus making perfect measurements theoretically impossible.
Looking at fig_chp2\fig2.21.pptx the effort to measure the unknown voltage Eo, is made theoretically impossible since the circuit is changed as soon as the voltmeter with its resistance Rm is connected.
64
Since the loading effect is going to change the output variable, the concept of input impedance will be used to characterize this effect.
Let qi1 be the input variable of primary interest as far as the instrument is concerned. Let there be an associated variable qi2 such that the product qi1 qi2has the dimensions of power and represents the instantaneous rate of energy withdrawal from the device. Then the generalized input impedance Zgi is defined as
2i
1igi q
qZ 65
This gives the power, P, drained by the voltmeter as
which shows the requirement of large input impedance to keep the power drain small.
For the voltmeter the input variable qi1=Em. An associated variable will be qi2=im. This will give Zgi=Em/im=Rm, the meter resistance.
For a general approach Thvenins theorem (without proof) will be used. If the input is a general load with input impedance Zl as shown in fig_chp2\fig2.22.pptx, the open circuit voltage Eo
gi
21i
ZqP =
66
is the one that is measured when the load is not connected and it is also possible to determine the impedance ZAB. Thvenins theorem states : If the load Zl is connected as shown in Fig.2. 22b, a current il will flow. This current will be the same as the current that flows in the fictitious equivalent circuit of Fig. 2. 22c. Thus the network contains only the output impedance ZAB and the single voltage source Eo.
Applying Thvenins theorem to the voltmeter connection of Fig.2.21a converted to Fig.2.21b
omab
mm ERR
RE+
=67
If the voltmeter is to indicate the true value Eo, then we must have Rm >> Rab.
For the generalized input, Zgi, and output impedance, Zgo, and for electrical and nonelectrical systems
Where qi1m=measured value of effort variableqi1u=undisturbed value of effort variable
High value of Zgi is required.Similar analysis for an ammeter shows the meter
resistance must be sufficiently low.
u1igigo
u1igigo
gim1i q1Z/Z
1qZZ
Zq
+=
+=
68
If the device is a power supply to the external load with Rm
For maximizing power (constant Eo, Rab) with respect to Rm (after differentiating) will require Rm=Rab.
Example 2.3example2.pptx
2
mab
m
m
2o
2
omab
m
mm
2m
RRR
REE
RRR
R1
REP
+
=
+
==
69
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