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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