2015 Chap.1 Lec.notes 2014-2015 Trim.3,EME2066 Chapter 1 Basic Concepts and Analysis of Data

EME 2066 Measurement and Instrumentation

FET 2004/2005 1

CHAPTER 1 BASIC CONCEPTS AND ANALYSIS OF DATA

1.0 Introduction

Many importance theories in science and engineering are come from experiments. To

collect useful data from experiment, engineers need to know how to measure the

importance physical variable (such as flow rate, temperature etc…) and what is the

suitable instrument to use. In order to get the best result, engineers need to know the

governing principles of measurement instruments.

1.0.1 Definition of terms

• Measurement – comparison between an unknown quantity and a predefined

standard

• Measurand – the unknown quantity to be measured.

• Instrument – physical device used to determine measurand numerically.

1.1 Standards

In order to compare the experimental results from others on a consistent basic, it is

necessary to establish certain standard units of length, weight, time and temperature

and electrical quantities.

However, the definition of these quantities has been changed several times.

The definition for 1 meter (m) is:

• before 1960

Length of platinum-iridium bar maintained at the International Bureau of

Weights and Measures in Sevres, France.

• at General conference on Weights and Measures, 1960

1,650,763.73 wavelengths of the orange-red light of a krypton-86 lamp

• 1982

the distance of light travels in 1/299,792,458ths of a second

The definition of 1 kilogram (kg) is the

Mass of platinum-iridium bar maintained at the International Bureau of

Weights and Measures in Sevres, France


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The definition of 1 second (s)

• in the past

1/86400 of a mean solar day

• Oct 1967 (the Thirteenth General Conference on Weights and Measures)

Duration of 9,192,631,770 periods of the radiation corresponding to the

transition between the two hyperfine levels of the fundamental state of the

atom of cesium-133.

1.2 Dimension and Units

• Dimension- defines some physical characteristics. Eg. length, volume, velocity,

heat and etc.

• Unit – is a standard or reference by which a dimension can be expressed

numerically

• SI unit – The international system of units

1.2.1 Fundamental and secondary units

• There are seven fundamental units or base units and two supplemental units. Table 1.1 Fundamental Quantity

Quantity Unit Symbol Length meter m Mass kilogram kg Time second s Electric current ampere A Temperature, kelvin K Luminous intensity candela cd Amount of substance mole mol

supplemental units Plane angle radian rad Solid angle steradian sr

• Table 1.2 showed some of the derived units. Secondary units /derived units are

the product of fundamental units

For eg : Area ( L2)- m2;


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Table 1.2 Derived Units

Quantity Unit Symbol Dimensions Area square meter m2 L2

Volume cubic meter m3 L3

Density kilogram per cubic meter kg/m3 ML-3 Velocity meter per second m/s LT-1 Charge Coulomb C IT Capacitance Farad F 2421 ITLM −− Inductance Henry H 222 −− ITML Potential Volt V 132 −− ITML Resistance Ohm Ω 232 −− ITML Energy Joule J 22 −TML Force Newton N 2−MLT Frequency Hertz f 1/T Power Watt W 32 −TML Pressure Pascal Pa 21 −− TML

Table 1.3 Metric Prefixes

Scientific notation Prefix Symbol Exa 1018 E peta 1015 P tera 1012 T giga 109 G mega 106 M kilo 103 k hecto 102 h deka 10 da deci 10-1 d centi 10-2 c milli 10-3 m micro 10-6 µ nano 10-9 n pico 10-12 p femto 10-15 f atto 10-18 a

1.3 Generalized Measurement System

Most measurement system may be divided into three parts:

1. A detector-transducer stage

-Detects the physical variable and performs mechanical/electrical

transformation to convert the signal into more useful form.


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2. Intermediate stage

-Modifies the direct signal by amplification, filtering, or other means so that a

desirable output is available.

3. Final or terminating stage

– Indicate, record, or control the variable being measured. The output

may also be digital or analog.

Figure 1.1 Schematic of the generalized measurement system

For example, in a bourdon-tube pressure gage (Figure 1.2)

Figure 1.2: Bourdon-tube pressure gage as the generalized measurement system.

A pressure meter. The deflecting torque is produced by the gas pressure which expanses the Bourdon tube. The pointer is deflected by a simple mechanical gear system.


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The first stage (detector-transducer stage):

-The bourdon tube converts the pressure signal into mechanical signal (displacement

of the tube)

The second stage (Intermediate stage):

-Amplifies the mechanical signal (small displacement of the tube into bigger

revolution of a gear) using a gearing system.

The third stage (Final or terminating stage):

-Calibrate the dial and pointer according to a known pressure input.

1.4 Basic Concept of Dynamic Measurement

1.4.1 Static vs Dynamic measurement

Static measurement- the input/output relationship is independent of the rate of change

of the input. (the input quantity is not change with time)

Dynamic measurement- input/output relationship depends on the rate of change of the

input.

The measurement process for dynamic measurement is much more difficult compare

to the static measurement.

The measurement system under dynamics conditions (transient) maybe described in

terms of a general variable x(t) written in differential equation:

)(... 011

1

1 tFxadtdxa

dtxda

dtxda n

n

nn

n

n =++++−

−

− (1.1)

1.4.2.Zero Order System

The zero order system may be express as

)(0 tFxa = (1.2)

-A zero order system is a system whose behaviour is independent of the time-

dependent characteristics of storage or inertia.

-The response of Zero order system is instantaneous response


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1.4.3 First Order System

The first order of the system may be express as

00

1 )(a

tFxdtdx

aa

=+ (1.3)

A first order system is a system characterised as having time-dependent storage or

dissipative ability but having no inertia.

If (1.3) is solved for the case of a sudden step input F(t)=A at time zero, then

F(t)=0 at t=0 and F(t) =A for t >0

along with the initial condition

x=x0 at t = 0

the solution to (1.3) is

τt

eaAx

aAtx

−

−+=

00

0

)( where τ =a1/a0 (1.4)

the first term of the rightè steady stage response

the second term of the rightètransient response

Designating the steady state value as x∞, (1.4) can be written in dimensionless form as

τt

exxxtx −

∞

∞ =−−

0

)( (1.5)

The term τt

e−

represents the error in achieving the steady-state value, x∞=A/ao.

The rise time is defined as the time required to achieve a response of 90% of the step

input.

For a first order system subjected to harmonic input with: -

Initial condition x = x0 at t = 0 and F(t)= A sin ωt for t > 0

The solution of 00

1 )(a

tFxdtdx

aa

=+ is

( )[ ]( )tωtω

ωτ

aA

Cetx τt 1

212

0 tansin1

)( −−−

++= (1.6)

where τ=a1/a0 is the time constant.

The phase shift angle φ is defined as

tωωφ 1tan)( −−= (1.7)

The time delay !t of the steady state response is defined by


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( )ωωφt =∆ (1.8)

A first order system will respond to a harmonic input in a harmonic fashion with the

same frequency, but with a phase shift and reduced amplitude. The larger the time

constant, the greater the phase lag and amplitude decrease.

Example 1.1

A certain thermometer has a time constant of 15s and an initial temperature of 20oC. It

is suddenly exposed to a temperature of 100oC. Determine the rise time, i.e., the time

to attain 90% of the steady-state value, and the temperature at this time.

Solution

The thermometer is a first order system,

T0=20oC= temperature at t=0

T∞=100oC= temperature at steady state

τ=15s=time constant

for the 90% rise time

1.0=−

τt

e

ln (0.1)=-t/15

t = 34.54s

1.010020100)(

=−−tT

T(t)= 92oC

Example 1.2

Suppose the thermometer in example 1.1 was subjected to a very slow harmonic

disturbance having a frequency of 0.01Hz. The time constant is still 15s. What is the

time delay in the response of the thermometer and how much does the steady state

amplitude response decrease?

Solution

ω =0.01Hz=0.06283 rad/s

τ =15s


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ωτ=(0.06283)(15)=0.9425

the phase angle is

φ(ω)= - tan-1(0.9425)=-43.3o=-0.756 rad

so the time delay is

sωωφt 03.12

06283.0756.0)(

−=−

==∆

The amplitude response decreases according to

( )[ ] ( )[ ]

7277.09425.01

1

1

12

12212

=+

=+ ωτ

1.4.4 Second Order System

A second order system would be governed by

00

12

2

0

2 )(a

tFxdtdx

aa

dtxd

aa

=++ (1.9)

A second order system is a system whose behaviour includes time-dependent inertia.

Examples of second order system are Pressure and Acceleration transducers.

Fig 1.3: Simple spring-mass damper system

Another example of second order system is a simple spring –mass damper system

showed in Fig 1.3. In the spring mass damper system,

x1(t) is the input displacement variable; x2(t) is the output displacement.

Assume that the damping force is proportional to velocity, from the Newton’s second

law of motion;

( ) 22

221

21 dtxdm

dtdx

dtdxcxxk =

−+− (1.10)

Rearrange the equation;

11

22

22

2

kxdtdxckx

dtdxc

dtxdm +=++ (1.11)


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Spring-mass damper system subjected to a force input

Fig 1.4 Spring mass damper system subjected to a force input

A force F (t)=F0 cos ω1t is impressed on the spring mass system shown in Fig 1.4, the

displacement of the mass x(t) need to be determined.

The differential equation for the system is

tωFkxdtdxc

dtxdm 102

2cos=++ (1.12)

The solution of the equation (1.12) is

( )

21

21

221

10

21

cos

+

−

−

=

ncn ωω

cc

ωω

φtωkF

x (1.13)

where

( ) ( )( )

112

1

2 / /tan

1 /c n

n

c c ω ωφ

ω ω−=

−

mkωn =

mkcc 2=

ϕ is called the phase angle, ωn is the natural frequency and cc is called the critical

damping coefficient.

x0 is the amplitude of the motion given by

21

21

221

0

0

21

+

−

=

ncn ωω

cc

ωω

kF

x (1.14)


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The ratio of output to input amplitude x0/(F0/k) is showed in Figure 1.5 to show the

frequency response of the system.

The phase angle ϕ is plotted in Fig 1.6 to show the phase-shift characteristics.

Fig 1.6 Phase-shift characteristics of the system in Fig 1.4

The rise time for a second order system is defined as the time to attain a value of 90%

of a step input. It may be reduced by reducing the damping ratio for values c/c2c

below about 0.7.

Fig 1.5 Frequency response of the system in Fig 1.4


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The ringing frequency ( )[ ] 2/12/1 cnr ccωω −= and the rise time are showed in Fig

1.7. The response time is the time taken for the system to settle within ±10% of the

steady-state value.

Fig 1.7 Effect of rise time and ringing on output response to a step input

In order for a system to have a good response, it must be treat all frequencies within

the range of application so that the ratio of output to input amplitude remains the same

over the frequency range desired (Linear frequency response). Further discussion of the second order system, higher order system and dynamic system

response/transient respond will be discussed in Control Engineering

.

1.4.5 Distortion

As a result of frequency response characteristics of a system, distortion happens in the

signal. Distortion is the variations of a signal from its true form. The distortion may

result from either poor frequency response or poor phase shift response.

1.5 Error in experimental data

The data from the experimental measurement need to be analyses in order to obtain

the useful information. In order to get a good result, experimentalists need to know

the validity of the data. (What are the experimental errors and how to handle them?)

• True value - Almost impossible to obtain in practice


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• Measured value – value indicated by an instrument or result estimated from a

number of quantities measured.

• It should be followed by uncertainty in measurement.

Example:

l = (1.5 ± 0.1) cm

• Nominal value – value of the quantity specified by the manufacturer

It is normally followed by a tolerance specification

Example: R= 10 kΩ ± 10 %

• Static error

The difference between the measured value and the true value of the quantity

tmδ AAAtruevaluevaluemeasurederror

−= (1.15)

• Relative static error

tr A

Aδε = (1.16)

1.5.1 Types of experimental errors

1.5.1.1 Gross Errors

• Refer to errors due to human mistake in reading instruments and recording and

calculating measurement results.

• Example i: read the temperature as 31.5oC while the actual reading may be 21.5oC

• Example ii: read 25.8oC and record as 28.5oC

• Prevention of the error: Read and record carefully, take several readings, look at

the set of readings and discard readings which can be identified as those caused by

gross error similar to those in the examples mentioned above.

1.5.1.2 Systematic Errors/ Bias Error

a) Instrument errors

i) Due to inherent shortcomings of instruments

• Inherent due to their mechanical structure.

• May be due to construction, manufacturing limitations, calibration, etc.


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• e.g., If the spring (used for producing controlling torque) of a permanent magnet

instrument has become weak, the instrument will always read high.

• Methods of overcoming the error

- re-calibrate the instrument carefully, as often as may be required (by comparison

with an instrument of higher accuracy).

- Apply correction factors after determining the instrument errors

ii) Due to misuse of instrument

• e.g. : failure to adjust the zero of instruments

iii) Due to loading effect of instruments

• Loading effect causes inaccuracy of measurement, even if there is no error in

the instrument readings

• This error can be minimised by using an appropriate instrument i.e., using a

voltmeter which has a relatively high resistance compared to the load resistance.

(b) Environmental Errors

Example: effects of temperature, pressure, humidity, dust, vibrations or external

magnetic or electrostatic fields.

Method of overcome this errors:

i) Keeping the conditions as nearly constant as possible.

E.g.: temperature can be kept constant by keeping the equipment in a temperature-

controlled enclosure.

ii) Use equipment that is immune to these effects

Example: variations of resistance with temperature can be minimized by using

resistance materials which have a very low resistance temperature co-efficient

iii) Employ techniques that eliminate the effects of disturbances

Example: effect of humidity & dust can be entirely eliminated by hermetically sealing

the equipment

iv) Apply computed correction


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(c) Observational Errors

i) Parallax error

ii) Reaction time

1.6 Uncertainty Analysis

In experimental measurement occurs some uncertainty, for example:

a certain pressure reading might be expressed as

p= 100 kPa ±1 kPa .

The ‘±’ notation is the uncertainty.

If a function R is depends on several independent variable (x1, x2, x3, …, xn),

R=R(x1, x2, x3, …, xn) (1.16)

then the uncertainty of the function R, wR is

21

22

33

2

22

2

11

....

∂∂

++

∂∂

+

∂∂

+

∂∂

= nn

R wxRw

xRw

xRw

xRw (1.17)

where w1, w2, w3, …, wn be the uncertainties in the independent variable.

Uncertainties for product functions

in some cases, the function is in the form of product function;

R= x1 a1x2

a2…xn an (1.16b)

and

ni an

ai

aa

ixxaxx

xR )...( 121

121−=

∂∂ (1.18)

dividing by R from (1.16b):

i

i

i xa

xR

R=

∂∂1 (1.19)

inserting into (1.17) gives

21

2

= ∑

i

xiiR

xwa

Rw

(1.20)

è this is the uncertainty for product function.


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Uncertainties for additive functions

For function in the form of additive function, R will be expressed as

∑=+++= iinn xaxaxaxaR ...2211 (1.21)

and

ii

axR

=∂∂ (1.22)

The uncertainty of additive function can be expressed as:

( )[ ] 212

21

22

∑∑ =

∂∂

= iixR wawxiRw

i (1.23)

Example 1.3

The resistance of a certain size of copper wire is given as

( )[ ]2010 −+= TαRR

where R0 = 6 Ω ± 0.3 percent is the resistance at 20oC, α =0.004oC-1 ±1 percent is the

temperature coefficient of resistance, and the temperature of the wire is T = 30±1oC.

Calculate the resistance of the wire and it’s uncertainly.

Solution

The nominal resistance is ( )[ ] Ω=−+= 24.6)2030(004.01)6(R

04.1)2030)(004.0(1)20(10

=−+=−+=∂∂ TαRR

)20(0 −=∂∂ TRαR

024.0)004.0)(6(0 ===∂∂ αRTR

Ω== 018.0)003.0)(6(0Rw

15104)01.0)(004.0( −−×== Coαw

Co1=Tw

The uncertainty in the resistance is

( ) ( )[ ] %.Ω or .WR 49003050)1()024.0()104()60(018.004.1 21

2225222 =+×+= −


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Example 1.4:

The two resistor R and Rs are connected in series as shown in the Fig 1.8. The voltage

drops across each resistor are measured as

%)1(1.010 VV ±=E

%)467.0(005.02.1 VV ±=sE

along with a value of %4/10066.0 ±Ω=SR

Determine the power dissipated in resistor R and its uncertainty

Fig 1.8 Resisitors in series

Solution

The power dissipated in resistor R is P=EI

The current through both resistors is I=Es/R, so that

s

s

REEP =

The nominal value of the power is therefore

P=(10)(1.2)/(0.0066)=1818.2W

( ) ( ) ( ) 011.00025.0)1(2.1

005.0110

1.012

1

222

22

2

21

222

=

−+

+

=

+

+

=

s

RR

s

EEEEP

Rwa

Ewa

Ewa

Pw ssss

then wp=(0.0111)(1818.2)= 20.18 W

1.7 Statistical Analysis of Experimental Data

When a set of readings is taken from an experiment. The individual reading will vary

from each other. If each reading is denoted by xi, and there are n readings, the

arithmetic mean is


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∑=

=n

iim x

nx

1

1 (1.24)

The deviation di for each reading is defined by

mii xxd −= (1.25)

the standard deviation or root-mean-square deviation is defined by

( )

−= ∑

=

n

imi xx

nσ

1

21 (1.26)

and the square of the standard deviation is called variance.

1.7.1 Method of Least Squares

Suppose we have a set of x (x1, x2, x3 …xn). The sum of the squares of their deviations

from some mean value is

( )∑=

−=n

imi xxS

1

2 (1.27)

and the mean value xm is

∑=

=n

iim x

nx

1

1 (1.28)

To minimize S with respect to the mean value xm,

−−==

∂∂ ∑

=

n

imi

mnxx

xS

1)(20 (1.29)

where n is the number of observations.

In experiment that involved of two variables x and y, an equation is used to

express their relationship. However, to get the best function from the scatter data is

not easy. The least squares method can use to obtain a better function.

For the equation y= ax + b

To minimize the quantity

[ ]∑=

+−=n

iii baxyS

1

2)( (1.30)

Setting the derivatives with respect to a and b equal to zero;

Further reading: EXAMPLE 3.7, page 64, Experimental Methods for Engineers, J.P. Holman


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∑ ∑=+ ii yxanb (1.31)

∑ ∑ ∑=+ iiii yxxaxb 2 (1.32)

Solving 1.31 and 1.32 gives

( )( )( )∑ ∑

∑ ∑∑−

−= 22

ii

iiii

xxn

yxyxna (1.33)

( )( ) ( )( )( )∑ ∑

∑∑∑∑−

−= 22

2

ii

iiiii

xxn

xyxxyb (1.34)

Designing the computed value of y as y ,

baxy +=ˆ

and the standard error of estimate of y for the data is

Standard error =( ) ( ) 2

122

12

22ˆ

−−−

=

−

−∑n

baxyn

yy iiii (1.35)

1.7.2 Regression Analysis

To consider the experimental uncertainties for xi and yi, the following should be take

note;

1) If the values of xi and yi are taken as the data value in y and the value of x on

the fitted curve for the same value of y, then there is a presumption that the

uncertainty in x is large compared with that in y. (regression of x on y)

2) If the values of xi and yi are taken as the data value in y and the value on the

fitted curve for the same value of x, the presumption is that the uncertainty in y

dominates. (regression of y on x)

3) If the uncertainly in xi and yi are believed to be of approximately equal

magnitude, a special averaging technique must be used.

Example 1.5

From the following data obtain y as a linear function of x using the method of least

square.

Further reading: The method of least square in higher order polynomial, page 92, Experimental Methods for Engineers, J.P. Holman


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Solution

To get a equation in the form of y=mx +c.

then ( )( )

( )540.022

=−

−=

∑ ∑∑ ∑∑

ii

iiii

xxn

yxyxna

and ( )( ) ( )( )

( )879.022

2

=−

−=

∑ ∑∑∑∑∑

ii

iiiii

xxn

xyxxyb

Thus, the equation is

y = 0.540 x + 0.879

1.8 Equation and Numbering

It is important to have a right concept to write an equation. For an example:

Y=MX + C

Where M and C are the constants. X and Y are the variables.

X is always refers as the input/ changes to the system. (Always put on the right side of

the equation.)

whereas, Y is always refers as the result/effect that cause by the changing of the X.

(Always put on the left side of the equation and also always as a single term)

# The same principles apply in plot a scientific graph. X-axis is always taken for the

independent variable (cause) and Y-axis is chosen for the dependent variable (result or

effect).

1.9 Graphical Analysis of Data

yi xi 1.2 1.0 2.0 1.6 2.4 3.4 3.5 4.0 3.5 5.2

∑yi=12.6 ∑xi=15.2

yi xi xiyi xi2

1.2 1.0 1.2 1.0 2.0 1.6 3.2 2.56 2.4 3.4 8.16 11.56 3.5 4.0 14.0 16.0 3.5 5.2 18.2 27.04

∑yi=12.6 ∑xi=15.2 ∑xiyi=44.76 ∑xi2=58.16


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The raw data obtained from an experiment need to be plotted into a graph for further

analysis. Therefore, curve-plotting technique is importance in order to give a good

understanding of the experimental results. The plotting methods for different type of

function are showed in Table 1.4. Table 1.4 Methods of plotting various functions to obtain straight lines

Further reading: page 108-109, Experimental Methods for Engineers, J.P. Holman


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1.9.1 Choice of graph format

To present the experimental data, several graph formats can be used; they are x-y

graphs, bar charts, column charts, pie charts etc. However, the x-y graph is the most

frequently used format in engineering. The variations of x-y graph are showed in the

Figure 1.9.

_______________________________________________________

References:

1. J. P. Holman, “Experimental Methods for Engineers”, McGraw-Hill International Edition.

2. E.O. Doebelin, “Measurement Systems: Application and Design”, McGraw-Hill International Edition.


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Fig 1.9: Choices of x-y graph formats.

a) with data marker, no connecting line b) with data marker, joined by smooth curve c) without data marker, joined by smooth curve d) with data marker, joined by straight line e) without data marker, joined by straight line f) with data marker, joined by correlation curve

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2015 Chap.1 Lec.notes 2014-2015 Trim.3,EME2066 Chapter 1 Basic Concepts and Analysis of Data

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