dynamic characteristics of measuring systems

7/22/2019 Dynamic Characteristics of Measuring Systems

1/102

DYNAMIC CHARACTERISTIC OFMEASURING INSTRUMENTS

The input varies time to time so does the output.The behavior of the system under such conditions

is described by its dynamic response.

When dynamic or time varying quantities are to be

measured, it is necessary to find the dynamic

response characteristics of the instrument being

used for measurement.


2/102

The periodic signal varies with time and repeats after a constantinterval. The input may be of harmonic or non-harmonic type.

The transient signal varies non cyclically with time. The signal is ofdefinite duration and becomes zero after a certain period of time.

The Random signal varies randomly with with time, with no definiteperiod and amplitude. This may be continuous, but not cyclic.

Dynamic Input

Periodic

Transient

Random

Harmonic Non-harmonic


3/102

..dynamic characteristic of measuring instruments

Speed of Response

Rapidity with which a measurement system responds tochanges in the measured quantity.

Measuring Lag

Delay in the response to a change in input. It may be due tocapacity, inertia or resistance.

retardation typeresponse of the system begins immediately after a changein measured quantity.

time delay type

response begin after a dead time after the application ofinput.


4/102

Dynamic System Response

Input/Output Model of Linear Dynamic Systems Time Response of Dynamic Systems

Solutions to Differential Equations

Transient and Steady State Response

Frequency Response of Dynamic Systems

Review of Complex Variables

Frequency Response Function

Gain and Phase Characteristics

System Integration


5/102

Linear Systems

Satisfies the Superposition Principal. Can be modeled by Linear Ordinary Differential Equations.

Input a sinusoidal signal of frequency f1, the output will be asinusoidal signal with the same frequency f1.

Linear SystemLinear SystemInput Output

x1(t) y1(t)

x2(t) y2(t)

A x1(t) + B x2(t)

Inputs

Simple

Input

dComplicate

Complicated

Output


6/102

Generalized Mathematical Model of a Measuring System

The most widely useful mathematical model for the studyof measurement-system dynamic response is theordinary differential equation with constant coefficients.

ioi

m

im

mm

im

m

ooo

n

o

n

nn

o

n

n

qbdt

dqb

dt

qdb

dt

qdb

qadt

dqa

dt

qda

dt

qda

++++=

++++

11

1

1

11

1

1

......

......

( ) ( ) ( )y t y t y tP H= +ParticularSolution

(Steady State Solution)

HomogeneousSolution

(Transient Solution)

123 123


7/102

( ) io11m1mmmoo1

1n

1n

n

n

qaDa....DaDa

qaDa....DaDa

++++=

++++

opiocfo qqq +=

With D operator method the complete solution is obtained in two parts

0aDa......DaDa o11n

1n

n

n =++++

The qocf

has n arbitrary constants; qopi

has none. These n

arbitrary constants may be evaluated numerically by

imposing n initial conditions. The solution qocf is obtained

by calculating the n roots of the characteristic equation


8/102

For each real unrepeated root s one term ofsolution is written as Cest, where C is anarbitrary constant. Thus, for example roots 1.5,+2.5 and 0 give a solution

Real Roots Unrepeated

3

t5.2

2

t5.1

1 CeCeC +++


9/102

Real Root Repeated

For each root s which appears p times, the solution

is written asst1p

1p

2

21o e)tC....tCtC(C

++++

( ) tCCetCtCCetCC 65t22

432

t

1o ++++++

Thus, if the roots are 1, -1, +2, +2, +2, 0 ,0, thesolution is written as.


10/102

Complex Root Unrepeated

A complex root has a general form a+ib . They are

always in pairs aib. For each such root pair, thecorresponding solution is

For example -2i4, 3i5 and 0i7 give a solution

)sin(btCe

at

+

( ) ( ) ( )2213

12 7sin5sin4sin +++++ tCteCteC to

to


11/102

Complex Root Repeated

( ) ( )

( )

( )1112

22

11

sin....

sin

sinsin

++

++

+++

patp

p

at

ato

ato

btetC

btetC

btteCbteC


12/102

Operational Transfer Function

o1

1n

1n

n

n

o1

1m

1m

m

m

i

o

aDa.............DaDa

bDb..........DbDb(D)

q

q

+++

+++=

Sinusoidal Transfer Function

( )

( ) ( )

( ) ( ) o11n

1nn

n

o11m

1mm

m

i

o

aia.......iaia

bib.....ibib

q

q

+++

+++

=

i


13/102

Operational transfer function

The frequency

response of a systemconsists of curves of

amplitude ratio and

phase shift as afunction of

frequency.


14/102

Zero Order Instruments


15/102

zeroth order instruments

No mathematical law can exactly represent

any system. There no pure resistance. Potentiometer shall

have some inductance or capacitance

Mechanical loading due to sliding contact.


16/102

FIRST ORDER INSTRUMENTS

ioooo

1 qbqadt

dqa =+

i

o

oo

o

o

1 qa

bq

dt

dq

a

a=+

ioi

m

im

mm

im

m

ooo

n

on

nn

on

n

qbdt

dqb

dt

qdb

dt

qdb

qadt

dqa

dt

qda

dt

qda

++++=

++++

11

1

1

11

1

1

......

......


17/102

First Order Instruments

1D

K

(D)q

q

i

o

+=

( ) io Kqq1D =+

seca

a)(constanttime

put)(output/ina

b(K)ysensitivitStatic

o

1

o

o

=

=


18/102

First Order Instruments

xo displacement from the reference mark

temperature of fluid in the bulb; Ttf=0when xo=0

expansion coefficient of thethermometer fluid and bulb glass

Vb volume of bulb

Ac cross section area of capillaryTi Temperature of measurand

Ab heat transfer area of the bulb

tf

c

bexo T

A

Vkx =


19/102

First order instruments: Thermometer

Heat in Heat out = energy stored

( ) tfbtfib CdTV0dtTTUA =

Assumptions The bulb wall and the fluid film on both sides is a pure

resistance to heat transfer with no heat storage capacity.

U is constant. The film coefficient and the bulb wallconductivity does not change with temperature.

Ab is constant but the change may take place by contractionor expansion.

No heat loss by thermometer by conduction upstream.

Mass of fluid inside the bulb is constant.

The specific heat of fluid is constant.


20/102

First order instruments: Thermometer

ibtfbft

b TUATUAdt

dTCV =+

ibo

bex

cbo

ex

c TUAxVK

AUA

dt

dx

K

CA=+

Cm/A

VKK o

c

bex=

sUA

CV

b

b

=


21/102

( ) iso Kqq1D =+

t

ocf Ceq

=

isopi Kqq =

is

t

o KqCeq +=

Step Response of First Order Instruments

initial conditions

qo= qi = 0 at t = 0

For t>=0; qo=qi

Time, t

qi


22/102

Step response of first order instruments

is

is

KqC

kqC0

=

+=

=

t

iso e1Kqq

which finally gives

Applying initial conditions we get

t

is

o e1kq

q =(in non-dimensional form)


23/102

Kqqeerror,tmeasuremen oim =

=

t

isism e1qqe

t

is

m eqe

=

1. the response is faster for a small value of time

constant

2. setting time is the time for the instrument to reach andstay within a tolerance band around the final value



24/102


time, t

qis

qi

1

2 1


25/102


t/

t/


26/102

=

=tq

0qq

is

o

i& 0t

0t

( ) tqKq1D iso &=+

Ramp Response of First Order Instruments

t

ocf Ceq

=

( ) ( ) ( )tqKdt

tqdtqKtqD1Kq isis

isisopi =

== &&


27/102

Ramp response of first order instruments

( )tqKCeq is

t

o +=

&

+= teqKq

t

iso&

{

ssm,

errorstatesteady

is

tm,

errortransient

t

iso

im

e

q

e

eq

K

qqe

&

43421 +

==

Measurement error

Applying initial conditions


28/102


29/102

Numerical-1

A thermometer is initially at a temperature of

20o

C and is suddenly plunged into a liquidbath, which is maintained at 150 oC. Thethermometer indicated 95 oC after the intervalof 3 seconds. Estimate the time constant ofthermometer. Also indicate temperature after5 time constants and comment upon thisresult.

Ans: 3.49 s, 149.1 oC


30/102

Numerical-2

A thermometer is sudenly subjected to a stepinput of 200 oC from 0 oC. Calculate the

temperature indicated by the thermometer aftera time of 1.5 seconds. The thermometer may beidealized as a first order system with a timeconstant of 2.5 seconds. Would there be anychange in the indicated temperature if thethermometer was initially held at 25 oC?

Ans: 90.2 oC; 103.95 oC


31/102

Numerical-3

A temperature sensitive transducer used to

measure the temperature of a furnace has beenidealized as a first order system subjected toramp input. Calculate the time constant of thetransducer if the furnace temperature increasesat a rate of 0.15 oC/s. The maximumpermissible error in temperature measurementis limited to 4.5 oC.

Ans: 30 s


32/102

Numerical-4

A weather balloon carrying a temperature sensingdevise with time constant of 10 s, rises through theatmosphere at 6 m/s. The balloon transmitsinformation about temperature and altitude throughradio signals. At 3000 m height, a temperature

indication of 35 oC has been received. Determine thetrue altitude at which 35 oC temperature occurs. Itmay be presumed that the temperature sensing

device is of the first order and that the temperaturevaries with altitude at a uniform rate of 0.01 oC/m.

Ans: 2940 m


33/102

( )T

KAKqqD io ==+ io KqqD

=t

o Ceq

Impulse Response of First Order Instruments


34/102

Impulse response of first order instruments

T

T

Te

e1KA

C

=

T

t

T

o

Te

ee1KA

q

=

by imposing initial conditions

&

For T 0 as the case is of impulse response

T

T

0Timt

t

T

T

0Timo0T

Te

e1KAee

Te

e1KAqlim

=

= LL


35/102


, an indeterminate form0

0

T

e1lim

T

0T =

( )

1

1

e

1

limT

e1lim

T

0T

T

0T ==

t

o e

KAq

=

Thus, finally for the impulse response of a first order-

instrument

Applying LHospitals rule


36/102


We take A=1 and T=0.01.

( ) tT

e

ee1K100

Tt0e1K100

q

01.0

t01.0

t

o

=

The shape of the pulse is immaterial for a short duration

ioo Kqq

dt

dq=+

++

=++ 0

0

i

0

0

o

0

0

o dtKqdtqdqKqq oo =++ 000

Kq

0o =+

(area under qi curve from t=0 to t=0+)

(area under impulse)


37/102


t/qo/(k/) qo/(k/)

( )

k

q o

t/


38/102


39/102

( ) ( )

( ) ( ) o11n

1n

n

n

o1

1m

1m

m

m

i

o

aia.......iaia

bib.....ibib)(i

q

q

+++

+++=

1

K)i(

q

q

A

A

22i

o

i

o

+

==

( )tanangle,phase 1 =

The ideal frequency response (zero order instrument) would have

( ) oi

o 0Kiq

q=

Frequency Response of First Order Instruments



40/102


qiis often combination of several sine waves of different frequency

t20sin3.0t2sin1q i +=

Suppose the is 0.2 sec. Since, this is a linear system,

we may use superposition principle to find qo.

( ) ( )

( ) ( ) o120i

o

o1

2

i

o

76K93.02.0x20tan116

Ki

q

q

8.21K93.02.0x2tan116.0

Ki

q

q

=+

=

=+

=

=

=

( )( ) ( ) ( )( ) ( )

( ) ( ) io

o

q76t20sin72.08.21t2sin93.0

K

q

76t20sinK24.03.08.21t2sinK93.01q

=+=

+=


41/102

Numerical 5

A signal prescribed by the following relation

is required to be measured by using firstorder system having a time constant of 0.1 s.Develop an expression for the corresponding

output. Comment on the result.

tti 10cos4.02sin3 +=



42/102



43/102

SECOND ORDER INSTRUMENTS

A second order instrument is one that follows the equation

ioooo

12

o2

2 qbqadt

dqadt

qda =++

o

o

a

bK,ysensitivitstatic =

a

afrequency,naturalundamped2

on =

aa2

aratio,damping2o

1=

Second Order Instruments


44/102


io

n

2

n

2

Kqq1

D2

Dgiveswhich =

++

1

D2

DK(D)

qqisfunctiontransferloperationa

n

2

n

2

i

o

++=



45/102

2

o

2

oso

idt

xdMxKdt

dxBf =

( ) ios2

fxKBDMD =++

s

K

1K=

M

K sn =

MK2

B

s

=



46/102

Step Response of Second Order Instruments

io

n

2

n

2

Kqq1

D2

D=

++

0tat0dt

dq

0tat0q

o

o

==

==+

+

isopi Kqq =

01

D2

D

n

2

n

2

=++

Step response of second order instruments


47/102

( ) 1et1Kq

q tn

is

o n

++=

critically damped ( real repeated)

( )21

n

2

2

tw

is

o

1sin

1t1sin1

eKqq

n

=

++

=

under damped (complex)


1e

12

1e

12

1

Kq

q t1

2

2t1

2

2

is

o n2

n2

+

+

+=

+

over damped ( real and unrepeated)



48/102

Nondimensional step-function response of second-order instrumentnt

S f d d 0


49/102

Step response of second order system n=10



50/102


Frequency of under damped oscillation

2

nd 1 =

Dynamic error in the measurement, em

( ) isn2

2

t

oi qt1sin

1e

Kqq

n

+=

Steady state error for the second order system for step input is zero.

0== mtss ee



51/102


The response of second order underdamped system

is sinusoid with a decaying amplitude. For = 0

tcos12

tsin1Kq

q

is

o

=

+=

Thus system has constant oscillations.

For>1 there are no oscillations but system is highly

sluggish in response.


52/102

Step response of

second orderinstruments



53/102


Time domain specifications

how fast the system moves to follow the applied input?

how oscillatory is the system? how long will it take the system to practically reach its

final steady state value?

Rise TimeTime required by the system to rise from 0 to 100 percent ofits final value.

( ) 11cost1sin1

e

Kq

qn

2

2

tw

is

on

=++

=



54/102


2

n

1

r

1

costtime,rise

=

Peak Time, tp

It is time required for the output to reach the peak of timeresponse or peak overshoot.Differentiate the equation and put derivative equal to zero

0t1sin 2n = ,...3,2,,0t1 2n =

2

n

p

1t

=



55/102


peak overshoot, Mp

is the difference of output and the input at tp

( ) 11t1sin1

e1

Kq

qn

2

2

Ptw

is

on

++

=

21

p eM

=



56/102


Settling Time

for 2% tolerance band 02.0

1

e

2

t sn

=

approximate solution is

n

s

4t =

for 5% band

n

s

3t

=

Terminated Ramp Response of


57/102

Terminated Ramp Response of


The devices with

extremely high

natural frequency

and very light

damping (


58/102

io

n

2

n

2

Kqq1D2D

=

+

+

=tT0.1

Tt0T

t

q i

0at t0dt

dqq oo

+===Since we are concerned with thelightly damped systems, weobtain the solution for only the

underdamped case

( )+

+

= t1sine

1T

1

T

2

T

t

K

qn

2t

2

nn

o n

Tt0

Terminated ramp response of second order instruments


59/102

p p

( )

( ) ( )( )

+

+

+

+=

Tt1sine1T1

T

21

T

t

t1sine1T

1

T

2

T

t

K

q

n2Tt

2

n

n

n

2t

2

nn

o

n

n

tT

=

21 1tan2


60/102

Terminated Ramp Response of Lightly Damped System

Terminated ramp response of second order instruments


61/102

Tt0For There is steady state error of size (2/nT)

The transient error can be no larger than 21/1 tn

Thus if=0 (no damping), the steady state error is zeroand transient error is a sustained sine wave ofAmplitude 1/(nT). Therefore ifn is sufficiently largerelative to 1/T. The transient error can be made very

small even if the damping is Practically nonexistent.

Ramp Response of Second-Order Instruments


62/102

p p

tqKq1D2D

iso

n

2

n

2

&=

+

+

+=== 0at t0

dt

dqq oo

overdamped

++

+

=

+

+

t1

2

22

t1

2

22

n

is

is

o

n2

n2

e14

1212

e14

12121

q2

tqK

q &

&

Ramp response of second-order instruments


63/102

+

= 2

t1e1

q2tq

K

q ntw

n

is

is

o n&

( )

+

=

t1sin12

e1

q2tq

K

qn

2

2

tw

n

isi

on&

&

12

12tan

2

2

=

critically damped

under damped



64/102

Ramp response of second order instrument



65/102

Nondimensional ramp response

Impulse Response of Second OrderInstruments


66/102

0q1D2D

o

n

2

n

2

=

+

+

=

+ t1t1

2n

o n2

n2

ee

12

1

KA

q

Over damped

Instruments

0at tdt

dq; 2 === n

o KAod

Impulse response of second order instruments


67/102

tn

n

o nteKA

q=

( )t1sine11

KA

qn

2t

2n

o n

=

critically damped

under damped

Impulse response of second order instruments


68/102

Non-dimensional Impulse Response of Second Order Instrument

Frequency Response of Second OrderInstruments


69/102

( )

1

i

2

i

Ki

q

q

n

2

n

i

o

+

+

=

+

=2

n

22

n

i

o

21

1)i(q

K

q

Instruments

n

n

1 2tan

=


70/102

Frequency response of second-order instrument

01M0 ===

n


71/102

-1800M

-9021M0

===

===

n

n

Frequency response characteristics of second order systems

Resonance Frequency


72/102

q y

The frequency at which M has the highest value is known

as resonant frequency, when

21frequencyresonant 2nr =

0

d

dM

n

=

Resonance Peak


73/102

1whenM,ofvalueMaximum =n

2

nr 21 ==

Band width


74/102

The frequency at which M = 0.707 is called cutoff

frequency c

Above this frequency the M reduces below 0.707 or(1/2)0.5. The frequency c represents half power

point. The band of frequency between zero to cutoff

frequency is called bandwidth of the system, b

Measurement systems are low pass filters as the

amplitude ratio is unity at =0. As the frequency ofinput signal increases the, the output gets attenuated.Bandwidth is ,therefore, indicative of the satisfactory

reproduction of output signal.

Dead Time Element


75/102

The dead time element is defined as a system in which

the output is exactly same form as input but occurs dt

seconds (dead time) later.

( ) ( )tkqtq dtio =

This type of delay is also known as pure delay or transport lag

HIGHER ORDER SYSTEMS


76/102

When the order of the governing equation of aninstrument of a combination of instruments is high,

it is convenient to plot the frequency response ofthe system by logarithmic plots, known as Bodediagram.

The advantage of this method is that the frequencyresponse of a complex system can be obtained by

adding the response due to various first andsecond order terms occurring in the transferfunction of the system.


77/102

Bode Plot


78/102

Usually, decibel notation is used to express Decibel value = 20 log M

321 MlogMlogMlogMlog ++=

( ) ( )321321i

o MMMiq

q++=

For the first order system, with governing equation

( ) io qqD1 =+

tAq ii sin=

Bode Plot


79/102

Bode Diagram ForFirst Order System

Bode Plot


80/102

1

K

A

AMratio,Amplitude22

i

o

+==

( )dB1log10Klog201

K20logdecibelsinM

22

22

+=

+=

-40dBM100

-20dB;M10-3dB;M1

0;M1,

=

==

=


81/102

Similarly the Bode diagram for a second order systemwith governing equation

tsinAqq1

D2Diio

n

2

n

2

==

++

( ) ( )

( ) ( ){ }222

222i

o

r2r1log10

r2r1

1log20AAlog20M

+=

+==


82/102


83/102


84/102

Figure For The Problem

PERIODIC INPUT


85/102

Non-harmonic Signals

Fourier series for a periodic function qi(t) is

++= nt

T

2sinbnt

T

2cosaa

2

1q nnoi

Where, T is the time period

( )dttqT

2a

2

T

2

T

io +

= ( ) dtntT2

costqT

2a in

=

( ) dtnt

T

2sintq

T

2b in

=

Non-harmonic signals


86/102

For the square wave signal

( )

20,

02T-,

T

tC

tCtqi


87/102

0dtntT

2cosCdtnt

T

2cosC

T

2a

0

2T

2

T

0

n =

+=

( )[ ]ncosncos2n

C

dtntT

2sinCdtntT

2sinCT

2b0

2

T

2

T

0

n

=

+=

Non-harmonic signals


88/102

2,4,6...nfor0,

...5,3,1,nfor,n

C4

==

==

( ) 5...3,1,n,Tnt2sin

nC4tq,Thus i ==

RANDOM INPUT


89/102

Random signal does not have a definite time periodor amplitude and has to be described statistically.

Only stationary random signals will be discussedhere. It is possible to describe such signalsstatistically over a certain period of time.

Random Signal

The statistical properties that are of rele ance are


90/102

The statistical properties that are of relevance are :(i) mean or average value of the random signal,(ii) rms value,(iii) mean square spectral density, and(iv)auto-correlation function.

( )

( ) dttqT2

1lim

tqsignalrandomofMean value

T

T

iT

i

+

=

Random Signal


91/102

The signal has to be fad in the filter which allows a

component of a signal center frequency c, with asmall bandwidth to pass through it. Ideally should be as small as possible

The filter output is then squared and averaged over a

certain time interval. Thus, we get the mean squarevalue of the component of a certain frequency c.

By varying the center frequency, the mean square

value corresponding to various frequencycomponents is obtained.

Random Signal


92/102

( )

=

2qS

S() represents thedensity, i.e. the amountper unit frequencyband width of the meansquare value.

From the mean square

spectral density S() isdefined as:

Random Signal

Consider an instrument with frequency response


93/102

Consider an instrument with frequency responsefunction M() and input signal

tsinqqi

=

Output response,

( ) ( ) tqMtq io sin=

Mean square value of input signal,

( ) =T

0

2i

2i dttsinq

T1tq

Random Signal

Mean square value of output signal


94/102

Mean square value of output signal,

( ) ( )[ ]

( )[ ] ( ){ }txM

dttsinqMT

1tq

22

T

0

22

i

22

o

=

=

the mean spectral sensitivity of the output signalSo() is given by

( ) ( )[ ] ( ) i2

o SMS =

COMPENSATION

In order to improve the dynamic characteristics of a


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In order to improve the dynamic characteristics of ameasuring system, compensation is employed. Thisinvolves use of additional elements.

First order system compensation

Governing equation for thermocouples is

( ) ( )tKqD io =+1

In order to reduce the effective value of time

constant, the voltage V1 can be applied to acircuit whose output is V2. The relationshipbetween V2 and V1 can be easily derived

First order compensation


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( )RiiRiV 2132 +==

o

211

R

VVi

=

( )dt

dD,VVCDi 212 ==

( ) RVVCDR

VVV 21

o

212

+

=


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Second Order System

The damping in the second order system affects the


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The damping in the second order system affects theoutput response considerably. Usually, damping inpractice is small and may be increased by additional

means, like use of compensation network.

Second order system


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i

CD

1LDRRV o1

+++=

o12 iRVV =

( )1

D2

D

1D

2D

1CDRRLCD1RCDLCD

VV

1

2

2

1

1

1

2

1

o

2

2

1

2

+

+

+

+

=+++

++=


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dynamic characteristics of measuring systems

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