Post on 10-Feb-2018
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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.
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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
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..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.
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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
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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
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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
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( ) 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
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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 +++
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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.
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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
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Complex Root Repeated
( ) ( )
( )
( )1112
22
11
sin....
sin
sinsin
++
++
+++
patp
p
at
ato
ato
btetC
btetC
btteCbteC
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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
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Operational transfer function
The frequency
response of a systemconsists of curves of
amplitude ratio and
phase shift as afunction of
frequency.
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Zero Order Instruments
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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.
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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
......
......
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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
=
=
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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 =
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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.
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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
=
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( ) 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
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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)
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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
Step response of first order instruments
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Step response of first order instruments
time, t
qis
qi
1
2 1
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Step response of first order instruments
t/
t/
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=
=tq
0qq
is
o
i& 0t
0t
( ) tqKq1D iso &=+
Ramp Response of First Order Instruments
t
ocf Ceq
=
( ) ( ) ( )tqKdt
tqdtqKtqD1Kq isis
isisopi =
== &&
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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
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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
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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
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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
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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
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( )T
KAKqqD io ==+ io KqqD
=t
o Ceq
Impulse Response of First Order Instruments
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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
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Impulse response of first order instruments
, 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
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Impulse response of first order instruments
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)
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Impulse response of first order instruments
t/qo/(k/) qo/(k/)
( )
k
q o
t/
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( ) ( )
( ) ( ) 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
Frequency Response of First Order Instruments
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Frequency Response of First Order Instruments
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
=+=
+=
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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 +=
Frequency Response of First Order Instruments
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Frequency Response of First Order Instruments
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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
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Second Order Instruments
io
n
2
n
2
Kqq1
D2
Dgiveswhich =
++
1
D2
DK(D)
qqisfunctiontransferloperationa
n
2
n
2
i
o
++=
Second Order Instruments
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2
o
2
oso
idt
xdMxKdt
dxBf =
( ) ios2
fxKBDMD =++
s
K
1K=
M
K sn =
MK2
B
s
=
Second Order Instruments
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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
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( ) 1et1Kq
q tn
is
o n
++=
critically damped ( real repeated)
( )21
n
2
2
tw
is
o
1sin
1t1sin1
eKqq
n
=
++
=
under damped (complex)
Step response of second order instruments
1e
12
1e
12
1
Kq
q t1
2
2t1
2
2
is
o n2
n2
+
+
+=
+
over damped ( real and unrepeated)
Step response of second order instruments
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Nondimensional step-function response of second-order instrumentnt
S f d d 0
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Step response of second order system n=10
Step response of second order instruments
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Step response of second order instruments
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
Step response of second order instruments
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Step response of second order instruments
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.
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Step response of
second orderinstruments
Step response of second order instruments
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Step response of second order instruments
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
=++
=
Step response of second order instruments
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Step response of second order instruments
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
=
Step response of second order instruments
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Step response of second order instruments
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
=
Step response of second order instruments
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Step response of second order instruments
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
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Terminated Ramp Response of
Second Order Instruments
The devices with
extremely high
natural frequency
and very light
damping (
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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
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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
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Terminated Ramp Response of Lightly Damped System
Terminated ramp response of second order instruments
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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
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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
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+
= 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
Ramp response of second-order instruments
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Ramp response of second order instrument
Ramp response of second-order instruments
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Nondimensional ramp response
Impulse Response of Second OrderInstruments
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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
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tn
n
o nteKA
q=
( )t1sine11
KA
qn
2t
2n
o n
=
critically damped
under damped
Impulse response of second order instruments
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Non-dimensional Impulse Response of Second Order Instrument
Frequency Response of Second OrderInstruments
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( )
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
=
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Frequency response of second-order instrument
01M0 ===
n
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-1800M
-9021M0
===
===
n
n
Frequency response characteristics of second order systems
Resonance Frequency
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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
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1whenM,ofvalueMaximum =n
2
nr 21 ==
Band width
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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
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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
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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.
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Bode Plot
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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
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Bode Diagram ForFirst Order System
Bode Plot
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1
K
A
AMratio,Amplitude22
i
o
+==
( )dB1log10Klog201
K20logdecibelsinM
22
22
+=
+=
-40dBM100
-20dB;M10-3dB;M1
0;M1,
=
==
=
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Similarly the Bode diagram for a second order systemwith governing equation
tsinAqq1
D2Diio
n
2
n
2
==
++
( ) ( )
( ) ( ){ }222
222i
o
r2r1log10
r2r1
1log20AAlog20M
+=
+==
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Figure For The Problem
PERIODIC INPUT
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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
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For the square wave signal
( )
20,
02T-,
T
tC
tCtqi
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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
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2,4,6...nfor0,
...5,3,1,nfor,n
C4
==
==
( ) 5...3,1,n,Tnt2sin
nC4tq,Thus i ==
RANDOM INPUT
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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
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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
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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
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( )
=
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
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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
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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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