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Dynamic Characteristics
Dynamic characteristics tell us about how well a sensor responds to changes in its input. For dynamic signals, the sensor or the measurement system must be able to respond fast enough to keep up with the input signals.
Sensor or
system
Input signalx(t)
Output signaly(t)
In many situations, we must use y(t) to infer x(t), therefore a qualitative understanding of the operation that the sensor or measurement system performs is imperative to understanding the input signal correctly.
General Model For A Measurement System
nth Order ordinary linear differential equation with constant coefficient
Where m ≤ ny(t) = output from the systemx(t) = input to the system
t = timea’s and b’s = system physical parameters, assumed constant
)()()()()()()()(011
1
1011
1
1 txbdt
tdxbdt
txdbdt
txdbtyadt
tdyadt
tydadt
tyda m
m
mm
m
mn
n
nn
n
n ++++=++++ −
−
−−
−
− LL
Measurement system
x(t) y(t)
y(0)
F(t) = forcing function
The solution
Where yocf = complementary-function part of solutionyopi = particular-integral part of solution
opiocf yyty +=)(
Complementary-Function Solution
The solution yocf is obtained by calculating the n roots of the algebraic characteristic equation
11 1 0... 0n n
n na D a D a D a−−+ + + + =Characteristic equation
Roots of the characteristic equation: 1 2, ,..., nD s s s=
1. Real roots, unrepeated:
2. Real roots, repeated: each root s which appear p times
3. Complex roots, unrepeated:the complex form: a ± ib
4. Complex roots, repeated:each pair of complex root which appear p times
stCe
( )2 10 1 2 1... p st
pC C t C t C t e−−+ + + +
sin( )atCe bt φ+
20 0 1 1 2 2
11 1
[ sin( ) sin( ) sin( )
... sin( )]p atp p
C bt C t bt C t bt
C t bt e
φ φ φ
φ−− −
+ + + + +
+ + +
Complementary-function solution:
Particular Solution
Method of undetermined coefficients:
•After a certain-order derivative, all higher derivatives are zero.•After a certain-order derivative, all higher derivatives have the same functional form as some lower-order derivatives.
•Upon repeated differentiation, new functional forms continue to arise.
Where f(t) = the function that describes input quantityA, B, C = constant which can be found by substituting yopi into ODEs
Important Notes
...)()()( +′′+′+= tfCtfBtAfyopi
Zero-order Systems
The behavior is characterized by its static sensitivity, K and remains constant regardless of input frequency (ideal dynamic characteristic).
A linear potentiometer used as position sensor is a zero-order sensor.
Vr
xm
x = 0y = V
-
+here, /r r m
m
xV V K V xx
= ⋅ =
Where 0 ≤ x ≤ xm and Vr is a reference voltage
All the a’s and b’s other than a0 and b0 are zero.
where K = static sensitivity = b0/a0)()( 00 txbtya = )()( tKxty =
Where K = b0/a0 is the static sensitivityτ = a1/a0 is the system’s time constant (dimension of time)
All the a’s and b’s other than a1, a0 and b0 are zero.
First-Order Systems
)()()(001 txbtya
dttdya =+
)()()( tKxtydt
tdy=+τ 1
)(+
=DKD
xy
τ
q
Sensorm, Ttf, C
Surfacearea A Ti
Thermometer based on a mass,m with specified heat, C
Consider a thermometer based on a mass m =ρVwith specified heat C (J/kg.K), heat transmission area A, and (convection heat transfer coefficient U(W/m2.K).
(Heat in) – (Heat out) = Energy stored
Assume no heat loss from the thermometer
( ) 0i tf tfUA T T dt VCdTρ− − =
tftf i
dTVC UAT UAT
dtρ + =
Therefore, we can immediately define K =1 and τ = ρ VC/UA
First-Order Systems
iifif TT
dtdT
=+τ
Time, t-1 0 1 2 3 4 5
U(t)
0
1
2
First-Order Systems: Step Response
The complete solution:
Assume for t < 0, y = y0 , at time = 0 the input quantity, x increases instantly by an amount A. Therefore t > 0
yocf yopi
Applying the initial condition, we get C = y0-KA, thus gives
τ/0 )()( teKAyKAty −−+=
>≤
==000
)()(tAt
tAUtx
)()()( tKAUtydt
tdy=+τ
Transient response
Steady state response
KACety t += − τ/)(
First-Order Systems: Step Response
t/τ
0 1 2 3 4 5
Out
put S
igna
l , (y
(t)-y
0)/(K
A-y 0)
0.0
.2
.4
.6
.8
1.0
0.632
t/τ
0 1 2 3 4 5
Erro
r fra
ctio
n, e
m
0.0
.2
.4
.6
.8
1.0
0.368
Non-dimensional step response of first-order instrument
τ/
)0()( te
KAyKAty −=
−−
τ/
0
0 1)( teyKAyty −−=
−−
Here, we define the term error fraction as
τ/
0 )()0()()()()( t
m eyyyty
KAyKAtyte −=
∞−∞−
=−−
=
Determination of Time constant
t
0 1 2 3 4 5
Erro
r fra
ctio
n, e
m
.001
.01
.1
1
τ/
)0()( te
KAyKAty −=
−−0.368
τtee mm −== log3.2ln
Slope = -1/τ
τ/
)0()( t
m eKAyKAtye −=
−−
=
First-Order Systems: Ramp Response
The complete solution:
Applying the initial condition, gives
>≤
=0 0 0
)(ttqt
txis&
Therefore
)()( / ττ τ −+= − teqKty tis&
Measurement error
Transient error
Steady state error
Assume that at initial condition, both y and x = 0, at time = 0, the input quantity start to change at a constant rate Thus, we haveisq&
)()()( ttUqKtydt
tdyis&=+τ
)()( / ττ −+= − tqKCety ist &
Transient response
Steady state response
ττ τis
tism qeq
Ktytxe && +−=−= − /)()(
First-Order Systems: Ramp Response
t/τ
0 2 4 6 8 10
Out
put s
igna
l , y/
K
0
2
4
6
8
10
Non-dimensional ramp response of first-order instrument
Steady state error = τisq&
τSteady state time lag =
Input x(t)
y(t)/K
First-Order Systems: Frequency Response
From the response of first-order system to sinusoidal inputs, we have
tKAydtdy ωτ sin=+
tAtx ωsin)( =
( ) tKAtyD ωτ sin)(1 =+
The complete solution: ( )ωτωωτ
τ 1
2
/ tansin)(1
)( −− −+
+= tKACety t
Transient response
Steady state response
If we do interest in only steady state response of the system, we can write the equation in general form
[ ])(sin)()( / ωφωωτ ++= − tBCety t
[ ] 2/12)(1)(
ωτω
+=
KAB
ωτωφ 1tan)( −−=
Where B(ω) = amplitude of the steady state response and φ(ω) = phase shift
Frequency response=
ωτ
.01 .1 1 10 100
Am
plitu
de ra
tio
0.0
.2
.4
.6
.8
1.0
1.2
Dec
ibel
s (d
B)
0
-2
-4-6-8-10
-20
ωτ
.01 .1 1 10 100Ph
ase
shift
, φ(ω
)-90
-80
-70
-60
-50
-40
-30
-20
-10
0
First-Order Systems: Frequency Response
The phase angle isThe amplitude ratio1)(
1)(2 +
=ωτ
ωM )(tan)( 1 ωτωφ −−=
( )[ ] 2/121
1)(ωτ
ω+
==KABM
Frequency response of the first order system
0.707 -3 dB
Cutoff frequency
Dynamic error, δ(ω) = M(ω) -1: a measure of an inability of a system to adequately reconstruct the amplitude of the input for a particular frequency
Dynamic error
First-Order Systems: Frequency Response
Ex: Inadequate frequency responseSuppose we want to measure
With a first-order instrument whose τ is 0.2 s and static sensitivity K
Superposition concept:
For ω = 2 rad/s:
For ω = 20 rad/s:
Therefore, we can write y(t) as
tttx 20sin3.02sin)( +=
oo 8.2193.08.21116.0
)rad/s 2( −∠=−∠+
= KKAB
oo 7624.076116
)rad/s 20( −∠=−∠+
= KKAB
)7620sin()24.0)(3.0()8.212sin()93.0)(1()( oo −+−= tKtKty
)7620sin(072.0)8.212sin(93.0)( oo −+−= tKtKty
x(t)
y(t)/K
The essential parameters
= the static sensitivity
= the damping ratio, dimensionless
= the natural angular frequency
0
0
bKa
=
0
2n
aa
ω =
1
0 22aa a
ζ =
Second-Order Systems
)()()()(0012
2
2 txbtyadt
tdyadt
tyda =++ )()(122
2
tKxtyDD
nn
=
++
ωζ
ω
In general, a second-order measurement system subjected to arbitrary input, x(t)
)()()(2)(12
2
2 tKxtydt
tdydt
tyd
nn
=++ωζ
ω
Consider the characteristic equation
Second-Order Systems
0121 22 =++ DD
nn ωζ
ωThis quadratic equation has two roots:
122,1 −±−= ζωζω nnS
Overdamped (ζ > 1):
Critically damped (ζ = 1):
Underdamped (ζ< 1): :
Depending on the value of ζ, three forms of complementary solutions are possible
( )Φ+−= − tCety nt
ocn 21sin)( ζωζω
tt
ocnn eCeCty
ωζζωζζ
−−−
−+−
+=1
2
1
1
22
)(
ttoc
nn teCeCty ωω −− += 21)(
t
yHtLtAe σ−
)sin( φω +td
Second-Order Systems
Case 2 Overdamped (ζ > 1):
Case 3 Critically damped (ζ = 1):
Case I Underdamped (ζ< 1):
d
nn
jS
ωσζωζω
±=
−±−=
12
2,1 ( ) nS ωζζ 122,1 −±−=
nS ω−=2,1
t
yHtL1=ζ
1>ζ
Second-order Systems
Example: The force-measuring spring
2
2o o
i s odx d xf B K x Mdt dt
− − =
forces=(mass)(acceleration)Σ
the second-order model:
consider a spring with spring constant Ks under applied force fiand the total mass M. At start, the scale is adjusted so that xo = 0 when fi = 0;
2( )s o iMD BD K x f+ + =
1 m/Ns
KK
=
rad/ssn
KM
ω =
2 s
BK M
ζ =
Second-order Systems: Step Response
For a step input x(t)
With the initial conditions: y = 0 at t = 0+, dy/dt = 0 at t = 0+
The complete solution:
Overdamped (ζ > 1):
Critically damped (ζ = 1):
Underdamped (ζ< 1): :
)(212
2
2 tKAUydtdy
dtyd
nn
=++ωζ
ω
112
112
1)( 1
2
21
2
2 22
+−
−−+
−
−+−=
−−−
−+− tt nn ee
KAty ωζζωζζ
ζ
ζζ
ζ
ζζ
)()(122
2
tKAUtyDD
nn
=
++
ωζ
ω
1)1()(++−= − t
nnet
KAty ωω
( ) 11sin1
)( 2
2++−
−−=
−
φωζζ
ζω
teKA
tyn
tn ( )21 1sin ζφ −= −
ωnt
0 2 4 6 8 10
Out
put s
igna
l, y(
t)/K
A
0.0
.5
1.0
1.5
2.0
Second-order Systems: Step Response
Non-dimensional step response of second-order instrument
ζ = 0
0.25
0.5
1.02.0
21 ζωω −= ndRinging frequency:
Rise time decreases ζ with but increases ringing
Optimum settling time can be obtained from ζ ~ 0.7
Practical systems use 0.6< ζ <0.8
ddT
ωπ2
=Ringing period
Time, t (s)0 5 10 15 20
Out
put s
igna
l , y(
t)/K
A
0.0
.2
.4
.6
.8
1.0
1.2
1.4
settling time
rise time
overshoot
100% ± 5%
Typical response of the 2nd order system
Second-order Systems: Step Response
Second-order Systems: Ramp Response
For a ramp input
With the initial conditions: y = dy/dt = 0 at t = 0+
The possible solutions:
Overdamped:
Critically damped:
Underdamped:
)(212
2
2 ttUqKydtdy
dtyd
isnn
&=++ωζ
ω)()( ttUqtx is&=
−
−−+−+
−
−−−+−=
−+−
−−−
t
t
n
isis
n
n
e
eqtqKty
ωζζ
ωζζ
ζζ
ζζζ
ζζ
ζζζωζ
1
2
22
1
2
22
2
2
141212
141212
12)( &&
+−−= − tn
n
isis
netqtqKty ωω
ω)
11(12)( &
&
( )
+−
−−−=
−
φωζζζω
ζ ζω
teqtqKty
n
t
n
isis
n2
21sin
1212)( &
&12
12tan 2
21
−−
= −
ζζζ
φ
Time, t (s)0 2 4 6 8 10
Out
put s
igna
l, y(
t)/K
0
2
4
6
8
10
Typical ramp response of second-order instrument
ζ = 0.30.6
1.02.0
Ramp input
Steady state error = n
isqωζ&2
Steady state time lag =
nωζ2
Second-order Systems: Ramp Response
where
Second-order Systems: Frequency Response
The response of a second-order to a sinusoidal input of the form x(t) = Asinωt
( )[ ] ( ){ } [ ])(sin/2/1
)()( 2/1222
ωφωωζωωω
++−
+= tKAtytynn
ocf
[ ])(sin)()(steady ωφωω += tBty
The steady state response of a second-order to a sinusoidal input
( )[ ] ( ){ } 2/1222 /2/1
)(nn
KABωζωωω
ω+−
=ωωωω
ζωφ//
2tan)( 1
nn −−= −
ωωωωζωφ
//2tan)( 1
nn −−= −
Where B(ω) = amplitude of the steady state response and φ(ω) = phase shift
( )[ ] ( ){ } 2/1222 /2/1
1)(nn
KABM
ωζωωωω
+−==
Second-order Systems: Frequency Response
Magnitude and Phase plot of second-order Instrumentω/ωn
.01 .1 1 10 100
Ampl
itude
ratio
0.0
.5
1.0
1.5
2.0
Dec
ibel
(dB
)
6
3
0
-3-6-10-15
ζ = 0.1
0.3
0.5
1.0
2.0
ω/ωn
.01 .1 1 10 100
Phas
e sh
ift, φ(ω)
-180
-160
-140
-120
-100
-80
-60
-40
-20
0 ζ0 = 0.1
0.3
0.5
1.0
2.0
The phase angleThe amplitude ratio
( )[ ] ( ){ } 2/1222 /2/1
1)(nn
Mωζωωω
ω+−
=ωωωω
ζωφ//
2tan)( 1
nn −−= −
Time, t (s)0 5 10 15 20
Out
put s
igna
l , y(
t)/K
A
0.0
.2
.4
.6
.8
1.0
1.2
1.4
For overdamped (ζ >1) or critical damped (ζ = 1), there is neither overshoot nor steady-state dynamic error in the response.
In an underdameped system (ζ < 1) the steady-state dynamic error is zero, but the speed and overshoot in the transient are related.
arctan( / )dr
d
t ω δω−
=
pd
t πω
=
( )2exp / 1pM πζ ζ= − −
Rise time:
Peak time:
Maximum overshoot:
2
12 1
rMζ ζ
=−
Resonanceamplitude:
2 2where = , 1 , and arcsin( 1 )n d nδ ζω ω ω ζ φ ζ= − = −
Resonancefrequency:
21 2r nω ω ζ= −
Second-order Systems
Td
settling time
rise time
overshoot
peak time
Dynamic CharacteristicsSpeed of response: indicates how fast the sensor (measurement system) reacts to changes in the input variable. (Step input)
Rise time: the length of time it takes the output to reach 90% of full response when a step is applied to the input
Time constant: (1st order system) the time for the output to change by 63.2% of itsmaximum possible change.
Settling time: the time it takes from the application of the input step until the output has settled within a specific band of the final value.
Transfer Function: a simple, concise and complete way of describing the sensor or system performance
H(s) = Y(s)/X(s) where Y(s) and X(s) are the Laplace Transforms of the input and output respectively. Sometimes, the transfer function is displayed graphically as magnitude and phase plots VS frequency (Bode plot).
Dynamic Characteristics
Cutoff frequency: the frequency at which the system response has fallen to 0.707 (-3 dB) of the stable low frequency.
cr f
t 35.0≈
Dynamic error, δ(ω) = M(ω) - 1 a measure of the inability of a system or sensor to adequately reconstruct the amplitude of the input for a particular frequency
Bandwidth the frequency band over which M(ω) ≥ 0.707 (-3 dB in decibel unit)
Frequency Response describe how the ratio of output and input changes with the input frequency. (sinusoidal input)
Dynamic CharacteristicsExample: A first order instrument is to measure signals with frequency content up to 100 Hz with an inaccuracy of 5%. What is the maximum allowable time constant? What will be the phase shift at 50 and 100 Hz?
Solution:1
1)(22 +
=τω
ωMDefine
( ) %10011
1%1001)(error Dynamic22
×
−
+=×−=
τωωM
From the condition |Dynamic error| < 5%, it implies that 05.11
195.022
≤+
≤τω
But for the first order system, the term can not be greater than 1 so that the constrain becomes
1/1 22 +τω
Solve this inequality give the range 33.00 ≤≤ωτThe largest allowable time constant for the input frequency 100 Hz is
The phase shift at 50 and 100 Hz can be found from ωτφ arctan−=
This give φ = -9.33o and = -18.19o at 50 and 100 Hz respectively
11
195.022
≤+
≤τω
ms 52.0Hz 1002
33.0==
πτ
Dynamic Characteristics
Am
plitu
de ra
tio M
(ω)
1.05
0.95
ωτ
M(ω) ≥ 0.95 region or δ(ω) ≤ 0.05 region
Dynamic CharacteristicsExample: A temperature measuring system, with a time constant 2 s, is used to measured temperature of a heating medium, which changes sinusoidal between 350 and 300oC with a periodic of 20 s. find the maximum and minimum values of temperature, as indicated by the measuring system and the time lag between the output and input signals
T
325oC
350oC
300oC
x(t)
t
T
325oC
350oC
300oC
x(t), y(t)
t
TL
Solution: C)56.0602sin(2.21325)( o−+= tty π
Dynamic CharacteristicsExample: The approximate time constant of a thermometer is determined by immersing it in a bath and noting the time it takes to reach 63% of the final reading. If the result is 28 s, determine the delay when measuring the temperature of a bath that is periodically changing 2 times per minute. s7.6=dt
Dynamic Characteristics
Example: A pressure transducer has a natural frequency of 30 rad/s, damping ratio of 0.1 and static sensitivity of 1.0 µV/Pa. A step pressure input of 8x105 N/m2 is applied. Determine the output of a transducer.
Example: A second order instrument is subjected to a sinusoidal input. Undamped natural frequency is 3 Hz and damping ratio is 0.5. Calculate the amplitude ratio and phase angle for an input frequency of 2 Hz.
Amplitude ratio y(t)/x(t) = 1.152 and phase shifts –50.2o.
Solution:
Solution:
V )]47.185.29sin(1[8.0)( 3 +−= − tety t
Dynamic CharacteristicsExample: An Accelerometer is to selected to measure a time-dependent motion. In particular, input signal frequencies below 100 Hz are of prime interest. Select a set of acceptable parameter specifications for the instrument, assuming a dynamic error of ±5% and damping ratio ζ =0.7
ωn ≥ 1047 rad/sSolution:
Am
plitu
de ra
tio M
(ω)
ω/ ωn
1.05
0.95
1.05
0.95A
mpl
itude
ratio
M(ω
)
ω/ ωn
Response of a General Form of System to a Periodic Input
The steady state response of any linear system to the complex periodic signal can be determined using the frequency response technique and principle of superposition.
( )∑∞
=
++=1
000 sincos)(n
nn tnBtnAAtx ωωLet x(t)
The frequency response of the measurement system
Where KM(ω) = Magnitude of the frequency response of the measurement system and φM(ω) = Phase shift = tan-1(An/Bn)
( ) ( ))()(sin)()( 00001
220 ωφωφωω nntnnKMBAKAty nM
nnn ++++= ∑
∞
=
Response of a General Form of System of a Periodic Input
x(t) Linear system
y(t)t
xHtLt
yHtL
|x(ω)|
ω
ω
φnφM
ω
ω
|KM(ω)|
X
+
|Y(ω)|
φY(ω)ω0 2ω0 3ω0 4ω0 5ω0
ω0 2ω0 3ω0 4ω0 5ω0
ω
ω
=
=
0.01 0.02 0.03 0.04t
-1
-0.5
0.5
1
qoHtL
0.01 0.02 0.03 0.04t
-1
-0.5
0.5
1
qoHtL
Example: If qi(t) as shown in Figure below is the input to a first-order system with a sensitivity of 1 and a time constant of 0.001 s, find Qo(iω) and qo(t) for the periodic steady state.
-1
-0.02
qi(t)
t, sec-0.01 0.01 0.02
+1 ( )21
4 1 1( ) 1
o nn
o
Q in n
ω φπ ω τ
∞
=
= ∠ +
∑
arctan( )n onφ ω τ= −( )2
1
4 1 1( ) sin( ) 1
o o nn
o
q t n tn n
ω φπ ω τ
∞
=
= + +
∑
Where n = odd number
and
n = 3n = 5n = 7 n = 25
Response of a General Form of System to a Periodic Input