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.
Input signal
x(t)
Sensor
or
system
Output signal
y(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
d m −1 x (t )
dx (t )
d n −1 y (t )
dy (t )
d m x (t )
d n y (t )
+
+
L
+
+
(
)
=
+
+
L
+
b
+ b0 x (t )
a
a
a
y
t
b
b
an
n
−
m
m
−
1
1
0
1
1
n
n −1
m
m −1
dt
dt
dt
dt
dt
dt
m≤n
y(t)
x(t)
t
a’s and b’s
F(t) = forcing function
Where
= output from the system
= input to the system
= time
= system physical parameters, assumed constant
y(0)
The solution
x(t)
Measurement
system
y(t)
y (t ) = y ocf + y opi
Where yocf = complementary-function part of solution
yopi = particular-integral part of solution
Complementary-Function Solution
The solution yocf is obtained by calculating the n roots of the algebraic characteristic
equation
Characteristic equation
an D n + an −1 D n −1 + ... + a1 D + a0 = 0
Roots of the characteristic equation:
D = s1 , s2 ,..., sn
Complementary-function solution:
Ce st
1. Real roots, unrepeated:
2. Real roots, repeated:
each root s which appear p times
3. Complex roots, unrepeated:
the complex form: a ± ib
(C
2
p −1
st
+
C
t
+
C
t
+
...
+
C
t
e
)
0
1
2
p −1
Ce at sin(bt + φ )
[C0 sin(bt + φ0 ) + C1t sin(bt + φ1 ) + C2t 2 sin(bt + φ2 )
4. Complex roots, repeated:
each pair of complex root which appear p times
+ ... + C p −1t p −1 sin(bt + φ p −1 )]e at
Particular Solution
Method of undetermined coefficients:
y opi = Af (t ) + B f ′(t ) + C f ′′(t ) + ...
Where f(t)
= the function that describes input quantity
A, B, C = constant which can be found by substituting yopi into ODEs
Important Notes
9 •After a certain-order derivative, all higher derivatives are zero.
a certain-order derivative, all higher derivatives have the same
9 •After
functional form as some lower-order derivatives.
2 •Upon repeated differentiation, new functional forms continue to arise.
Zero-order Systems
All the a’s and b’s other than a0 and b0 are zero.
y (t ) = Kx (t )
a0 y (t ) = b0 x(t )
where K = static sensitivity = b0/a0
The behavior is characterized by its static sensitivity, K and remains
constant regardless of input frequency (ideal dynamic characteristic).
xm
Vr
+
x=0
y=V
-
x
here, K = Vr / xm
V = Vr ⋅
xm
Where 0 ≤ x ≤ xm and Vr is a reference voltage
A linear potentiometer used as position
sensor is a zero-order sensor.
First-Order Systems
All the a’s and b’s other than a1, a0 and b0 are zero.
a1
dy (t )
+ a 0 y (t ) = b0 x (t )
dt
τ
dy (t )
+ y (t ) = Kx (t )
dt
y
K
(D) =
x
τD + 1
Where K = b0/a0 is the static sensitivity
τ = a1/a0 is the system’s time constant (dimension of time)
First-Order Systems
Surface
area A
Ti
Consider a thermometer based on a mass m =ρV
with 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
q
Assume no heat loss from the thermometer
Sensor
m, Ttf, C
UA(Ti − Ttf )dt − 0 = ρVCdTtf
ρVC
Thermometer based on a mass,m with
specified heat, C
dTtf
dt
+ UATtf = UATi
Therefore, we can immediately define K =1 and τ = ρ VC/UA
τ
dTif
dt
+ Tif = Ti
First-Order Systems: Step Response
Assume for t < 0, y = y0 , at time = 0 the input quantity, x increases instantly
by an amount A. Therefore t > 0
0 t ≤ 0
x(t ) = AU (t ) = 
A t > 0
τ
dy (t )
+ y (t ) = KAU (t )
dt
y (t ) = Ce − t /τ + KA
The complete solution:
2
U(t)
yocf
Transient
response
1
yopi
Steady state
response
Applying the initial condition, we get C = y0-KA, thus
gives
0
-1
0
1
2
Time, t
3
4
5
y (t ) = KA + ( y0 − KA)e − t /τ
First-Order Systems: Step Response
Here, we define the term error fraction as
y (t ) − KA y (t ) − y (∞)
=
= e −t / τ
y0 − KA
y ( 0) − y ( ∞ )
1.0
1.0
.8
.8
Error fraction, em
Output Signal, (y(t)-y0)/(KA-y0)
em (t ) =
0.632
.6
y (t ) − y0
= 1 − e −t / τ
KA − y0
.4
.6
.4
0.368
.2
.2
0.0
y (t ) − KA
= e −t /τ
y (0) − KA
0
1
2
3
t/τ
4
5
0.0
0
1
2
3
t/τ
Non-dimensional step response of first-order instrument
4
5
Determination of Time constant
y (t ) − KA
= e −t /τ
y (0) − KA
ln em = 2.3 log em = −
t
τ
1
y (t ) − KA
= e −t /τ
y (0) − KA
0.368
Error fraction, em
em =
.1
Slope = -1/τ
.01
.001
0
1
2
3
t
4
5
First-Order Systems: Ramp Response
Assume that at initial condition, both y and x = 0, at time = 0, the input quantity
start to change at a constant rate q&is Thus, we have
Therefore
The complete solution:
t≤0
0
x(t ) = 
q&is t t > 0
dy (t )
+ y (t ) = Kq&is tU (t )
τ
dt
y (t ) = Ce − t /τ + Kq&is (t − τ )
Transient Steady state
response response
Applying the initial condition, gives
Measurement error
em = x(t ) −
y (t ) = Kq&is (τe − t /τ + t − τ )
y (t )
= − q&isτe −t /τ + q&isτ
K
Transient
error
Steady
state error
First-Order Systems: Ramp Response
10
Input x(t)
Output signal, y/K
8
y(t)/K
6
Steady state
time lag = τ
4
Steady state
error = q&isτ
2
0
0
2
4
6
8
10
t/τ
Non-dimensional ramp response of first-order instrument
First-Order Systems: Frequency Response
From the response of first-order system to sinusoidal inputs, x(t ) = A sin ωt
we have
τ
dy
+ y = KA sin ωt
dt
The complete solution:
(τD + 1) y(t ) = KA sin ωt
y (t ) = Ce −t /τ +
KA
1 + (ωτ )
Transient
response
2
(
sin ωt − tan −1 ωτ
Steady state
response
=
)
Frequency
response
If we do interest in only steady state response of the system, we can write the
equation in general form
y (t ) = Ce −t /τ + B(ω ) sin[ωt + φ (ω )]
B(ω ) =
KA
[1 + (ωτ ) ]
2 1/ 2
φ (ω ) = − tan −1 ωτ
Where B(ω) = amplitude of the steady state response and φ(ω) = phase shift
First-Order Systems: Frequency Response
M (ω ) =
The amplitude ratio
[
(ωτ ) 2 + 1
-2
-3 dB
.6
-4
.4
-6
-8
-10
Cutoff frequency
.1
-20
1
ωτ
10
100
-20
Phase shift, φ(ω)
.8
0.707
φ (ω ) = − tan −1 (ωτ )
-10
Decibels (dB)
0
0.0
.01
The phase angle is
0
1.0
.2
]
Dynamic error
1.2
Amplitude ratio
1
M (ω ) =
B
1
=
KA 1 + (ωτ )2 1/ 2
-30
-40
-50
-60
-70
-80
-90
.01
.1
Frequency response of the first order system
1
10
100
ωτ
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
First-Order Systems: Frequency Response
Ex: Inadequate frequency response
Suppose we want to measure
x(t ) = sin 2t + 0.3 sin 20t
x(t)
With a first-order instrument whose τ is 0.2 s and
static sensitivity K
Superposition concept:
For ω = 2 rad/s: B (2 rad/s) =
A
y(t)/K
For ω = 20 rad/s: B (20 rad/s) =
A
K
∠ − 21.8o = 0.93K∠ − 21.8o
0.16 + 1
K
∠ − 76o = 0.24 K∠ − 76o
16 + 1
Therefore, we can write y(t) as
y (t ) = (1)(0.93K ) sin(2t − 21.8o ) + (0.3)(0.24 K ) sin(20t − 76o )
y (t ) = 0.93K sin(2t − 21.8o ) + 0.072K sin(20t − 76o )
Second-Order Systems
In general, a second-order measurement system subjected to arbitrary input, x(t)
2
a2
d y (t )
dy (t )
+
+ a0 y (t ) = b0 x(t )
a
1
2
dt
dt
 D 2 2ζ

 2 +
D + 1 y (t ) = Kx(t )
 ωn ωn

1 d 2 y (t ) 2ζ dy (t )
+
+ y (t ) = Kx(t )
2
2
ωn dt
ωn dt
The essential parameters
K=
ζ =
b0
a0
a1
2 a0 a2
ωn =
a0
a2
= the static sensitivity
= the damping ratio, dimensionless
= the natural angular frequency
Second-Order Systems
Consider the characteristic equation
1 2 2ζ
D +
D +1 = 0
2
ωn
ωn
This quadratic equation has two roots:
S1, 2 = −ζω n ± ωn ζ 2 − 1
Depending on the value of ζ, three forms of complementary solutions are possible
 −ζ + ζ 2 −1 ω t

 n
Overdamped (ζ > 1):
yoc (t ) = C1e
Critically damped (ζ = 1):
yoc (t ) = C1e −ωnt + C2te −ωnt
Underdamped (ζ< 1): :
yoc (t ) = Ce −ζω nt sin ωn 1 − ζ 2 t + Φ
(
+ C2 e
 −ζ − ζ 2 −1 ω t

 n
)
Second-Order Systems
Case I Underdamped (ζ< 1):
Case 2 Overdamped (ζ > 1):
yHtL
)
(
S1, 2 = −ζω n ± ωn ζ 2 − 1
S1, 2 = − ζ ± ζ 2 − 1 ωn
= σ ± jωd
Case 3 Critically damped (ζ = 1):
Ae
S1, 2 = −ωn
−σt
t
sin(ωd t + φ )
yHtL
ζ =1
ζ >1
t
Second-order Systems
Example: The force-measuring spring
consider a spring with spring constant Ks under applied force fi
and the total mass M. At start, the scale is adjusted so that xo = 0
when fi = 0;
Σforces=(mass)(acceleration)
dxo
d 2 xo
fi − B
− K s xo = M
dt
dt 2
( MD 2 + BD + K s ) xo = f i
the second-order model:
K=
1
Ks
m/N
Ks
rad/s
M
B
ζ =
2 Ks M
ωn =
Second-order Systems: Step Response
For a step input x(t)
 D 2 2ζ

 2 +
D + 1 y (t ) = KAU (t )
 ωn ωn

1 d 2 y 2ζ dy
+
+ y = KAU (t )
2
2
ωn dt
ωn dt
With the initial conditions: y = 0 at t = 0+, dy/dt = 0 at t = 0+
The complete solution:
Overdamped (ζ > 1):
ζ + ζ 2 − 1  −ζ +
y (t )
=−
e
2
KA
2 ζ −1
Critically damped (ζ = 1):
y (t )
= −(1 + ωnt )e −ωnt + 1
KA
Underdamped (ζ< 1): :
y (t )
e −ζω nt
=−
sin 1 − ζ 2 ωnt + φ + 1
KA
1− ζ 2
(
ζ 2 −1 ω n t

+
ζ − ζ 2 −1
2 ζ −1
2
)
e
 −ζ − ζ 2 −1 ω t

 n
(
+1
φ = sin −1 1 − ζ 2
)
Second-order Systems: Step Response
Ringing period
Output signal, y(t)/KA
2.0
Ringing frequency:
ζ=0
0.25
1.5
2π
ωd
ωd = ωn 1 − ζ 2
Rise time decreases ζ with but
increases ringing
0.5
1.0
Optimum settling time can be obtained
from ζ ~ 0.7
.5
1.0
Practical systems use 0.6< ζ <0.8
2.0
0.0
Td =
0
2
4
6
8
10
ωnt
Non-dimensional step response of second-order instrument
Second-order Systems: Step Response
1.4
overshoot
Output signal, y(t)/KA
1.2
100% ± 5%
1.0
.8
.6
.4
settling
time
.2
0.0
rise time
0
5
10
15
Time, t (s)
Typical response of the 2nd order system
20
Second-order Systems: Ramp Response
For a ramp input x(t ) = q&is tU (t )
1 d 2 y 2ζ dy
+
+ y = Kq&is tU (t )
2
2
ωn dt
ωn dt
With the initial conditions: y = dy/dt = 0 at t = 0+
The possible solutions:
Overdamped:
2ζq&is
y (t )
= q&is t −
ωn
K
 2ζ 2 − 1 − 2ζ ζ 2 − 1  −ζ −
1 +
e

4ζ ζ 2 − 1

+
Critically damped:
Underdamped:
2q&
y (t )
= q&is t − is
ωn
K
2ζq&is
y (t )
= q&is t −
K
ωn
− 2ζ 2 + 1 − 2ζ ζ 2 − 1
4ζ ζ − 1
2
e
ζ 2 −1 ω n t

 −ζ + ζ 2 −1 ω t

 n




ωnt −ωnt 

1
(
1
)e 
−
+

1


(

e −ζω nt
sin 1 − ζ 2 ωnt + φ
1 −
 2ζ 1 − ζ 2
)



2ζ 1 − ζ 2
φ = tan
2ζ 2 − 1
−1
Second-order Systems: Ramp Response
Steady state error =
Output signal, y(t)/K
10
8
ωn
6
4
0
ζ = 0.3
1.0
2.0
0
2
4
ωn
Steady state
time lag = 2ζ
Ramp input
2
2q&isζ
6
0.6
8
10
Time, t (s)
Typical ramp response of second-order instrument
Second-order Systems: Frequency Response
The response of a second-order to a sinusoidal input of the form x(t) = Asinωt
KA
y (t ) = yocf (t ) +
sin[ωt + φ (ω )]
1/ 2
2 2
2
1 − (ω / ωn ) + (2ζω / ωn )
{[
where
]
}
φ (ω ) = − tan −1
2ζ
ω / ωn − ωn / ω
The steady state response of a second-order to a sinusoidal input
ysteady (t ) = B (ω ) sin[ωt + φ (ω )]
B(ω ) =
KA
{[1 − (ω / ω ) ] + (2ζω / ω ) }
2 2
n
2
1/ 2
φ (ω ) = − tan −1
n
2ζ
ω / ωn − ωn / ω
Where B(ω) = amplitude of the steady state response and φ(ω) = phase shift
M (ω ) =
B
1
=
KA 1 − (ω / ω )2 2 + (2ζω / ω )2 1/ 2
n
n
{[
]
}
Second-order Systems: Frequency Response
The phase angle
The amplitude ratio
M (ω ) =
1
φ (ω ) = − tan −1
{[1 − (ω / ω ) ] + (2ζω / ω ) }
2 2
2
n
1/ 2
n
2ζ
ω / ωn − ωn / ω
0
0.3
1.5
0.5
1.0
.5
3
0
-3
-6
-10
-15
1.0
2.0
0.0
.01
.1
1
ω/ωn
10
100
ζ0 = 0.1
-20
Phase shift, φ(ω)
ζ = 0.1
6
Decibel (dB)
Amplitude ratio
2.0
0.3
-40
-60
-80
0.5
1.0
2.0
-100
-120
-140
-160
-180
.01
.1
1
ω/ωn
Magnitude and Phase plot of second-order Instrument
10
100
Second-order Systems
For overdamped (ζ >1) or critical damped (ζ = 1), there is neither overshoot nor steadystate 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.
1.4
arctan(−ω d / δ )
Rise time:
tr =
Maximum
overshoot:
M p = exp −πζ / 1 − ζ 2
Peak time:
tp =
ωd
π
ωd
Resonance
frequency:
ω r = ω n 1 − 2ζ 2
Resonance
amplitude:
Mr =
1
2
)
Output signal, y(t)/KA
1.2
(
2ζ 1 − ζ
Td
overshoot
1.0
.8
peak
time
.6
.4
settling
time
.2
rise time
0.0
0
where δ =ζω n , ω d = ω n 1 − ζ 2 , and φ = arcsin( 1 − ζ 2 )
5
10
Time, t (s)
15
20
Dynamic Characteristics
Speed 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 its
maximum 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
Frequency Response describe how the ratio of output and input changes
with the input frequency. (sinusoidal input)
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)
Cutoff frequency: the frequency at which the system response has fallen to
0.707 (-3 dB) of the stable low frequency.
tr ≈
0.35
fc
Dynamic Characteristics
Example: 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: Define
M (ω ) =
1
ω 2τ 2 + 1


1

− 1 ×100%
Dynamic error = (M (ω ) − 1)× 100% = 
2 2
 ω τ +1 
1
0
.
95
≤
≤ 1.05
From the condition |Dynamic error| < 5%, it implies that
2 2
ω τ +1
But for the first order system, the term 1 / ω 2τ 2 + 1 can not be greater than 1 so that the
constrain becomes
1
0.95 ≤
Solve this inequality give the range
≤1
ω τ +1
0 ≤ ωτ ≤ 0.33
2 2
The largest allowable time constant for the input frequency 100 Hz is τ =
The phase shift at 50 and 100 Hz can be found from
φ = − arctan ωτ
This give φ = -9.33o and = -18.19o at 50 and 100 Hz respectively
0.33
= 0.52 ms
2π 100 Hz
Dynamic Characteristics
Amplitude ratio M(ω)
1.05
0.95
M(ω) ≥ 0.95 region or
δ(ω) ≤ 0.05 region
ωτ
Dynamic Characteristics
Example: 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
2π
Solution: y (t ) = 325 + 21.2 sin( t − 0.56) o C
60
x(t), y(t)
x(t)
350oC
350oC
325oC
325oC
300oC
300oC
T
TL
t
T
t
Dynamic Characteristics
Example: 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.
t d = 6.7s
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.
Solution:
y (t ) = 0.8[1 − e −3t sin( 29.85t + 1.47)] V
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.
Solution:
Amplitude ratio y(t)/x(t) = 1.152 and phase shifts –50.2o.
Dynamic Characteristics
Example: 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/s
Amplitude ratio M(ω)
Amplitude ratio M(ω)
Solution:
1.05
0.95
ω/ ωn
1.05
0.95
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.
Let x(t)
∞
x(t ) = A0 + ∑ ( An cos nω0t + Bn sin nω0t )
n =1
The frequency response of the measurement system
∞
y (t ) = KA0 + ∑
n =1
(
)
An2 + Bn2 KM (nω0 ) sin (nω0t + φM (nω0 ) + φn (nω0 ) )
Where KM(ω) = Magnitude of the frequency response of the measurement
system and
φM(ω) = Phase shift = tan-1(An/Bn)
Response of a General Form of System of a Periodic Input
yHt L
xHt L
x(t)
t
Linear
system
y(t)
|KM(ω)|
|x(ω)|
|Y(ω)|
=
X
ω0
φn
2ω0 3ω0 4ω0 5ω0
ω
φM
ω0
ω
ω
2ω0 3ω0 4ω0 5ω0
φY(ω)
=
+
ω
t
ω
ω
Response of a General Form of System to a Periodic Input
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.
qi(t)

1
Qo (iω ) = ∑ 
π n =1  n

4
+1
-0.02
-0.01
0.01
0.02
t, sec
-1

1
qo (t ) = ∑ 
π n =1  n

4
qoH tL
n=7
n=5
n=3
0.5
0.01
-1
∞

∠φn 
2
( nω oτ ) + 1 
1
Where n = odd number

sin(nω ot + φn )  and φ n
2

( nω oτ ) + 1

1
= − arctan(nω oτ )
qoH tL
1
-0.5
∞
0.02
0.03
0.04
1
n = 25
0.5
t
0.01
-0.5
-1
0.02
0.03
0.04
t