2141-375
Measurement and Instrumentation
Measurement System
Behavior
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 n y (t )
d n −1 y (t )
dy (t )
d m x (t )
d m −1 x (t )
dx (t )
+
+
L
+
+
(
)
=
+
+
L
+
+ b0 x ( t )
an
a
a
a
y
t
b
b
b
n
−
1
1
0
m
m
−
1
1
n
n −1
m
m −1
dt
dt
dt
dt
dt
dt
F(t) = forcing function
Where
m≤n
y(t)
x(t)
t
a’s and b’s
= 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
0
+ C1t + C2t 2 + ... + C p −1t p −1 ) e st
Ceat 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
Complementary-Function Solution
Case 1: Real roots, unrepeated:
ex: -1.7, 3.2, 0
Case 2: Real roots, repeated:
ex: -1, -1, 2, 2, 2, 0, 0
Case 3: Complex number, unrepeated:
ex: -3 ± j4, 2 ± j5, 0 ± j7
Case 4: Complex number, repeated:
ex: -3 ± j2, -3 ± j2, -3 ± j2
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
•After a certain-order derivative, all higher derivatives are zero.
a certain-order derivative, all higher derivatives have the same
•After
functional form as some lower-order derivatives.
•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
V = Vr ⋅
here, K = Vr / xm
xm
+
x=0
y=V
-
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
Where K = b0/a0 is the static sensitivity
τ = a1/a0 is the system’s time constant (dimension of time)
First-Order Systems
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).
Surface
area A
Ti
q
(Heat in) – (Heat out) = Energy stored
Assume no heat loss from the thermometer
UA(Ti − Ttf )dt − 0 = ρVCdTtf
Sensor
m, Ttf, C
ρVC
Thermometer based on a mass,m with
specified heat, C
dTtf
dt
+ UATtf = UATi
ρVC dTtf
UA dt
Therefore, we can immediately define K =1 and τ = ρ VC/UA
τ
dTtf
dt
+ Ttf = Ti
+ Ttf = 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
τ
dy (t )
+ y (t ) = KAU (t )
dt
y (t ) = Ce −t /τ + KA
The complete solution
2
U(t)
yocf
Transient
response
yopi
Steady state
response
1
Applying the initial condition, we get C = y0-KA, thus
gives
y (t ) = KA + ( y0 − KA)e − t /τ
0
-1
0
1
2
Time, t
3
4
5
t>0
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, Γ
Output Signal, (y(t)-y0)/(KA-y0)
Γ=
0.632
.6
y (t ) − y0
= 1 − e −t /τ
KA − y0
.4
y (t ) − KA
= e −t / τ
y (0) − KA
.6
.4
0.368
.2
.2
0.0
0.0
0
1
2
3
t/τ
4
5
0
1
2
3
t/τ
Non-dimensional step response of first-order instrument
4
5
First-Order Systems: Step Response
Time constant, τ : the time required for a system to response
63.2% of a step change
Rise time, tr : the time required for a system to response 90% of a
step change. (sometimes defined time for 10 to 90% of a step
change) ~ 2.3 τ
First-order system response
t/ττ
% response
1
63.2
2
86.5
2.3
90.0
3
95.0
5
99.3
Determination of Time constant
Γ=
y (t ) − KA
= e −t /τ
y (0) − KA
ln Γ = 2.3 log Γ = −
t
τ
1
Error fraction, em
Advantage
- Check whether the system is
1st order
- Avoid the influence of an
error in any one data point
- eliminate the need to
accurately determine the 0.368
point
y (t ) − KA
= e −t /τ
y (0) − KA
0.368
.1
Slope = -1/2.3τ
.01
.001
0
1
2
3
t/τ
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 )
τ
+ y (t ) = Kq&is tU (t )
dt
Initial condition: y(0) = 0
y (t ) = Ce − t /τ + Kq&is (t − τ )
Transient Steady state
response response
Applying the initial condition, gives
Measurement error
em =
y (t ) = Kq&is (τe −t /τ + t − τ )
y (t )
− x (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
The complete solution:
dy
+ y = KA sin ωt
dt
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
x(t ) = A sin ωt
y (t ) = Ce −t /τ + B(ω ) sin[ωt + φ (ω )]
Input
Output
Input x(t) and Output y(t)
2.0
1.5
1.0
.5
B
A
0.0
-.5
-1.0
-1.5
-2.0
0.00
td
.01
.02
.03
Time (sec)
.04
.05
.06
First-Order Systems: Frequency Response
M (ω ) =
The amplitude ratio
1
M (ω ) =
B
1
=
KA 1 + (ωτ )2 1/ 2
[
(ωτ ) 2 + 1
]
The phase angle is
Dynamic error
1.2
φ (ω ) = − tan −1 (ωτ )
0
.8
-2
-3 dB
0.707
.6
-4
.4
-6
-8
-10
.2
Cutoff frequency
-20
0.0
.01
.1
1
ωτ
10
100
-20
Phase shift, φ(ω)
0
Decibels (dB)
Amplitude ratio
-10
1.0
-30
-40
-50
-60
-70
-80
-90
.01
.1
1
10
100
ωτ
Frequency response of the first order system
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:
y(t)/K
For ω = 2 rad/s:
M (2 rad/s) =
For ω = 20 rad/s:
M (20 rad/s) =
1
∠ − 21.8o = 0.93∠ − 21.8o
0.16 + 1
1
∠ − 76o = 0.24∠ − 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 ) + 0.072 K sin(20t − 76o )
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
Dynamic error = (M (ω ) − 1)× 100% = 
− 1 × 100%
2 2
 ω τ +1 
1
≤ 1.05
From the condition |Dynamic error| < 5%, it implies that 0.95 ≤
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
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
τ=
0.33
= 0.52 ms
2π 100 Hz
Dynamic Characteristics
Amplitude ratio M(ω
ω)
1.05
0.95
M(ω
ω) ≥ 0.95 region or
δ(ω
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 sinusoidally 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
Solution:
 2π

y (t ) = 325 + 21.3 sin 
t − 31.36 o 
 20

t d = 1.75 s
x(t), y(t)
x(t)
350oC
350oC
325oC
325oC
300oC
300oC
T
T
td
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.
Solution:
t d = 6.69 s
Second-Order Systems
In general, a second-order measurement system subjected to arbitrary input, x(t)
d 2 y (t )
dy (t )
a2
+
a
+ a0 y (t ) = b0 x(t )
1
dt 2
dt
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:
D1, 2 = −ζω n ± ωn ζ 2 − 1
Depending on the value of ζ, three forms of complementary solutions are possible
Overdamped (ζ
ζ > 1):
yoc (t ) = C1e
 −ζ + ζ 2 −1 ω t

 n
+ C2 e
 −ζ − ζ 2 −1 ω t

 n
−ω n t
+ C2te −ωnt
Critically damped (ζ
ζ = 1): yoc (t ) = C1e
Underdamped (ζ
ζ< 1): :
(
yoc (t ) = Ce −ζω nt sin ωn 1 − ζ 2 t + Φ
Unrepeated real roots
Repeated real roots
)
Complex roots
Second-Order Systems
Case I Underdamped (ζ
ζ< 1):
Case 2 Overdamped (ζ
ζ > 1):
)
(
D1, 2 = −ζω n ± ωn ζ 2 − 1
D1, 2 = − ζ ± ζ 2 − 1 ωn
= −σ ± jωd
ζ = 1):
Case 3 Critically damped (ζ
y(t)
HL
Ae
D1, 2 = −ωn
−σt
y(t)
HL
t
sin(ωd t + φ )
ζ =1
ζ >1
t
The complementary function solution determines the transient responses of the system
Second-order Systems: Step Response
For a step input x(t)
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
ζ > 1):
 ζ + ζ 2 − 1  −ζ +
y (t ) = KA−
e
2
 2 ζ − 1
ζ 2 −1 ω n t

+
ζ − ζ 2 −1
2 ζ −1
Critically damped (ζ
ζ = 1):
y (t ) = − KA(1 + ωn t )e −ωnt + KA
Underdamped (ζ
ζ< 1): :
 e −ζω nt
y (t ) = − KA
sin 1 − ζ 2 ωn t + φ
 1 − ζ 2
(
2
)

 + KA

e
 −ζ − ζ 2 −1 ω t

 n

 + KA

(
φ = sin −1 1 − ζ 2
)
Second-order Systems: Step Response
Ringing period
Output signal, y(t)/KA
2.0
1.5
Td =
2π
ωd
ωd = ωn 1 − ζ 2
ζ=0
Ringing frequency:
0.25
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
0
2
4
6
8
10
ωnt
Non-dimensional step response of second-order instrument
Second-Order Systems
1.4
overshoot
Output signal, y(t)/KA
1.2
100% ± 5%
1.0
.8
.6
.4
settling
time
.2
rise time
0.0
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:
2ζKq&is  2ζ 2 − 1 − 2ζ ζ 2 − 1  −ζ −
1+
y (t ) = Kq&is t −
e
2

ωn 
4ζ ζ − 1
Overdamped:
+
Critically damped:
Underdamped:
y (t ) = Kq&is t −
− 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




2 Kq&is 
ωnt −ωnt 
1
−
(
1
+
)e 
1
ωn 

(
2ζKq&is 
e −ζω nt
y (t ) = Kq&is t −
sin 1 − ζ 2 ωn t + φ
1 −
ωn  2ζ 1 − ζ 2
)

2
ζ
ζ
2
1
−
−
1
 φ = tan

2ζ 2 − 1
Second-order Systems: Step Response
Steady state error =
Output signal, y(t)/K
10
8
ωn
6
4
ζ = 0.3
0.6
1.0
2.0
0
0
2
4
ωn
Steady state
time lag = 2ζ
Ramp input
2
2q&isζ
6
8
10
Time, t (s)
Typical ramp response of second-order instrument
Second-order Instrument: Frequency Response
The response of a second-order to a sinusoidal input of the form x(t) = Asinω
ωt
KA
y (t ) = yoc (t ) +
sin[ωt + φ (ω )]
1/ 2
2 2
2
1 − (ω / ωn ) + (2ζω / ωn )
{[
}
]


2ζ

where φ (ω ) = tan −1  −
 ω / ωn − ω n / 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
n
1/ 2


2ζ


φ (ω ) = tan  −
 ω / ωn − ω n / ω 
−1
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
2.0
6
-20
3
0.5
1.0
0
-3
-6
-10
-15
1.0
.5
2.0
0.0
.01
.1
1
ω/ωn
10
100
Phase shift, φ(ω)
1.5
Decibel (dB)
Amplitude ratio
ζ = 0.1
0.3
ζ0 = 0.1
0.3
-40
0.5
-60
-80
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. (Time response)
In an underdameped system (ζ < 1) the steady-state dynamic error is zero, but the speed
and overshoot in the transient are related. (Time response)
1.4
; 0 < ζ <1
Rise time:
tr ~
Maximum
overshoot:
M p = exp −πζ / 1 − ζ 2
Peak time:
tp =
Resonance
frequency:
Resonance
amplitude:
ωn
(
π
ωd
ω r = ω n 1 − 2ζ 2
Mr =
1
2ζ 1 − ζ
)
Td
overshoot
1.2
Output signal, y(t)/KA
0.8 + 2.5ζ
1.0
.8
peak
time
.6
.4
settling
time
.2
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
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 M(ω)= 1.152 and phase shifts = -0.875 rad =–50.2o.
Dynamic Characteristics
Example: An Accelerometer is 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
ω)
Amplitude ratio M(ω
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
∞
x(t ) = A0 + ∑ An2 + Bn2 sin (nω0t + φn (nω0 ))
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
ω) = Magnitude of the frequency response of the measurement
system and
φM(nω
ω0) = Phase shift of the measurement system at nω
ω0
φn(nω
ω0) = tan-1(An/Bn)
Response of a General Form of System of a Periodic Input
HL
HL
x(t)
t
Linear
system
y(t)
t
|KM(ω
ω)|
|x(ω)|
|Y(ω
ω)|
=
X
ω0
2ω0 3ω0 4ω0 5ω0
ω
ω0
ω
φM
φn
φY(ω
ω)
=
+
ω
ω
2ω0 3ω0 4ω0 5ω0
ω
ω
Response of a General Form of System to a Periodic Input
Example: If x(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 y(t) for the periodic steady state.
x(t)
1
y (iω ) = ∑ 
π n =1  n

4
+1
1
y (t ) = ∑ 
π n =1  n

4
-0.02
-0.01
0.01
0.02
t, sec
∞

∠φ (nωo ) where n = odd number
2

(nωoτ ) + 1

1

sin( nωot + φ (nωo ) and φ (nωo ) = − arctan (nωoτ )
2

(nωoτ ) + 1

∞
1
-1
y(t)
y(t)
qoH tL
HL
n=7
n=5
n=3
1
0.5
0.01
-0.5
-1
0.02
0.03
0.04
1
n = 25
0.5
t
0.01
-0.5
-1
0.02
0.03
0.04
t