Dynamic Performance
Class 4
Dynamic Performance
 The dynamic characteristics of a
measuring instrument describe its
behavior between the time a
measured quantity changes value
and the time when the instrument
output attains a steady value in
response.
 Because dynamic signals vary with
time, the measurement system
must be able to respond fast
enough to keep up with the input
signal. Further, we need to
understand how the input signal is
applied to the sensor because that
plays a role in system response.
Mechanical Zero-Order Systems
l1
 The simplest model of a measurement
system is the zero-order system model.
This is represented by the zero-order
differential equation:
ao x  f (t )
x 
1
f ( t )  Kf ( t )
ao
 K is the static sensitivity or steady gain of
the system. It is a measure of the
amount of change in the output in
response to the change in the input.
l2
f(t)
x(t)
x(t)/l2 = f(t)/l1
K = l2/l1
Mechanical Zero-Order Systems
l1
 In a zero-order system, the
output responds to the
input signal
instantaneously.
 If an input signal of
magnitude f(t) = A were
applied, the instrument
would indicate KA. The
scale of the measuring
device would normally be
calibrated to indicate A
directly.
l2
f(t)
x(t)
x(t)/l2 = f(t)/l1
K = l2/l1
ao x  f (t )
x 
1
ao
f ( t )  Kf ( t )
Electrical Zero-Order Systems
 In a zero-order system, the
output responds to the
input signal
instantaneously.
vi
R1
R2
Vo
v o  iR 1 
vo 
vi
R1  R 2
R1
R1  R 2
vi
R1
A Unity Gain Zero-Order system
xi
l
xo(t) = xi(t)
K=1
xo
Non-zero Order Systems
 Measurement systems that contain
storage or dissipative elements do not
respond instantaneously to changes in
input. In the bulb thermometer, when
the ambient temperature changes, the
liquid inside the bulb will need to
store a certain amount of energy in
order for it to reach the temperature
of the environment. The temperature
of the bulb sensor changes with time
until this equilibrium is reached, which
accounts physically for its lag in
response.
 In general, systems with a storage or
dissipative capability but negligible
inertial forces may be modeled using a
first-order differential equation.
xo
xi
c
m
k
Non-zero Order Systems
 Measurement systems that contain
storage or dissipative elements do not
respond instantaneously to changes in
input.
 In the bulb thermometer, for
example, when the ambient
temperature changes, the liquid inside
the bulb will need to store a certain
amount of energy in order for it to
reach the temperature of the
environment. The temperature of the
bulb sensor changes with time until
this equilibrium is reached, which
accounts physically for its lag in
response.
First Order Systems
 Consider the time response of a bulb
thermometer for measuring body
temperature. The thermometer, initially at
room temperature, is placed under the
tongue. Body temperature itself is constant
(static) during the measurement, but the
input signal to the thermometer is suddenly
changed from room temperature to body
temperature. This, is a step change in the
measured signal.
 The thermometer must gain energy from its
new environment to reach thermal
equilibrium, and this takes a finite amount of
time. The ability of any measurement system
to follow dynamic signals is a characteristic
of the measuring system components.
Body
Temperature
Room
Temperature
t
time
First Order Systems
 Suppose a bulb thermometer originally at
temperature To is suddenly exposed to a fluid
temperature T∞. Develop a model that
simulates the thermometer output response.
 Rate of energy stored = Rate of energy in
E stored  Q in
mc
mc
dT
p
dt
dT
p
mc
hA
p
 hA T   T
 hAT  hAT
dt
dT
dt

Body
Temperature

 T  T
Room
Temperature
t
 The ratio mcp/hA has a units of seconds and
is called the time constant
time
First Order Systems
mc
p
hA

dT
dt
dT
dt
 T  T
 T  T
 The ratio mcp/hA has a units of seconds and
is called the time constant, τ. If the time
constant is much less then 1 second, the
system may be approximated by a unity gain
zero-order system.
Body
Temperature
Room
Temperature
t
time
1st Order Systems
 Examples:
 Bulb Thermometer
 RC Circuits
 Terminal velocity
 Mathematical Model:
𝜏
τ:
𝑑𝑥
+𝑥 =𝑓 𝑡
𝑑𝑡
Time constant
𝑓 𝑡 : Input (quantity to be measured)
𝑥:
Output (instrument response)
1st Order Systems with Step Input
𝑓 𝑡 = 𝐾𝑢(𝑡)
𝑢 𝑡 =
0 𝑡<0
1 𝑡≥0
ds
𝑑𝑥
𝜏
+ 𝑥 = 𝐾𝑢(𝑡)
𝑑𝑡
𝑥 0 = 𝑥0
Second Order Systems
 In the system shown, the input
displacement, xi, will cause a
deflection in the spring, and
some time will be needed for
the output displacement xo to
reach the input displacement.
xo
xi
c
m
k
Second Order Systems
F
 m xo
k  x i  x o   c  x i  x o   m xo
xi
xo
c
m xo  c x o  kx o  c x i  kx i
m
k
xo 
c
k
x o  x o 
c
k
x i  x i
m
k
 If m/k << 1 s2 and c/k << 1 s, the system may be approximated as a zero
order system with unity gain.
 If, on the other hand, m/k << 1 s2 , but c/k is not, the system may be
approximated by a first order system. Systems with a storage and
dissipative capability but negligible inertial may be modeled using a
first-order differential equation.
Example – Automobile Accelerometer
 Consider the accelerometer used in
seismic and vibration engineering to
determine the motion of large bodies to
which the accelerometer is attached.
 The acceleration of the large body places
the piezoelectric crystal into compression
or tension, causing a surface charge to
develop on the crystal. The charge is
proportional to the motion. As the large
body moves, the mass of the
accelerometer will move with an inertial
response. The stiffness of the spring, k,
provides a restoring force to move the
accelerometer mass back to equilibrium
while internal frictional damping, c,
opposes any displacement away from
equilibrium.
Piezoelectric
crystal
xi
xo
c
m
k
Zero-Order systems
 Can we model the system below as a zero-order system? If the mass,
stiffness, and damping coefficient satisfy certain conditions, we may.
xo
xi
c
m
k
F
 m xo
k  x i  x o   c  x i  x o   m xo
k  x i  x o   c  x i  x o   m  xi  xo   m xi
k   c   m   m xi
First Order Systems
 Measurement systems that contain
storage elements do not respond
instantaneously to changes in input. The
bulb thermometer is a good example.
When the ambient temperature changes,
the liquid inside the bulb will need to
store a certain amount of energy in order
for it to reach the temperature of the
environment. The temperature of the
bulb sensor changes with time until this
equilibrium is reached, which accounts
physically for its lag in response.
 In general, systems with a storage or
dissipative capability but negligible inertial
forces may be modeled using a first-order
differential equation.
1st Order Systems with Step Input
error ratio =
excitation ratio =
current error current deviation from final value
=
starting error starting deviation from final value
current excitation
current deviation from initial value
=
desired (input) excitation
input deviation from initial value
𝜏
𝑑𝑥
+ 𝑥 = 𝐾𝑢(𝑡)
𝑑𝑡
𝑥 0 = 𝑥0
𝑥 𝑡 = 𝐾 + 𝑥0 − 𝐾 𝑒 −𝑡/𝜏
Error Ratio
𝑥 𝑡 −𝐾
= 𝑒 −𝑡/𝜏
𝑥0 − 𝐾
Excitation
Ratio
𝑥 𝑡 − 𝑥0
= 1 − 𝑒 −𝑡/𝜏
𝐾 − 𝑥0
Excitation ratio may also be called response ratio
= current response / desired response
 Note that the excitation ratio also represents
the system response in case of x0=0 and K=1
Example 1
 A bulb thermometer with a time constant τ
=100 s. is subjected to a step change in the
input temperature. Find the time needed
for the response ratio to reach 90%
Example 1 Solution
 A bulb thermometer with a time constant τ =100 s. is subjected to a
step change in the input temperature. Find the time needed for the
response ratio to reach 90%
𝑥 𝑡 − 𝑥0
= 1 − 𝑒 −𝑡/𝜏 = 0.9
𝐾 − 𝑥0
𝑥 𝑡 −𝐾
= 𝑒 −𝑡/𝜏 = 0.1
𝑥0 − 𝐾
𝑡 𝜏 = ln 10 = 2.3
𝑡 = 2.3 × 𝜏 = 230 𝑠.
≈ 4 minutes
1st Order Systems with Ramp Input
𝜏
𝑑𝑥
+ 𝑥 = 𝑥0 + 𝐾𝑟 𝑡𝑢(𝑡)
𝑑𝑡
𝑥 0 = 𝑥0
excitation ratio =
𝑥 𝑡 = 𝑥0 + 𝐾𝑟 𝑡 − 𝐾𝑟 𝜏(1 − 𝑒 −𝑡/𝜏 )
Error = 𝑥 𝑡 − 𝑓 𝑡 = −𝐾𝑟 𝜏(1 − 𝑒 −𝑡/𝜏 )
Steady State Error
𝑆. 𝑆. 𝐸 = lim 𝑥(𝑡) − 𝑓(𝑡)
𝑡→∞
𝑆. 𝑆. 𝐸 = 𝐾𝑟 𝜏
𝐸𝑥𝑐𝑖𝑡𝑎𝑡𝑖𝑜𝑛 𝑅𝑎𝑡𝑖𝑜 =
𝑥 𝑡 − 𝑥0
𝐾𝑟 𝑡
(1 − 𝑒 −𝑡/𝜏 )
𝐸𝑥𝑐𝑖𝑡𝑎𝑡𝑖𝑜𝑛 𝑅𝑎𝑡𝑖𝑜 = 1 −
𝑡 𝜏
Note that using L’Hospital rule
lim
𝑡 𝜏 →0
= lim
𝑡 𝜏 →0
1−
(1 − 𝑒 −𝑡/𝜏 )
𝑡 𝜏
1 − lim
𝑡 𝜏 →0
1 − 𝑒 −𝑡/𝜏
𝑡 𝜏
𝑒 −𝑡/𝜏
= 1 − lim
= 1−1= 0
𝑡 𝜏 →0 1
excitation ratio =
current excitation
desired (input) excitation
current deviation from initial value
input deviation from initial value
1st Order Systems with Harmonic Input
𝜏
𝑑𝑥
+ 𝑥 = 𝐹 cos⁡
(ωt)
𝑑𝑡
𝑥 𝑡 = 𝐶𝑒 −𝑡/𝜏 + 𝑋cos⁡
(ωt − φ)
C depends on the initial conditions and the
exponential term will vanish with time. We
are interested in the particular steady
solution 𝑋cos⁡
(ωt − φ). Solving for 𝑋 and φ,
we find
𝑋=
𝐹
1 + 𝜏𝜔
φ = tan−1 𝜏𝜔
2
1st Order Systems with Harmonic Input
Define the amplitude ratio 𝐴𝑟 = 𝑋 𝐹 and
the time ratio 𝑇𝑟 = 𝜏 𝑇 where 𝑇 = 2𝜋 𝜔 is
the period of the excitation function,
𝐴𝑟 =
𝑋
=
𝐹
1
1 + 𝜏𝜔
2
=
1
1 + 4𝜋 2 𝑇𝑟 2
φ = tan−1 𝜏𝜔 = tan−1 2𝜋𝑇𝑟
1st Order Systems with Harmonic Input
 The amplitude ratio, Ar(ω), and the
corresponding phase shift, ϕ, are
plotted vs. ωτ. The effects of τ and
ω on frequency response are
shown.
 For those values of ωτ for which the
system responds with Ar near unity,
the measurement system transfers
all or nearly all of the input signal
amplitude to the output and with
very little time delay; that is, X will
be nearly equal to F in magnitude
and ϕ will be near zero degrees.
1st Order Systems with Harmonic Input
 At large values of ωτ the measurement system filters out any frequency
information of the input signal by responding with very small amplitudes,
which is seen by the small Ar(ω) , and by large time delays, as evidenced by
increasingly nonzero ϕ.
1st Order Systems with Harmonic Input
 Any equal product of ω
and τ produces the
same results. If we
wanted to measure
signals with highfrequency content, then
we would need a system
having a small τ.
 On the other hand,
systems of large τ may
be adequate to measure
signals of low-frequency
content. Often the
trade-offs compete
available technology
against cost.
dB = 20 log Ar(ω)
1st Order Systems with Harmonic Input
 The dynamic error,δ(ω), of a system is defined as
δ(ω) = (X(ω) – F)/F
δ(ω) = Ar(ω) –1
It is a measure of the inability of a system to adequately reconstruct the
amplitude of the input signal for a particular input frequency. We normally
want measurement systems to have an amplitude ratio at or near unity over
the anticipated frequency band of the input signal to minimize δ(ω) .
 As perfect reproduction of the input signal is not possible, some dynamic error
is inevitable. We need some way to quantify this. For a first-order system, we
define a frequency bandwidth as the frequency band over which Ar(ω) >
0.707; in terms of the decibel defined as
dB = 20 log Ar(ω)
This is the band of frequencies within which Ar(ω) remains above 3 dB
Example 2
A temperature sensor is to be selected to measure temperature within a
reaction vessel. It is suspected that the temperature will behave as a simple
periodic waveform with a frequency somewhere between 1 and 5 Hz. Sensors
of several sizes are available, each with a known time constant. Based on time
constant, select a suitable sensor, assuming that a dynamic error of 2% is
acceptable.
Example 2. Solution
 A temperature sensor is to be
selected to measure temperature
within a reaction vessel. It is
suspected that the temperature will
behave as a simple periodic
waveform with a frequency
somewhere between 1 and 5 Hz.
Sensors of several sizes are
available, each with a known time
constant. Based on time constant,
select a suitable sensor, assuming
that an absolute value for the
dynamic error of 2% is acceptable.
 Accordingly, a sensor having a time
constant of 6.4 ms or less will work.
𝛿 ω
≤ 0.02
−0.02 ≤ 𝛿 ω ≤ 0.02
−0.02 ≤ 𝐴𝑟 − 1 ≤ 0.02
0.98 ≤ 𝐴𝑟 ≤ 1.02
0.98 ≤
1
1 + 𝜏𝜔
2
≤ 1.0
0 ≤ 𝜏𝜔 ≤ 0.2
The smallest value of 𝐴𝑟 will occur at
the largest frequency
𝜔 = 2𝜋𝑓 = 2𝜋(5)
0 ≤ 2𝜏𝜋(5) ≤ 0.2
𝜏 ≤ 6.4 × 10−3 s.
Example 2. Solution
 A temperature sensor is to be
selected to measure temperature
within a reaction vessel. It is
suspected that the temperature will
behave as a simple periodic
waveform with a frequency
somewhere between 1 and 5 Hz.
Sensors of several sizes are
available, each with a known time
constant. Based on time constant,
select a suitable sensor, assuming
that an absolute value for the
dynamic error of 2% is acceptable.
 Accordingly, a sensor having a time
constant of 6.4 ms or less will work.
𝛿 ω
≤ 0.02
−0.02 ≤ 𝛿 ω ≤ 0.02
−0.02 ≤ 𝐴𝑟 − 1 ≤ 0.02
0.98 ≤ 𝐴𝑟 ≤ 1.02
0.98 ≤
1
1 + 𝜏𝜔
2
≤ 1.0
0 ≤ 𝜏𝜔 ≤ 0.2
The smallest value of 𝐴𝑟 will occur at
the largest frequency
𝜔 = 2𝜋𝑓 = 2𝜋(5)
0 ≤ 2𝜏𝜋(5) ≤ 0.2
𝜏 ≤ 6.4 × 10−3 s.
2nd Order Systems
 Example:
 Spring – mass damper
 RLC Circuits
 Accelerometers
 Mathematical Model:
𝑑2 𝑥
𝑑𝑥
+
2𝜁𝜔
+ 𝜔𝑛 2 𝑥 = 𝑓 𝑡
𝑛
2
𝑑𝑡
𝑑𝑡
𝜁
Damping ratio (dimensionless)
𝜔𝑛
Natural frequency (1/s)
𝑓 𝑡 : Input (quantity to be measured)
𝑥:
Output (instrument response)
2nd Order Systems with step input
𝑓 𝑡 = 𝐾𝑢(𝑡)
𝑢 𝑡 =
0 𝑡<0
1 𝑡≥0
ds
𝑑2 𝑥
𝑑𝑥
2
+
2𝜁𝜔
+
𝜔
𝑥 = 𝐴𝑓 𝑡
𝑛
𝑛
𝑑𝑡 2
𝑑𝑡
𝜁
Damping ratio (dimensionless)
𝜔𝑛
Natural frequency (1/s)
𝑓 𝑡 : Input (quantity to be measured)
𝑥:
Output (instrument response)
𝐴:
Arbitrary constant
2nd Order Systems with step input
Correction to Figliola’s Book
Theory and Design for Mechanical Measurements
The sign should be – not +


2
2
2




1




1
2  1

0
0
y ( 0 )  KA  KA 
e 
e   KA  KA
0
2
2
 2  2 1

2


1
2

1


2nd Order Systems with step input
2nd Order Systems with periodic input
2nd Order Systems with step input