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Class 4
Dynamic Performance
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
The simplest model of a measurement system is the zero-order system
model. This is represented by the zero-order differential equation:
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.
)()(1
)(
tKftfa
x
tfxa
o
o
l1 l
2
f(t)x(t)
x(t)/l2
= f(t)/l
1
K = l2
/l1
Mechanical Zero-Order Systems
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.
)()(1
)(
tKftfa
x
tfxa
o
o
l1 l
2
f(t)x(t)
x(t)/l2
= f(t)/l
1
K = l2
/l1
Electrical Zero-Order Systems
In a zero-order system, the output responds to
the input signal instantaneously.
vi
R1
R2
Vo
io
io
vRR
Rv
RRR
viRv
21
1
121
1
A Unity Gain Zero-Order system
l
xi
xo
xo
(t) = xi(t)
K = 1
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.
m
k
c
xi
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, 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 T
o 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
The ratio mcp/hA has a units of seconds and is called the time constant
Body Temperature
Room Temperature
t
time
TTdt
dT
hA
mc
hAThATdt
dTmc
TThAdt
dTmc
QE
p
p
p
instored
First Order Systems
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
TTdt
dT
TTdt
dT
hA
mcp
1st Order Systems Examples:
Bulb Thermometer
RC Circuits
Terminal velocity
Mathematical Model:
𝜏𝑑𝑥𝑑𝑡 + 𝑥= 𝑓ሺ𝑡ሻ τ: Time constant 𝑓ሺ𝑡ሻ: Input (quantity to be measured) 𝑥: Output (instrument response)
1st Order Systems with Step Input
𝑓ሺ𝑡ሻ= 𝐾𝑢(𝑡)
𝑢ሺ𝑡ሻ= ቄ0 𝑡 < 01 𝑡 ≥ 0
ds
𝜏𝑑𝑥𝑑𝑡 + 𝑥= 𝐾𝑢(𝑡)
𝑥ሺ0ሻ= 𝑥0
Second Order Systems In the system shown, the input displacement, x
i, will
cause a deflection in the spring, and some time will be
needed for the output displacement xo
to reach the
input displacement.
m
k
c
xi
xo
Second Order Systems
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.
m
k
c
xi x
o
iiooo
iiooo
ooioi
o
xxk
cxx
k
cx
k
m
kxxckxxcxm
xmxxcxxk
xmF
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.
m
k
c
xi x
o
Piezoelectric crystal
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.
m
k
c
xi
xo
i
ioioioi
ooioi
o
xmmck
xmxxmxxcxxk
xmxxcxxk
xmF
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.
𝜏𝑑𝑥𝑑𝑡 + 𝑥= 𝐾𝑢(𝑡)
𝑥ሺ0ሻ= 𝑥0 𝑥ሺ𝑡ሻ= 𝐾+ሺ𝑥0 − 𝐾ሻ𝑒−𝑡/𝜏 𝑥ሺ𝑡ሻ− 𝐾𝑥0 − 𝐾 = 𝑒−𝑡/𝜏
𝑥ሺ𝑡ሻ− 𝑥0𝐾− 𝑥0 = 1− 𝑒−𝑡/𝜏
1st Order Systems with Step Input
Error Ratio
Excitation Ratio
Note that the excitation ratio also represents the system response in case of
x0=0 and K=1
error ratio = current errorstarting error = current deviation from final valuestarting deviation from final value
excitation ratio = current excitationdesired (input) excitation= current deviation from initial valueinput deviation from initial value
Excitation ratio may also be called response ratio = current response / desired
response
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%
𝑥ሺ𝑡ሻ− 𝑥0𝐾− 𝑥0 = 1− 𝑒−𝑡/𝜏 = 0.9
𝑥ሺ𝑡ሻ− 𝐾𝑥0 − 𝐾 = 𝑒−𝑡/𝜏 = 0.1
𝑡 𝜏= lnሺ10ሻ= 2.3Τ
𝑡 = 2.3× 𝜏= 230 𝑠.≈ 4 minutes
1st Order Systems with Ramp Input excitation ratio = current excitationdesired (input) excitation
excitation ratio = current deviation from initial valueinput deviation from initial value
𝜏𝑑𝑥𝑑𝑡 + 𝑥= 𝑥0 + 𝐾𝑟𝑡𝑢(𝑡)
𝑥ሺ0ሻ= 𝑥0 𝑥ሺ𝑡ሻ= 𝑥0 + 𝐾𝑟𝑡− 𝐾𝑟𝜏(1− 𝑒−𝑡/𝜏)
𝑆.𝑆.𝐸= lim𝑡→∞ሺ𝑥(𝑡) − 𝑓(𝑡)ሻ 𝑆.𝑆.𝐸= 𝐾𝑟𝜏
𝐸𝑥𝑐𝑖𝑡𝑎𝑡𝑖𝑜𝑛 𝑅𝑎𝑡𝑖𝑜 = 𝑥ሺ𝑡ሻ− 𝑥0𝐾𝑟𝑡
𝐸𝑥𝑐𝑖𝑡𝑎𝑡𝑖𝑜𝑛 𝑅𝑎𝑡𝑖𝑜 = 1− (1− 𝑒−𝑡/𝜏)𝑡 𝜏Τ
limሺ𝑡 𝜏Τ ሻ→0ቆ1− (1− 𝑒−𝑡/𝜏)𝑡 𝜏Τ ቇ
= limሺ𝑡 𝜏Τ ሻ→0ሺ1ሻ− lim
ሺ𝑡 𝜏Τ ሻ→0൫1− 𝑒−𝑡/𝜏൯𝑡 𝜏Τ
= 1− limሺ𝑡 𝜏Τ ሻ→0𝑒−𝑡/𝜏1 = 1− 1 = 0
Error = 𝑥ሺ𝑡ሻ− 𝑓ሺ𝑡ሻ= −𝐾𝑟𝜏(1− 𝑒−𝑡/𝜏)
Steady State Error
Note that using L’Hospital rule
1st Order Systems with Harmonic Input
𝜏𝑑𝑥𝑑𝑡 + 𝑥= 𝐹 cos (ωt) 𝑥ሺ𝑡ሻ= 𝐶𝑒−𝑡/𝜏 + 𝑋cos (ωt − φ)
𝑋= 𝐹ඥ1+ሺ𝜏𝜔ሻ2
φ = tan−1ሺ𝜏𝜔ሻ
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
1st Order Systems with Harmonic Input
𝐴𝑟 = 𝑋𝐹= 1ඥ1+ሺ𝜏𝜔ሻ2 = 1
ඥ1+ 4𝜋2𝑇𝑟2
φ = tan−1ሺ𝜏𝜔ሻ=tan−1ሺ2𝜋𝑇𝑟ሻ
Define the amplitude ratio 𝐴𝑟 = 𝑋 𝐹Τ and the time ratio 𝑇𝑟 = 𝜏 𝑇Τ where 𝑇= 2𝜋 𝜔Τ is the period of the excitation function,
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 high-frequency 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.
aC𝛿ሺωሻaC≤ 0.02 −0.02 ≤ 𝛿ሺωሻ≤ 0.02 −0.02 ≤ 𝐴𝑟 − 1 ≤ 0.02 0.98 ≤ 𝐴𝑟 ≤ 1.02
0.98 ≤ 1ඥ1+ሺ𝜏𝜔ሻ2 ≤ 1.0
0 ≤ 𝜏𝜔≤ 0.2
0 ≤ 2𝜏𝜋(5) ≤ 0.2
The smallest value of 𝐴𝑟 will occur at the largest frequency
𝜔= 2𝜋𝑓= 2𝜋(5)
𝜏≤ 6.4× 10−3s.
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.
aC𝛿ሺωሻaC≤ 0.02 −0.02 ≤ 𝛿ሺωሻ≤ 0.02 −0.02 ≤ 𝐴𝑟 − 1 ≤ 0.02 0.98 ≤ 𝐴𝑟 ≤ 1.02
0.98 ≤ 1ඥ1+ሺ𝜏𝜔ሻ2 ≤ 1.0
0 ≤ 𝜏𝜔≤ 0.2
0 ≤ 2𝜏𝜋(5) ≤ 0.2
The smallest value of 𝐴𝑟 will occur at the largest frequency
𝜔= 2𝜋𝑓= 2𝜋(5)
𝜏≤ 6.4× 10−3s.
2nd Order Systems Example:
Spring – mass damper
RLC Circuits
Accelerometers
Mathematical Model:
𝑑2𝑥𝑑𝑡2 + 2𝜁𝜔𝑛 𝑑𝑥𝑑𝑡 + 𝜔𝑛2𝑥= 𝑓ሺ𝑡ሻ 𝜁 Damping ratio (dimensionless) 𝜔𝑛 Natural frequency (1/s) 𝑓ሺ𝑡ሻ: Input (quantity to be measured) 𝑥: Output (instrument response)
2nd Order Systems with step input
𝑓ሺ𝑡ሻ= 𝐾𝑢(𝑡)
𝑢ሺ𝑡ሻ= ቄ0 𝑡 < 01 𝑡 ≥ 0
ds
𝑑2𝑥𝑑𝑡2 + 2𝜁𝜔𝑛 𝑑𝑥𝑑𝑡 + 𝜔𝑛2𝑥= 𝐴𝑓ሺ𝑡ሻ 𝜁 Damping ratio (dimensionless) 𝜔𝑛 Natural frequency (1/s) 𝑓ሺ𝑡ሻ: 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 +
012
12
12
1
12
1)0(
2
20
2
20
2
2
KAKAeeKAKAy
2nd Order Systems with step input
2nd Order Systems with periodic input
2nd Order Systems with step input