Class 4 - Dynamic Performance Characteristics
Class 4 - Dynamic Performance Characteristics
Class 4 - Dynamic Performance Characteristics
Class 4
Dynamic Performance
The dynamic characteristics of a measuring instrument
in system response.
Mechanical Zero-Order Systems
l l
The simplest model of a measurement system is the zero-order system 1 2
model. This is represented by the zero-order differential equation:
ao x f (t ) f(t)
x(t)
1 x(t)/l = f(t)/l
x f (t ) Kf (t )
K is the static sensitivity or steady gain of the system. It is a measure of 2 1
the amountaofochange in the output in response to the change in the
input.
K = l /l
2 1
Mechanical Zero-Order Systems
l l
In a zero-order system, the output responds to 1 2
the input signal instantaneously.
x(t)/l = f(t)/l
2 1
K = l /l
2 1
ao x f (t )
1
x f (t ) Kf (t )
ao
Electrical Zero-Order Systems
In a zero-order system, the output responds to v
i
the input signal instantaneously.
R R
1 2
V
o
vi
vo iR1 R1
R1 R2
R1
vo vi
R1 R2
A Unity Gain Zero-Order system
l
x x
i o
x (t) = x (t)
o i
K=1
Non-zero Order Systems
Measurement systems that contain storage or dissipative elements
response.
k
Non-zero Order Systems
Measurement systems that contain storage or dissipative elements
temperature changes, the liquid inside the bulb will need to store
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
Room Temperature
time
First Order Systems
Suppose a bulb thermometer originally at temperature T is suddenly
o
exposed to a fluid temperature T . Develop a model that simulates the
∞
thermometer output response.
E stored Q in
dT
mc p hA T T Body Temperature
dt
dT
mc p hAT hAT
dt
Room Temperature
mc p dT
The ratio mcp/hA has a T and is called the time constant
Tof seconds
units
hA dt
t
time
First Order Systems
mc p dT
T T
hA dt
dT
T T
dt
The ratio mcp/hA has a units of seconds and is called the time constant, τ. If
Body Temperature
the time constant is much less then 1 second, the system may be
Room Temperature
time
1st Order Systems
Examples:
Bulb Thermometer
RC Circuits
Terminal velocity
Mathematical Model:
𝑑𝑥
𝜏 + 𝑥 = 𝑓ሺ𝑡ሻ
𝑑𝑡
τ: Time constant
𝑓ሺ𝑡ሻ: Input (quantity to be measured)
𝑥: Output (instrument response)
1st Order Systems with Step Input
𝑓ሺ𝑡ሻ = 𝐾𝑢(𝑡)
0 𝑡<0
𝑢ሺ𝑡ሻ = ቄ
1 𝑡≥0
ds
𝑑𝑥
𝜏 + 𝑥 = 𝐾𝑢(𝑡)
𝑑𝑡
𝑥 ሺ0ሻ = 𝑥0
Second Order Systems
In the system shown, the input displacement, x , will
i
cause a deflection in the spring, and some time will be
x x
i o
c
k
Second Order Systems
F mx o x
k xi xo c x i x o mxo
i x
o
c
m c
xo x o xo cx i kxi k
k 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
attached.
The acceleration of the large body places the piezoelectric crystal into
x x
i
o
c
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.
x x
i o
c
F mx o
k xi xo c x i x o mxo
k xi xo c x i x o m xi xo mxi
k c m mx i
First Order Systems
Measurement systems that contain storage elements do not respond
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
𝑑𝑥
𝜏 + 𝑥 = 𝐾𝑢(𝑡)
𝑑𝑡
Excitation ratio may also be called response ratio = current response / desired
𝑥ሺ0ሻ = 𝑥0
response
Note that the excitation ratio also represents the system response in case of
change in the input temperature. Find the time needed for the response
𝑥ሺ𝑡ሻ− 𝑥0
= 1 − 𝑒 −𝑡/𝜏 = 0.9
𝐾 − 𝑥0
𝑥ሺ𝑡 ሻ− 𝐾
= 𝑒 −𝑡/𝜏 = 0.1
𝑥0 − 𝐾
𝑡Τ𝜏 = lnሺ10ሻ = 2.3
𝑡 = 2.3 × 𝜏 = 230 𝑠.
≈ 4 minutes
1st Order Systems with Ramp Input
𝑑𝑥
𝜏 + 𝑥 = 𝑥0 + 𝐾𝑟 𝑡𝑢(𝑡)
𝑑𝑡
current excitation
𝑥ሺ0ሻ = 𝑥0 excitation ratio =
desired (input) excitation
𝑥 ሺ𝑡ሻ = 𝑥0 + 𝐾𝑟 𝑡 − 𝐾𝑟 𝜏(1 − 𝑒 −𝑡/𝜏 )
Error = 𝑥ሺ𝑡ሻ− 𝑓ሺ𝑡ሻ= −𝐾𝑟 𝜏(1 − 𝑒 −𝑡/𝜏 ) current deviation from initial value
excitation ratio =
Steady State Error input deviation from initial value
𝑆. 𝑆. 𝐸 = 𝐾𝑟 𝜏
𝑥ሺ𝑡ሻ− 𝑥0
𝐸𝑥𝑐𝑖𝑡𝑎𝑡𝑖𝑜𝑛 𝑅𝑎𝑡𝑖𝑜 =
𝐾𝑟 𝑡
(1 − 𝑒 −𝑡/𝜏 )
𝐸𝑥𝑐𝑖𝑡𝑎𝑡𝑖𝑜𝑛 𝑅𝑎𝑡𝑖𝑜 = 1 −
𝑡 Τ𝜏
Note that using L’Hospital rule
(1 − 𝑒 −𝑡/𝜏 )
lim ቆ1 − ቇ
ሺ𝑡 Τ𝜏 ሻ→0 𝑡 Τ𝜏
൫1 − 𝑒 −𝑡/𝜏 ൯
= lim ሺ1ሻ− lim
ሺ𝑡 Τ𝜏 ሻ→0 ሺ𝑡 Τ𝜏 ሻ→0 𝑡Τ𝜏
𝑒 −𝑡/𝜏
= 1 − lim = 1−1= 0
ሺ𝑡 Τ𝜏 ሻ→0 1
1st Order Systems with Harmonic Input
𝑑𝑥
𝜏 + 𝑥 = 𝐹 cos(ωt)
𝑑𝑡
𝑥 ሺ𝑡ሻ = 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
𝐹
𝑋=
ඥ1 + ሺ𝜏𝜔ሻ2
φ = tan−1 ሺ𝜏𝜔ሻ
1st Order Systems with Harmonic Input
Define the amplitude ratio 𝐴𝑟 = 𝑋 Τ𝐹 and
the time ratio 𝑇𝑟 = 𝜏Τ𝑇 where 𝑇 = 2𝜋Τ𝜔 is
the period of the excitation function,
𝑋 1 1
𝐴𝑟 = = =
𝐹 ඥ1 + ሺ𝜏𝜔 ሻ2 ඥ1 + 4𝜋 2 𝑇𝑟 2
of the input signal amplitude to the output and with very little
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
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) = 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
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)
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 ȁ𝛿 ሺωሻȁ ≤ 0.02
temperature within a reaction vessel. It is suspected that the
−0.02 ≤ 𝛿 ሺωሻ ≤ 0.02
temperature will behave as a simple periodic waveform with a
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 ȁ𝛿 ሺωሻȁ ≤ 0.02
temperature within a reaction vessel. It is suspected that the
−0.02 ≤ 𝛿 ሺωሻ ≤ 0.02
temperature will behave as a simple periodic waveform with a
𝜔 = 2𝜋𝑓 = 2𝜋(5)
0 ≤ 2𝜏𝜋(5) ≤ 0.2
𝜏 ≤ 6.4 × 10−3 s.
2nd Order Systems
Example:
RLC Circuits
Accelerometers
Mathematical Model:
𝑑2 𝑥 𝑑𝑥
2
+ 2𝜁𝜔𝑛 + 𝜔𝑛 2 𝑥 = 𝑓 ሺ𝑡ሻ
𝑑𝑡 𝑑𝑡
𝜁 Damping ratio (dimensionless)
𝜔𝑛 Natural frequency (1/s)
𝑓 ሺ𝑡ሻ: Input (quantity to be measured)
𝑥: Output (instrument response)
2nd Order Systems with step input
𝑓ሺ𝑡ሻ = 𝐾𝑢(𝑡)
𝑑2 𝑥 𝑑𝑥
+ 2𝜁𝜔𝑛 + 𝜔𝑛
2
𝑥 = 𝐴𝑓ሺ𝑡ሻ
𝑑𝑡 2 𝑑𝑡
0 𝑡<0
𝑢ሺ𝑡ሻ = ቄ 𝜁 Damping ratio (dimensionless)
1 𝑡≥0
𝜔𝑛 Natural frequency (1/s)
ds 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
2 2
1 0 1 0 2 2 1
y (0) KA KA e e KA KA 0
2 2 1 2 2
1 2 2
1
2nd Order Systems with step input
2nd Order Systems with periodic input
2nd Order Systems with step input