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MAE 334 - INTRODUCTION TO COMPUTERS AND INSTRUMENTATION

3 Dynamic Behavior of Measurement Systems


Order of a Dynamic Measurement System
Every measurement system responds to inputs in a unique way. For
example, your ability to hear high frequency sounds will probably degrade
as you age and will never be as keen as most dogs hearing. Sound
pressure waves are a dynamic signal and the sensing of these pressure
waves by a flexible membrane (like your ear drum) can be mathematically
modeled and therefore simulated.
Our goal in this section is to apply our understanding of the physics
involved in sensing a signal and build a mathematical model that could be
used to describe the response of the measurement system to a dynamic
signal. In prior sections we described the response of a measurement
system to a static signal and built a mathematical model which described
that response. The process of characterizing that response is referred to
as a static calibration and the resulting mathematical model is called the
static calibration curve.
In the first lab you will perform both a static and dynamic calibration of a
temperature sensor and determine the corresponding static and dynamic
models which describe the sensor response. In the case of a signal that is
changing with time (dynamic) a sensor that can keep up, or is fast
enough, is needed to accurately detect the change. In the case of the
temperature sensors used in the first lab both the sensor and the
environment being sensed must be at the same temperature to make an
accurate measurement. If the sensor is initially at a different temperature
then some amount of time is required for the sensor and the environment
to become the same temperature. There has been a dynamic change in
the sensor temperature in response to a dynamic change in the input
temperature signal.
In this example we understand that heat must be transferred from the
environment to the sensor. The physics of that heat transfer might be
modeled based on our understanding of conduction, convection, radiation
or possibly some combination thereof. In general we could reason that
the temperature sensor performs some mathematical operation on the
input signal and outputs the result.
In fact most measurement systems can be modeled using a differential
equation that describes the relationship between the input signal and the
output signal. In the first lab you will find the linear equation that describes
the response to a static input (a static calibration) and the first order
differential equation that describes the conductive heat transfer to and
from the sensor (a dynamic calibration).

MeasurementSystemBehaviorNotes.docx 1 of 12 9/11/2009
Scott H Woodward
MAE 334 - INTRODUCTION TO COMPUTERS AND INSTRUMENTATION

Figure 3.2 Measurement system operation on an input signal, F(t), provides the
output signal, y(t).

Measurement System Model


If the measurement system operation performed on the input signal, F(t),
in figure 3.2 is an nth-order linear differential equation then the output
signal, y(t), can be represented with the equation:

dny d n 1y dy
an n an 1 n 1  a1 a0 y F (t ) (3.1)
dt dt dt
where the coefficients, a0, a1, a2, …, an represent the physical system
parameters whose properties and values will depend on the
measurement system itself. The forcing function, F(t), can also be
generalized into an mth-order equation of the form:

dmx d m 1x dx
F (t ) bm m
bm 1 m 1
 b1 b0 x m n
dt dt dt
where b0, b1,…, bm also represent physical system parameters. The
nature of these equations should reflect the governing equations of the
pertinent fundamental physical laws of nature that are relevant to the
measurement system.

Zero-Order System
If all the derivatives in Equation 3.1 are zero then the most basic model of
a measurement system is obtained, the zero-order differential equation:

a0 y F (t )

From this equation it is easy to see that any input, F(t), is instantly
reflected in the output y with only a factor, a0, modification. If the input is a
dynamically varying signal b0x then y = b0/a0x or y = Kx. The factor K is

MeasurementSystemBehaviorNotes.docx 2 of 12 9/11/2009
Scott H Woodward
MAE 334 - INTRODUCTION TO COMPUTERS AND INSTRUMENTATION

often times referred to as the static sensitivity found during a static


calibration.

First-Order System
A linear time-invariant (LTI) first-order system contains a single mode of
energy storage. A simple Resister-Capacitor circuit is a first order system.

Here the underlying physics is described by the equation


dVout
RC Vout Vin
dt
This circuit is called a single pole low-pass RC filter and will be discussed
in greater detail in subsequent sections on signal conditioning and filters.
Systems with thermal capacity like a bulb thermometer or thermocouple
require heat transfer, Q, from their environment to effect a sensor
temperature change. The change in energy, E, with respect to time is
described by the first-order equation.
dE dT
Q mCv hAs T (t ) Ts (t )
dt dt
where m is the sensor’s mass, Cv is the sensor’s specific heat, h is the
convective heat transfer coefficient, As is the surface area of the sensor,
T is the temperature of the surrounding material and Ts is the
temperature of the sensor. This can be rearranged as

dT
mCv hAsT (t ) hAsTs (t )
dt
dT
mCv hAsTs (t ) hAs F (t )
dt
This can obviously be represented as a first-order differential equation in
the form of equation 3.1 as
dy
a1 a0 y F (t )
dt
To help clarify the underlying physics the equation can be recast by
dividing through by a0 and setting y dy dt .

MeasurementSystemBehaviorNotes.docx 3 of 12 9/11/2009
Scott H Woodward
MAE 334 - INTRODUCTION TO COMPUTERS AND INSTRUMENTATION

y y KF (t )

where a1 a0 . The parameter is called the time constant of the


system. Reflecting back it is easy to see that the time constant of a
single-pole low-pass RC filter is 1/RC and that of a temperature sensor is
based on the mass, specific heat, heat transfer coefficient, and the
surface area of the sensor, mCv hAs .

It is essential that you grasp the insight that the time constant of
such systems (LTI) or sensors is based on properties that do not
change (under normal operating conditions). I.e. a bulb
thermometer does not change in mass when subjected to a
temperature change nor does its specific heat, surface area or heat
transfer coefficient change therefore its time constant remains
constant.

Dynamic Calibration of a First-Order System


Like a static calibration, the sensor is subjected to a known input and
the resulting sensor output is recorded. For a dynamic calibration a
dynamic input is needed and the ability to measure a time varying
signal is required. Being engineers is makes sense that we would
start with equations that model the physics and find a simple solution
to them. If the input function, F(t), is a unit step function, U(t), then
y y KF (t ) can be recast as, y + y = y which has the solution:

y = y + [ y0 y ]e-t /
Recall that the unit step function, U(t), is zero for all time prior to t0
and 1 for all time thereafter. In practice U(t) usually has an
amplitude, A, other than 1. The difference between the input and
the output is often referred to as the error. With a simple
rearrangement of terms that error is clearly shown to be an
exponential function

y y -t /
e
[ y0 y ]
When t the error function is e-1 = 0.368 or y = 0.632(KA - y0).
By taking the natural log of the error function or when plotted in
semi-log coordinates the equation assumes a linear form.

y y
ln ln t/
[ y0 y ]

MeasurementSystemBehaviorNotes.docx 4 of 12 9/11/2009
Scott H Woodward
MAE 334 - INTRODUCTION TO COMPUTERS AND INSTRUMENTATION

Figure 3.8. The error function plotted in semi-log coordinates.

The slope of the linearized error function is -1/ . Finding the slope of a
line is less sensitive to errors than finding a point on a curve (at a
value of y = 0.632(KA - y0).)

Dynamic Calibration of Thermocouple


In the first lab you will be performing a dynamic calibration of a
thermocouple by subjecting it to a sudden change in temperature (i.e.
moving it from cold water to warm water).

dT
The governing equation from above mCv hAsTs (t ) hAs F (t ) can
dt
easily be recast in the more familiar form

mCv dT
Ts (t ) F (t )
hAs dt
mCv
where the time constant is defined by the physical constants . Here
hAs
the natural log of the error function is plotted in linear coordinates and a
line is fit to a portion of the data from 5 to 15 seconds. The slope of that
line, -0.208, is -1/time constant or = 4.8 seconds.

MeasurementSystemBehaviorNotes.docx 5 of 12 9/11/2009
Scott H Woodward
MAE 334 - INTRODUCTION TO COMPUTERS AND INSTRUMENTATION

Large Step Input Thermocouple


Dynamic Calibration in Water

90
80
T∞ = 75 C
70
Temperature (C)

60
50 ~2/3 Step = (0.632)(T∞-T0)+T0 ≈ 50 C
40
30
Time Constant ≈ (7 - t0) = 5 seconds
20
10
T0 = 5 C
0
0 5 10 15 20 25 30
t0 = 2 seconds Time (sec)

Linearized Thermocouple Step Input Response


Error Function, (t)
1
Linear Regression Fit
y = -0.208t + .482
0
τ = -1/-0.208 = 4.8 seconds

-1

-2
ln( (t))

-3

-4

-5
0 5 10 15 20 25 30
Time (sec)

MeasurementSystemBehaviorNotes.docx 6 of 12 9/11/2009
Scott H Woodward
MAE 334 - INTRODUCTION TO COMPUTERS AND INSTRUMENTATION

An example data set using the same thermocouple containing


considerable noise and quantization error is plotted below. Of note is the
difficulty with which ~2/3 of the step response could be determined. In
contrast the time constant determined from the linearized error function is
relatively insensitive to the noise and quantization errors.

Small Step Input Thermocouple


Dynamic Calibration in Water
12
11
10
Temperature (C)

9
8
7
6
5
4
0 5 10 15 20 25 30
Time (sec)

Linearized Thermocouple Step Input Response


Error Function, (t)
1

0.5

-0.5

-1
ln( (t))

-1.5

-2

-2.5

-3
0 5 10 15 20 25 30
Time (sec)

The line above is fit to only the first 2 seconds (~3 to 5 seconds) of the
linearized error function of the step input response and yields a = 4.8 s.

MeasurementSystemBehaviorNotes.docx 7 of 12 9/11/2009
Scott H Woodward
MAE 334 - INTRODUCTION TO COMPUTERS AND INSTRUMENTATION

Frequency Response of a First-Order System


The determination of the static sensitivity and time constant of a first-
order system transforms the black box “Measurement system operation”
in figure 3.2 into a known function. That implies that for any given output
which is correctly recorded the input that produced it could be
ascertained. This is possible because the differential equation describing
the physics of the measurement system is known and is solvable with
relative ease.
An intuitive description of the relationship between a system input and
output requires an understanding of the user, their intent as well as the
measurement system. Most users of thermometers are not interested in
dynamic temperature measurement. It is fair to say that few have the
training needed to relate the math to the physics and then apply this
knowledge to understand a dynamic phenomenon or solve a problem.
To facilitate an understanding of a system’s dynamic response we will
start by finding a solution to a simple sine wave input.

dy
+ y = KA sin(t )
dt
dy
We already know that the complementary equation, dt + y = 0 has a
t/
decaying exponential solution of the form y(t ) Ce . By applying
appropriate initial conditions a particular solution at a single frequency,
1, can be found in the form of y(t)= B1 sin[ 1t + ( 1 )] .

MeasurementSystemBehaviorNotes.docx 8 of 12 9/11/2009
Scott H Woodward
MAE 334 - INTRODUCTION TO COMPUTERS AND INSTRUMENTATION

For an input A1 sin( 1t) the output B1 sin[ 1t + ( 1 )] is produced. There


is an amplitude reduction from A1 to B1 and a delay in time, 1, reflected
in the phase shift of ( 1 ) .

The complete solution is


t/
y(t ) = B( )sin[ t + ( )]+Ce
1

)2
2
where B ( ) KA / 1 ( and ( ) tan 1 ( ) . This solution
depends only on the static sensitivity, K, and the time constant, . The
time constant is the only system characteristic that affects the frequency
response. This solution provides a relationship between the input and
output for all frequencies. The ratio of output/input magnitude would
therefore be M ( ) B( ) / KA . Figure 3.12 is plot of the magnitude ratio
versus the normalized frequency . Note that at the magnitude is
1

)2
2
0.707 which can be derived from M ( ) 1/ 1 ( 1/ 2 .

Figure 3.12 is plotted with the magnitude ratio expressed in decibels


below. A decibel, dB, is unit of power defined as 20 log10(M( )).
The frequency response of a first-order sensor is defined based on the
time constant as . This can be stated in terms of the magnitude
ratio as the -3 dB point or the point when the magnitude ratio is 0.707.
This definition also carries over to other non-first-order systems even
though they are based on a different mathematical model (physics).

MeasurementSystemBehaviorNotes.docx 9 of 12 9/11/2009
Scott H Woodward
MAE 334 - INTRODUCTION TO COMPUTERS AND INSTRUMENTATION

Bode Plot of Single Pole Low Pass Filter

3
0
-3
-6
Amplitude Ratio (dB)

-9
-12
-15
-18
-21
-24
-27
-30
0.001 0.010 0.100 1.000 10.000
log(f/fc) or

A first-order system can be thought of as a low pass filter. They attenuate


higher frequencies and pass lower frequencies with little attenuation.
A first-order system always delays the input signal in time. That delay
results in a phase shift as evidenced by the ( ) term in the complete
solution. The time delay, , in figure 3.11 above can be solved for as

( 1) tan 1 ( 1 )
1
1 1

By removing the particular frequency, , from the above equation a plot


similar to the magnitude ratio can be generated for all frequencies.

MeasurementSystemBehaviorNotes.docx 10 of 12 9/11/2009
Scott H Woodward
MAE 334 - INTRODUCTION TO COMPUTERS AND INSTRUMENTATION

Example

Suppose I want to measure a temperature which fluctuates with a


frequency of 0.1 Hz with a minimum of 98% amplitude reduction.

ASSUMPTIONS: basic first-order temperature sensor like a thermocouple

FIND: Magnitude ratio of at least 0.98

M ( ) 0.98, or dB = 20log 0.98 = 0.175

B 1
M( ) 1
KA 1 ( )2
2

rearranging gives
1/2
= 1/ M ( ) 2 1
so for M ( ) 98%, 0.2
or, 0.2 / = 0.2 / 2 f = 0.2 / 2 3.142 0.1
0.31sec
Problem 3.7

A thermocouple has a time constant of 20 ms. Determine its 90% rise


time.

MeasurementSystemBehaviorNotes.docx 11 of 12 9/11/2009
Scott H Woodward
MAE 334 - INTRODUCTION TO COMPUTERS AND INSTRUMENTATION

Example 3.3

Suppose a bulb thermometer originally


indicating 20ºC is suddenly exposed to
a fluid temperature of 37 ºC. Develop
a simple model to simulate the
thermometer output response.
KNOWN:
T0 = 20ºC
T∞ = 37ºC
F(t) = [T∞ - T0]U(t)
ASSUMPTIONS:
Normal first-order response
FIND: T(t)
SOLUTION:
The rate at which energy is exchanged
between the sensor and the environment through convection, , must be
balanced by the storage of energy within the thermometer, dE/dt.

For a constant mass temperature sensor,

This can be written in the form

dividing by hAs

Therefore:

The thermometer response is therefore:

[ºC]

MeasurementSystemBehaviorNotes.docx 12 of 12 9/11/2009
Scott H Woodward

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