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Measurement System Behavior
Most measurement system dynamic behavior can be characterized by a linear ordinary differential equation of order n:
n n-1
n n-1 1 on n-1
m m-1
m m-1 om-1
y y dyd d + ( ) + + + y = F(t )a a a a
dtdt dt
where : (3.1)
x xd dF(t) = + + + x m nb b b
dt dt
Figure 3.2 Measurement system operation on an input signal, F(t), provides the output signal, y(t).
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Zero-Order System The simplest model of a system is the zero-order system, for which all the derivatives drop out: y = K F(t) (3.2) K is measured by static calibration. Most instruments do not exactly "follow" dynamic changes and hence do not behave as zero-order systems. We will use one instrument in the lab that is very close to a zero-order instrument - the linear position transducer. In our experiment, its behavior is very close to that predicted by Equation 3.2.
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First-Order Systems Consider the thermocouple we will use in Experiment 2.
Following example 3.3, we calculate the rate of temperature change:
where convective heat transfer
coefficient, surface area
sv
s
Q = dE/dt = m dT(t )/dt = h [ (t ) - T(t )]C A T
h =
= A
This can be rearranged:
v s
s s
mc dT(t)/dt +hA [T(t)-T(O)]
= hA [ (t)-T(O)] = hA F(t)T
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This is a first order linear differential equation. Suppose now F(t) is a step function, U(t): F(t) = O for t < O F(t) = A for t > O
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This equation can be written in the form:
which has the solution:
where
dT(t)
+T(t) = Tdt
0[ ] -t/T(t)= + T T eT
sv = m / h c A
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We can rewrite this equation in the form:
Taking the natural log of both sides gives
so a plot of the right side of the equation vs t will have a slope of -1/τ.
0
-t/ T(t)-T = e
T -T
0
lnT(t) -T
-t/ = T -T
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First Order Systems Step Response For the general first order equation
the solution is:
where A is the height of the step and U(t) is a unit step.
dy+ y = KA U(t)
dt
-t /
oy (t ) = K A + ( - K A) y e
Figure 3.6 in 2nd
and 3rd
Edition
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First Order System Frequency Response
We will now look at the response to a sinusoidal signal
First look at the complementary equation
This has the solution:
A particular solution can be found in the form
The complete solution is
sin( )
dy+ y = KA t
dt
0dy
+ y =dt
/tY(t)= Ce
sin[ ( )]y(t)= B t +
/
2 1/ 2
( ) ( )sin[ ( )]
where ( ) /[ ( ) ]
( ) ( )tan
t
-1
y t = B t + +Ce
B KA 1+
=
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We can therefore describe the entire frequency response characteristics in terms of a magnitude ratio and a phase shift
Note: τ is the only system characteristic which
affects the frequency response
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The amplitude is usually expressed in terms of the decibel dB = 20 log10 M(ω). The frequency bandwidth of an instrument is defined as the frequency below which M(ω)=0.707, or dB = -3 ("3 dB down"). First order systems act as low pass filters, in other words they attenuate high frequencies.
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A useful measure of the phase shift is the time delay of the signal:
tan
-1
1
( ) ( ) = =
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Example Suppose I want to measure a temperature which fluctuates with a frequency of 0.1 Hz with a minimum of 98% amplitude reduction. I require
2 1/ 2
0.98, or 20log 0.98 0.175
1 1
M( ) db = =
M( )= B/(KA)= /[ +( ) ]
rearranging
1/ 221/ ( ) 1
so for 98%, 0.2
or, 0.2 0.2 2 0.2 2 3.142 0.1
0.31sec
= M
M ( )
/ = / f = /( )
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.
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KNOWN: T(0) = 20ºC
T∞ = 37ºC
F(t) = [T∞ - T(0)]U(t)
ASSUMPTIONS... 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]