a a a a + ( ) + + + y = F(t ) ddy y dy dt dt dt F(t) = + + + x m nb b b dt dt · 2007-09-14 · FirstOrderSystems.docx 9/14/2007 1:03 PM Page 2 Zero-Order System The simplest model

FirstOrderSystems.docx 9/14/2007 1:03 PM Page 1

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).


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.



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


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


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


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


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


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+

=


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


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.


A useful measure of the phase shift is the time delay of the signal:

tan

-1

1

( ) ( ) = =


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.


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]

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a a a a + ( ) + + + y = F(t ) ddy y dy dt dt dt F(t) = + + + x m nb b b dt dt · 2007-09-14 · FirstOrderSystems.docx 9/14/2007 1:03 PM Page 2 Zero-Order System The simplest model