Dynamic Characteristics
Lecture04 SME3242 Instrumentation
1
Static transfer function – how the output related to input if the input is constant
Dynamic transfer function – also called time response
Lecture04 SME3242 Instrumentation
2
FIGURE 1.27 The dynamic transfer function specifies how a
sensor output varies when the input changes instantaneously in
time (i.e., a step change).
Curtis Johnson
Process Control Instrumentation Technology, 8e]
Lecture04 SME3242 Instrumentation
Copyright ©2006 by Pearson Education, Inc.
Upper Saddle River, New Jersey 307458
All rights reserved.
2.1.3: Mathematical model structure
Input and output relationship of a linear measurement
system - ordinary differential equation (ODE):
n1
n
dy
d y
an n  an1 n1  a1  a0 y
dt
dt
dt
d y
bm
m
d u
dt m
 bm1
m1
u  du 
b0 u
 b1
m1
dt
dt
d
where,
u = input, y = output;
u and y varies with t
n>m
a, b = constant coefficients
Lecture04 SME3242 Instrumentation
4
: Transfer
Function of An Accelerometer
xi
xo
k
m
Fs
Applying 2nd Newton’s
Law: F = ma
ma
=
FD
c
Differential equation:
2
dxi dxo
d xo

)m 2
k ( xi  xo )  c(
dt
dt
dt
2
d xo c dxo k
c dxi k

 xo 
 xi
2
dt
m dt m
m dt m
Lecture04 SME3242 Instrumentation
5
Dynamic characteristic
- the output response of the instrument against
time when the input is varied
- the relation between any input and output for
nth order system can be written as:
n
n1
d y
dy
d y
an n an1 n1 a1 a0 y  b0 u
dt
dt
dt
- 3 types of response: zero order response, first
order response and second order response
Lecture04 SME3242 Instrumentation
6
Dynamic response of zero order instrument
(i.e. n = 0)
- the zero order instrument is represented by
a0y=b0u or y=Ku
or
y/u = K
(y=output, u=input, K=b0/a0=static sensitivity)
- the output responses linearly to the input
Lecture04 SME3242 Instrumentation
7
Eg: potentiometer
Lecture04 SME3242 Instrumentation
8
Dynamic response of first order instrument
(i.e. n = 1)
dy
 a 0 y  b0u
a1
dt
- dividing the equation by a0, and apply D-operator
b0
a 1 dy 

u ; ( TD  1 ) y  Ku
y
a0
a 0 dt
- T=a1/a0=time constant, K=b0/a0=static sensitivity
Lecture04 SME3242 Instrumentation
9
- the operational transfer function
K
y

u (1  TD)
u(t)
Sensor
y(t)
- The time constant; T, represents the time taken for
the output to reach 63% of the final value and it
reaches its final value (99%) at around 5T.
Lecture04 SME3242 Instrumentation
10
Eg.: Thermocouple
Lecture04 SME3242 Instrumentation
11
Characteristic first-order time response of
a sensor.
t
T
2T
3T
4T
5T
10T
y(t)
0.63212
0.86466
0.95021
0.98168
0.99326
0.99995
Lecture04 SME3242 Instrumentation
%
63.212
86.466
95.021
98.168
99.326
99.995
12
Characteristic first-order time response of
a sensor.
Lecture04 SME3242 Instrumentation
13
General equation as function of time following a step input is given as:
y (t )  yi  ( y f  yi )[1  e
 t /T
]
where,
yi = initial output from static transfer function and initial input
yf = final output from static transfer function and final input
T = time constant = 63% time
Lecture04 SME3242 Instrumentation
14
Characteristic first‐order exponential time response of a sensor
to a step change of input.
y
yf
yi
Curtis Johnson
Process Control Instrumentation Technology,
Lecture04 SME3242 Instrumentation
8e]
Copyright ©2006 by Pearson
Education, Inc.
Upper Saddle River, New Jersey
07458
All rights reserved.
15
A sensor measures temperature linearly with a static
transfer function of 33 mV/0C and has a 1.5‐s time
constant. Find the output 0.75 s after input changes
from 200C to 410C. Find the error in temperature this
represents.
* Time response analysis always applied to the output
of the sensor because it is only the output of the
sensor that lagged
Lecture04 SME3242 Instrumentation
16
Given static transfer function:
V = (33mV/ºC)T
Hence, initial and final output of the sensor are:
yi = (33mV/ºC)(20ºC)
= 660mV
yf = (33mV/ºC)(41ºC)
= 1353mV
Lecture04 SME3242 Instrumentation
17
Time response of first-order system,
t / T
=
y (t ) yi  (y f  yi )[1  e ]
Substitute the value of yi and yf,
y 0.75
= 660  (1353 - 660)[1  e  0.75 /1.5 ]
= 932.7 mV
Lecture04 SME3242 Instrumentation
18
The corresponding temperature for sensor output of
932.7 mV,
932 .7 mV
T
33 mV / C
 28 .3 C
Since the actual temperature is 41ºC, hence the error
in temperature is:
error = (true value – instrument reading)
= (41ºC – 28.3ºC)
= 12.7ºC
Lecture04 SME3242 Instrumentation
19
When t = 5T i.e. t = 5(1.5) = 7.5 s,
y 7.5 = 660  (1353 - 660)[1  e
 7.5 /1.5
]
= 1348.3 mV
The corresponding temperature for sensor output of
1348.3 mV is:
mV

1348.3


T
41 C

33 mV / C
which is the exact measured temperature
Lecture04 SME3242 Instrumentation
20
Dynamic response of second-order instrument
(i.e. n = 2)
a2
d2y
dy
 a1
 a 0 y  b0u
2
dt
dt
Applying D operator
y 
b0u
(a 0  a1D  a 2 D 2 )
Lecture04 SME3242 Instrumentation
21
Applying Laplace Transform (with all initial
conditions equal to zero) and rearranging the
equation:
2

K n
y  2
s  2  n s  
2
n
u
where,
 = damping ratio
n= natural frequency
Lecture04 SME3242 Instrumentation
22
The time response is given as:
q0(t)  qe-atsin(nt)
where q=amplitude and a= n is output
damping ratio
Eg.: Accelerometer
Lecture04 SME3242 Instrumentation
23
Lecture04 SME3242 Instrumentation
24
FIGURE 1.29 Characteristic second-order oscillatory time response
of a sensor.
Curtis Johnson
Process Control Instrumentation
Lecture04 SME3242 Instrumentation
Technology, 8e]
Copyright ©2006 by Pearson
Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.25
undamped (=0)
under damped (1>>0)
over damped (1)
Lecture04 SME3242 Instrumentation
26
Lecture04 SME3242 Instrumentation
27

Compact I/O module for DeviceNet™ - TURCK

Lecture 3: Nonhomogeneous Linear Systems of ODEs

70001 Testing of metal adhesive bondings