lecture04 dynamic characteristicsmohamed/ukur inst notes/lecture 04.pdf · 2012. 10. 2. · dynamic...

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



FIGURE 1.27 The dynamic transfer function specifies how asensor output varies when the input changes instantaneously intime (i.e., a step change).

Curtis JohnsonProcess Control Instrumentation Technology, 8e] Copyright ©2006 by Pearson Education, Inc.

Upper Saddle River, New Jersey 07458All rights reserved.


Input and output relationship of a linear measurement system - ordinary differential equation (ODE):

where,u = input, y = output; u and y varies with tn > ma, b = constant coefficients

2.1.3: Mathematical model structure

d dyy yadt

adtda

dta n

n

nn

n

n 011

1

1

dtdtdtd du ubbdbub m

m

mm

m

m 011

1

1

y

u


: Transfer Function of An Accelerometer

Differential equation:

xoxi

mk

c FD

=Fs ma

dxdx o d2

2

)()(dt

xmdtdt

cxxk oioi

kk ccx dx ii

ooo x

mdtdx

mx

mdtmdtd

2

2

Applying 2nd Newton’s Law: F = ma


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:

- 3 types of response: zero order response, firstorder response and second order response

y d dy ubyadt

adt

adtda n

n

nn

n

n 0011

1

1

y


Dynamic response of zero order instrument (i.e. n = 0)

- the zero order instrument is represented bya0y=b0u or y=Ku

or

(y=output, u=input, K=b0/a0=static sensitivity)

- the output responses linearly to the input

y/u = K


Eg: potentiometer


Dynamic response of first order instrument (i.e. n = 1)

- dividing the equation by a0, and apply D-operator

- T=a1/a0=time constant, K=b0/a0=static sensitivity

1 KuyTDuabydt

dyaa )1(;

0

0

0

ubyadtdya 001


Sensoru(t) y(t)

- the operational transfer function

y)1( TD

Ku

- The time constant; T, represents the time taken forthe output to reach 63% of the final value and itreaches its final value (99%) at around 5T.


Eg.: Thermocouple


Characteristic first-order time response of a sensor.

t y(t) %T 0.63212 63.2122T 0.86466 86.4663T 0.95021 95.0214T 0.98168 98.1685T 0.99326 99.32610T 0.99995 99.995


Characteristic first-order time response of a sensor.

General equation as function of time following a step input is given as:

where,yi = initial output from static transfer

function and initial inputyf = final output from static transfer function

and final inputT = time constant = 63% time


]1)[()( /Ttifi eyyyty


Characteristic first‐order exponential time response of a sensorto a step change of input.

Curtis JohnsonProcess Control Instrumentation Technology, 8e]

Copyright ©2006 by Pearson Education, Inc.

Upper Saddle River, New Jersey 07458

All rights reserved.

yf

yi

y

A sensor measures temperature linearly with a statictransfer function of 33 mV/0C and has a 1.5‐s timeconstant. Find the output 0.75 s after input changesfrom 200C to 410C. Find the error in temperature thisrepresents.

* Time response analysis always applied to the outputof the sensor because it is only the output of thesensor that lagged


Given static transfer function:V = (33mV/ºC)T

Hence, initial and final output of the sensor are:yi = (33mV/ºC)(20ºC)

= 660mVyf = (33mV/ºC)(41ºC)

= 1353mV


Substitute the value of yi and yf,


0.75y 660 )[ ]1 /1.50.75e(1353 - 660=

= 932.7 mV

]1)[()( /Ttifi eyyyty =

Time response of first-order system,


The corresponding temperature for sensor output of932.7 mV,

C3.28

mVCmV

T/33

7.932

Since the actual temperature is 41ºC, hence the errorin temperature is:

error = (true value – instrument reading)= (41ºC – 28.3ºC)= 12.7ºC


When t = 5T i.e. t = 5(1.5) = 7.5 s,

7.5y 660 )[ ]1 /1.57.5e (1353 - 660=

= 1348.3 mV

The corresponding temperature for sensor output of1348.3 mV is:

C41mVCmV

T/33

1348.3

which is the exact measured temperature


Dynamic response of second-order instrument(i.e. n = 2)

Applying D operator

dtdty dy ubyaada 0012

2

2

)( 2210

0

DaDaauby

Applying Laplace Transform (with all initialconditions equal to zero) and rearranging theequation:

where, = damping ration= natural frequency


uss

Ky

nn

n22

2

2


The time response is given as:q0(t) qe-atsin(nt)

where q=amplitude and a= n is output damping ratio

Eg.: Accelerometer


FIGURE 1.29 Characteristic second-order oscillatory time responseof a sensor.

Curtis JohnsonProcess Control Instrumentation Technology, 8e]

Copyright ©2006 by Pearson Education, Inc.

Upper Saddle River, New Jersey 07458All rights reserved.


undamped (=0)

under damped (1>>0)

over damped (1)

lecture04 dynamic characteristicsmohamed/ukur inst notes/lecture 04.pdf · 2012. 10. 2. · dynamic...

Documents