Dynamic Characteristics. Lecture04 SME3242 Instrumentation

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1 Dynamic Characteristics Lecture04 SME3242 Instrumentation 1

2 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

3 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] Copyright 2006 by Pearson Education, Inc. Lecture04 SME3242 Instrumentation Upper Saddle River, New Jersey All rights reserved.

4 2.1.3: Mathematical model structure Input and output relationship of a linear measurement system - ordinary differential equation (ODE): a d n dt y a n1 d y dy 1 a a n 1 1 dt dt n n n 0 b d dt m u b m1 y d u du b 1 b m 1 1 dt dt m m m 0 u where, u = input, y = output; u and y varies with t n > m a, b = constant coefficients Lecture04 SME3242 Instrumentation 4

5 : Transfer Function of An Accelerometer x i k c m x o Applying 2 nd Newton s Law: F = ma F s ma = F D Differential equation: 2 dxi dxo d x k( x x ) c( ) i o m 2 dt dt dt 2 d xo c dx k c k o dx i xo x 2 i dt m dt m m dt m Lecture04 SME3242 Instrumentation 5 o

6 Dynamic characteristic - the output response of the instrument against time when the input is varied - the relation between any input and output for n th order system can be written as: a n n1 d y d dy a y a a y b u n n1 n dt dt dt n 0-3 types of response: zero order response, first order response and second order response Lecture04 SME3242 Instrumentation 6

7 Dynamic response of zero order instrument (i.e. n = 0) - the zero order instrument is represented by a 0 y=b 0 u or y=ku or y/u = K (y=output, u=input, K=b 0 /a 0 =static sensitivity) - the output responses linearly to the input Lecture04 SME3242 Instrumentation 7

8 Eg: potentiometer Lecture04 SME3242 Instrumentation 8

9 Dynamic response of first order instrument (i.e. n = 1) - dividing the equation by a 0, and apply D-operator a 1 dy b y 0 u ; ( TD 1) y Ku a dt a 0 dy a a y 1 0 b 0 u dt 0 - T=a 1 /a 0 =time constant, K=b 0 /a 0 =static sensitivity Lecture04 SME3242 Instrumentation 9

10 - the operational transfer function y u K ( 1TD) 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

11 Eg.: Thermocouple Lecture04 SME3242 Instrumentation 11

12 Characteristic first-order time response of a sensor. t y(t) % T T T T T T Lecture04 SME3242 Instrumentation 12

13 Characteristic first-order time response of a sensor. Lecture04 SME3242 Instrumentation 13

14 General equation as function of time following a step input is given as: y( t) y i ( y f y i )[1 e t /T ] where, y i = initial output from static transfer function and initial input y f = final output from static transfer function and final input T = time constant = 63% time Lecture04 SME3242 Instrumentation 14

15 Characteristic first order exponential time response of a sensor to a step change of input. y y f y i Curtis Johnson Process Control Instrumentation Technology, 8e] Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Lecture04 SME3242 Instrumentation 15

16 A sensor measures temperature linearly with a static transfer function of 33 mv/ 0 C and has a 1.5 s time constant. Find the output 0.75 s after input changes from 20 0 Cto41 0 C. 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

17 Given static transfer function: V = (33mV/ºC)T Hence, initial and final output of the sensor are: y i = (33mV/ºC)(20ºC) = 660mV y f = (33mV/ºC)(41ºC) = 1353mV Lecture04 SME3242 Instrumentation 17

18 Time response of first-order system, y( t) = y i ( y f y i )[1 e t /T ] Substitute the value of y i and y f, y 660 )[ 1 e 0.75/1.5 = ( ] 0.75 = mv Lecture04 SME3242 Instrumentation 18

19 The corresponding temperature for sensor output of mv, 932.7mV T 33mV/ 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

20 When t =5T i.e. t =5(1.5)=7.5s, y )[ 1 e 7.5/1.5 = ( ] = mv The corresponding temperature for sensor output of mv is: mv T C 33mV/ C which is the exact measured temperature Lecture04 SME3242 Instrumentation 20

21 Dynamic response of second-order instrument (i.e. n = 2) 2 d y dy a a a y b 0u 2 dt dt Applying D operator y ( a 0 a 1 b 0u D a 2 D 2 ) Lecture04 SME3242 Instrumentation 21

22 Applying Laplace Transform (with all initial conditions equal to zero) and rearranging the equation: y s 2 2 K n 2 s n 2 n u where, = damping ratio n = natural frequency Lecture04 SME3242 Instrumentation 22

23 The time response is given as: q 0 (t) qe -at sin( n t) where q=amplitude and a= n is output damping ratio Eg.: Accelerometer Lecture04 SME3242 Instrumentation 23

24 Lecture04 SME3242 Instrumentation 24

25 FIGURE 1.29 Characteristic second-order oscillatory time response of a sensor. Copyright 2006 by Pearson Curtis Johnson Education, Inc. Process Control Instrumentation Upper Saddle River, New Jersey Technology, 8e] Lecture04 SME3242 Instrumentation All rights reserved. 25

26 undamped ( =0) under damped (1> >0) over damped ( 1) Lecture04 SME3242 Instrumentation 26

27 Lecture04 SME3242 Instrumentation 27

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