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] Copyright 2006 by Pearson Education, Inc. Lecture04 SME3242 Instrumentation Upper Saddle River, New Jersey 3 07458 All rights reserved.
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
: 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
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 1 1 0 dt dt dt n 0-3 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 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
Eg: potentiometer Lecture04 SME3242 Instrumentation 8
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
- 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
Eg.: Thermocouple Lecture04 SME3242 Instrumentation 11
Characteristic first-order time response of a sensor. t y(t) % T 0.63212 63.212 2T 0.86466 86.466 3T 0.95021 95.021 4T 0.98168 98.168 5T 0.99326 99.326 10T 0.99995 99.995 Lecture04 SME3242 Instrumentation 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) 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
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 07458 All rights reserved. Lecture04 SME3242 Instrumentation 15
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
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
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 = (1353-660 ] 0.75 = 932.7 mv Lecture04 SME3242 Instrumentation 18
The corresponding temperature for sensor output of 932.7 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
When t =5T i.e. t =5(1.5)=7.5s, y7.5 660 )[ 1 e 7.5/1.5 = (1353-660 ] = 1348.3 mv The corresponding temperature for sensor output of 1348.3 mv is: mv T 1348.3 41 C 33mV/ C which is the exact measured temperature Lecture04 SME3242 Instrumentation 20
Dynamic response of second-order instrument (i.e. n = 2) 2 d y dy a a a y 2 1 0 b 0u 2 dt dt Applying D operator y ( a 0 a 1 b 0u D a 2 D 2 ) Lecture04 SME3242 Instrumentation 21
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
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
Lecture04 SME3242 Instrumentation 24
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 07458 Technology, 8e] Lecture04 SME3242 Instrumentation All rights reserved. 25
undamped ( =0) under damped (1> >0) over damped ( 1) Lecture04 SME3242 Instrumentation 26
Lecture04 SME3242 Instrumentation 27