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-1n n-1 1 on n-1m m-1m m-1 om-1y y dydd + ( ) + + + y = F(t )a a a adtdt dt where : (3.1)xxddF(t) = + + + x m nb b bdtdt Figure 3.2 Measurement system operation on an input signal, F(t), provides the output signal, y(t).FirstOrderSystems.docx 9/14/2007 1:03 PM Page 2 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.FirstOrderSystems.docx 9/14/2007 1:03 PM Page 3FirstOrderSystems.docx 9/14/2007 1:03 PM Page 4 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 transfercoefficient, surface areasvsQ = dE/dt = m dT(t )/dt = h [ (t ) - T(t )]CAT h = = A This can be rearranged: vsssmc dT(t)/dt+hA [T(t)-T(O)]= hA [ (t)-T(O)] = hA F(t)TFirstOrderSystems.docx 9/14/2007 1:03 PM Page 5 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 > OFirstOrderSystems.docx 9/14/2007 1:03 PM Page 6 This equation can be written in the form: which has the solution: where dT(t) +T(t) = Tdt 0[]-t/T(t)= + T TeT sv = m / h cAFirstOrderSystems.docx 9/14/2007 1:03 PM Page 7 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 = eT-T 0lnT(t)-T-t/ = T-TFirstOrderSystems.docx 9/14/2007 1:03 PM Page 8 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) ye Figure 3.6 in 2nd and 3rd EditionFirstOrderSystems.docx 9/14/2007 1:03 PM Page 9 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 tdt 0dy+ y=dt /tY(t)=Ce sin[ ( )]y(t)= B t+ /2 1/2( ) ( )sin[ ( )]where ( ) /[ ( ) ]( ) ( )tant-1y t = B t+ +Ce B KA 1+=FirstOrderSystems.docx 9/14/2007 1:03 PM Page 10 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 responseFirstOrderSystems.docx 9/14/2007 1:03 PM Page 11 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.FirstOrderSystems.docx 9/14/2007 1:03 PM Page 12 A useful measure of the phase shift is the time delay of the signal: tan-11 ( ) ( ) = =FirstOrderSystems.docx 9/14/2007 1:03 PM Page 13 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 21/20.98, or 20log0.98 0.17511M( ) db= =M( )= B/(KA)= /[ +( )] rearranging 1/221/ ( ) 1so for 98%, 0.2or, 0.2 0.2 2 0.2 2 3.142 0.10.31sec = 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.FirstOrderSystems.docx 9/14/2007 1:03 PM Page 14 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: