# AOSC 634 Air Sampling and Analysis Lecture 3 Measurement Theory Performance Characteristics of Instruments Dynamic Performance of Sensor Systems Response.

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AOSC 634 Air Sampling and Analysis Lecture 3 Measurement Theory Performance Characteristics of Instruments Dynamic Performance of Sensor Systems Response of a second order system to A step change A ramp change Copyright Brock et al. 1984; Dickerson 2015 1

Dynamic Response Sensor output in response to changing input. Dynamic Characteristics of Second Order Systems EQ I Where  n is the undamped natural frequency, a constant (s -1 ).  is the damping ratio, a unitless constant. We must solve an initial value problem. 2

Solving the differential equation 3 We will use the technique of variation of parameters to find complementary solutions. We must assume a time dependence of the form e rt and substitute this into Eq I. The characteristic equation is: Each root gives rise to a solution; there are four.

Four roots of the characteristic equation  leads to free oscillations  X c (t) = C sin(  n t +  ) 2. 0 <  leads to damped oscillations X c (t) = C exp (−  n  n t) sin(  m t +  ) Where  m =  n (1 –  2 ) ½ {  n within 5% for  < 0.3}  leads to critically damped  X c (t) = exp (−  n t) (At + B) Where A and B are constants.  1 leads to an overdamped solution. 4

5  1 leads to an overdamped solution. Where  m = 1/  m And the characteristic time is 1/  m In dimensionless time  n t = t” Critically damped systems are an ideal; in the real world only overdamped and underdamped systems exist. We will focus on underdamped systems such as the dew pointer (or a car with bad shocks). Overdamped systems lead to a “double first order”.

Time Response of second order systems. Start with a system at rest where both the input and output are zero. X(0) = X I (0) = X 0 Their first derivatives are likewise zero at time zero. We will proceed as with the first order system assuming a step change. Using the dimension- less form.  = 0, no damping. X’(t”) = 1 – cos (t”) 6

Time Response of second order systems.  < 1.0, underdamped.  = 0, critically damped. X’(t”) = 1 – e -t” (t” + 1) 7

Time Response of second order systems.  > 1.0, overdamped. The damping number is  With  see Figure 2-11 of Brock et al. 8

Response of a second order system to an a step increase of input. 9 Undamped Underdamped Critical Overdamped Dimensionless time t” =  n t Output 

An example with doubly normalized time 10

Notes on Figure 2-11. For all  the final state is X I (t”) for t” > 0 – The slope is real, continuous, and near zero where t” << 1.0. Contrast with first order. For  undamped systems,  there is free oscillation at  n. For 0 <  there is damping at a frequency of: The modified (damped) natural freq in Hz. For  there is large overshoot and a long time lag.  11

Notes on Figure 2-11, continued. For  there is large overshoot and a long time lag.  – The amplitude of the oscillations decreases exponentially with a time constant of  -1. The extrema can be found: Where the sub e represents extrema. The extrema come on time at  t”. Where n is a positive integer. 12

From the amplitude of the first extreme (assume here a maximum) we can calculate the damping ratio  : 13 Practical application

From the time (in units of t” of the first extreme (assume here a maximum) we can calculate the natural undamped frequency  n : 14

Note, the closer to  is to unity, and the smaller  n, the faster X’(t”) approaches X I. Example using Figure 2-11. Try this yourself with a mm ruler. Let’s check the curve with  = 0.10 for the first maximum. Looking at a paper copy, X’(t”) max = 60 mm X’(t) final = X I = 35 mm Close to the 0.100 value in the book. If the max amplitude is twice the input then (2/1 – 1) is 1 and  =0. 15

Example using Figure 2-11, continued. Let’s look for the natural frequency,  n. Let the time of the first max be 30 s, an arbitrary value. To get within e -1 of the final value requires: t” = 1/  = 10 =  n t = 0.1t and t = 100 s! In general, the time to e -1 is (  n  ) -1 for  < 0.3. For  > 0.3, use  m. 16

Summary Although less intuitive than first order systems, second order systems lend themselves to analysis of performance characteristics. A step change is in some ways a worst case scenario for overshoot. Any second order systems provide perfectly adequate temporal response in the real world where geophysical variables tend to show wave structure. 17

References MacCready and Henry, J. Appl Meteor., 1964. Determination of the Dynamic Response of a Nitric Oxide Detector, K. L. Civerolo, J. W. Stehr, and R. R. Dickerson, Rev. Sci. Instrum., 70(10), 4078-4080, 1999. 18

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