#### Transcript Lecture 3. Dynamic Response Second Order Systems

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 d X dX 2 2 + 2Vw n + wn X = wn XI 2 dt dt 2 EQ I • Where wn is the undamped natural frequency, a constant (s-1). • z is the damping ratio, a unitless constant. • We must solve an initial value problem. 2 Solving the differential equation We will use the technique of variation of parameters to find complementary solutions. We must assume a time dependence of the form ert and substitute this into Eq I. The characteristic equation is: r + 2Vwnr + w = 0 2 2 n Each root gives rise to a solution; there are four. 3 Four roots of the characteristic equation 1. z = 0 2. 0 < z < 1 3. z = 1 4. z > 1 leads to free oscillations L Xc(t) = C sin(wnt + q) leads to damped oscillations Xc(t) = C exp (−wnzn t) sin(wmt + q) Where wm = wn(1 – z2)½ {= wn within 5% for z < 0.3} leads to critically damped. J Xc(t) = exp (−wn t) (At + B) Where A and B are constants. leads to an overdamped solution. XC (t) = e-Vwnt éë Ae+t/t m + Be-t/t m ùû 4 4. z > 1 leads to an overdamped solution. XC (t) = e -Vwnt éë Ae+t/t m + Be-t/t m ùû Where tm = 1/wm And the characteristic time is 1/wm In dimensionless time wnt = 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”. 5 Time Response of second order systems. • Start with a system at rest where both the input and output are zero. X(0) = XI(0) = X0 Their first derivatives are likewise zero at time zero. dX dX I dt = dt =0 We will proceed as with the first order system assuming a step change. Using the dimensionless form. 1. z = 0, no damping. X’(t”) = 1 – cos (t”) 6 Time Response of second order systems. 2. 0 < z < 1.0, underdamped. X ' (t" ) = 1 e t " 1 cos 1 1 where q = cos 1 2 1/ 2 2 1/ 2 2 1/ 2 t"q 3. z = 0, critically damped. X’(t”) = 1 – e-t”(t” + 1) 7 Time Response of second order systems. 4. z > 1.0, overdamped. 1 n 1 / 2t " X ' (t" ) = 1 (n 1) e n n 1 e n 1 / 2t " 1 (1 2 )1/ 2 where v = 1 (1 2 )1/ 2 The damping number is n. n +1 With z = see Figure 2-11 of Brock et al. 2 v 8 Output Response of a second order system to an a step increase of input. Undamped Underdamped Dimensionless time t” = wnt 9 An example with doubly normalized time 10 Notes on Figure 2-11. • For all z > 0 the final state is XI(t”) for t” > 0 – The slope is real, continuous, and near zero where t” << 1.0. Contrast with first order. • For z = 0, undamped systems, there is free oscillation at wn. • For 0 < z < 1, there is damping at a frequency of: wn (1- V 2 )1/2 = wm The modified (damped) natural freq in Hz. • For z << 1, there is large overshoot and a long time lag. L 11 Notes on Figure 2-11, continued. • For z << 1, there is large overshoot and a long time lag. L – The amplitude of the oscillations decreases exponentially with a time constant of z-1. The extrema can be found: Where the sub e represents extrema. The extrema come on time at pt”. t"e = np +1 1- V 2 Where n is a positive integer. 12 Practical application • From the amplitude of the first extreme (assume here a maximum) we can calculate the damping ratio z: X 'max - Vp X 'e (t") = = 1+ e XI -1 æ X 'max ö V = ln ç -1÷ p è XI ø 13 • From the time (in units of t” of the first extreme (assume here a maximum) we can calculate the natural undamped frequency wn: tmax = p t " º w nt wn = p tmax 14 • Note, the closer to z is to unity, and the smaller wn, the faster X’(t”) approaches XI. • Example using Figure 2-11. Try this yourself with a mm ruler. Let’s check the curve with z = 0.10 for the first maximum. Looking at a paper copy, X’(t”)max = 60 mm X’(t)final = XI = 35 mm æ 60 ö V = ln ç -1÷ = 0.107 p è 35 ø -1 Close to the 0.100 value in the book. If the max amplitude is twice the input then (2/1 – 1) is 1 and z =0. 15 • Example using Figure 2-11, continued. Let’s look for the natural frequency, wn. Let the time of the first max be 30 s, an arbitrary value. p wn = » 0.10s-1 or 0.1 Hz. 30s To get within e-1 of the final value requires: t” = 1/z = 10 = wnt = 0.1t and t = 100 s! In general, the time to e-1 is (wnz)-1 for z < 0.3. For z > 0.3, use wm. 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