General Response of Second Order System

Transcription

1 General Response of Second Order System Slide 1

2 Learning Objectives Learn to analyze a general second order system and to obtain the general solution Identify the over-damped, under-damped, and critically damped solutions Convert complex solution to real solution Suspended mass-spring-damper equivalent system Slide

3 Second order RC circuit System with state variables Described by two coupled first-order differential equations i c i L C L R - + States Voltage across the capacitor - V 1 Current through the inductor - i L What to obtain state equations of the form: x = Ax Need to obtain expression for d /dt in terms of V 1 and i L Need to obtain expression for di l /dt in terms of V 1 and i L Slide 3

4 State Differential Equations For the capacitor and inductor we have, d dt = i c C, di L dt = v L L = L Above immediately gives us an expression for di l /dt in terms of V 1 Need to obtain expression for d /dt in terms of V 1 and i L Write node equation at to obtain: i c + i L + / R = 0 Hence the state equations are given by: (1)!!! d dt =!i L C! RC Slide 4 ()!!! di L dt = L

5 Solving the state differential equations d dt! " i L $!'1 / RC & = '1 / C $! " 1 / L 0 &! " $ " A! [ ] = s + 1 / RC 1 / C i L $ &!!! $! si ' A " '1 / L s & ((s) =!characteristic!eqn!=!det(si ' A) = 0 det(si ' A) = s + s RC + 1 LC = 0 solve!quadratic!equation!to!obtain :!!!!!!!!!!s 1 = Slide 5 ' 1 RC + ) 1, * + RC -. ' 4 LC,!!!s = ' 1 RC ' ) 1, * + RC -. ' 4 LC!!!

6 The natural response of RLC circuits Three cases Over-damped response: Characteristic equation has two (negative) real roots Response is a decaying exponential No oscillation (hence the name over-damped, because the resistor damps out the frequency of oscillation) Under-damped response: Characteristic equation has two distinct complex roots Response is a decaying exponential that oscillates Critically-damped response: Characteristic equation has two read, distinct roots Solution no longer a pure exponential Response is on the verge of oscillation Analogy to oscillating suspended spring-mass-damper system; where energy is stored in the spring and mass Slide 6

7 Over-damped response 1 (RC) > 4 LC Characteristic equation has two distinct real roots; s 1, s Both s 1 > 0, s > 0 (why?) Solution of the form: s = ae 1 1 e s 1t s + be 1 e s t s i l = ae 1 e s 1t s + be e s t Over-damped response Decaying exponential response Slide 7

8 Critically-damped response 1 (RC) = 4 LC Characteristic equation has two real repeated roots; s 1, s Both s 1 = s = -1/RC Solution no longer a pure exponential defective eigen-values only one independent eigen-vector Cannot solve for (two) initial conditions on inductor and capacity However, solution can still be found and is of the form: = a 1 e st + b 1 te st i l = a e s 1t + b te s t See chapter 5 of Edwards and Penny; or notes Slide 8

9 Critically-damped response, cont. = ae!t / RC!t / RC + bte Response is on the verge of oscillation Known as critically damped critically-damped response Slide 9

10 Under-damped response 1 (RC) < 4 LC Characteristic equation has two distinct complex roots; s 1, s s 1 =!" + j,!!s = s 1 * =!"! j " = 1 / RC, = 1 / (RC)! 4 / LC Two distinct eigenvectors, V 1 and V 1 * Complex eigenvalues and eigenvectors Solution of the form: = a 1 e s 1t i l = a e s 1t + a 1 * e s t + a * e s t Slide 10

11 Under-damped response, cont. = a 1 e (!" + j )t + a * (!"! j )t 1 e Euler 's formula :!!e! + j" = e! (cos(") + j sin(")) = a 1 e $!t (cos("t) + j sin("t)) + a 1 * e $!t (cos($"t) + j sin($"t)) cos($") = cos("),!!sin($") = $sin(") = e $!t [ Re[a 1 ]cos("t) $ Im[a 1 ]sin("t)] Under-damped response Decaying oscillation between -Re(a 1 ) and + Re(a 1 ) or - Im(a 1 ) and + Im(a 1 ) Slide 11

12 Example of over-damped response (complex eigen-values and eigen-vectors) Consider the same circuit with the following values C=1/ F, L=1/5 H, R=1 Initial conditions, V 1 (0)=1v, i c (0)=1A d dt! " i L! si ' A $!'1 / RC & = '1 / C $! " 1 / L 0 &! " $ " A! [ ] = s + " $ '5 s & i L $!' '$! &!!!= " 5 0 & " i L $ &!! Slide 1! s + $ " '5 s & = s + s + 10 = 0 s 1 = '1+ 3j,!!!s = '1 ' 3j

13 Example, continued: Eigen-vectors Now find the eigen-vectors 1+ 3j & s 1 =!1 + 3j " si! A = $!5!1+ 3j ( ' $ "!5V s1 1 + (!1 + 3j)V s1 = 0 " V s1 s1 1 = [!1 / / 5 j]v " V s1 = V 1 $ s1 V s1 & ( =!1 / / 5 j & ' $ 1 ( ' V 1 s1 V s1 & ( = 0 ' Slide 13 s = s 1 * =!1! 3j V s = V 1 $ s1 V s1 & ( ' * =!1 / / 5 j & $ 1 ( ' * =!1 / 5! 3 / 5 j & $ 1 ( '

14 The total solution V 1 (t) = a 1 (!1 / / 5 j)e (!1+ 3 j )t (!1! 3 j )t + a (!1 / 513 / 5 j)e i c (t) = a 1 (1)e (!1+ 3 j )t (!1! 3 j )t + a (1)e Use!V 1 (0) = 1,!!i c (0) = 1!!to!obtain :!!a 1 = 1 /! j,!!a = 1 / + j V 1 (t) = (1 / + 1 / j)e (!1+ 3 j )t (!1! 3 j )t + (1 /! 1 / j)e i c (t) = (1 /! j)e (!1+ 3 j )t (!1! 3 j )t + (1 / + j)e Slide 14 to!express!as!real!values!use, Euler's!formula:!!e " +j = e " (cos() + j sin()) ' some!algebra $ V 1 (t) = [ cos(3t)! sin(3t) ]e!t & (' i c (t) = [ cos(3t) + sin(3t) ]e!t

15 nd order mechanical systems mass-spring-damper spring mass M Damper (dashpot) X=0 X>0 frictionless support Force exerted by spring is proportional to the displacement (x) of the mass from its equilibrium position and acts in the opposite direction of the displacement Fs = -kx Fs < 0 if x > 0 (I.e., spring stretched) Fs > 0 if x < 0 (spring compressed) Damper force is proportional to velocity (v = x ) and acts in opposite direction to motion Fr = -bv = -bx Slide 15

16 Spring-mass-damper system If Fr and Fs are the only forces acting on the mass m, then F = ma where a = dv/dt = x F = m x = Fs + Fr = -kx - bx So the differential equation for the mass position, x, is given by: mx + bx + kx = 0 This is a second order system Guessing x = Ae st for the solution we get, mas e st + base st + kae st = 0 The characteristic equation is thus s + (b/m)s + k/m = 0 Slide 16 s 1 = (!b / m) + (b / m)! 4(k / m),!!!s = (!b / m)! (b / m)! 4(k / m)

17 Response of Spring-mass-damper system x(t) = A 1 e s 1t + A e s t where!a 1!and!A!are!chosen!to!satisfy!initial!conditions Note that for this system the state can be described by Position, x(t), Velocity, x (t) Hence, the initial conditions would be x(0) and x (0) Note similarity to RLC circuit response:!!!!s 1 =!1 RC + ( 1 ) RC! 4 LC,!!!s =!1 RC! 1 RC ( )! 4 LC Slide 17 Notice relationship between 1/R in RLC circuit and damping factor (b) in spring-mass-damper system B ~ 0 un-damped system oscillation This is the basis for the terminology, over-damped, under-damped, etc. Over-damped system damping factor is large and system does not oscillate (just exponential decay)