nlyzing Dmped Oscilltions Prolem (Medor, exmple 2-18, pp 44-48): Determine the eqution of the following grph. The eqution is ssumed to e of the following form f ( t) = K 1 u( t) + K 2 e!"t sin (#t + $ )u( t) (1) where u t! ( ) = 1 t > 0 " # 0 t < 0 is the unit step function. (2) The solution we will find is Solution: f ( t) = 10u( t) + 12.6e!0.207t sin( 0.469t + 0.328)u( t) (3) We need to figure out the following prmeters: K 1, K 2,!,!, nd!. Pge 1 of 10
Find K 1 : We note tht the second term goes to zero for lrge time due to the exponentil decy of the envelope function, f t ( ) = 10u( t) + 12.6e!0.207t u( t). We see tht lim f t t!" ( ) = 10 so K 1 = 10. Find! : We note tht the grph of f t those points, tke the derivtive of f t t>0. d dt f t ( ) hs mxim nd minim. To find the vlues of the function t ( ) = K 1 + K 2 e!"t sin (#t + $ ) nd set it equl to zero with { } = 0 (4) ( ) = K 2e!"t!" sin (#t + $ ) + # cos (#t + $ ) We rerrnge to find tht t the extrem ( ) =! # tn!t mx/min + " (5) Since the period of the tngent function is!, the phse ngle etween the first two extrem is!. Note the right hnd side of eqution (5) is constnt. Therefore 8.5! + " = 1.8! + " + # (6) Solving for! we get Pge 2 of 10
! = " = 0.469 rd/sec (7) 8.5 # 1.8 Find! : To find the phse ngle,!, we look t the first "zero". f t ( )! K 1 = K 2 e!"t sin (#t + $ ) = 0 (8) This vlue occurs when!t + " = 0,#,2#, Since the sine is descending t t=6, we see! * 6 + " = # (9) Since we now know! = 0.469, we get! = " # 6 * 0.469 = 0.328 rd (10) Find! : ( ) t t=1.8 nd 8.5. ( )! 10 = K 2 e!1.8" sin( 0.469 *1.8 + 0.328) ( )! 10 = K 2 e!8.5" sin( 0.469 * 8.5 + 0.328) We use the knowledge of t nd f t f 1.8 f 8.5 or, equivlently, 8 = 0.922K 2 e!1.8" (11)!2 =!0.922K 2 e!8.5" (11) Divide these two equtions to hve the K 2 cncel ech other thus we determine!. 4 = e 6.7! (12) Solve for!.! = ln 4 = 0.207 (13) 6.7 Find K 2 : We now know ll ut K 2. We find K 2 y evluting the function t known point. 18 = 10 + K 2 e!0.207*1.8 sin 0.469 *1.8 + 0.328 We find K 2 = 12.6. In summry, we find the eqution of the originl grph to e Pge 3 of 10 ( ) (14) f ( t) = 10u( t) + 12.6e!2.07t sin( 0.469t + 0.328)u( t) (15)
qundry Finding!. When finding!, we found tn (!t mx/min + ") =! # Therefore,! = " tn "t mx/min + # ( ) = 0.469 tn( 0.469 *1.8 + 0.328) = 0.198 (16) However, this is not the sme nswer we got efore which ws 0.207. Wht's going on? It turns out tht our grph reding is not quite right. Look t some interesting points from the TI-89 from grphing eqution (15) t f(t) 1.7636 18.0015 (mximum) 1.8 18.000148 5.9991 10. 8.462 8.000204 (minimum) 8.5 8.00058 The smll numer of significnt figures in t is ffecting the nswer. Hence, you need to red the grph with greter ccurcy, which my or my not e possile. However, this method is good enough to estimte prmeters to within two significnt figures, the ccurcy of the time dt. 0.207! 0.198 The error in! is = 0.0045 " 0.5% which is certinly within the error uilt into 2 the clcultion. Connection to the dmped hrmonic oscilltor For n undriven spring of spring constnt k, mss m, nd dmping constnt, we hve m d 2 x dt + c dx + kx = 0 (17) 2 dt Divide y m. d 2 x dt + c 2 m dx dt + k m x = 0 (18) Now we trnsition to more informtive prmeters. Define! 0 2 = k m nd 2!" 0 = c m (19) Now eqution (18) ecomes Pge 4 of 10
d 2 x dt + 2!" dx 2 0 dt + " 2 0 x = 0 (20) ssume solution of the form x = Ce!t where C is constnt nd! is to e determined so tht this solution is ctully solution of eqution (20). When we plug it into eqution (20), Ce!t is in every term so fctors out (it is never 0) nd we get the following eqution, known s the chrcteristic eqution.! 2 + 2"# 0! + # 2 0 = 0 (21) The solution to this qudrtic eqution is { } (22)! = "# 0 $ ± $ 2 " 1 When! <1 we get { } (23)! = "# 0 $ ± j 1" $ 2 We use Euler's reltionship e ± jx = cos x ± j sin x (24) We therefore hve the solution x = e!" 0#t e j"t + e! j"t { } (25) Other forms re (see ppendices nd B) x = e!" 0#t C 1 cos "t { ( ) + C 2 sin ("t)} (26) ( ) (27) x = e!" 0#t sin "t + $ where! =! 0 1" # 2 (28) We now relte the hrmonic oscilltor to the originl prolem.! =! 0 1" # 2 = 0.469 (29)! = " 0 # = 0.207 (30) s we did efore, we divide the two equtions! " = 1 # $ 2 $ Rewriting we get Solve for!. = 0.469 = 2.266 (31) 0.207 1! " 2 = 2.266" (32) Pge 5 of 10
! = 0.404 (33) We now use eqution (30) to solve for! 0.! 0 = 0.207 = 0.207 = 0.512 rd/sec (34) " 0.404 We cn now ck out of the ove c m nd k m. k m =! 0 = 0.263 (35) c m = 2!" = 2( 0.404) 0 ( ) = 0.414 (36) In summry, our system hs the following prmeter vlues: Prmeter Vlue Interreltionship k 0.263 m c m 0.414! 0 0.512 rd/sec! 0 = k m! 0.404 c c! = = 2m" 0 2 km Note tht ll these prmeters involve rtios of hrdwre vlues, m, c, nd k. The performnce of the system depends on the scled prmeters,! 0 nd!, s shown in column 3 ove. Some questions rise with this system. 1) How do we get to criticl dmping? Tht occurs when! = 1 so c crit = 2m! 0 = 2 km (37) The ove reltionship provides wy to determine the effect of chnging t lest one of the physicl prmeters to chieve criticl dmping. We need to choose m, c, nd k so tht eqution (37) holds nd we chieve criticl dmping. Or, we cn use eqution (37) to sty wy from criticl dmping mke sure tht m, c, nd k re chosen such tht our system is fr from the stisfying eqution (37). 2) If, insted, you wish to specify the period of oscilltion, T. Pge 6 of 10
You re then you re choosing m, c, nd k such tht eqution (38) provides the desired vlue of T.! =! 0 1" # 2 =! 2 0 " (! 0 #) 2 = $ k ' $ c ' % & m( ) " % & 2m( ) 2 = 2* T (38) 3) If you wish to specify the dmping time constnt, ", then you re specifying!" 0 = c 2m = 1 # We need to choose c nd m so tht eqution (39) holds for the chosen vlue of ". (39) So, nlyzing the grph provides dt on your system, which you cn then chnge to your desired specifictions. Connection to RLC circuits For series RLC circuit the Kirchhoff Voltge Lw (KVL) with no voltge source is L d 2 Q dt 2 or d 2 Q dt 2 + R L + R dq dt + 1 C Q = E t ( ) = 0 (40) With the pproprite definitions, dq dt + 1 LC Q = E t ( ) = 0 (41)! 0 = 1 LC nd! = R 2L" 0 (42) the dmped spring nlysis pplies. Note tht ccording to Pisno, the cnonicl form for electricl prolems is d 2 x dt +! n dx 2 Q dt +! 2 n x = 0 (43) Compre this eqution with eqution (20), the cnonicl form for mechnicl systems. d 2 x dt + 2!" dx 2 0 dt + " 2 0 x = 0 (20) Pge 7 of 10
ppendix Hrmonic Identities Derrick nd Grossmn pge 130 Identity #1: " t + sin" t = cos( " t #! ) cos (1) With + 2 2 =, cos! =,, sin! = hence tn! = When determining! y inverse trig functions, you know which qudrnt is the correct qudrnt since you know the sign of oth the sine nd the cosine. The inverse sin nd tn return ngles from -90º to + 90º nd the inverse cos from 0º to 180º. To get n ngle in Qudrnt III tke the inverse tn then dd or sutrct 180º. Be creful of rdins vs. degrees s! t is usully in rdins. Qudrnt ngles cos! = sin! = I 0º to 90º + + II 90º to 180º # + III 180º to 270º # # IV 270º to 360º + # Proof: Use cos( " t #! ) = cos" t cos! + sin" t sin! So cos( " t #! ) = cos" t( cos! ) + sin" t( sin! ) Compre with cos (! t # " ) = cos! t + sin! t Identity #2: " t + sin" t = sin( " t +!) cos (2) With + 2 2 =, cos! =, sin! =, hence tn! = When determining! y inverse trig functions, you know which qudrnt is the correct qudrnt since you know the sign of oth the sine nd the cosine. The inverse sin nd tn return ngles from -90º to + 90º nd the inverse cos from 0º to 180º. To get n ngle in Qudrnt III tke the inverse tn then dd or sutrct 180º. Be creful of rdins vs. degrees s! t is usully in rdins. Qudrnt ngles cos! = Pge 8 of 10 sin! = I 0º to 90º + + II 90º to 180º # +
III 180º to 270º # # IV 270º to 360º + # Proof: Use sin( " t +!) = cos" t sin! + sin" t cos! So sin( " t +!) = cos" t( sin! ) + sin" t( cos! ) Compre with sin (! t + ") = cos! t + sin! t TI-89: F2/Trig/tExpnd nd F2/Trig/tCollect will pply the trig identity. To find the phse ngle, use vectors: [ cos!, sin! ]!Polr. i.e. [,]!Polr with similr result for identity 1. Note: cos! e j!t = cos (!t) + j sin!t Therefore e j!t + e " j!t = C 1 cos (!t) + C 2 sin (!t) ppendix B Euler's Identity ( ), e! j"t = cos "t ( )! j sin ("t) # ( ) = sin! + " $ % 2 & ' ( ppendix C Period of the Tngent Function From the grph elow, you cn see tht the period of the tngent is!. Pge 9 of 10
ppendix D References Don.Medor, nlog Signl Processing nd Lplce Trnsforms nd ctive Filter Design, Delmr, 2002. lert Pisno, MEC 219-EEC246, Lecture 8 ccelerometry, 2005, www.me.erkeley.edu/me219/lectures/l08me219.pdf Derrick nd Grossmn, Elementry Differentil Equtions, 4 th ed., ddison Wesley, 1997. Pge 10 of 10