0 for t < 0 1 for t > 0


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1 8.0 Sep nd del funcions Auhor: Jeremy Orloff The uni Sep Funcion We define he uni sep funcion by u() = 0 for < 0 for > 0 I is clled he uni sep funcion becuse i kes uni sep = 0. I is someimes clled he Heviside funcion. The grph of u() is simple. u() Inegrls of u () From clculus we know u () d = u() For exmple: 5 2 u () d = u(5) u( 2) =, nd b u () d = u(b) u(). u () d = u() u() = 0, 5 u () d = u( ) u( 5) = 0. In fc, he inegrl over ny inervl [, b] conining 0 equls nd over for ny inervl no conining 0 i equls 0. Le 0 + be infiniesimlly o he lef of 0 nd 0 + infiniesimlly o he righ of 0. Th is, 0 < 0 < 0 +. We hve more exmples of inegrls of u : u () d =, 0 u () d = 0, 0 u () d =. Generlized funcions nd derivives Of course u() is no coninuous funcion, so in he 8.0 sense is derivive doesn exis. Noneheless we sw h we could mke sense of is inegrls. We cll u () he generlized derivive of u(). You cn do everyhing wih u () you cn do wih n ordinry funcion, bu i cn go nywhere we hve n inpu funcion in 8.0. Le s delve lile deeper ino u (). I s cler u () = 0 if 0. A = 0 he slope is infinie so u (0) =. We define δ() = u ()
2 8.0 Sep nd Del Funcions 2 nd cll i he del funcion or he uni impulse funcion. We hve seen he following properies of δ(). 0 if 0. δ() = if = δ() d = u(), δ() d = δ() d =. 0 We cn grph δ() s n infinie spike he origin. The inegrls show h he re under his grph equls nd i is ll concenred he origin. δ() δ( ) We lso show δ( ) which is jus δ() shifed o he righ. Del funcions re your friend Recll how pinful inegrion could be. In conrs, inegrls wih del funcions re lwys esy nd involve no echniques of inegrion. Suppose we scle δ(): he inegrls re jus scled. 5 5 δ() d =, The inegrl b 5 δ() d = 0, f()δ() d is lso esy. We hve b f() d = This jus mouns o evluing f. 0 δ() d =, f(0) if [, b] conins 0 0 if [, b] does no conin δ() d = 0. Reson: δ() is 0 everywhere excep = 0 so f()δ() is 0 everywhere excep = 0 nd scles he del funcion by f(0) 0. Technicliy: We mus hve f() coninuous = 0. If we shif by we hve, e.g., Prcice wih δ δ()2e 42 d = 2, δ( 2)2e 42 d = 2e 6, δ()2e 42 d = 0, 5 0 δ( 2)2e n2 () d = 2e n2 (8). δ( ) = f(). δ( 2)2e 42 d = 0, δ()2e 42 d = 2, 0 0 δ()2e n2 () d =. δ( 2)2e 42 d = 2e 6 0
3 8.0 Sep nd Del Funcions Tol inpu s he inegrl of inpu re Consider he re equion dx + k f(). To be specific, ssume x is in unis of d kilogrms. The derivive x nd he inpu f() re in unis of kg/ime. So, he ol moun inpu from ime 0 o ime is 4 Exmple : If f() = /4 0 f(u) du. or f() = /8 hen (in eiher cse) he ol inpu over ll ime is kg. Suppose we coninue o shoren he ime inervl in which we inpu ol of kg. In he limiing cse we dump kg in ll once. This is n insnneous jump in he moun x, clled n impulse. In generl, for firs order sysem n impulse is n insnneous jump in he moun. Now consider he second order sysem mx + bx + k F (). Here, F () is force i chnges momemum over ime. The ol momemum dded in he inervl [0, ] is 0 F (u) du. Similr o he firs order cse, n impulse is when we dd ll he momemum once. In his cse, we ge jump in he momemum. For our ypicl fixed mss sysem his is he sme s jump in he momenum. Th is, jump in mx. Pre nd posiniil condiions. Becuse n impulse cuses n insnneous jump in some vlue we hve o consider he condiions jus before nd jus fer he impulse. Assume he impulse occures = 0 hen A = 0 : A = 0 + : he condiions re preiniil condiions he condiions re posiniil condiions Del funcion s uni impulse An impulse o sysem cuses n insnneous jump. To hrmonic oscillor kick cuses jump in momenum. To firs order sysem, e.g. dumping lemmings in wildlife refuge, n impulse is sudden ddiion of lod of lemmings. A uni impulse cuses uni chnge. I is modeled by he δ funcion: For exmple, consider he equion x + k δ() wih inpu δ.. The ol inpu (inegrl of δ()) is. 2. All he inpu occurs = 0. Exmple 2: Solve x + k δ; res IC. nswer: This is firs order exponenil decy sysem. The uni impulse he = 0 cuses n insnneous jump of in he vlue of x. The preiniil condiion is x(0 ) = 0, so he posic is x(0 + ) =. Afer h he sysem is homogeneous. So, 8
4 8.0 Sep nd Del Funcions 4 for > 0 we hve x + k 0, x(0 + ) =. The soluion for > 0 is x() = e k (check he vlue 0 + ). The full soluion is 0 for < 0 e k for > 0. x Response from res o inpu = δ Exmple : Solve mx + bx + k δ; res IC. nswer: Res IC mens h x() = 0 for < 0. So our preic re x(0 ) = x (0 ) = 0. The uni impulse cuses uni jump in momenum so he posic re x(0 + ) = 0, mx(0 + ) = x(0 + ) = /m. For > 0 we hve he homogeneous equion mx + bx + k 0. We won grind hrough he symbolic lgebr. Insed le s work our sndrd exmple Exmple 4: Solve x + 8x + 7 δ; res IC. nswer: Here m = so for > 0 we hve o solve x + 8x + 7 0, x(0 + ) = 0, x (zp) =. We know c e + c 2 e 7. The posic deermine he coefficiens nd we ge 0 for < 0 6 e 6 e 7 for > 0 δ s limi of owers We define uni box of widh h funcion by u h () = h for 0 < < h 0 for h < (See grphs below.) The ol re under he grph of u h () is lwys. So, if we use u h () s inpu he ol inpu is. As h 0 he grph ges ller nd hinner nd looks more like spike. Also s h 0 he inpu is pplied over shorer nd shorer ime nd cs more like uni impulse. Th is, δ = lim h 0 u h 2 2
5 8.0 Sep nd Del Funcions 5 The funcions u, u /2, u / limiing o δ(). Exmple 5: Le s compre uni box (u h ()) inpu wih uni impulse (δ()) inpu: Solve x + k u h wih res IC. (physicl resoning:) This models rdiocive dumping. u h = re mer dded over ime ol moun dded = h 0 u h =. The grph shows he inpu u h s blck box. The response is grphed in red. If here were no decy we would see x increse linerly from 0 o over he inervl [0, h]. Since here is decy, he end of he inervl we hve x <. Afer ime = h here is no more inpu nd he response shows exponenil decy. As h 0 he inpu becomes he uni impulse δ(). The ol moun dded = 0 δ() d = nd i is dumped in ll once. The response curve shows jump of = 0. I is wrong o pu he re nd moun on he sme xes since hey hve differen unis. A he very les we should use lef vericl xis wih unis of x nd righ vericl xis wih unis of x. / 2 /2 / Inpu nd response o sequence of box funcions limiing o δ(). For compleeness we give he exc soluion o he IVP x + k u h, res IC. ( hk e k ) hk (ekh )e k for 0 < < h for h < Jus s expeced, s h 0 he inpu becomes δ nd he oupu becomes e k e kh (i.e. lim = ) h 0 hk Exmple 6: Solve x + δ() wih res IC. nswer: Preinil condiions re 0. The posiniil condiions ( = 0 + ) re x(0 + ) = 0, x (0 + ) =. For > 0 he DE becomes x + 0. This DE is esy o solve: For > 0 : x() = c cos() + c 2 sin(). We find c nd c 2 using he posiniil condiions: c = 0, c 2 =. The complee soluion is 0 for < 0 x() = sin() for > 0.
6 8.0 Sep nd Del Funcions 6 Physicl explnion: = 0 n impulse kicks he simple hrmonic oscillor ino moion. Afer h inpu is 0 nd he sysem is in simple hrmonic moion. The jump in momenum = corner in grph 0. x Exmple 7: Solve x + δ( ) wih res IC. The physicl explnion nd grph re he sme s in he previous exmple shifed by. x Exmple 8: (Resonnce) Solve x + f; res IC. Where f = blow every 2π seconds, mgniude 2 in he posiive direcion. nswer: We hve f = 2δ() + 2δ( 2π) + 2δ( 4π) +... Using he previous exmple nd superposiion we ge 2 sin for 0 < < 2π 4 sin for 2π < < 4π 6 sin for 4π < < 6π The deils showing his re no hrd. Keep in mind h he response o δ( 2nπ) is 0 before = 2nπ nd 2 sin( 2nπ) = 2 sin() fer = 2nπ. This mens h every 2π seconds we dd noher copy of 2 sin() o he oupu. This is resonnce he blows come he nurl frequence (every 2π seconds) of he sysem. Derivive of squre wve The grphs below re of funcion sq() (clled squre wve) nd is derivive. sq() lernes every π seconds beween ±. The derivive sq () is clerly 0 everywhere excep he jumps. A jump of +2 gives (generlized) derivive of 2δ nd jump of 2 gives (generlized) derivive of 2δ. Thus we hve sq () = δ( + 2π) 2δ( + π) + 2δ() 2δ( π) + 2δ( 2π) 2δ( π) +...
7 8.0 Sep nd Del Funcions π 2π π π 2π π 4π Grph of sq() = squre wve π π π π 2π 2π 4π Grph of sq () = impulse rin End opic sep nd del noes
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