1.0 Electrical Systems


 Caitlin Haynes
 6 years ago
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1 . Elecricl Sysems The ypes of dynmicl sysems we will e sudying cn e modeled in erms of lgeric equions, differenil equions, or inegrl equions. We will egin y looking fmilir mhemicl models of idel resisors, idel cpciors, nd idel inducors. Then we will egin puing hese models ogeher o develop models for L nd C circuis. Finlly, we will review soluion echniques for he firs order differenil equion we derive o model he sysems.. Idel esisors The governing equion for resisor wih resisnce is given y Ohm s lw, v () = i () where v () is he volge cross he resisor nd i () is he curren hrough he resisor. Here is mesured in Ohms, v () is mesured in vols, nd i () is mesured in mps. The enire expression mus e in vols, so we ge he uni expression. Idel Cpciors [vols] = [Ohms][mps] The governing equion for cpcior wih cpcince C is given y dv() i () = C d HereC is mesured in frds, nd gin v () is mesured in vols nd i () is mesured in mps. This expression lso helps us wih he unis. The enire expression mus e in erms of curren, so looking he differenil relionship we cn deermine he uni expression [mps] = [frds][vols]/[seconds] We cn inegre his equion from n iniil ime dv() i () = C d up o he curren ime s follows: i() d = dv() C Nex, since we wn o inegre up o finl ime, we need o use dummy vrile in he inegrl h is no. This is n imporn hi o ge ino do no use s he dummy vrile of inegrion if we expec funcion of ime s he oupu! Here we Copyrigh 9 oer D. Throne
2 hve chosen o use he dummy vrile λ. Also we incorpore he fc h ime volge is v ( ), while ime he volge is v () v ( ) v( ) C i( λ) dλ = d λ v( o ) he Crrying ou he inegrion we ge C which we cn rerrnge s i( λ) dλ = v ( ) v ( v () = v ( ) i( C λ) dλ This expression ells us here re wo componens o he volge cross cpcior, he iniil volge v ( ) nd he pr due o ny curren flowing hrough he cpcior fer h ime, C i( λ) dλ Finlly, hese expressions help us deermine some imporn chrcerisics of our idel cpcior: If he volge cross he cpcior is consn, hen he curren hrough he cpcior mus e zero since he curren is proporionl o he re of chnge of he volge. Hence, cpcior is n open circui o dc. I is no possile o chnge he volge cross cpcior in zero ime.the volge cross cpcior mus e coninuous funcion of ime, oherwise n infinie moun of curren would e required..3 Idel Inducors The governing equion for n inducor wih inducnce L is given y v () = L di() d Here L is mesured in henrys, nd gin v () is mesured in vols nd i () is mesured in mps. This expression lso helps us wih he unis. The enire expression mus e in erms of volge, so looking he differenil relionship we cn deermine he uni expression [vols] = [henrys][mps]/[seconds] ) We cn inegre his equion from n iniil ime up o he curren ime s follows: Copyrigh 9 oer D. Throne
3 di() v () = L d vd () = di () L Nex, since we wn o inegre up o finl ime, so we gin hve chosen o use he dummy vrile λ. Also we incorpore he fc h ime he curren is, i ( ) while ime he curren is i (). Crrying ou he inegrion we ge L which we cn rerrnge s L i ( ) v( λ) dλ = di( λ) i( o ) v( λ) dλ = i ( ) i ( = ) L i () i ( v( λ) dλ This expression ells us here re wo componens o he curren hrough n inducor, he iniil curren i ( ) nd he pr due o ny volge cross he inducor fer h ime, L v( λ) dλ. Finlly, hese expressions help us deermine some imporn chrcerisics of our idel inducor: If he curren hough n inducor is consn, hen he volge cross he inducor mus e zero since he volge is proporionl o he re of chnge of he curren. Hence, n inducor is shor circui o dc. I is no possile o chnge he curren hrough n inducor in zero ime.the curren hrough n inducor mus e coninuous funcion of ime, oherwise n infinie moun of volge would e required. ) Copyrigh 9 oer D. Throne 3
4 . Firs Order Circuis A firs order circui is circui wih one effecive energy sorge elemen, eiher n inducor or cpcior. (In some circuis i my e possile o comine muliple cpciors or inducors ino one equivlen cpcior or inducor. ) We egin his secion wih he derivion of he governing differenil equion for vrious firs order circuis. We will hen pu he firs order equion ino sndrd form h llows us o esily deermine physicl chrcerisics of he circui. Nex we show n lernive mehod for checking some prs of he governing differenil equions. We hen solve he differenil equions for he cse of piecewise consn inpus, nd finish he secion wih n lernive mehod of solving he differenil equions using inegring fcors.. Governing Differenil Equions for Firs Order Circuis In his secion we derive he governing differenil equions h model vrious L nd C circuis. We hen pu he governing firs order differenil equions ino sndrd form, which llows us o red off descripive informion ou he sysem very esily. The sndrd form we will use is dy() y () = Kx () d Here we ssume he sysem inpu is x() nd he sysem oupu is y (). is he sysem ime consn, which indices how long i will ke he sysem o rech sedy se for sep (consn) inpu. K is he sic gin of he sysem. For consn inpu of mpliude dy() A ( x() = Au(), where u() is he uni sep funcion), in sedy se we hve = d nd y () = Kx () = KA. Hence he sic gin les us esily compue he sedy se vlue of he oupu. For circuis wih cpciors he differenil equion will in generl e in erms of volge (he oupu y () will e volge), while for circuis wih inducors he differenil equion will in generl e in erms of curren (he oupu y () will e curren ). Exmple... Consider he C circui shown in Figure.. The volge source is vs ( ). We sr o derive he governing differenil equion y deermining he single curren in he loop vs() vc() dvc() i() = = ic() = C d or dvc() vs() vc() C = d where v ( c ) is he volge cross he cpcior nd he curren in he loop is equl o he curren hrough he resisor i ( ) nd he curren hrough he cpcior ic ( ). We cn pu his ino more sndrd form y rerrnging he erms Copyrigh 9 oer D. Throne 4
5 dvc () C vc() = vs() d If we define he ime consn = C, hen we hve Here he sic gin K =. dvc () vc() = vs() d vs () v ()  c C  Figure.. Circui for Exmple... Exmple... Consider he C circui shown in Figure.. Agin he volge source is vs (). We gin sr o derive he governing differenil equion y deermining he curren hrough resisor, vs() vc() i() = This curren mus e equl o he sum of he currens hrough he cpcior nd, vc ( ) dv ( ) i( ) C c = d Equing hese we ge he governing differenil equion: errnging erms we ge i ( v () v () vc () dv ( ) = = C d ) s c c Copyrigh 9 oer D. Throne 5
6 dvc () C vc() = vs() d dvc () C vc () = vs () d or C dvc () vc () = vs () d C Wih ime consn = nd sic gin K = we ge dvc () vc() = Kvs() d v v () s ()  c C  Figure.. Circui used in Exmple... Exmple..3. Consider he operionlmplifier circui shown in Figure.3. The inpu volge is gin vs ( ) nd he oupu volge (he volge cross he lod resisor L ) is he sme s he volge cross he cpcior (since he erminl of he op mp is ssumed o e grounded). We will ssume n idel op mp, which implies he condiions i () = i () = v () = v () Le s look he currens flowing ino he negive (feedck) erminl of he opmp using he idel opmp model. Since for our exmple he noninvering erminl is ied o ground we hve v ( ) =. Wih hese ssumpions our governing differenil equion ecomes vs () v () dv () = C d c c Copyrigh 9 oer D. Throne 6
7 errnging his gives dvc() vc() vs () C = d or dvc () C vc () = vs() d Seing he ime consn = C nd sic gin K = we finlly hve dvc () vc() = Kvs() d C   vc () vs ()  L Figure.3. Circui for Exmple..3. Exmple..4. Consider he L circui shown in Figure.4. The single curren in he loop is given y vs() vl() i () = where d ( ) vl () = L i d Comining nd rerrnging we ge di() L i() = vs () d Copyrigh 9 oer D. Throne 7
8 or Ldi () i () = vs () d L Wih ime consn = nd sic gin K = he governing differenil equion is di() i ( ) = Kvs ( ) d vs () v ()  L  L Figure.4. Circui for Exmple..4. Exmple..5. Consider he C circui shown in Figure.5. The single curren source mus e divided eween he curren flowing hrough resisor nd he curren flowing hrough he cpciorc, vc() dvc() is () = C d errnging we ge dvc () C vc() = i s() d Wih ime consn = C nd sic gin K = he governing differenil equion is dvc () vc() = Kis() d Copyrigh 9 oer D. Throne 8
9 is( ) vc ()  C Figure.5. Circui used in Exmple..5.. Thevenin esisnce, Time Consns, nd Sic Gin Alhough we re focusing our enion on deriving he governing equions for firs order circuis, i is useful nd very convenien o e le o check our equions s much s possile. Firs of ll, for firs order C circuis he ime consns will e of he form = hceq where h is he Thevenin resisnce seen from he pors of he equivlen cpcior, C eq. Leq For firs order L circuis he ime consns will e of he form = where h is he h Thevenin resisnce seen from he pors of he equivlen inducor, L eq. ecll h when deermining he Thevenin resisnce ll independen volge sources re reed s shor circuis, nd ll independen curren sources re reed s open circuis. Secondly, if we re looking consn inpus, hen we use he fc h cpcior is n open circui o dc nd n inducor is shor circui o dc. In ddiion, for consn inpus in sedy se ll of he ime derivives re zero (in sedy se nohing chnges in ime). Exmple... Consider he circui shown in Figure. (Exmple..). The Thevenin resisnce seen from he cpcior is equl o, so he ime consn is = C. For dc inpu, he cpcior looks like n open circui, so in sedy se he volge cross he cpcior is equl o v s, he inpu volge, so he sic gin is K =. These resuls mch our previous resuls. Exmple... Consider he circui shown in Figure. (Exmple..). The Thevenin resisnce seen from he cpcior is h = =, so he ime consn is C = C h =. For dc inpu, he cpcior looks like n open circui, so in sedy Copyrigh 9 oer D. Throne 9
10 se he volge cross he cpcior is given y he volge divider relionship vc = v s, so he sic gin is K =. These resuls mch our previous resuls. Exmple..3. Consider he circui shown in Figure.3 (Exmple..3). The Thevenin resisnce seen y he cpcior is lile more difficul o deermine, nd o do i correcly is eyond he scope of his course. For dc inpu, he cpcior looks like n open circui, so summing he currens ino he negive erminl of he op mp we hve v c v s =, or in sedy se vc = vs Hence he sic gin is K =. Exmple..4. Consider he circui shown in Figure.4 (Exmple..4). The Thevenin resisnce seen y he inducor is h =. For dc inpu, he inducor looks like shor circui. Hence he sedy se curren flowing in he circui for dc inpu is i = v s, so he sic gin is K =. Exmple..5. Consider he circui shown in Figure.5 (Exmple..5). The Thevenin resisnce seen y he cpcior is h = so he ime consn is = C. For dc inpu he cpcior looks like n open circui, so in sedy se vc = i, so he sic gin is K =..3 Solving Firs Order Differenil Equions In his secion we will go over wo mehods for solving firs order differenil equions. We will iniilly solve he equions y reking he soluion ino he nurl response (he response wih no inpu) nd hen he forced response (he response when he inpu is urned on). We will pply his mehod o prolems where he inpu is consn vlue, or is swiched eween consn vlues. This mehod will lso work wih ny inpu, nd we will exmine he resuls for sinusoidl inpu ler. In he ls secion we will go over differen mehod of soluion using inegring fcors, which will work for ny ype of inpu, nd is n imporn mehod in helping us chrcerize how sysem will respond o ny ype of inpu..3. Soluion using Nurl nd Forced esponses Consider sysem descried y he firs order differenil equion dy() y () = Kx () d In his equion, is he ime consn nd K is he sic gin. We will solve his equion in wo prs. We will firs deermine he nurl response, y ( ). The nurl n Copyrigh 9 oer D. Throne
11 response is he response due only o iniil condiions when no inpus re presen. Then we will deermine he forced response, y (). The forced response is he response due o f he inpu only, ssuming ll iniil condiions re zero. The ol response is hen he sum of he nurl nd forced responses, y () = y() y (). Nurl esponse: To deermine he nurl response we ssume here is no inpu in he sysem, so we hve he equion dy () n yn d () = r Le s ssume soluion of he form yn () = ce, where c nd r re prmeers o e deermined. Susiuing his ssumpion ino he differenil equion we ge n r r r rce ce = ce [ r ] = If c = hen we re done, nd he nurl response will e y ( ) =. This soluion r cerinly sisfies he differenil equion. However, if c, nd since e cn never e zero, we mus hve r =, or r =. In his cse he nurl response will e / y () = ce. n f n Forced esponse: To deermine he forced response we mus know he sysem inpu, x(). We will iniilly ssume n inpu h is zero efore = nd hen hs consn mpliude A for, < x () = A Then for we hve he equion dy f () y f () = KA d Since his is liner ordinry differenil equion we only need o find one soluion. One dy f () ovious soluion o his equion is he soluion in sedy se, when =. In d sedy se we hve y () = KA f Noe h for consn inpu, he sedy se oupu is he produc of he sic gin nd he mpliude of he inpu. Tol Soluion: The ol soluion o he prolem is he sum of hese wo soluions / y () = yn() yf() = ce KA Now ssume he iniil ime is = nd he sysem is iniilly res, i.e. here is no energy sored in he sysem so y() =. Susiuing his ino our equion we hve y() = = c K A, or c= KA, nd our ol soluion is Copyrigh 9 oer D. Throne
12 / y () = KA( e ) For simpliciy, le s wrie our sedy se vlue explicily, so y( ) = KA nd we hve he soluion / y () = y( )( e ) Finlly, le s deermine more generl form of he soluion for y(). Then we hve or so he ol soluion is y() = c KA= c y( ) c = y() y( ) Significnce of he Time Consn / [ ] y () = y() y( ) e y( ) In much of wh we do, we will e concerned wih he ime consns of sysem in one wy or noher. Le s look he response of our firs order sysem ssuming he sysem is iniilly res ( y () = ) nd he finl vlue is one ( y( ) = ). Le s look he response of our sysem s he ime kes on he vlues of ineger numer of ime consns: Time ( ) / / y() = e Figure.6 show his resul grphiclly, The wy his informion is usully inerpreed is h sysem is wihin 5% of is finl vlue in 3 ime consns, wihin % of is finl vlue in 4 ime consns, nd wihin % of is finl vlue in 5 ime consns. Hence he use of ime consns gives us quick wy o descrie one spec of he ehvior of sysem. As we will see, s he sysems ecome more complex, he use of ime consns indices which pr of he soluion is he mos imporn nd how he sysem responds o periodic inpus (sines nd cosines). Copyrigh 9 oer D. Throne
13 Exmple.3.. Consider he circui in Figure. (Exmple..). Le s firs ssume = = kωnd C = μ F. Then = kω, = ms, nd K =.5. Nex we will h ssume he iniil volge on he cpcior is zero ( vc( ) = vc() = ) nd he inpu is s follows: < 8 vs () = 8 < 6 > y() Numer of Time Consns / Figure.6. Grph of y() = e for = up o = 7. y () is wihin 5% of is finl vlue in 3 ime consn, wihin % of is finl vlue in 4 ime consns, nd wihin % of is finl vlue in 5 ime consns. Here he inpu is in vols nd he ime is mesured in milliseconds. We now wn o deermine he oupu. We will do his y looking he iniil nd finl vlues for ech ime inervl, where he ime inervls re deermined y he imes during which he inpu volge is consn. The differenil equion is Copyrigh 9 oer D. Throne 3
14 dvc () vc() = Kvs() d Clerly y() = v () nd x() = v (). The soluion in ech inervl will e of he form c s / [ ] y () = y() y( ) e y( ) A his poin we jus need o e le o deermine wh inervl. y () nd y( ) men for ech Firs inervl ( <8 ms) : We hve he iniil vlue in his inervl y() = v c () = vols. To deermine he finl vlue, we use he sic gin nd he mpliude of he inpu for his inervl. vc ( ) = y( ) = Ki= i = Hence for his inervl, we hve he soluion v () = y() = e = e c / /. Before we go on o he nex inervl we need o figure ou he vlue of y () he end of his inervl, his vlue will e he iniil poin during he nex inervl. A he end of he inervl we will hve.8/. 8 y(.8) = e = e = Second inervl (8 < 6 ms) : The iniil vlue for his inervl will e he end poin of he previous inervl, so y () =. To deermine he finl vlue we gin use he sic gin y( ) = Ki( ) = i ( ) = We now hve lmos everyhing we need, however, our soluion ssumed ime of zero ws mesured he eginning of he inervl. Hence o use our previous soluion we need o surc he ime he eginning of he inervl from our cul ime in our form of he soluion, so our ime will e mesured from he eginning of he inervl. Our soluion for his inervl is hen y() = [ ( ) ( ) (.8)/ (.8)/. ] e = e A he end of his inervl we will hve y (.6.8)/. 8 ( = e = e = ). Third inervl ( >6 ms) : The iniil vlue for his inervl will e he end poin of he previous inervl, so y () =. To deermine he finl vlue we hve Copyrigh 9 oer D. Throne 4
15 y( ) = Ki = K = Agin we mus scle our soluion so ime is mesured from he eginning of he inervl, so we hve (.6)/ (.6)/. y ( ) = [.5] e. 5=.5 e. 5 Tol soluion: To ge he ol soluion, we lis he soluions during ech ime inervl: < /. e < 8ms vc () = y() = (.8)/. e 8 < 6ms (.6)/..5e.5 6ms dvc () To ge he curren hrough he cpcior, we use he relionship ic () = C for ech d ime inervl ove. Doing his we ge < /. dv (). 8 c e < ms ic () = C = (.8)/. d.e 8 < 6 ms (.6)/..5e 6ms Here i ( c ) is mesured in mps. Figure.7 shows he inpu volge, he volge cross he cpcior, nd he curren hrough he cpcior s funcion of ime. Noe h he volge cross he cpcior is coninuous, s i mus e. However for his inpu, which is disconinuous, he curren hrough he cpcior is disconinuous. Le s lso look he nswer o see if we cn check our resuls nd if he nswer mkes sense. When he source volge is iniilly urned on, he volge cross he cpcior is zero nd ll of he volge genered y he source is equl o he volge cross resisor. If here were ny volge drop cross he iniil ime, here would lso e volge drop cross he cpcior since hey re in prllel. In sedy se, he cpcior looks like n open circui, so here is no curren flowing hrough he cpcior nd he mximum possile volge his ime is hlf he volge of he source, which grees wih our resuls. In his exmple he inpu ws held consn for n equivlen of eigh ime consns, so he volge cross he cpcior hd essenilly reched sedy se. Finlly is useful o poin ou h if he volge cross he cpcior is descried y he relionship v = v v e v / c() [ c() c( )] c( ) Then he curren hrough he cpcior is given y Copyrigh 9 oer D. Throne 5
16 dvc () C ic() = C = [ vc() vc( )] e d Wh his mens is h if he volge cross cpcior is growing exponenilly, hen he curren hrough he cpcior is decresing exponenilly. Similrly, if he volge cross cpcior is decresing exponenilly, he curren hrough he cpcior will e growing exponenilly. This is lso ehvior our resuls show. Similr resuls lso hold for inducors. / V s () (vols) V c () (vols) i c () (ma) Time (ms) Figure.7. esuls for Exmple.3.. Copyrigh 9 oer D. Throne 6
17 Exmple.3.. Consider he circui in Figure.4 (Exmple..4). Le s firs ssume L. = h = Ωnd L = mh. Then = = =. = μs nd K=.. Nex we will ssume he iniil curren hrough he inducor is i() = mand he inpu is s follows: < <. vs () = 3. < Here he inpu is in vols nd he ime is mesured in milliseconds. We now wn o deermine he oupu. We will do his y looking he iniil nd finl vlues for ech ime inervl, where he ime inervls re deermined y he imes during which he inpu volge is consn. The differenil equion for his sysem is gin di() i () = Kvs () d Clerly y () = i () nd x() = v (). The soluion in ech inervl will e of he form s / [ ] y () = y() y( ) e y( ) A his poin we jus need o e le o deermine wh inervl. y () nd y( ) men for ech Firs inervl ( <.ms) : We hve he iniil vlue y() = i() =.mps in his inervl. To deermine he finl vlue, we use he sic gin nd he mpliude of he inpu for his inervl i( ) = y( ) = Ki= i =. Hence for his inervl, we hve he soluion / / /. [ ] y ( ) = y() y( ) e y( ) = [..] e. =.e. Before we go on o he nex inervl we need o figure ou he vlue of y () he end of his inervl, his vlue will e he iniil poin during he nex inervl. A he end of he inervl we will hve y e e./. (.) =.. =.. =.63 Copyrigh 9 oer D. Throne 7
18 Third inervl (. <. 5 ms) : The iniil vlue for his inervl will e he end poin of he previous inervl, so y () =.63. To deermine he finl vlue we gin use he sic gin y( ) = K i( 3) =.i ( 3) =.3 We gin need o surc he ime he eginning of he inervl from our cul ime in our form of he soluion, so our ime will e mesured from he eginning of he inervl. Our soluion for his inervl is hen y ( ) = [.63 (.3)] e (.3) =.463e.3 (.)/ (.)/. A he end of his inervl we will hve y (.5.)/..5 (.5) =.463e.3 =.463e.3 =.966 Fourh inervl (.5 ms) : The iniil vlue for his inervl will e he end poin of he previous inervl, so y() =.966. To deermine he finl vlue we hve y( ) = Ki 4=.4 Agin we mus scle our soluion so ime is mesured from he eginning of he inervl, so we hve (.5)/ (.5)/. y ( ) = [.966.4] e. 4 =.5966e.4 Tol soluion: To ge he ol soluion, we lis he soluions during ech ime inervl: < /..e. <. ms il () = y() = (.)/..463e.3. <.5 ms (.5)/..5966e.4.5 ms To ge he volge cross he inducor we use he relionship () di () L vl = L nd d compue he volge for ech ime inervl. Doing his we ge < /. di (). L e < ms vl() = L = (.)/. d 4.63e. <.5 ms (.5)/ e.5 ms Figure.8 shows he inpu volge, he curren hrough he inducor, nd he volge cross he inducor s funcion of ime. Noe h he curren hrough he inducor is Copyrigh 9 oer D. Throne 8
19 coninuous, s i mus e, while in his cse he volge cross he inducor is no coninuous. Agin le s look our soluion o see if i mkes sense. Firs of ll, he volge/curren relionships for he inducor re consisen wih wh we expec. The iniil curren in he inducor is ma, s we require, nd he iniil volge from he source is vols. Applying Kirchhoff s lws round he loop, we expec he iniil volge drop cross he inducor o e given y vs () i() = (.)() = vols, which is wh we hve. In sedy se he inducor looks like shor circui, so here should e no volge drop cross he inducor once he sysem reches sedy se, which gin mches our resuls. Noe h he sysem only reches sedy se ner.7 or.8 ms. In ddiion, in sedy se he volge drop cross he resisor mus mch he volge supplied y he source, or vs ( ) i( ) = 4 (.4)() = vols, which gin mches our resuls. Le s look he resuls one oher convenien poin in ime, sy =. ms. Using he equions we derived ove (nd he known inpu) we hve v (.) = 3 vols il (.) =.96 ma vl(.) =.7 vols Applying Kirchhoff s lws round he loop we hve s vs(.) is(.) vl(.) = 3 (.96)() (.7). We cn oviously check s mny poins in ime s we wn in his wy. This ype of checking does no gurnee our nswer is correc, u i does help find ovious errors..3. Soluion Using Inegring Fcors An lernive mehod of soluion of firs order differenil equions is y he use of inegring fcors. This mehod of soluion is imporn o undersnd ecuse s we sr o nlyze differen ypes of sysems, we need o e le o undersnd how we would solve for he oupu when we don cully know wh he inpu is. This helps us chrcerize sysems independen of he cul (specific) inpu. The use of inegring fcors for solving firs order differenil equions is sed on he fc h when we differenie n exponenil, we ge he sme exponenil ck () muliplied y some oher erm. For exmple, if x() = e φ, hen d d φ() φ() dφ() dφ() x () = e = e = x () d d d d In wh follows, he mehod looks firly lenghy, u wih prcice mos of he seps cn e done in your hed. Le s pply his ide o our equion dy() y () = Kx () d Copyrigh 9 oer D. Throne 9
20 4 V s () (vols) i L () (ma) v L () (vols) Time (ms) Figure.8. esuls for Exmple.3.. This mehod will work eer if we rerrnge our equion i o he form dy() K y () = x () d () Nex, we look differeniing he produc y() e φ, where φ () will e deermined y he differenil equion we re rying o solve. This leds o he equion d φ() dy() φ() dφ() φ() φ() dy() dφ( ) ye ( ) e y() e e y ( ) d = = d d d d Nex, we eque he erm in rckes o he lef hnd side of our originl differenil equion, Copyrigh 9 oer D. Throne
21 dy() dφ() dy() y () = y () d d d Clerly his mens h dφ( ) = d Solving his simple equion we ge φ() = Now we pu his ck ino our equion ove o ge d / dy() / / / dy() ye () e y() e e y () d = = d d The erm on he fr righ is he sme s he lef hnd side of our differenil equion / muliplied y e, so his mus equl he righ hnd side of our differenil equion muliplied y he sme hing, d / / dy() / K ye () e y () e x () d = = d Nex we elimine he middle erm o ge he exc differenil we wn d / / K ye () e x () d = Finlly we inegre from nd iniil ime wih iniil vlue vlue y (), d λ / λ / K y( λ) e dλ e x( λ) d dλ = The lef hnd side cn e inegred s λ y( ) o finl ime wih d / / / y( / e dλ = = K λ) e λ dλ y( ) e y( ) e λ x( λ) dλ or ( )/ ( λ)/ K () = ( ) ( λ) y y e e x dλ This is he generl soluion, for ny inpu x(). Copyrigh 9 oer D. Throne
22 Exmple.3.. Le s now look he sme inpu s efore, x() = A for wih iniil condiion = nd y( ) = y(). The soluion o he differenil equion ecomes / ( λ)/ K y () y() e e Adλ = / / / K λ y () = y() e e e Adλ / / λ/ λ = = e λ = y () y() e KA e y e K / / / ( ) = y() e A e Wih he susiuion y( ) = KA, we ge = / / y ( ) y() e KA e or he sme soluion s efore. / / () ( ) y = y e y( ) e / [ ] y () = y() y( ) e y( ) Exmple.3.. Le s use inegrion fcors o deermine he soluion o he differenil equion dy() = y() x() d The firs hing we need o do is pu ll of he y erms on he lef hnd side, Then we need or Then we hve dy() y() = x() d dφ() = d φ () = d ye () e = x () d Inegring oh sides we ge Copyrigh 9 oer D. Throne
23 or d λ ( ) ( ) ( ) λ λ λ ( λ) dλ = = e y d e y e y e x dλ ( ) () = ( ) λ ( λ) y e y e e x dλ Exmple.3.3. Le s use inegrion fcors o deermine he soluion o he differenil equion dy() y() = x() d Then we need dφ() = d or φ () = Then we hve d y () e = e x () d Inegring oh sides we ge e y( λ) dλ e y( ) e y( ) e x( λ) d λ λ d = = dλ λ or λ ( ) y () = e y ( ) e e x( λ) dλ Copyrigh 9 oer D. Throne 3
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