s10_lec

# s10_lec - Steps in solving a linear dynamic network(for now...

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Unformatted text preview: Steps in solving a linear dynamic network (for now): 1. Choose to use node method or loop method, based on circuit topology and types of energy storage .. elemehts. 2. Write down node or loop equations, which yields differential and/or integral equations for variables. 3. Assume exponential solution, so that all variables are proportional to e“. 4. Rewrite node or loop equations in terms of amplitudes. 5. Write equations in matrix form, e.g., M(s)l_3= Q. N.B.: Can skip steps (2), (3), and (4), and do this step directly. 6. Find roots of equation det[M(s)] = 0; for each root si, find E. By superposition, general solution is QU‘) : Zai EieSit i From this solution, find the voltage or current in each capacitor or inductor-.- Solve for the ai that give the correct " initial conditions. Lac-PM 5’0 From /a§-’~ +ime I IVA—e: C; S J ,1, are, “IL-Q. Cohdug‘l-ahLei L4 5 TLL/ awe. cal/CA ’/adm}HalqceSU qu ) l aN- Ma. res)s+am¢es C; S IA TL‘QJ «A called IM/Oedahceéu TD/uj 214 Com/300,614+ Vo/ues: 4.5 + __5_‘. -J. L )2 g E, __‘ ___L J. ~4— __L_ E2, 6 Q 25 ”HQ-E ”:— O .L. .5: _/ \ 25 ‘32. /‘ \$515) : I " l/L I/c""/zs : 251+ 4' 5- 2° 245 TL»: “(errqck/isﬁg vanes" are .\ - S,:-l 62¢ sac—Vam' Seld’io‘n Procedure—.1 For each cLa/qc‘kV/S'HC VaWSL.) ginok oquaa-lreng‘Hp Vick! __E__ Ha? sanisC-Pes eﬁooHoa-d, /5<.4M u SalquhsJ duck 56+ coaé+qa4 SucL I #4- ;m—aqx mamm art SqHspiﬁd. [VII-ﬂ): [”224 'l) f‘gi ”VG | 1 "/L Va +2211) .— "‘ 7,2 “’I/L # l/C -1/3 2am reckon/.6 N (‘1) CD —/12. ‘ ®__Q:Z;Z- CD >< (n) ® -—//4 _ “”3 \ 6» Q @+’/<’7® Reduczd wa+f)¥ l5 ﬁ:[l 1] SuﬂDDSQ. ‘HAL )wH-ial Candi-Hons 6H. +laa+ U;{:=)=Z 720,4” Lq/b)= I awn/m C092 we use- +sz. 7% {ind 4,12? >15 buvl— [E is d1‘#/CHH‘. Express 3716)) (L466) [91 +€VMS Q4 44 = J— 387.5%)d'6 L4 -5/ t 3 J— (ae'£+fbe 1) Z. -—t ~57U= Nah: 77425 5k}; ‘15 a Pain) QSPQcc'aHJ 1-?- 5; am cowP)-Q><. 79%st wH/L a‘odaroadn Q) Cow‘olich-ti, («Al‘Ha Mang Sk/OS © EzoaLiﬂnS (M 355)) QM ma“ {H +€Vm\$ a! :Fundqmn4Q‘ Varioglu" mm anA 91¢) => wqu (HG?— a/ojar‘ooxcL Haml— QWPI/Mji e5 role. O‘C ‘guwciaMQm-La( Tam/10A, m ggmgfﬂ; g—C “Si-3+6" TLo. 544%. O-C a sjg-lem is q 5e4- OL VGNQLRS (a! swaHeH— Poss/Hg, Si )+lrw+3 thH—UL (AWL am art/3644‘s +° ‘HW .5 54cm) [5 .Sq'H-icaew‘ ‘I’Q Emu—4' + ﬁuhbe. (wee/\$001+) elmvior 01C HAL 5j5+em, ‘ Each 54—«4Q var/Webb; L‘s assocéa‘ki \ \\ // MM'HA mummy ‘3 Eacl. s+q+€ VaHCJoLa. C5 asjoCZal—edl w‘l‘HA an EMF/”(‘4‘ Cond/Hoa " S+A+C Vat/(awe; are O'Heu, agocia‘kl com/x €142er 5+0/‘U a. [(4 our exam/3’9 'H/Q 344+t5 are. The Definition of “State” The state of a system is a set of variables (of smallest possible size) that, together with any inputs to the system, is sufficient to predict the future (important) behavior of the system. ...
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