Homework6 - P14.4—1 3 3 3 3 A BS C F[5}=—3 =—‘ =_j.5 33 6.5 4(3 1(3 1}‘ 3 3 1.5" 23 4 Where A =3(s 1]‘ 3 s=—1 3 Then 2 3 B C I w

Info iconThis preview shows pages 1–16. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 12
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 14
Background image of page 15

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 16
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: P14.4—1 3+3 3+3 A BS+C F[5}=—3 , =—‘ =_+_j .5' +33'+6.5'+4 (3+1) [(3+1}‘ + 3] 3+1 .5"+23+4 Where A =3 (s+1]‘+3 s=—1 3 Then 2 +3 _ B C I w. 4 3 %=i+ 1L2}(s+3]=(—+B) s" +[—+B+C]s+—+C [5+1] (3 +23+4) S+l S'+23+4 5 3 3 Equating etc-efficient yields '1 '3‘ 53: 0=:+B => B=—: 3 3 '3 s: 12i—:+C :: C:1 3 3 3 Then '} "r a '1 1 : _:S+l : _: (3+1) T 3 1 3—! — l 64 flea-45! + 1 3 3 E8 _r~:.i11-..gr. r33 0 P14.4—4 1 1 A BS+C F(s)= 1. = 1' =—+ 3 (s+l](3 +23+2] (3+1][(3+1)'+1] 3+1 (3+1) +1 1 when: A = ‘- =1 S_+1.S'+25=_1 "Next 1 = 1 + BHC z} 1=sl+zs+2+(35+cj(s+1) (3+1) (32 +23+2) 3+1 .5'2 +23+2 2:: 1={B+1]33+[B+C+1}S+C+2 Equating coefficients: 330:3“ 2» B=—l 3:D=B+C+2 := C=—1 Finally 17(3) = i — [Si—{EH 2} y {r} = [3" _ 3-! cm. I)“ (r) P14.4—5 F[) 3(3+3] 1 + —[S+l] + 2 S =—=_ —,_ [3+1}(32+Es+5) 3+1 (MU—+4 (5+1]1+4 f{i‘]=[€._: —e_’ cos(1r]+e" sin[2r]]u(r] P14.4—fi 1 5+5 F(3]= ( J = A + B + C s[s+l](s+2} .5' 3+1 3+3 when: 2 3 2 3 A=SF[S) _ =[S—Jr) =3 B—(3+1)F{s}| = {3+} =—4 5—D [3+1](3+2]:=0 ='1 S(.5'+2]:=—1 and 2 3 (3+1]F[3)l = (3+ ] =C=1 “— S[.5'+1)5=—2 Finally 3 —4 1 1314.43; FLY] = ca+{ea}—ma:) = 1[ c+jat + c—jai ] {3+a)'+m- 2 3+fi—JW S+H+Jw l majfi mafia 3 3 —1 d =—[—+—]uhara m=1fe +d . fl=tan A 2 aha—jag S+a+jra .'.f[r)=a"” [ceos afi—dsi11wt]u(t]= a”![-vt'aalfii"E e0s[mr+3)]u[r) = m 3—“? eos(mr+6)u(t) P1454 Initial value: 5' s+16 31 + 165' 110) =JJ'1115V[3) = lintzl—] = ljn12—= 1 H“? Hens +43+12 Hens +4S+12 Final value: 15 I 2 16 v[oo) =li1u .9 = Lim ZS+—S = 0 H“ S +4S+12_ 5-?” .5' +4S+12 (Cheek: PIS} is stable because Re{pl.} <1 0 since p,- =—2 i 2.828}. “Te expect the final value to exist.) P14.5—3 Initial value: 31 +1.0 5 19(D)=li1113V(3} = lint fi = 0 Ha H33 33 +23 +3 Final value: .5' S+10 11:0] = lint 3V [5] = lint #:10 H0 5"“ .9 [35‘ +23+1) (Cheek: PIS} is stable because pi = —0.333 t 0.4T1 j. ‘We expect the final value to exist.) P14.5—4 Initial value: I I _253 _143 f[0)= h111$F(3]=1u111—=—2 Hm HMS —23+10 Final Value: F(S) is net stable because Re {pf} > O sincepf =1i 3}. No final value 0f f ( I] exists. P145545 Afler the switch opens, applj,r KCL and KVL to get / R1 f{t}+C%v{t}]+v{t]= s; K Apply KVL to get 1 _ i. . }{I]—LJE![I]+R2H[E'_} Substituting v{t } into the first equation gives a, :[rj+oi[sir{:j+az:[rjnwiqmagr{:_}=t; K d: dt at then reload—Ertzrht'ttlores+L}£:[rj+q’al+aa}:[rt=t»: o‘t2 ' ' a: - - v * Dividing to}; R1 CL: 3 “a CR +5“ “a +3 ; d—ijzth —1 2 iron 1 3 or]: IS {jib I? " R R1 d3 5 t' - . . _- _. ,_. = _ _ Eiththegtvenvalues drlrm+25drt[t]|+lfio_2)t[t] 12) Talo'ng the Laplace transform: '3 I/o' , _ 125 5'1 5 — —t' 11+ +51f 0+ +25 5] 5 —t' 13+ +136251 5 =— [amt yd] [Hit] [)5 We need the initial conditions. For t =:: {1, the switch is closed and the circuit is at steady state. At steady state the capacitor acts like an open circuit and the inductor acts like a short circuit. Using voltage division v[o—t=;2o=i4_ts4 v ' 9+[is H]. Then, using current division ' = 0.323 A K 4 ‘|V[D—~I The capacitor voltage and inductor current are continuous so v [ ID +} = v [ ID — ] and hit] +} = r' [0—] . After the switch opens v[o+t aim] 14.154 stuns] '+ = +— 9.4 :14 0.4 = 29.508 d. . d. v[r]=LE:[r]+Rli[r] :3 Eiifl+j_ Substituting these initial conditions into the Laplace transformed differential equation gives [:3 r[st—[aasoswsassjhasst{E]—o.3zs]+iss_zs rm: 12—3" , _ _ , 5 r 7' t (r +255+1ss.25]r[5) =L—3+[29.sos+o_32353+ as (cars; . .5 " so E1323 51 +[29_5fl3 + 25 [GEESJ] +125 f s = I: 1' 451955455251 0.32351+{29.5D8+25(fl.323i]+125414?] 23.5 {is 5{.-:+12.5]- 5+12-3 [s+12.5]‘ 5 Taking the inverse Laplace transform MEI]=D.E+o'”"'{lifit—flfil} A. forms 50 _ x 0.3233 fortSEJ ’[I»'=io.s+e-“-5*[asst—0.4m A fort-ED P1 1.6—? 4 H KCL: T+i=?e_fl‘ 3 . K‘s-FL: 4fl+3i—v._ =IEI :} 1IL=4fl+3f ?fl_brtitr], A 0 dr iii (if . 4—+3t (3,} 35 Then ml +i= Tie—‘5' :3 —+2i=—g"“ 3 d: 4 Taking the Laplace lmnsfoim of the dfierential equation: 35 1 3' 1 sitsi—iifli+lflsl=:m EHSFTDW Where we have used iii-ti) = ID . Next. we pen-form partial fiaction expansion. +=i+i what-e A: 1 =1 afld3=L{ =_l {s+2]{s+fi} 5+2 5+6 5+6 4 s+2§__5 4 .l' -—1 Then ;{5)=EL_E; 2;, f{r}=3_3g'3'—3—3e'“‘ A, EEC! 1155+? lfis+fi 15 15 FILE-8 201;! L1 Apply KCL at node a to get o’v 191—13 {111' I i_1=i l :3. 2v]+—]=2‘L' v.2 48 cit 24 cit Apply KCL at node is to get vz—fiflcoslr vi—v] v3 1 din dv: —+ +—+— 1 ‘: 2E] 24 3E] 24 d: cit =fifleos2t Take the Laplace transforms ofthese equation; using v] (G) =10 V and 13(0): 25 U: to get 255] + 505 +199 [2+s‘li’Tlfisj—2Ktjsj = 10 and —T«’l{s}+[3+s]i’1{sj= 1 a - s ‘ s‘+4 Solve these equations using Gunner‘s rule to get I/ 1' 2551+fifl5+1flfl If, Ema 5+4 f“: {2+5}{2551+505+1m}+10[51+4} 4' 5 =~—=Ifl—~ ‘1‘ ] [2+5] {3+5}—1 [5‘+4]|{5+1][5+4’I _ 2553+12953+22fl5+24fl [51+4]|{5+1][5+4] Next, partial fraction expansion gives V45]: A_ + A_ +i+£ ' 5+J2 5—J2 5+1 5+4 where 2553+12053+2205+240 —24fl—J24fl _ :1 =—_ =—=fi+_:fi [5+1]|{5+4][5—J2] 5.1;: —”1-{} A'=fi—j6 B =2553+12fl53+2205+24fl =11_3=§ [252+4jl[5++:| 5-—l 15 3 C _ 2553+12053+2205+24IJ _ — 20 _E {51+4][5+1] 5—4 —I5fl 3 Than ' _' ' x ij=fi+5fi+fi 56+23fi+lfil3 5+J2 5—32 5+1 5+4 Finally 121(3) = llcnsir+125in25+%e"+%e" if 2'20 f}':} 4“ Node equations: F ‘—.T:’ F" 5 —'[5*' C[SJ+ EIII‘5]=1 2:: Paris]: 6 I’c[5|+—fi 5 I5 5 5+6 ’ 5+5 5 I, 6 , x fr F.3[5]—— Ictsl- —T'.:[5I+—] 1 5 3 5+3+M+_+_VCEJ~I__=Q 4 5 5 4 ' '2 After quite a bite of algebra: If: [[5]: f fi5'\+:afi5+f132_ {5+2’|{5+3]1L5+3-] Partial fiactiml expansion: fl 1 V13): W=i_i+i ‘3 [5+3][5+'2j|[5+3] 5+2 5+3 5+5 Iuvcrse Laplace: transform: 1: {2'} = fle'l' — 99'“ + ie's' V. r 3; G C 3 3 P1 LT—fi '.|II'II§' Ciilll'hl’l [cg um: wit I'J 'n 'u"; Write a node equation in the fiequency domain: R R, E I; ~ I; x %+55 ml; 5—103- i+ “(S'I—5C+ °l[”={} 2:» I’L[5]=—1—~,=— 1+—11L R1 R3 — 55+ 1 5 5+ Cs REC; REC [marge Laplacc mfmm: RH -” i=2:a _ — w(r]=—_1u =+Es—m—-]e’=’fi=c]=_. 10—551mm] V f“””' PILT-D 2&1 HID fiequency domain Wfltmg a mash Equation: 5 2“ K *. ’4+55U[sj+3€1+£=0 : j{jj=fi=_ 3+ 3 IL ' 5 K 4) 5 4 5|5+7 5+: K 3 x 3.: Take the luvcrse Laplacc mfmm: :[r]=—3[l+e‘”'“'} A forr>fl P1 LT—II} Steady—state for t<::{}: From the equation for v.33“): _'J. t'o[oe]=ti+12£ "(301:6 U From the circuit: t 3 t v at: =— 13 o[ J R+3[ 1 Therefore: 6:3 R+3 || Ch. :3 {13) :3 rt Ittst=iftst+E=i ' Cit ' 5 (35 Taking the inverse Laplace transform: 918V 31.! Steady-state for t.'-={}: vc{3}= mug-"3“ V for r p a 1 Comparing this to the given equation for veg): we see that 2 = E :3 C' = £1.25 P. Phi—13 For I -=:: E]. the switch is open and the circuit is at steady state. and the circuit is at stead}r state. At steady state. the capacitor acts like an open circuit. .. . A A H‘th ER and v.3[r] 2 Consequently. ., A A Itfl—]=fi and VC[D—]=E iC{CI—]=D The capacitor voltage and inductor current are continuous so v.3 [flfl = ta: {-0—} and :‘L {43+} =r._ {cl—j. For I ::= ID. the voltage source voltage is 1'2 V. Represent the circuit in the frequency domain using the Laplace transform as shown. IL [5} and EC [5] are nJesh currents. 1|Writing a mesh equations gives l —I 5 C5 cl: J _ Dcinniatrixforin . “AL :1 K _ ‘1 _ _ Ls+R R1 {IL—{5] ER 5 —R rt+a RICH] _ _i “ K 25 ‘I. u“ \. {Ls+Rj[—_i +R i is“): sx KER s = 2L -“ 1 * ~ 2 n i [Ls+R]LR+aJ—R s +RCs+LC 11.}When R=3fl= L=2H, C=fif and :1=1'2"v', 3 3 I [5]_;_;_i_i C 51+Es+12 {s+2}{s+fi] 5+2 5+3- Taking the inverse Laplace transfonn gives '~. :E{i]= gash—i “‘ Jam] A b.) R=2fl= L=2H, C=éF and.-1=1'2V, 3 3 I =1—= 1 a“) s'+4s+4 [5+2]' Taking the inverse Laplace transfonn gives it {2'}=3Ie'2'u[r} A c.}R=1C| fl. L=2H__ C=$ F andA=12V 3 3 3 4 Icli51|=.—=—.=—X—a s‘+4s+lfl [5+2]‘+1fi 4 [s+2]‘+115 Taking the inverse Laplace transfonn gives fc[r]=§e'3‘sin[4f}u[r] A FLU—14 For I-=:: CI. The input is 1'2 V. At steady state. the capacitor acts like an open circuit. Notice that viii]: is a node voltage. Express the controlling voltage of the dependent source as a function of the node voltage: vi = —v(r} Writing a node equation: “’12—vvi'" 1;. *_ k 8 I|+ 4 +L 4v[I]J|—fl —12+v{t}+2v[r]—fiv[t]=fl : v[t]=—l'v' For H: CI. represent the circuit in the fiequencv 0 7'73 lit-i] domain using the Laplace transform as shown. E15 II is a node voltage. Express the controlling voltage of the depeth source in terms of the node voltages I’a{s]=—T»’{s] Writing a node equation gives PM] 15 7— .v ‘ K x —“‘+ [5'l+3—5 I’lisHi =0.i‘5l«’[s] 8 4 4|!)K ' s Solvinggives _’J / {5-5‘fl’{5]=E_4 2:, Iris]: 10F _i=_‘+i_i_=_2 1+; ' E s[r—)] 5—5 5 5—5 5—) k5 5—5 Taking the inverse Laplace transform gives v{t‘}=—2[l+ei'] V fori'EEJ This voltage becomes very large as time goes on. P14.S—2 Let 21 = R + CL and E1 = R + La then the input impedance is .3 1 ' 3 L' 22 R+— (R+Ls] LES + RC+— 3+1 2(a) = 11 =—?- =R —. R- al+z2 R+_+R+LS LCS +2RC3 +1 CS wL‘s—ow require : RC + i = ERC 2} L = REC then 2 = R P 14.8—5 The transfer function can also be calculated form the circuit itself. The circuit can be represented in the frequency domain as 10kg Fi' "We can save ourselves some work be noticing that the 10000 olnu resistor. the resistor labeled R and the op amp comprise a non-inverting amplifier. Thus R Vatsl=ll+mlmsl Now. writing node equations. —VC{S]—VE{S]+CSPL[S:I:D 0nd —VO{S}—P;(S] P1113} 1000 Solving these node eqlmtions gives 1 [ R ]5000 1000 1+ 10000 Hm: C L I 1 5000] 3+ S + 1000C L Comparing These two equations for the Transfer function gives [3-+ 1 ] (34-2000) or {s-— 1 ]==(34-5000) 1000:? . l 5000] 5+ L (s——2000) or [s-F———— 1 [1 .R 'J5000 + 100001 10000. L The solution isn‘t unique. but there are only two possibilities. ICline of These possibilities is 1 1000C F+ }=0+2mm)=0 C=0j0r 5000 3+ L ={3+5000] :> L=1H 1 .0 5000 -—6[1+—}—:15x10ESP :11» 3:51.31 1000fllfixl0 ] 10000 1 P 14.8—5 The Transfer functic-n of the circuit is R2 1 1+R203 RIC Hm:— R =——1 1 3+ RIC The give atcp rcapcngc is VD (r) = 4(1—3'350’)u(r) V . The correspond n‘anafcr function ii. calclflntcd are. ELF)=2:_4(1_€—250r]u(r)}=_[i_ 4 J: —1DI[){J : H“): —lU'U'U S .9 S+250_ S[S+ESD) S+250 Comparing Thcic rcsrfl‘rs giver- 1 =50 2:: RE: 1 =—]L _ =4cm REC ZSDC 250(D.1><10 6) l =1CJU'U' 2} R1: 1 l =lflkfl RIC 1000c —1{]DD[D.1:<1|D_6) ...
View Full Document

This note was uploaded on 01/24/2010 for the course ECE 2025 taught by Professor Juang during the Spring '08 term at Georgia Institute of Technology.

Page1 / 16

Homework6 - P14.4—1 3 3 3 3 A BS C F[5}=—3 =—‘ =_j.5 33 6.5 4(3 1(3 1}‘ 3 3 1.5" 23 4 Where A =3(s 1]‘ 3 s=—1 3 Then 2 3 B C I w

This preview shows document pages 1 - 16. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online