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Homework6

# Homework6 - P14.4—1 3 3 3 3 A BS C F[5}=—3 =—‘...

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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) ...
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