RLC Circuits - Engineering Electrical Circuits - Lecture Slides, Slides of Electrical Circuit Analysis

Download RLC Circuits - Engineering Electrical Circuits - Lecture Slides and more Slides Electrical Circuit Analysis in PDF only on Docsity! 2nd Order RLC Circuits Docsity.com Cap & Ind Physics Summary • Under Steady-State (DC) Conditions – Caps act as OPEN Circuits – Inds act as SHORT Circuits • Under Transient (time-varying) Conditions – Cap VOLTAGE can NOT Change Instantly • Resists Changes in Voltage Across it – Ind CURRENT can NOT change Instantly • Resists Changes in Curring Thru it Docsity.com Second Order Circuits  Single Node-Pair Ri Li Ci 0=+++− CLRS iiii )();()(1;)( 0 0 t dt dvCitidxxv L i R tvi CL t t LR =+== ∫ SL t t it dt dvCtidxxv LR v =+++ ∫ )()()( 1 0 0 • By KCL −+ Rv −+ Cv − + Lv • By KVL 0=+++− LCRS vvvv )();()(1; 0 0 t dt diLvtvdxxi C vRiv LC t t CR =+== ∫ SC t t vt dt diLtvdxxi C Ri =+++ ∫ )()()( 1 0 0  Single Loop Differentiating dt di L v dt dv Rdt vdC S=++ 12 2 dt dv C i dt diR dt idL S=++2 2 Docsity.com Second Order Circuits  Single Node-Pair Ri Li Ci • By KCL Obtained −+ Rv −+ Cv − + Lv • By KVL Obtained  Single Loop dt di L v dt dv Rdt vdC S=++ 12 2 dt dv C i dt diR dt idL S=++2 2  Make CoEfficient of 2nd Order Term = 1 1∙(2nd Order Term) dt di CLC v dt dv RCdt vd S11 2 2 =++ dt dv L i LCdt di L R dt id S11 2 2 =++ Docsity.com ODE for iL(t) in SNP • Single-Node Ckt • By KCL • Note That • Use Ohm & Cap Laws • Recall v-i Relation for Inductors • Sub Out vL in above Ri Li Ci SCLR iiii =++ ( ) ( )tvtv L= S L L L i dt dvCi R v =++ dt diLv LL = S L L L i dt diL dt dCi dt diL R =     ++     1 Docsity.com Illustration • Write The Differential Eqn for v(t) & i(t) Respectively    > < = 0 00 )( tI t ti S S dt di L v dt dv Rdt vdC S=++ 12 2 0;0)( >= tt dt diS    > < = 00 0 )( t tV tv SS dt dv C i dt diR dt idL S=++2 2 0;0)( >= tt dt dvS Si − + Sv  The Forcing Function  Parallel RLC Model  In This Case  So 012 2 =++ L v dt dv Rdt vdC  The Forcing Function  Series RLC Model  In This Case  So 02 2 =++ C i dt diR dt idL Docsity.com 2nd Order Response Equation • Need Solutions to the 2nd Order ODE  As Before The Solution Should Take This form  If the Forcing Fcn is a Constant, A, Then Discern a Particular Soln  Verify xp )()()()(1 212 2 tftxat dt dxat dt xd =++⋅ )()()( txtxtx cp +=  Where • xp ≡ Particular Solution • xc ≡ Complementary Solution )( 2a AxAtf p =⇒= A a Aaxa dt xd dt dx a Ax p pp p ==⇒ ==⇒= 2 22 2 2 2 0  For Any const Forcing Fcn, f(t) = A )()( 2 tx a Atx c+=  Docsity.com The Complementary Solution • The Complementary Solution Satisfies the HOMOGENOUS Eqn  Nomenclature • α ≡ Damping Coefficient • ζ ≡ Damping Ratio • ω0 ≡ Undamped (or Resonant) Frequency  Need xc So That the “0th”, 1st & 2nd Derivatives Have the same form so they will CANCEL in the Homogeneous Eqn  Look for Solution of the form  ReWrite in Std form 0)()()( 212 2 =++ txat dt dxat dt xd 0)()(2)( 202 2 =++ txt dt dxt dt xd ωα  Where • a1 ≡ 2α = 2ζω0 • a2 ≡ ω02 stKetx =)( Docsity.com Complementary Solution cont.3 • Example Cont.  Before Moving On, Verify that Kest is a Solution To The Homogenous Eqn  Then  K=0 is the TRIVIAL Solution • We need More 042 0)(4)(2)( 2 2 2 =++ =++ ss txt dt dxt dt xd UnitLess)(5.0 2 1 1222 2/2 44 0 0 1 0 2 0 2 0 ==∴ =⇒== =≡∴ =⇒= − ζ ω ζαζω ω ωω SS stst Kes dt xdsKe dt dx 2 2 2 ; == 0)2( or 0)()(2)( 2 0 2 2 02 2 =++ =++ stKess txt dt dxt dt xd ωα ωα Docsity.com Complementary Solution cont.4 • If Kest is a Solution Then Need  Solve By Completing the Square • The CHARACTERISTIC Equation  Solve For s by One of • Quadratic Eqn • Completing The Square • Factoring (if we’re REALLY Lucky)  The Solution for s Generates 3 Cases 1. ζ>1 2. ζ<1 3. ζ=1 02 20 2 =++ ωαss 1 1 0)()( 2 002,1 2 0 2 02,1 2 0 2 22 0 2 −±−= −±−= ⇒−±−= =−++ ζωζω ωαωα ωαα αωα s s s s Docsity.com Aside: Completing the Square • Start with: • ReArrange: • Add Zero → 0 = y−y: • ReArrange: • Grouping – The First Group is a PERFECT Square • ReWriting: 02 20 2 =++ ωαss 02 20 2 =++ ωαss ( ) 02 20222 =+−++ ωαααss 02 220 22 =−+++ αωααss ( ) ( ) 02 22022 =−+++ αωααss ( ) ( ) 02202 =−++ αωαs Docsity.com Case 2: ζ<1 → UNDERdamped • Since ζ<1 The Characteristic Eqn Yields COMPLEX Roots as Complex Conjugates – So with j=√(-1)  Then The UnderDamped UnForced (Natural) Response Equation  Where • ωn ≡ Damped natural Oscillation Frequency • α ≡ Damping Coefficient n n jjs jjs ωαζωζω ωαζωζω −−=−−−= +−=−+−= 2 002 2 001 1 1 ( ) ( )tAtAetx nntc ωωα sincos 21 += − Docsity.com UnderDamped Eqn Development • Start w/ Soln to Homogeneous Eqn  From Appendix-A; The Euler Identity  Since K1 & K2 are Arbitrary Constants, Replace with NEW Arbitrary Constants tsts c eKeKtx 21 21)( += tjte nn tj n ωωω sincos ±=±  Then ( ) [ ] [ ]( )tKKjtKKe tjKtKtjKtKe eeKeeK eKeKtx nn t nnnn t tjttjt tjtj c ndn nn ωω ωωωω α α ωαωα ωαωα sincos sincossincos )( 2121 2211 21 )( 2 )( 1 −++= −++= += += − − −−− +−+− [ ]212 211 KKjA KKA −= +=  Sub A1 & A2 to Obtain ( )tAtAetx nntc ωωα sincos)( 21 += −  Docsity.com UnderDamped IC’s • Find Under Damped Constants A1 & A2 • Given “Zero Order” IC  With xp = D (const) then at t=0 for total solution  Now dx/dt at any t 0)0( Xx =  For 1st-Order IC ( ) DXA DAAe Xx nn −=∴ ++ == − 01 21 0 0 0sin0cos )0( ωωα 10 Xdtdx t = = ( ) ( ) ( ) ( ) 121 21 120 21 12 )0( 0sin 0cos )0( sin cos AAX dt dx AA AA e dt dx tAA tAA e dt dx n nd nn nn nnt αω ωαω ωαω ωαω ωαω α α −==       +− − =       +− − = − −  Arrive at Two Eqns in Two Unknowns • But MUST have a Number for X1 Docsity.com Example: Case Analyses cont. • For Char. Eqn Complete the Square  Then the Solution  The Roots are Complex and Unequal → an Underdamped (Case 2) System • Find the Damped Parameters ( ) 31 3)1( 03)1( 31242 2 2 22 js s s ssss ±−=⇒ −=+∴ =++= +++=++ 122 0 12 0 1 0 0 0 314 325.0121 122 2122 24 − − − =−=−= =−=−= === =⇒= == S S S n n αωω ζωω ζωα ζζω ω ( ) ( )tAtAetx tAtAetx t nn t 3sin3cos)( sincos)( 21 21 += += − − ωωα Docsity.com UnderDamped Parallel RLC Exmpl • Find Damping Ratio and Undamped Natural Frequency given – R =1 Ω – L = 2 H – C = 2 F • The Homogeneous Eqn from KCL (1-node Pair)  Or, In Std From 012 2 =++ L v dt dv Rdt vdC 0 21 12 2 2 =++ v dt dv dt vd 0 42 1 2 2 =++ v dt dv dt vd  Recognize Parameters 2 1 2 12; 2 1 4 1 00 =⇒=== ζζωω Docsity.com Parallel RLC Example cont • Then: Damping Factor, Damped Frequency 4 3 4 11 2 11 20 =−=−= ζωωn 4 1 0 == ζωα  Then The Response Equation       += − tAtAetv t c 4 3sin 4 3cos)( 214  If: v(0)=10 V, and dv(0)/dt = 0 V/S, Then Find: 31010 21 == AA  Plot on Next Slide       += − ttetv t c 4 3sin 3 1 4 3cos10)( 4 Docsity.com KEY to 2nd Order → [dx/dt]t=0+ • Most Confusion in 2nd Order Ckts comes in the from of the First-Derivative IC  If x = iL, Then Find vL  MUST Find at t=0+ vL or iC  Note that THESE Quantities CAN Change Instantaneously • iC (but NOT vC) • vL (but NOT iL) ( ) ( ) LvX v dt diL L L t L += += += 0or 0 1 0  If x = vC, Then Find iC 10 Xdtdx t = = ( ) ( ) CiX i dt dvC C C t C += += += 0or 0 1 0 Docsity.com [dx/dt]t=0+ → Find iC(0+) & vL(0+) • If this is needed • Then Find a CAP and determine the Current through it • If this is needed • Then Find an IND and determine the Voltage through it +=0tdt dv ( ) C i dt dv t + += = 0 0 +=0tdt di ( ) L v dt di t + += = 0 0 Docsity.com —W§3-lvo nr 60) 2H WhiteBoard > Find v,(t) i,(t) + @ 27r=0 18 G) tf 24V v,(t) 4V 12V Docsity.com _\arnGz7¥3 8 | Engineering Instructor — 2 > os Chabot College = 7HvS Da = Ac ch Ue 25655 Hesperian Blvd g at Hayward, CA 94545 O eMail: bmayer@chabotcollege.édu At =Ao= CLUs D8 l4*4G ABOVE /v KVL EGN LehUe ~/ONCAR + Ue = -8BV ole “ak CorTd K=2e € CHMReE Z MUe 4 /Olx/F AUe + Ver-8V ee at Be At v FROM PBOVE P/scenn pA2TICO AL S6lunl 7 Vie = —8V NOW THE HoMomavecos Gow Gat eageeve Fe LOD 4 FLUE 4 [8UL =O @ werrgeeenne @ 2¢ 7272/4 | Oven 3 4 THE CHAR SGN: 82+ 95778 +0 FACTINNG D> (Bt3)(Ste) =O Docsity.com THEN THe [C9078 : gs = 7-3 ok So-G THEN THE HOMOG SOLV Dee = Ke == tke ante THUS THE TOTAC SOCN = VectUp *—-8V “ Vee Let tee NOU US& ZS 7 FO &, ke UW. COT) = OV te OV= K, e°+Ki. 2° av OR K, +Ky =tAV Ca) NOW (KE 4ce= Cebhvcfat CH Ya SK lt=ot THOS Yel = ACO ap ay BE ot x “Ss = £Cot) =A, (o*) = ZA 4 2ZEFCRZL2 |g? $3 SI Docsity.com So AN = BeYyY- /AaY (ae E> $4 (Ries 8") ks -3ke**-6h eo * Cl Bor kbve /ebt leno => /&@VK Sb oS (6 ~>-BsK, A -Chee OL 2K, +64 =-7@ KOurs Salve fot & # KE. US HQ Gors (1) €@) BK. ¢€Gkr2=-%@ (ep tke = 4+8)-3 7A#-os Ke = -¥Y3 = SEV PREN (2, = BV—K, = BY-CIeV)=#2, SQ Ve=t2vVe 14 VE eV & Numerical Example cont • Steady State for t<0  The Analysis at t = 0+  Then Find The Constants from ICs  Then di0/dt by vL = LdiL/dt  Solving for K1 and K2 Ω Ω )0(Li − + )0(CvV24 0)0( =Cv A VViL 5.0186 1224)0( = Ω+Ω − = )(5.0)0()0( Aii Lo =+=+ )0()0()0( +=+=+ dt diL dt diLv oLL 012)0(18)0(4 =+++++− LL iv ( ) ( )( ) VAvL 175.0181240 −=Ω−−=+  KVL at t=0+ (vc(0+) = 0) HV dt dio 2/17)0( −=+ 21 21 A 5.0)0( 63 S A 2 17)0( KKi KK dt di o o +==+ −−=−=+ A 6 14;A 6 11 21 =−= KK Docsity.com Numerical Example cont.2 • Return to the ODE  Yields Char. Eqn Roots Ω Ω V24 63 21 −=−= ss 0)(18)(9)(2 2 =++ tit dt dit dt id  Write Soln for i0 0; 6 14 6 11)( 63 >+−= −− teeti tto  And Recall io & vo reln )(12)(18)( Vtitv oo += 0;124233)( 63 >++−= −− tVeetv tto 0;)( 62 3 1 >+= −− teKeKti tto ( )( )630 0189 2 ++= =++ ss ss  So Finally Docsity.com General Ckt Solution Strategy • Apply KCL or KVL depending on Nature of ckt (single: node-pair? loop?) • Convert between V↔I using • Ohm’s Law • Cap Law • Ind Law ( ) ( )0 0 1 tvdxxi C v dt dvCi c t t cc c c ∫ += = Rvi Riv RR RR = = ( ) ( )0 0 1 tidxxv L i dt diLv L t t LL L L ∫ += =  Solve Resulting Ckt Analytical-Model using Any & All MATH Methods Docsity.com 2nd Order ODE SuperSUMMARY-3 • Then the TOTAL Solution: x = xc + xp • All TOTAL Solutions for x(t) include 2 UnKnown Constants • Use the Two INITIAL Conditions to generate two Eqns for the 2 unknowns • Solve for the 2 Unknowns to Complete the Solution Process ( ) ( ) pnnt p st p tsts xtAtAex xbmtex xeKeKx ++= ++= ++= − ωωα sincos 21 21 21 Docsity.com All Done for Today 2nd Order IC is Critical! ( )+=    += 0 Case Series 0 L t L v dt diL ( )+=    += 0 Case Parallel 0 C t C i dt dvC Docsity.com Complete the Square -1 • Consider the General 2nd Order Polynomial – a.k.a; the Quadratic Eqn • Where a, b, c are CONSTANTS  Solve This Eqn for x by Completing the Square  First; isolate the Terms involving x  Next, Divide by “a” to give the second order term the coefficient of 1  Now add to both Sides of the eqn a “quadratic supplement” of (b/2a)2 02 =++ cbxax cbxax −=+2 a cx a bx −=+2 a cababx a bx −     =     ++ 22 2 22 Docsity.com Derive Quadratic Eqn -2 • Note that Denom is, itself, a PERFECT SQ  Next, Isolate x  But this the Renowned QUADRATIC FORMULA  Note That it was DERIVED by COMPLETING the SQUARE a acb a bx a acb a bx 2 4 2 4 4 2 2 2 2 − ±=+ − ±=+ a acb a bx 2 4 2 2 − ±−= a acbbx 2 42 −±− =  Now Combine over Common Denom Docsity.com E 9 SOME BY COMALCENYER SOUMZCE Docs 6™=p2. Cus ~ide > og ote 2 S20S =-Wwe- e u y S APP Q@UADRLANC S LPPLCEMENT J Cine TD BOM SIOES 2 5242 Sia,s+Ftoe = Yua*- wet Boe AMS (S$ PERMTEPT SF . _ (st Se.) °= wé[S% 1 Y TPKE SQG-BT OF Bot 31063 Se Foye WN 5*-/ Complete the Squa LSOLATIING § ge Sato sso FEVET | @