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

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Engineering 43 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 L i Ci 0 CLRS iiii )();()( 1 ; )( 0 0 t dt dv Citidxxv L i R tv i CL t t LR   SL t t it dt dv Ctidxxv LR v   )()()( 1 0 0 • By KCL  Rv  Cv   Lv • By KVL 0 LCRS vvvv )();()( 1 ; 0 0 t dt di Lvtvdxxi C vRiv LC t t CR   SC t t vt dt di Ltvdxxi C Ri   )()()( 1 0 0  Single Loop Differentiating dt di L v dt dv Rdt vd C S 1 2 2 dt dv C i dt di R dt id L S 2 2 Docsity.com Second Order Circuits  Single Node-Pair Ri L i Ci • By KCL Obtained  Rv  Cv   Lv • By KVL Obtained  Single Loop dt di L v dt dv Rdt vd C S 1 2 2 dt dv C i dt di R dt id L 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 L i Ci SCLR iiii     tvtv L S L L L i dt dv Ci R v  dt di Lv LL  S L L L i dt di L dt d Ci dt di L 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 vd C S 1 2 2 0;0)(  tt dt diS       00 0 )( t tV tv S S dt dv C i dt di R dt id L S 2 2 0;0)(  tt dt dvS Si   Sv  The Forcing Function  Parallel RLC Model  In This Case  So 0 1 2 2  L v dt dv Rdt vd C  The Forcing Function  Series RLC Model  In This Case  So 0 2 2  C i dt di R dt id L 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 dx at dt xd  )()()( txtxtx cp   Where • xp  Particular Solution • xc  Complementary Solution )( 2a A xAtf p  A a A axa dt xd dt dx a A x p pp p   2 22 2 2 2 0  For Any const Forcing Fcn, f(t) = A )()( 2 tx a A tx c  Docsity.com The Complementary Solution • The Complementary Solution Satisfies the HOMOGENOUS Eqn  Nomenclature • α  Damping Coefficient •   Damping Ratio • 0  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 dx at dt xd 0)()(2)( 202 2  txt dt dx t dt xd   Where • a1  2α = 20 • a2  0 2 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 dx t dt xd UnitLess)(5.0 2 1 1 222 2/2 44 0 0 1 0 2 0 2 0           SS stst Kes dt xd sKe dt dx 2 2 2 ;  0)2( or 0)()(2)( 2 0 2 2 02 2   stKess txt dt dx t dt xd   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. >1 2. <1 3. =1 02 20 2  ss 1 1 0)()( 2 002,1 2 0 2 02,1 2 0 2 22 0 2         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  ss 02 20 2  ss   02 20222  ss 02 220 22  ss     02 22022  ss     0220 2  s Docsity.com Case 2: <1 → UNDERdamped • Since <1 The Characteristic Eqn Yields COMPLEX Roots as Complex Conjugates – So with j=(-1)  Then The UnderDamped UnForced (Natural) Response Equation  Where • n  Damped natural Oscillation Frequency • α  Damping Coefficient n n jjs jjs     2 002 2 001 1 1    tAtAetx nn t c   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  sincos   Then       tKKjtKKe tjKtKtjKtKe eeKeeK eKeKtx nn t nnnn t tjttjt tjtj c ndn nn       sincos sincossincos )( 2121 2211 21 )( 2 )( 1          212 211 KKjA KKA    Sub A1 & A2 to Obtain  tAtAetx nn t c   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(  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                              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         tAtAetx tAtAetx t nn t 3sin3cos)( sincos)( 21 21      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 0 1 2 2  L v dt dv Rdt vd C 0 21 1 2 2 2  v dt dv dt vd 0 42 1 2 2  v dt dv dt vd  Recognize Parameters 2 1 2 1 2; 2 1 4 1 00   Docsity.com Parallel RLC Example cont • Then: Damping Factor, Damped Frequency 4 3 4 1 1 2 1 1 20  n 4 1 0    Then The Response Equation           tAtAetv t c 4 3 sin 4 3 cos)( 21 4  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 3 sin 3 1 4 3 cos10)( 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 di L L L t L    0or 0 1 0  If x = vC, Then Find iC 10 Xdtdx t       CiX i dt dv C 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 Numerical Example • For The Given 2nd Order Ckt Find for t>0 – io(t), vo(t) • From Ckt Diagram Recognize by Ohm’s Law  KVL at t>0  The Char Eqn & Roots   V24 )(12)(18)( 00 Vtitv  012)(18)(2)0()( 36/1 1 4 0         titdt di vdxxi t C 0)(36)(18)(2 2 2  tit dt di t dt id    6,3 :roots REAL 630 0189 : Eq. Ch. 2    s ss ss 0;)( 62 3 1   teKeKti tto  Taking d(KVL)/dt → ODE  The Solution Model KVL 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 dv Ci c t t cc c c    Rvi Riv RR RR      0 0 1 tidxxv L i dt di Lv L t t LL L L     Solve Resulting Ckt Analytical-Model using Any & All MATH Methods Docsity.com 2nd Order ODE SuperSUMMARY-1 • Find ANY Particular Solution to the ODE, xp (often a CONSTANT) • Homogenize ODE → set RHS = 0 • Assume xc = Ke st; Sub into ODE • Find Characteristic Eqn for xc  a 2nd order Polynomial dt di L v dt dv Rdt vd C S 1 2 2 dt dv C i dt di R dt id L S 2 2 Differentiating Docsity.com 2nd Order ODE SuperSUMMARY-2 • Find Roots to Char Eqn Using Quadratic Formula (or Sq-Completion) • Examine Nature of Roots to Reveal form of the Eqn for the Complementary Solution: – Real & Unequal Roots → xc = Decaying Constants – Real & Equal Roots → xc = Decaying Line – Complex Roots → xc = Decaying Sinusoid Docsity.com Complete the Square -2 • Now the Left-Hand-Side (LHS) is a PERFECT Square  Solve for x; but first let  Use the Perfect Sq Expression  Finally Find the Roots of the Quadratic Eqn a c a b a b x a cabab x a b x                           22 22 2 22 22 Fab DRHS a cab        2 2 2   DFx a c a b a b x               2 22 or 22   DFxx DFx DFx    21 2 , or Docsity.com Derive Quadratic Eqn -1 • Start with the PERFECT SQUARE Expression  Take the Square Root of Both Sides  Combine Terms inside the Radical over a Common Denom a c a b a b x              22 22 a c a b a b x        2 22 2 2 2 2 2 2 4 4 2 4 4 42 42 a acb a b x aa ac a b a b x a c a b a b x     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 b x a acb a b x 2 4 2 4 4 2 2 2 2     a acb a b x 2 4 2 2   a acbb x 2 42    Now Combine over Common Denom Docsity.com