Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege

1. Engineering 43 2nd OrderRLC Circuits Bruce Mayer, PE Registered Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. 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

3. ReCall RLC VI Relationships Inductor Resistor Capacitor

4. Transient Response • The VI Relations can be Combined with KCL and/or KVL to solve for the Transient (time-varying) Response of RL, RC, and RLC circuits • Kirchoff’s Current Law • Kirchoff’s Voltage Law • The sum of all Currents entering any Circuit-Node is equal to Zero • The sum of all the Voltage-Drops around any Closed Circuit-Loop is equal to Zero

5. Second Order Circuits • Single Node-Pair • Single Loop • By KVL • By KCL Differentiating

6. Second Order Circuits • Single Node-Pair • Single Loop • By KVL Obtained • By KCL Obtained • Make CoEfficient of 2nd Order Term = 1 1∙(2nd Order Term)

7. ODE for iL(t) in SNP • Single-Node Ckt • By KCL • Note That • Use Ohm & Cap Laws • Recall v-iRelation for Inductors • Sub Out vL in above

8. ODE Derivation Alternative • Take Derivative and ReArrange • Make CoEff of 2nd Order Term = 1 • Use Similar Method to vC(t) for Single LOOP Circuit • Then

9. ODE for vC(t) in SNP • Single LOOP Ckt • Importantly • Thus the KVL eqn • Cleaning Up

10. Illustration • Write The Differential Eqn for v(t) & i(t) Respectively • The Forcing Function • The Forcing Function • Parallel RLC Model • Series RLC Model • In This Case • In This Case • So • So

11. Need Solutions to the 2nd Order ODE 2nd Order Response Equation • If the Forcing Fcn is a Constant, A, Then Discern a Particular Soln • As Before The Solution Should Take This form • Verify xp  • Where • xp Particular Solution • xc  Complementary Solution • For Any const Forcing Fcn, f(t) = A

12. The Complementary Solution Satisfies the HOMOGENOUS Eqn The Complementary Solution • 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 • Where • a1 2α = 20 • a2  02

13. Sub Assumed Solution (x = Kest) into the Homogenous Eqn Complementary Solution cont • A value for “s” That SATISFIES the CHARACTERISTIC Eqn ensures that Kestis a SOLUTION to the Homogeneous Eqn • Units Analysis • Canceling Kest • The Above is Called the Characteristic Equation

14. Recall Homog. Eqn. Complementary Solution cont.2 • Short Example: Given Homogenous Eqn Determine • Characteristic Eqn • Damping Ratio,  • Natural frequency, 0 • Given Homog. Eqn • Discern Units after Canceling Amps • Coefficient of 2nd Order Term MUST be 1

15. Example Cont. Complementary Solution cont.3 • Before Moving On, Verify that Kest is a Solution To The Homogenous Eqn • Then • K=0 is the TRIVIAL Solution • We need More

16. If Kest is a Solution Then Need Complementary Solution cont.4 • 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 • =1

17. Aside: Completing the Square • Start with: • ReArrange: • Add Zero → 0 = y−y: • ReArrange: • Grouping • The First Group is a PERFECT Square • ReWriting:

18. Summarize the TOTAL solution for f(t) = const, A Two Eqns in Two Unknowns Initial Conditions • Find K1 and K2 From INITIAL CONDITIONS x(0) AND (this is important) [dx/dt]t=0; e.g.; • Must Somehow find a NUMBER for

19. Case 1: >1 → OVERdamped • The Damped Natural Frequencies, s1 and s2, are REAL and UNequal • The Natural Response Described by the Relation • The TOTAL Natural Response is thus a Decaying Exponential plus a Constant

20. Case 2: <1 → UNDERdamped • Since <1 The Characteristic Eqn Yields COMPLEX Roots as Complex Conjugates • So with j=(-1) • Where • n Damped natural Oscillation Frequency • α Damping Coefficient • Then The UnderDampedUnForced (Natural) Response Equation

21. Start w/ Soln to Homogeneous Eqn UnderDampedEqn Development • Since K1 & K2 are Arbitrary Constants, Replace with NEW Arbitrary Constants • From Appendix-A; The Euler Identity • Sub A1 & A2 to Obtain • Then 

22. Find Under Damped Constants A1 & A2 Given “Zero Order” IC UnderDamped IC’s • Now dx/dt at any t • With xp = D (const) then at t=0 for total solution • Arrive at Two Eqns in Two Unknowns • But MUST have aNumber for X1 • For 1st-Order IC

23. Case 3: =1 →CRITICALLY damped • The Damped Natural Frequencies, s1 and s2, are REAL and EQUAL • The Natural Response Described by Relation • Find Constants from Initial Conditions and TOTAL response • The Natural Response is a Decaying Exponential against The Sum of a CONSTANT and a LINEAR Term • EXERCISE • VERIFY that the Above IS a solution to the Homogenous Equation

24. Example: Case Analyses • Determine The General Form Of The Solution • Characteristic Eqn • Recast To Std Form • Then The Undamped Frequency and Damping Ratio • Factor The Char. Eqn • Real, Equal Roots → Critically Damped (C3)

25. 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

26. 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) UnderDamped Parallel RLC Exmpl • Or, In Std From • Recognize Parameters

27. Then: Damping Factor, Damped Frequency Parallel RLC Example cont • Then The Response Equation • If: v(0)=10 V, and dv(0)/dt = 0 V/S, Then Find: • Plot on Next Slide

29. Standardized form of the ODE Including the FORCING FCN “A” Determine Constants Using ICs • Case-2 → UnderDamped • Case-1 → OverDamped • Case-3 → Crit. Damping

30. Most Confusion in 2nd Order Ckts comes in the from of the First-Derivative IC KEY to 2nd Order → [dx/dt]t=0+ • If x = vC, Then Find iC • MUST Find at t=0+ vL or iC • Note that THESE Quantities CAN Change Instantaneously • iC (but NOT vC) • vL (but NOT iL) • If x = iL, Then Find vL

31. [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

32. WhiteBoard → Find vO(t)

39. For The Given 2nd Order Ckt Find for t>0 io(t), vo(t) From Ckt Diagram Recognize by Ohm’s Law KVL Numerical Example • The Char Eqn & Roots • KVL at t>0 • Taking d(KVL)/dt → ODE • The Solution Model

40. Steady State for t<0 Numerical Example cont • The Analysis at t = 0+ • Then Find The Constants from ICs • KVL at t=0+ (vc(0+) = 0) • Then di0/dt by vL = LdiL/dt • Solving for K1 and K2

41. Return to the ODE Numerical Example cont.2 • Yields Char. Eqn Roots • And Recall io & vo reln • Write Soln for i0 • So Finally

42. 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 • Solve Resulting Ckt Analytical-Model using Any & All MATH Methods

43. 2nd Order ODE SuperSUMMARY-1 • Find ANY Particular Solution to the ODE, xp (often a CONSTANT) • Homogenize ODE → set RHS = 0 • Assume xc = Kest; Sub into ODE • Find Characteristic Eqn for xc a 2nd order Polynomial Differentiating

44. 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

45. 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

46. All Done for Today 2nd OrderIC is Critical!

47. Consider the General 2nd Order Polynomial a.k.a; the Quadratic Eqn Complete the Square -1 • Next, Divide by “a” to give the second order term the coefficient of 1 • Where a, b, c are CONSTANTS • Solve This Eqn for x by Completing the Square • First; isolate the Terms involving x • Now add to both Sides of the eqn a “quadratic supplement” of (b/2a)2

48. Now the Left-Hand-Side (LHS) is a PERFECT Square Complete the Square -2 • Use the Perfect Sq Expression • Finally Find the Roots of the Quadratic Eqn • Solve for x; but first let

49. Start with the PERFECT SQUARE Expression Derive Quadratic Eqn -1 • Combine Terms inside the Radical over a Common Denom • Take the Square Root of Both Sides

50. Note that Denom is, itself, a PERFECT SQ Derive Quadratic Eqn -2 • Now Combine over Common Denom • But this the Renowned QUADRATIC FORMULA • Note That it was DERIVED by COMPLETING theSQUARE • Next, Isolate x