1. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 1 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 43 2nd Order RLC Circuits

2. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 2 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 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. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 3 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis ReCall RLC VI Relationships ResistorCapacitor Inductor

4. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 4 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 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  The sum of all Currents entering any Circuit-Node is equal to Zero  Kirchoff’s Voltage Law  The sum of all the Voltage- Drops around any Closed Circuit-Loop is equal to Zero

5. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 5 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Second Order Circuits  Single Node-Pair By KCL By KVL  Single Loop Differentiating

6. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 6 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Second Order Circuits  Single Node-Pair By KCL Obtained By KVL Obtained  Single Loop  Make CoEfficient of 2 nd Order Term = 1 1∙(2 nd Order Term)

7. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 7 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis ODE for i L (t) in SNP  Single-Node Ckt  By KCL  Note That  Use Ohm & Cap Laws  Recall v-i Relation for Inductors  Sub Out v L in above

8. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 8 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis ODE Derivation Alternative  Take Derivative and ReArrange  Make CoEff of 2 nd Order Term = 1

9. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 9 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis  Single LOOP Ckt  Importantly  Thus the KVL eqn  Cleaning Up

10. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 10 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Illustration  Write The Differential Eqn for v(t) & i(t) Respectively  The Forcing Function  Parallel RLC Model  In This Case  So  The Forcing Function  Series RLC Model  In This Case  So

11. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 11 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 2 nd Order Response Equation  Need Solutions to the 2 nd Order ODE  As Before The Solution Should to this Linear Eqn Takes This form  If the Forcing Fcn is a Constant, A, Then Discern a Particular Soln  Verify x p  Where x p  Particular Solution x c  Complementary Solution  For Any const Forcing Fcn, f(t) = A 

12. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 12 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis The Complementary Solution  The Complementary Solution Satisfies the HOMOGENOUS Eqn  Nomenclature α  Damping Coefficient   Damping Ratio  0  Undamped (or Resonant) Frequency  Need x c So That the “0 th ”, 1 st & 2 nd Derivatives Have the same form so they will CANCEL in the Homogeneous Eqn  Look for Solution of the form  ReWrite in Std form  Where a 1  2α = 2  0 a 2   0 2

13. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 13 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Complementary Solution cont  Sub Assumed Solution (x = Ke st ) into the Homogenous Eqn  A value for “s” That SATISFIES the CHARACTERISTIC Eqn ensures that Ke st is a SOLUTION to the Homogeneous Eqn  Units Analysis  Canceling Ke st  The Above is Called the Characteristic Equation

14. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 14 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Complementary Solution cont.2  Recall Homog. Eqn.  Short Example: Given Homogenous Eqn Determine Characteristic Eqn Damping Ratio,  Natural frequency,  0  Given Homog. Eqn  Discern Units after Canceling Amps  Coefficient of 2 nd Order Term MUST be 1

15. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 15 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Complementary Solution cont.3  Example Cont.  Before Moving On, Verify that Ke st is a Solution To The Homogenous Eqn  Then  K=0 is the TRIVIAL Solution We need More

16. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 16 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Complementary Solution cont.4  If Ke st is a Solution Then Need  Solve By Completing the Square  The Solution for s Generates 3 Cases 1.  >1 2.  <1 3.  =1

17. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 17 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Aside: Completing the Square  Start with:  ReArrange:  Add Zero → 0 = y−y:  ReArrange:  Grouping The First Group is a PERFECT Square  ReWriting:

18. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 18 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Initial Conditions  Summarize the TOTAL solution for f(t) = const, A  Find K 1 and K 2 From INITIAL CONDITIONS x(0) AND (this is important) [dx/dt] t=0 ; e.g.; Two Eqns in Two Unknowns  Must Somehow find a NUMBER for

19. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 19 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Case 1:  >1 → OVERdamped  The Damped Natural Frequencies, s 1 and s 2, are REAL and UNequal  The Natural Response Described by the Relation  The TOTAL Natural Response is thus a Decaying Exponential plus a Constant

20. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 20 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Case 2:  <1 → UNDERdamped  Then The UnderDamped UnForced (Natural) Response Equation  Where  n  Damped natural Oscillation Frequency α  Damping Coefficient

21. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 21 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis UnderDamped Eqn Development  Start w/ Soln to Homogeneous Eqn  From Appendix-A; The Euler Identity  Since K 1 & K 2 are Arbitrary Constants, Replace with NEW Arbitrary Constants  Then  Sub A 1 & A 2 to Obtain 

22. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 22 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis UnderDamped IC’s  Find Under Damped Constants A 1 & A 2  Given “Zero Order” IC  With x p = D (const) then at t=0 for total solution  Now dx/dt at any t  For 1 st -Order IC  Arrive at Two Eqns in Two Unknowns But MUST have a Number for X 1

23. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 23 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Case 3:  =1 →CRITICALLY damped  The Natural Response is a Decaying Exponential against The the Equation of a LINE  Find Constants from Initial Conditions and TOTAL response  The Damped Natural Frequencies, s 1 and s 2, are REAL and EQUAL  The Natural Response Described by Relation  EXERCISE VERIFY that the Above IS a solution to the Homogenous Equation

24. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 24 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example: Case Analyses  Recast To Std Form  Characteristic Eqn  Determine The General Form Of The Solution  Factor The Char. Eqn  Real, Equal Roots → Critically Damped (C3)  Then The Undamped Frequency and Damping Ratio

25. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 25 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example: Case Analyses cont.  Then the Solution  The Roots are Complex and Unequal → an Underdamped (Case 2) System Find the Damped Parameters  For Char. Eqn Complete the Square

26. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 26 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 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  Recognize Parameters

27. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 27 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Parallel RLC Example cont  Then: Damping Factor, Damped Frequency  Then The Response Equation  If: v(0)=10 V, and dv(0)/dt = 0 V/S, Then Find:  Plot on Next Slide

28. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 28 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

29. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 29 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Determine Constants Using ICs  Standardized form of the ODE Including the FORCING FCN “A”  Case-1 → OverDamped  Case-2 → UnderDamped  Case-3 → Crit. Damping

30. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 30 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis KEY to 2 nd Order → [dx/dt] t=0+  Most Confusion in 2 nd Order Ckts comes in the from of the First-Derivative IC  If x = i L, Then Find v L  MUST Find at t=0 + v L or i C  Note that THESE Quantities CAN Change Instantaneously i C (but NOT v C ) v L (but NOT i L )  If x = v C, Then Find i C

31. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 31 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis [dx/dt] t=0+ [dx/dt] t=0+ → Find i C (0 + ) i C (0 + ) & v L (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. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 32 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis WhiteBoard → Find v O (t)

33. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 33 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

34. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 34 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

35. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 35 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

36. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 36 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

37. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 37 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

38. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 38 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

39. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 39 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Numerical Example  For The Given 2 nd Order Ckt Find for t>0 i o (t), v o (t)  From Ckt Diagram Recognize by Ohm’s Law  KVL at t>0  The Char Eqn & Roots  Taking d(KVL)/dt → ODE  The Solution Model KVL

40. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 40 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Numerical Example cont  Steady State for t<0  The Analysis at t = 0+  Then Find The Constants from ICs  Then di 0 /dt by v L = Ldi L /dt  Solving for K 1 and K 2  KVL at t=0+ (v c (0+) = 0)

41. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 41 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Numerical Example cont.2  Return to the ODE  Yields Char. Eqn Roots  Write Soln for i 0  And Recall i o & v o reln  So Finally

42. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 42 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis General Ckt Solution Strategy  Apply KCL or KVL depending on Nature of ckt (single: node-pair? loop?)  Convert between V  I using Ohm’s LawCap LawInd Law  Solve Resulting Ckt Analytical-Model using Any & All MATH Methods

43. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 43 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 2 nd Order ODE SuperSUMMARY-1  Find ANY Particular Solution to the ODE, x p (often a CONSTANT)  Homogenize ODE → set RHS = 0  Assume x c = Ke st ; Sub into ODE  Find Characteristic Eqn for x c  a 2 nd order Polynomial Differentiating

44. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 44 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 2 nd 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 → x c = Decaying Constants Real & Equal Roots → x c = Decaying Line Complex Roots → x c = Decaying Sinusoid

45. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 45 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 2 nd Order ODE SuperSUMMARY-3  Then the TOTAL Solution: x = x c + x p  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. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 46 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis All Done for Today 2 nd Order IC is Critical!

47. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 47 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Complete the Square -1  Consider the General 2 nd 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

48. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 48 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 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

49. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 49 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 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

50. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 50 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 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  Now Combine over Common Denom

51. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 51 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Complete the Square