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 Ckts & MATLAB

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

3. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 3 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Second Order Circuits  Single Node-Pair By KCL By KVL  Single Loop Differentiating

4. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 4 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 form 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

5. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 5 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Hand Example Simple No.s  Solve this Equation for i o  Do on WhiteBoard, Plot with MSExcel  Some Findings

6. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 6 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis The Single-Node Pair  Assume that the SWITCH is a BETTER Short-Circuit than the INDUCTOR for the t=0 − Steady-State Analysis  Start WhiteBoard Work  Start WhiteBoard Work using Above as Template

7. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 7 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

8. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 8 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

9. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 9 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

10. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 10 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

11. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 11 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

12. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 12 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

13. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 13 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

14. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 14 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 804.7 mS 133.75 mA i O,max

15. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 15 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

16. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 16 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis i 0,max = 133.7 mA

17. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 17 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis MATLAB Example Real No.s  Solve this Equation for i o  The Parameter values Again Assume that the Switch is a Better DC-Short than the Inductor R 1 = 5.6 kΩR 2 = 8.2 kΩK = 50mA L = 10 mHC = 22 nFα = 2/s

18. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 18 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

19. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 19 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

20. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 20 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

21. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 21 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

22. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 22 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

23. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 23 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

24. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 24 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Solve Using MATLAB MuPAD  See MATLAB file: E43_Chp4_2nd_Order_D epSrc_Parrallel_LCR_Ex ample_1107.mn

25. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 25 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis The Answer for i o (t)

26. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 26 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

27. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 27 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

28. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 28 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

29. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 29 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 2 nd Order ODE SuperSUMMARY-3  Then the Complete 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

30. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 30 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis All Done for Today 2 nd Order IC is Critical!

31. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 31 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis WhiteBoard Work  Let’s Work This Prob Some Findings

32. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 32 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

33. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 33 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

34. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 34 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

35. BMayer@ChabotCollege.edu ENGR-43_Lec-04b_2nd_Order_Ckts.pptx 35 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

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