1. Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu
Engineering 43
Chp 7 – 2nd
Order Ckts
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer

2. 2nd Order  Basic Ckt Eqn Parallel Series Single Node-Pair → KCL
Single Loop → KVL
At DC Steady-State: for Node-Pair v(t) = 0 as IND is a SHORT, for Loop i(t) = 0 as CAP is OPEN
Differentiating

3. 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

4. 2nd Order Response Equation
Need Solutions to the 2nd Order ODE
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

5. The Complementary Solution
The Complementary Solution Satisfies the HOMOGENOUS Eqn
Nomenclature
  “Damping Ratio”
0  Undamped “Natural” 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
Squiggle greek ltr => zeta
Where
a1  20
a2  02

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

7. 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 2nd Order Term MUST be 1

8. 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

9. 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
The Solution for s Generates 3 Cases
>1
<1
=1

10. Initial Conditions Summarize the Complementary solution
Find K1 and K2 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

11. Case 1: >1 → OVERdamped
The Damped Natural Frequencies, s1 and s2, are REAL and UNequal
The Natural Response Described by the Relation
Find Constants from Initial Conditions
The Response is a Decaying Exponential

12. Case 2: <1 → UNDERdamped
Since <1 The Characteristic Eqn Yields COMPLEX Roots as Complex Conjugates
So with j=(-1)
Where
d  Damped Oscillation Frequency
  Damping Factor
Then The Underdamped Response Equation

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

14. Underdamped IC’s Find Under Damped Constants A1 & A2
Given “Zero Order” IC
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 a Number for X1
For 1st-Order IC

15. 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
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

16. 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)

17. 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

18. Parallel RLC Example
Find Damping Ratio and Undamped Natural Frequency given
R =1 Ω
L = 2 H
C = 2 F
The Homogeneous Eqn from KCL
Or, In Std From
Recognize Parameters

19. 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

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

22. 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
The Char Eqn & Roots
KVL at t>0
Taking d(KVL)/dt → ODE
The Solution Model

23. Numerical Example cont
Steady State for t<0
The Analysis at t = 0+
Then Find The Constants from ICs
KVL at t=0+ (vc(0+) = 0)
Note that VOLTAGE across ind CAN change instantaneously
Then di0/dt by vL = LdiL/dt
Solving for K1 and K2

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

25. Complete the Square -1 Consider the General 2nd Order Polynomial
a.k.a; the Quadratic Eqn
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

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

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

28. Derive Quadratic Eqn -2 Note that Denom is, itself, a PERFECT SQ
Now Combine over Common Denom
But this the Renowned QUADRATIC FORMULA
Note That it was DERIVED by COMPLETING the SQUARE
Next, Isolate x
The quadratic formula was derived BEFORE Phythagorus

29. KEY to 2nd Order → [dx/dt]t=0+
Most Confusion in 2nd Order Ckts comes in the form of the First-Derivative IC
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

30. 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

31. 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

32. 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

33. 2nd Order ODE SuperSUMMARY-3
Then the Complete Solution: x = xc + xp
All Complete 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

34. WhiteBoard Work Let’s Work This Prob Some Findings
7e P6.69 => see 2nd_Order_ODE_7e_6-69_0804.ppt

35. The Single-Node Pair
Assume that the SWITCH is a BETTER Short-Circuit than the INDUCTOR for t = 0− Steady-State Analysis

36. iO,max mA
804.7 mS

38. i0,max = mA

39. Complete the Square