bruce mayer, pe registered electrical & mechanical engineer bmayer@chabotcollege

BMayer@ChabotCollege.edu • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx1

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PERegistered Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Engineering 43

2nd OrderRLC

Circuits

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

ReCall RLC VI RelationshipsResistor Capacitor

CC vduuiC

tv tiRtv RR dttdiLtv L

dttdvCti C

C tvGti RR 01

LL iduuvL

Inductor

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

Second Order Circuits Single Node-Pair

Ri Li Ci

0 CLRS iiii

)();()(1;)(0

tdtdvCitidxxv

Rtvi CL

itdtdvCtidxxv

)()()(10

• By KCL

• By KVL0 LCRS vvvv

)();()(1; 00

tdtdiLvtvdxxi

CvRiv LC

vtdtdiLtvdxxi

CRi )()()(1

Single Loop

Differentiatingdtdi

RdtvdC S 12

dtidL S2

Second Order Circuits Single Node-Pair

Ri Li Ci

• By KCL Obtained

• By KVL Obtained

Single Loop

RdtvdC S 12

dtidL S2

Make CoEfficient of 2nd Order Term = 1

1∙(2nd Order Term)dtdi

RCdtvd S112

LCdtdi

dtid S112

ODE for iL(t) in SNP Single-Node Ckt

By KCL

Note That Use Ohm & Cap

Recall v-i Relation for Inductors

Sub Out vL in above

Ri Li Ci

SCLR iiii

tvtv L

dtdvCi

dtdiLv L

ODE Derivation Alternative

Take Derivative and ReArrange

Make CoEff of 2nd Order Term = 1

Use Similar Method to vC(t) for Single LOOP Circuit

SLLL iidtdi

dtidLC 2

LCdtdi

RCdtid S

iivvvv

dtdvCi

vdtdiLvRi

ODE for vC(t) in SNP Single LOOP Ckt

Importantly

Thus the KVL eqn

Cleaning Up

vdtdvC

dtdvCR

vdtdiLvRi

dtdvCi C

C LCvv

LCdtdv

vdtvdLCv

dtdvRC

Illustration Write The Differential Eqn for v(t) & i(t) Respectively

RdtvdC S

0;0)( ttdtdiS

dtidL S2

0;0)( ttdtdvS

The Forcing Function

Parallel RLC Model

In This Case So 01

RdtvdC

The Forcing Function

Series RLC Model

In This Case So 02

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

tftxatdtdxat

)()()( txtxtx cp Where

• xp Particular Solution• xc Complementary

Solution

)(2aAxAtf p

AaAaxa

For Any const Forcing Fcn, f(t) = A

txaAtx c

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

txatdtdxat

0)()(2)( 202

txtdtdxt

Where• a1 2α = 20

• a2 02

stKetx )(

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

Unitless Also

A/SA/S;

[A] Amps)()(Let

dtiddtdi

2 ststst KesKeKes

0)()(2)( 202

txtdtdxt

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

0)()(2)( 202

titdtdit

UnitLess1

11)(1secradians/1

1)(1/11

SameUnits

0)(16)(8)(4 2

txtdtdxt

Coefficient of 2nd Order Term MUST be 1

0)(4)(2)(2

txtdtdxt

Complementary Solution cont.3 Example Cont. Before Moving On,

Verify that Kest is a Solution To The Homogenous Eqn

K=0 is the TRIVIAL Solution• We need More

0)(4)(2)(

txtdtdxt

UnitLess)(5.021

stst KesdtxdsKe

dtdx 2

0)()(2)(

stKess

txtdtdxt

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 Cases1. >12. <13. =1

2002,1

Aside: Completing the Square Start with: ReArrange: Add Zero → 0 = y−y:

ReArrange: Grouping• The First Group is a PERFECT Square

ReWriting:

222 ss

02 220

Initial Conditions Summarize the TOTAL solution for f(t) = const, A

Find K1 and K2 From INITIAL CONDITIONS x(0) AND (this is important) [dx/dt]t=0; e.g.;

21)()()( AeKeKtxtxtx tstspcTOT

)0()(and

then)( 21

KsKsdt

dxdttdx

AeKeKtx

Two Eqns in Two Unknowns

Must Somehow find a NUMBER for 0

tdttdx

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

tstsc eKeKtx 21

txtxtx

202,1 and as s

Case 2: <1 → UNDERdamped

Then The UnderDamped UnForced (Natural) Response Equation

Where• n Damped natural

Oscillation Frequency• α Damping

Coefficient

Since <1 The Characteristic Eqn Yields COMPLEX Roots as Complex Conjugates• So with j=(-1)

tAtAetx nnt

c sincos 21

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

tstsc eKeKtx 21

tjte nntj n sincos

tKKjtKKe

tjKtKtjKtKe

eeKeeK

eKeKtx

tjttjt

sincos

sincossincos

KKjAKKA

Sub A1 & A2 to Obtain

tAtAetx nnt

c sincos)( 21

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

0sin0cos

10Xdtdx

0sin0cos

sincos

AAXdtdx

tAAtAA

Arrive at Two Eqns in Two Unknowns• But MUST have a

Number for X1

Case 3: =1 →CRITICALLY damped

The Natural Response is a Decaying Exponential against The Sum of a CONSTANT and a LINEAR Term

Find Constants from Initial Conditions and TOTAL response

The Damped Natural Frequencies, s1 and s2, are REAL and EQUAL

The Natural Response Described by Relation

tttc etBBteBeBtx 2121

EXERCISE• VERIFY that the Above

IS a solution to the Homogenous Equation

Example: Case Analyses

Recast To Std Form Characteristic Eqn

Determine The General Form Of The Solution

Factor The Char. Eqn

0)(4)(4)()1( 2

txtdtdxt

0442 ss

0)2(044 22 sss

24 020

1 & 2422 0

etBBtx

etBBtx2

Real, Equal Roots → Critically Damped (C3)

0)(16)(8)(4)2( 2

txtdtdxt

0)(4)(2)(2

txtdtdxt

Then The Undamped Frequency and Damping Ratio

20 24 S

5.022 0

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

325.0121

212224

tAtAetx

tAtAetxt

3sin3cos)(

sincos)(

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

RdtvdC

Recognize Parameters

Parallel RLC Example cont Then: Damping Factor,

Damped Frequency

Then The Response Equation

tAtAetv

c 43sin

43cos)( 21

If: v(0)=10 V, and dv(0)/dt = 0 V/S, Then Find:31010 21 AA

Plot on Next Slide

c 43sin

43cos10)( 4

Underdamped Parallel RLC Circuit Response

0 2 4 6 8 10 12 14 16 18 20

Time (s)

Envelope (top)v(t) (V)Envelope (bot)

file = Engr44_Lec_06-1_Last_example_Fall03..xls

c 43sin

43cosV10)( 4

Determine Constants Using ICs Standardized form of

the ODE Including the FORCING FCN “A”

Case-1 → OverDamped

Case-2 → UnderDamped

Atxtdtdxt

)()(2)( 202

tsts eKeKAtx 21212

2211)0( KsKsdtdx

)0( KKAx

tAtAeAtx nnt

sincos)( 212

)0( AAx

21)0( AAdtdx

Case-3 → Crit. Damping

tetBBAtx

)0( BAx

21)0( BBdtdx

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)

vdtdiL

If x = vC, Then Find iC

10Xdtdx

idtdvC

[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

0tdtdv

0tdtdi

WhiteBoard → Find vO(t)

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

)(12)(18)( 00 Vtitv

012)(18)(2)0()(36/114

dtdivdxxi

0)(36)(18)(2 2

titdtdit

6,3 :roots REAL

6300189 : Eq. Ch. 2

0;)( 62

31 teKeKti tt

Taking d(KVL)/dt → ODE The Solution Model

Numerical Example cont Steady State for t<0

The Analysis at t = 0+

Then Find The Constants from ICs

Then di0/dt by vL = LdiL/dt Solving for K1 and K2

)0(CvV24

0)0( Cv AVViL 5.01861224)0(

)(5.0)0()0( Aii Lo

)0()0()0( dtdiL

dtdiLv oL

012)0(18)0(4 LL iv VAvL 175.0181240

KVL at t=0+ (vc(0+) = 0)

HVdtdio 2/17)0(

A 5.0)0(

217)0(

KKdtdi

1121 KK

Numerical Example cont.2 Return to the ODE

Yields Char. Eqn Roots

63 21 ss

0)(18)(9)(2

titdtdit

Write Soln for i0

11)( 63 teeti tto

And Recall io & vo reln

)(12)(18)( Vtitv oo

0;124233)( 63 tVeetv tto

0;)( 62

31 teKeKti tt

6300189 2

So Finally

General Ckt Solution Strategy Apply KCL or KVL depending on Nature

of ckt (single: node-pair? loop?) Convert between VI using• Ohm’s Law • Cap Law • Ind Law

1 tvdxxiC

dtdvCi

1 tidxxvL

dtdiLv

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

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

RdtvdC S

dtidL S2

Differentiating

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

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

xtAtAexxbmtexxeKeKx

sincos 21

All Done for Today

2nd OrderIC is

Critical!

0CaseSeries

L vdtdiL

0CaseParallel

C idtdvC

Complete the Square -1 Consider the General

2nd 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

02 cbxax

cbxax 2

acababx

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

acababx

DRHSacab

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

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 theSQUARE

aacbbx

Now Combine over Common Denom

Complete the Square

bruce mayer, pe registered electrical & mechanical engineer bmayer@chabotcollege

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