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

[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx1

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

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

01

0C

t

CC vduuiC

tv tiRtv RR dttdiLtv L

L

dttdvCti C

C tvGti RR 01

0L

t

LL iduuvL

ti

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

N

kk ti

1

0

N

kk tv

1

0

http://en.wikipedia.org/wiki/File:KCL.png

http://en.wikipedia.org/wiki/File:KVL.png



Second Order Circuits Single Node-Pair

Ri Li Ci

0 CLRS iiii

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

0

tdtdvCitidxxv

Li

Rtvi CL

t

tLR

SL

t

t

itdtdvCtidxxv

LRv

)()()(10

0

• By KCL

Rv

Cv

Lv

• By KVL0 LCRS vvvv

)();()(1; 00

tdtdiLvtvdxxi

CvRiv LC

t

tCR

SC

t

t

vtdtdiLtvdxxi

CRi )()()(1

00

Single Loop

Differentiatingdtdi

Lv

dtdv

RdtvdC S 12

2

dtdv

Ci

dtdiR

dtidL S2

2



Second Order Circuits Single Node-Pair

Ri Li Ci

• By KCL Obtained

Rv

Cv

Lv

• By KVL Obtained

Single Loop

dtdi

Lv

dtdv

RdtvdC S 12

2

dtdv

Ci

dtdiR

dtidL S2

2

Make CoEfficient of 2nd Order Term = 1

1∙(2nd Order Term)dtdi

CLCv

dtdv

RCdtvd S112

2

dtdv

Li

LCdtdi

LR

dtid S112

2



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

By KCL

Note That Use Ohm & Cap

Laws

Recall v-i Relation for Inductors

Sub Out vL in above

Ri Li Ci

SCLR iiii

tvtv L

SL

LL i

dtdvCi

Rv

dtdiLv L

L

SL

LL i

dtdiL

dtdCi

dtdiL

R

1



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

SL

LL i

dtdiL

dtdCi

dtdiL

R

1

SLLL iidtdi

RL

dtidLC 2

2

LCii

LCdtdi

RCdtid S

LLL

112

2

C

SLCR

iivvvv

and

dtdvCi

vdtdiLvRi

CC

SC

CC

but



ODE for vC(t) in SNP Single LOOP Ckt

Importantly

Thus the KVL eqn

Cleaning Up

Rv

Cv

LvS

CC

C

SC

CC

vdtdvC

dtdLv

dtdvCR

vdtdiLvRi

dtdvCi C

C LCvv

LCdtdv

LR

dtvd

vdtvdLCv

dtdvRC

SC

CC

SC

CC

12

2

2

2



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

000

)(tIt

tiS

S

dtdi

Lv

dtdv

RdtvdC S

12

2

0;0)( ttdtdiS

000

)(ttV

tv SS

dtdv

Ci

dtdiR

dtidL S2

2

0;0)( ttdtdvS

Si

Sv

The Forcing Function

Parallel RLC Model

In This Case So 01

2

2

Lv

dtdv

RdtvdC

The Forcing Function

Series RLC Model

In This Case So 02

2

Ci

dtdiR

dtidL



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

2

tftxatdtdxat

dtxd

)()()( txtxtx cp Where

• xp Particular Solution• xc Complementary

Solution

)(2aAxAtf p

AaAaxa

dtxd

dtdx

aAx

p

ppp

222

2

2

2

0

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

)()(2

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

2

txatdtdxat

dtxd

0)()(2)( 202

2

txtdtdxt

dtxd

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

1-

222

S

Unitless Also

A/SA/S;

[A] Amps)()(Let

s

ste

dtiddtdi

titx

st

02 20

2 ststst KesKeKes

0)()(2)( 202

2

txtdtdxt

dtxd

02 20

2 ss



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

2

titdtdit

dtid

UnitLess1

11)(1secradians/1

1)(1/11

1

002

10

220

202

1

SS

dtdtt

dt

SS

Stdt

SSsst

SameUnits

SameUnits

SameUnits

0)(16)(8)(4 2

2

txtdtdxt

dtxd

Coefficient of 2nd Order Term MUST be 1

0)(4)(2)(2

2

txtdtdxt

dtxd



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

042

0)(4)(2)(

2

2

2

ss

txtdtdxt

dtxd

UnitLess)(5.021

1222

2/2

44

00

10

20

20

SS

stst KesdtxdsKe

dtdx 2

2

2

;

0)2(

or

0)()(2)(

20

2

202

2

stKess

txtdtdxt

dtxd



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

02 20

2 ss

1

1

0)()(

2002,1

20

202,1

20

2

220

2

s

s

s

s



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

ReArrange: Grouping• The First Group is a PERFECT Square

ReWriting:

02 20

2 ss

02 20

2 ss

02 20

222 ss

02 220

22 ss

02 220

22 ss

0220

2 s



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.;

2021

21)()()( AeKeKtxtxtx tstspcTOT

22110

2021

2021

)0()(and

)0(

then)( 21

KsKsdt

dxdttdx

AKKx

AeKeKtx

t

tsts

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

21)(

20

1

2

1

1

200

200

)()(

AeKeK

txtxtx

tt

pcTOT

020

20

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)

n

n

jjs

jjs

2

002

2001

1

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

21)(

tjte nntj n sincos

Then

tKKjtKKe

tjKtKtjKtKe

eeKeeK

eKeKtx

nnt

nnnnt

tjttjt

tjtjc

ndn

nn

sincos

sincossincos

)(

2121

2211

21

)(2

)(1

212

211

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

DXA

DAAe

Xx

nn

01

210

0

0sin0cos

)0(

10Xdtdx

t

121

21

120

21

12

)0(

0sin0cos

)0(

sincos

AAXdtdx

AAAA

edtdx

tAAtAA

edtdx

n

nd

nn

nn

nnt

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

00)(

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

2

txtdtdxt

dtxd

0442 ss

0)2(044 22 sss

24 020

1 & 2422 0

t

st

etBBtx

etBBtx2

21

21

)()(

)()(

Real, Equal Roots → Critically Damped (C3)

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

2

txtdtdxt

dtxd

0)(4)(2)(2

2

txtdtdxt

dtxd

Then The Undamped Frequency and Damping Ratio

10

20 24 S

122

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

31

3)1(

03)1(

31242

2

2

22

js

s

s

ssss

1220

120

10

0

0

314

325.0121

122

212224

S

S

S

n

n

tAtAetx

tAtAetxt

nnt

3sin3cos)(

sincos)(

21

21



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

012

2

Lv

dtdv

RdtvdC

021

12 2

2

v

dtdv

dtvd

042

12

2

v

dtdv

dtvd

Recognize Parameters

21

212;

21

41

00



Parallel RLC Example cont Then: Damping Factor,

Damped Frequency

43

411

211 2

0 n

41

0

Then The Response Equation

tAtAetv

t

c 43sin

43cos)( 21

4

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

Plot on Next Slide

ttetv

t

c 43sin

31

43cos10)( 4



Underdamped Parallel RLC Circuit Response

-10.0

-7.5

-5.0

-2.5

0.0

2.5

5.0

7.5

10.0

0 2 4 6 8 10 12 14 16 18 20

Time (s)

vO (V

)

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

file = Engr44_Lec_06-1_Last_example_Fall03..xls

ttetv

t

c 43sin

31

43cosV10)( 4



Determine Constants Using ICs Standardized form of

the ODE Including the FORCING FCN “A”

Case-1 → OverDamped

Case-2 → UnderDamped

Atxtdtdxt

dtxd

)()(2)( 202

2

tsts eKeKAtx 21212

0

)(

2211)0( KsKsdtdx

2120

)0( KKAx

tAtAeAtx nnt

sincos)( 212

0

120

)0( AAx

21)0( AAdtdx

n

Case-3 → Crit. Damping

tetBBAtx

212

0

)(

120

)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)

LvX

vdtdiL

L

Lt

L

0or

0

1

0

If x = vC, Then Find iC

10Xdtdx

t

CiX

idtdvC

C

Ct

C

0or

0

1

0



[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

Ci

dtdv

t

0

0

0tdtdi

L

vdtdi

t

0

0



WhiteBoard → Find vO(t)

V24



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

V24

)(12)(18)( 00 Vtitv

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

0

tit

dtdivdxxi

t

C

0)(36)(18)(2 2

2

titdtdit

dtid

6,3 :roots REAL

6300189 : Eq. Ch. 2

sss

ss

0;)( 62

31 teKeKti tt

o

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

KVL



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(Li

)0(CvV24

0)0( Cv AVViL 5.01861224)0(

)(5.0)0()0( Aii Lo

)0()0()0( dtdiL

dtdiLv oL

L

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

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

HVdtdio 2/17)0(

21

21

A 5.0)0(

63SA

217)0(

KKi

KKdtdi

o

o

A6

14;A6

1121 KK



Numerical Example cont.2 Return to the ODE

Yields Char. Eqn Roots

V24

63 21 ss

0)(18)(9)(2

2

titdtdit

dtid

Write Soln for i0

0;6

146

11)( 63 teeti tto

And Recall io & vo reln

)(12)(18)( Vtitv oo

0;124233)( 63 tVeetv tto

0;)( 62

31 teKeKti tt

o

6300189 2

ssss

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

00

1 tvdxxiC

v

dtdvCi

c

t

t cc

cc

Rvi

Riv

RR

RR

00

1 tidxxvL

i

dtdiLv

L

t

t LL

LL

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

dtdi

Lv

dtdv

RdtvdC S

12

2

dtdv

Ci

dtdiR

dtidL S2

2

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

pnnt

pst

ptsts

xtAtAexxbmtexxeKeKx

sincos 21

2121



All Done for Today

2nd OrderIC is

Critical!

0CaseSeries

0L

t

L vdtdiL

0CaseParallel

0C

t

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

acx

abx 2

acababx

abx

222

22



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

ac

ab

abx

acababx

abx

22

222

22

22

Fab

DRHSacab

22

2

DFx

ac

ab

abx

2

22

or

22

DFxx

DFx

DFx

21

2

,

or



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

ac

ab

abx

22

22

ac

ab

abx

2

22 2

2

2

2

2

2

44

2

44

42

42

aacb

abx

aaac

ab

abx

ac

ab

abx



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

aacb

abx

aacb

abx

24

2

44

22

2

2

aacb

abx

24

2

2

aacbbx

242

Now Combine over Common Denom



Complete the Square

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

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