SECOND-ORDER CIRCUITS

SECOND-ORDER CIRCUITS

THE BASIC CIRCUIT EQUATION

Single Node-pair: Use KCL

Ri Li Ci

0 CLRS iiii

)();()(1

;)(

0

0

tdt

dvCitidxxv

Li

R

tvi CL

t

tLR

SL

t

t

itdt

dvCtidxxv

LR

v )()()(

10

0

Differentiating

dt

di

L

v

dt

dv

Rdt

vdC S 12

2

Single Loop: Use KVL

Rv

Cv

Lv

0 LCRS vvvv

)();()(1

; 0

0

tdt

diLvtvdxxi

CvRiv LC

t

tCR

dt

dv

C

i

dt

diR

dt

idL S2

2

SC

t

t

vtdt

diLtvdxxi

CRi )()()(

10

0

LEARNING BY DOING

LYRESPECTIVE FOR EQUATIONAL DIFFERENTI THE WRITE ),(),( titv

0

00)(

tI

tti

SS

Si

MODEL FOR RLC PARALLEL

dt

di

L

v

dt

dv

Rdt

vdC S 12

2

0;0)( ttdt

diS

01

2

2

L

v

dt

dv

Rdt

vdC

Sv

00

0)(

t

tVtv S

S

MODEL FOR RLC SERIES

dt

dv

C

i

dt

diR

dt

idL S2

2

0;0)( ttdt

dvS

02

2

C

i

dt

diR

dt

idL

THE RESPONSE EQUATION

)()()()( 212

2

tftxatdt

dxat

dt

xd

EQUATION

THE FOR SOLUTIONS THESTUDY WE

solutionary complement

solution particular

:KNOWN

c

p

cp

x

x

txtxtx )()()(

0)()()( 212

2

txatdt

dxat

dt

xdc

cc

SATISIFES

SOLUTIONARY COMPLEMENT THE

IF THE FORCING FUNCTION IS A CONSTANT

solution particular a is 2

)(a

AxAtf p

Axadt

xd

dt

dx

a

Ax p

ppp 22

2

2

0 :PROOF

)()(

)(

2

txa

Atx

Atf

c

FUNCTION FORCING ANY FOR

0)(4)(2)(2

2

txtdt

dxt

dt

xd

LEARNING BY DOING

0)(16)(8)(4 2

2

txtdt

dxt

dt

xd

FREQUENCYNATURAL

ANDRATIO DAMPING EQUATION,

STICCHARACTERI THE DETERMINE

COEFFICIENT OF SECOND DERIVATIVEMUST BE ONE

0)(4)(2)(2

2

txtdt

dxt

dt

xd

2nn2

0422 ss

EQUATION STICCHARACTERI

DAMPING RATIO, NATURAL FREQUENCY

2 n

5.0

THE HOMOGENEOUS EQUATION

0)()()( 212

2

txatdt

dxat

dt

xd

0)()(2)( 22

2

txtdt

dxt

dt

xdnn

FORM NORMALIZED

2

11

22

2

22

a

aa

aa

n

nn

ratio damping

frequency natural (undamped)

n

02 22 nnss


ANALYSIS OF THE HOMOGENEOUS EQUATION

0)()(2)( 22

2

txtdt

dxt

dt

xdnn

FORM NORMALIZED

02

)(22

nn

st

ss

Ketx

iff solution a is

Iff s is solution of the characteristicequation

stst Kesdt

xdsKet

dt

dx 22

2

;)( :PROOF

stnnnn Kesstxt

dt

dxt

dt

xd)2()()(2)( 222

2

2

02 22 nnss


roots)distinct and (real :1 CASE 1tsts eKeKtx 21

21)( roots) conjugate(complex :2 CASE 1

d

nn

js

js

21

tAtAetx ddt sincos)( 21

tjttjst dndn eeee )( :HINTtjte dd

tj d sincos

roots) equal and (real 1 :3 CASE ns

tnetBBtx 21)(

)022()02( 22 nnn

st

sANDss

te

iff solution is :HINT

tsts eKeKtx 2121)(

frequency noscillatio dampedd

*12)( KKtx real

tj

d

deKtxjs

KK )(1

*12 Re2)(

2/)( 211 jAAK ASSUME

1

0)()(

2

222

2222

nn

nnn

nnn

s

s

s

(modes of the system)

factor damping

LEARNING EXTENSIONS

0)(4)(4)(2

2

txtdt

dxt

dt

xd

0442 ss


0)2(044 22 sss

242 nn 142 n

system 3) (case damped critically a is this

t

st

etBBtx

etBBtx2

21

21

)()(

)()(

DETERMINE THE GENERAL FORM OF THE SOLUTION

0)(16)(8)(4 2

2

txtdt

dxt

dt

xd

0)(4)(2)(2

2

txtdt

dxt

dt

xd

Divide by coefficient of second derivative

242 nn 5.022 n

system 2) (case dunderdampe

325.0121;1 2 ndn

tAtAetx

tAtAetxt

ddt

3sin3cos)(

sincos)(

21

21

3103)1(42 22 jssss

Roots are real and equal Roots are complex conjugated

LEARNING EXTENSIONS

FCHLR

RLC

2,2,1 WITH CIRCUIT PARALLEL

01

2

2

L

v

dt

dv

Rdt

vdC 02

2

C

i

dt

diR

dt

idL

042

1

02

2

2

2

2

2

v

dt

dv

dt

vd

v

dt

dv

dt

vd

016

3)4

1(

4

1

222 s

ss

2

1

4

1;2

1 nn

tAtAetv

t

c 4

3sin

4

3cos)( 21

4

4

3

4

11

2

11 2 nd

FFFCHLR

RLC

2,1,5.0,1;2 WITH CIRCUIT SERIES

HOMOGENEOUS EQUATION

022

2

C

i

dt

di

dt

id

valuesreplace& L/:

CC nn 22;1

C=0.5 underdampedC=1.0 critically dampedC=2.0 overdamped

Classify the responses forthe given values of C

Form of the solution

4

1

C

44 ntdiscrimina

THE NETWORK RESPONSE

DETERMINING THE CONSTANTS

Atxtdt

dxt

dt

xdnn )()(2)( 2

2

2

FORM NORMALIZED

tsts

n

eKeKA

tx 21212)(

212)0( KKA

xn

2211)0( KsKsdt

dx

tAtAeA

tx ddt

n

n

sincos)( 212

12)0( AA

xn

21)0( AAdt

dxdn

t

n

netBBA

tx

212)(

12)0( BA

xn

21)0( BBdt

dxn

LEARNING EXAMPLE FCHLR 51,5,2 VvAi CL 4)0(,1)0(

t

L dt

dvCidxxv

LR

v

0

0)0()(1

011

2

2

vLCdt

dv

RCdt

vd

015.22 ss


5.1;1 n

2

5.15.2

2

4)5.2(5.2 2

s

tt eKeKtv 5.02

21)(

To determine the constants we need

)0();0( dt

dvv

Vvvv CC 4)0()0()0( 0t KCL AT

0)0()0()0(

dt

dvCi

R

vL

C

5)5/1(

)1(

)5/1(2

4)0(

dt

dv

2;255.02

421

21

21

KKKK

KK

0;22)( 5.02 teetv tt

RiLi Ci

0 CLR iii

STEP 1 MODEL

STEP 2

STEP 3ROOTS

STEP 4FORM OFSOLUTION

STEP 5: FIND CONSTANTS

)0(),0( LC iv FIND GIVEN NOT IF

ANALYZECIRCUIT ATt=0+

%script6p7.m%plots the response in Example 6.7%v(t)=2exp(-2t)+2exp(-0.5t); t>0t=linspace(0,20,1000);v=2*exp(-2*t)+2*exp(-0.5*t);plot(t,v,'mo'), grid, xlabel('time(sec)'), ylabel('V(Volts)')title('RESPONSE OF OVERDAMPED PARALLEL RLC CIRCUIT')

USING MATLAB TO VISUALIZE THE RESPONSE

LEARNING EXAMPLE FCHLR 04.0,1,6 VvAi CL 4)0(;4)0(

t

CvdxxiC

tdt

diLtRi

0

0)0()(1

)()(

0)(1

)(2

2

tiLC

tdt

di

L

R

dt

id

0)(25)(62

2

titdt

di

dt

id

02562 ss :Eq. Ch.6.062

5252

n

nn

432

100366js :roots d

)4sin4cos()( 213 tAtAeti t

Aii L 4)0()0(

)0( dt

di COMPUTE TO )()( t

dt

diLtvL

)0()0()0( CvRidt

diL 20)0(

dt

di

41 A

)4cos44sin4()(3)( 213 tAtAetit

dt

di t

24)4(320:0@ 22 AAt

0];)[4sin24cos4()( 3 tAtteti t

Rv Lv

Cv0 CLR vvv

t

CC dxxiC

vtdt

diLtRitv

0

)(1

)0()()()(

0];)[4sin224cos4()( 3 tVttetv tC

NO SWITCHING OR DISCONTINUITY AT t=0. USE t=0 OR t=0+

model

Form:

USING MATLAB TO VISUALIZE THE RESPONSE%script6p8.m%displays the function i(t)=exp(-3t)(4cos(4t)-2sin(4t))% and the function vc(t)=exp(-3t)(-4cos(4t)+22sin(4t))% use a simle algorithm to estimate display timetau=1/3;tend=10*tau;t=linspace(0,tend,350);it=exp(-3*t).*(4*cos(4*t)-2*sin(4*t));vc=exp(-3*t).*(-4*cos(4*t)+22*sin(4*t));plot(t,it,'ro',t,vc,'bd'),grid,xlabel('Time(s)'),ylabel('Voltage/Current')title('CURRENT AND CAPACITOR VOLTAGE')legend('CURRENT(A)','CAPACITOR VOLTAGE(V)')

LEARNING EXAMPLE HLFCRR 2,1,8,10 21 AiVv LC 5.0)0(,1)0(

KVL

0)()()( 1 tvtiRtdt

diL

KCL

)()(

)(2

tdt

dvC

R

tvti

0)()()(

)(1

212

2

2

tvt

dt

dvC

R

tvR

dt

vdCt

dt

dv

RL

0)()(1

)(2

211

22

2

tv

LCR

RRt

dt

dv

L

R

CRt

dt

vd

0)(9)(6)(2

2

tvtdt

dvt

dt

vd 0962 ss :Eq. Ch.

162,3 nn

22 )3(096 sss :Eq. Ch.

tBBetv t21

3)(

Vvv c 1)0()0(

)0()0(

)0()0(

0

2 dt

dvC

R

vii

t

L

KCL AT

3)0( dt

dv

11)0( Bv

63)0(3)0( 22 BBvdt

dv

0;61)( 3 ttetv t

NO SWITCHING OR DISCONTINUITYAT t=0. USE t=0 OR t=0+

USING MATLAB TO VISUALIZE RESPONSE

%script6p9.m%displays the function v(t)=exp(-3t)(1+6t)tau=1/3;tend=ceil(10*tau);t=linspace(0,tend,400);vt=exp(-3*t).*(1+6*t);plot(t,vt,'rx'),grid, xlabel('Time(s)'), ylabel('Voltage(V)')title('CAPACITOR VOLTAGE')

LEARNING EXTENSION 0),( tti FIND

Once the switch opens the circuit is RLCseries

0)()0()(2)(30

t

C dxxivtdt

diti

0)(2

1)(

2

3)(2

2

titdt

dit

dt

id

5.0,1

05.05.12

s

ss

:roots

:Ch.Eq.

0;)( 221 teKeKtit

t

To find initial conditions use steady stateanalysis for t<0

dt

diCvC

AiL 2)0(

VvC 0)0(

0)0( dt

di

21

21

2

10

2

KK

KK

0;42)( 2 teetit

t

Cv

Rv

Lv

And analyze circuit at t=0+

0tat KCL Ai 2)0(

=0=2

LEARNING EXTENSION 0),(0 ttv FIND

For t>0 the circuit is RLC series

)(2)(0 titv

)(ti

KVL

0)(2)0()(3/2

1)(

2

1

0

tivdxxitdt

diC

t

0)(3)(4)(2

2

titdt

dit

dt

id

3,1

0342

s

ss

:roots

: Eq. Ch.

0;)( 321 teKeKti tt

To find initial conditions we use steadystate analysis for t<0

AiL 2)0(

0)0(Cv

And analyze circuit at t=0+

Ai 2)0( 0)0()0( dt

diCvC

030)0(

20)0(

21

21

KKdt

di

KKi

3

1

1

2

K

K

0;32)(

0;3)(3

0

3

teetv

teetitt

tt

LEARNING EXTENSION 0);(),( 00 ttvti DETERMINE )(12)(18)( 00 Vtitv

KVL

012)(18)(2)0()(36/1

14

0

titdt

divdxxi

t

C

0)(18)(9)(2

2

titdt

dit

dt

id

6,3

01892

s

ss

:roots

: Eq. Ch.

0;)( 62

310 teKeKti tt

0)0( Cv AiL 5.0)0(

Analysis at t=0+

)(5.0)0()0(0 Aii L

)0()0()0( 0 dt

diL

dt

diLv L

L )0(Lv

0)0(Cv

012)0(18)0(4 LL iv

210

210

5.0)0(

632/17)0(

KKi

KKdt

di

Steady state t<0

)0(Li

)0(CvV24

17)0( Lv

0;6

14

6

11)( 63

0 teeti tt

6

14;6

1121 KK

SecondOrder

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SECOND-ORDER CIRCUITS