Download pdf - Elec 3202 Chap 7

UAE University Department of Electrical Engineering Dr.Hazem N.Nounou

Chapter (7)Second Order Transient Circuits

•Here , we consider circuits with both capacitor , inductor , andresistors (RLC)•We expect the circuit to be descried by a second order differential equation.•Consider the parallel RLC circuit

iS (t)

iR (t) iL (t) iC (t)

L CR

+

-

VC (t)


Take first derivative

( ) (t)idt

(t)dvCdττv

L1)(ti

R(t)v

sc

t

tc0L

c

0

=+++∴ ∫

dt(t)di

dt(t)vd

C(t)vL1

dt(t)dv

R1 s

2c

2

cc =++

Let’s assume that the voltage across the capacitor is VC(t)

(t)i(t)i(t)i(t)i CLRS ++=KCL :


dt(t)di

C1(t)v

LC1

dt(t)dv

CR1

dt(t)vd s

cc

2c

2

=++

0dt

(t)diS =

0(t)vLC1

dt(t)dv

CR1

dt(t)vd

cc

2c

2

=++

If IS(t) = constant (DC)

Consider the series RLC :

VS(t)

+-

R L

Ci


KVL :

( ) (t)VdττiC1)(tV

dtdi(t)Li(t)R

0V(t)V(t)V(t)V

S

t

t0C

CLRS

0

=+++

=+++−

∫

dt(t)dV

(t)iC1

dti(t)dL

dtdi(t)R S

2

2

=++

0dt

(t)dVS =

Take first derivative :

If VS = constant

0(t)iCL

1dt

di(t)LR

dti(t)d

2

2

=++ Series RLC


Hence, let’s assume that the differential equation we wish to solve is

A(t)xadt

dx(t)adt

(t)d212

X2

=++

It is known that the solution x (t) can be expressed as

(t)x(t)xx(t) CP +=

xP(t) : particular ( forced solution )xC(t) : complementary ( natural solution)

To find xP(t)

The only solution is the constant xP(t) = k0


( ) ( )

2P

20

020102

2

aA(t)x

aAk

Akakdtdak

dtd

=⇒

=

=++

0(t)xadt

(t)dxadt

(t)dc2

c12

cX2

=++

To find xC(t) : ( natural solution )

For simplicity let’s define :

202

01

a

ξ2a

ω

ω

=

=


Where :ξ : damping ratio.w0 : underdamped natural frequency

0(t)xdt

(t)dxξ2dt

(t)dc

20

c02

cX2

=++ ωω

Assume xc(t) = k e s t

[ ] [ ] [ ]

[ ] 0sξ2sek

0ekeskξ2esk

0ekekdtdξ2ek

dtd

200

2ts

ts20

ts0

ts2

ts20

ts0

ts2

2

=++

=++

=++

ωω

ωω

ωω


k : non-zeroest : non-zero

0sξ2s 200

2 =++ ωω

1ξξs2

4ξ4ξ2s

200

20

20

20

−±−=

−±−=

ωω

ωωω

Characteristic polynomial

The roots :

1ξξs

1ξξs2

002

2001

−−−=

−+−=

ωω

ωω Complex frequencies[ rad / sec]


ts2

ts1c

21 ekek(t)x +=⇒

ts2

ts10

CP

21 ekekkx(t)

(t)x(t)xx(t)

++=

+=Q

210 kkkx(0) ++=

k1 , k2 can be found from initial conditions

From x(0):

From dx(0) / dt:

2211

ts22

ts11

skskdt

dx(0)

eskeskdt

dx(t)21

+=

+=


Note :There are 3 types of responses based on the values of (S)

1. Overdamped response ( ξ > 1)

S1 and S2 are real and distinct .

20

ts2

ts10

aAk

ekekkx(t) 21

=

++=

k1 and k2 can be found from initial conditions

2211

210

skskdt

dx(0)kkkx(0)

+=

++=


2. Underdamped response ( ξ < 1 )

S1 & S2 are complex conjugate

2001,2

2001,2

2001,2

ξ-1ωjωξS

1ξ-1ωωξS

1ξωωξS

m

m

m

−=

−−=⇒

−−=⇒

−≡

≡2

0d

0

ξ1ωω

ωξσLet Damped radian frequency

[ rad / sec ]

d1,2 ωjσS m−=⇒


( ) ( )

( )( ) ( )ΘsinjΘcose

ekekek

ekekk

ekekkx(t)

Θj

tωj2

tωj1

tσ0

tωjσ2

tωjσ10

ts2

ts10

dd

dd

21

±=

++=

++=

++=

±

−−

+−−−

Q

( ) ( )( ) ( ) ( )( )[ ]( ) ( ) ( )[ ]tωsinkjkjtωcos)k(kek

ωsinωcoskωsinωcoskekx(t)

d21d21tσ

0

dd2dd1tσ

0

−+++=

−+++=−

− tjttjt

( )212

211

kkjAkkALet−=

+=


( ) ( )

20

dtσ

2dtσ

10

aAk

tωsinetωcoseAkx(t)

=

++= −− A

A1 & A2 can be found from initial conditions :

( ) ( )

( ) ( )

2d1

tσ-2dd

tσd2

tσ1

tσddd

tσ1

10

AωAσdt

dx(0)σeAtωsinωetωcosA

eσAetωcosωtωsineAdt

dx(0)Akx(0)

+−=

−+

−=

+=

−

−−


+−=

+=

=

⇒

2d1

10

20

AωAσdt

dx(0)Akx(0)

aAk

021

2001,2

ωξSS1ξωωξSSince

−==⇒

−−= m

3. Critical damped response ( ξ = 1 )

*0(t)xαdt

(t)dxα2dt

(t)xd

ωξαLet

0(t)xωdt

(t)dxωξ2dt

(t)xd

c2

0c

2c

20

c20

c02

c2

=++⇒

≡

=++⇒


Which means thattα

2tα

1c ekek(t)x −− +=

tα2

tα1c ekek(t)x −− += t

It is impossible for this solution to satisfy two initial conditions :

It can be shown that the following solution also satisfies the differential equation:

tα2

tα10

cP

etkekkx(t)

(t)x(t)xx(t)−− ++=

+=⇒


21

tα2

tα2

tα1

10

20

kαkdt

dx(0)

ekeαtkeαkdt

dx(t)kkx(0)

aAk

+−=

+−−=

+=

=

−−−

+−=

+=

=

⇒

21

10

20

kαkdt

dx(0)kkx(0)

aAk


V (0)

+

-

RLC

Example :

A3(0)i,V12v(0)AssumeF1C

H0.5L

Ω31R

L ====

=

Find v (t) ?


0CL

1sCR

1s

0(t)vLC1

dt(t)dv

CR1

dt(t)vd

2

cc

2c

2

=

++

=++

Characteristic polynomial is :

11.0622

3ω23ξ

3ωξ22ω2ω

0CL

1SCR

1S

0

0

02

0

2

>===

==⇒=

=++

Overdamped

From before we know


A363

112(0)i12v(0)Since

(1)kk12v(0)ekekv(t)

ekekv(t)

2s,1s

R

21

t2-2

t-1

tS2

tS1

21

21

==⇒=

+==+=

+=

−=−=

LL

39dt

dv(0)

0dt

dv(0)139

0dt

dv(0)C336

0(0)i(0)i(0)i CLR

−=

=+

=++

=++


t2t2

1

21

21

t22

t1

e27e15-v(t)

27k15-k

(2)39k2k

39k2kdt

dv(0)

ek2ekdt

dv(t)

−−

−−

+=

==

=+

−=−−=

−−=

KK


24 V

+-

RL

Ct = 0

F4.0CH0.1LΩ802R

µ===

Find vc (t) , t > 0 ?

Example :

If vc(0) = 0iL(0) = 0


KVL :

( )dττiC1(0)Vi(t)R

dtdi(t)L24

V(t)V(t)V24t

tC

CRL

0

∫+++=

++=

0(t)i10*25dt

di(t)2800dt

i(t)d

0(t)iCL

1dt

di(t)LR

dti(t)d

(t)iC1

dtdi(t)R

dti(t)dL0

62

2

2

2

2

2

=++

=++

++=


Char. Poly. is

128.0100002800

ω228008002ωξ2

rad/sec0005ω10*52ω

010*52s8002s

00

062

0

62

<===⇒=

=⇒=

=++

ξ

Underdamped

4800j1400ωjσ

ξ-1wjwξs

d

20021,

m

m

m

−=−=

−=


( ) ( )0k

tωsineAtωcoseAki(t)

0

dtσ

2dtσ

10

=++= −−

( ) ( )t4800sinet4800coseAi(t) t14002

t14001

−− += A

A1 & A2 can be found from initial conditions

420R(0)i(0)V42(0)V(0)V(0)V

L

CRL

=++=++

i(0) = A1 = 0

Since homogenous


t)sin(4800e0.05i(t)

0.05AA4800Aω402

4800ω

AωAσ402dt

di(0)

42dt

di(0)L(0)v

t1400-2

22d

d

2d1

L

=

===⇒

=

+−==

==

dtdi(t)0.1i(t)28042(t)v

(t)Vi(t)Rdt

di(t)L24

V(t)V(t)V24

c

C

CRL

−−=

++=

++=From KVL


( )[ ]( )

( ) ( )

+−

−=

−

−

−

t1400

t1400

t1400c

e1400-0.05*t4800sin4800t4800cose0.05

0.1

t4800sine0.0528042(t)v

( )( ) ( )

( ) ( )t4800cose42t4800sine742(t)vt4800sine7t4800cose42

t4800sine1442(t)v

t1400t1400c

t1400t1400

t1400c

−−

−−

−

−−=

+−

−=


+

-

R

L Ci L(0) vc(0)

Example :

F91C

H1LΩ6R

=

==

Find vc (t)?

vc(0) = 1 V , iL(0) = 0

0(t)i9dt

di(t)6dt

i(t)d

0(t)iCL

1dt

di(t)LR

dti(t)d

2

2

2

2

=++

=++


α−==⇒−==⇒=+

=++

21

212

2

ss3ss03)(s

09s6s

0ki(0)etkeki(t)

1

t32

t31

==+= −−

1dt

di(0)

010dt

di(0)

01R(0)idt

di(0)L

0(0)V(0)V(0)V CRL

−=

=++

=++

=++From KVL :


t32

21

eti(t)

1k

1kkdt

di(0)

−−=

−=

−=+−= α

dtdi(t)L(t)iR(t)V

0(t)V(t)V(t)V

C

CRL

−−=

=++From KVL :

Now, we need to find Vc(t):


[ ] ( ) ( )[ ]

t3t3c

t3t3t3

t3t3t3c

eet3(t)veet3et6

1e3et-et-6(t)v

−−

−−−

−−−

+=

+−=

−+−−−=