Lecture06 - Higher Order Transient Anaysis

8/8/2019 Lecture06 - Higher Order Transient Anaysis

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Elmer R. Magsino MS EECOE Revised 2009 Education for a Fast Changing World

CIRCUITS 1 p1

Transient Analysis of Transient Analysis of HigherHigher -- Order Netw orksOrder Netw orks

BY: ArtemioBY: Artemio P. MagaboP. MagaboProfessor of Electrical Engineering, UP EEEIProfessor of Electrical Engineering, UP EEEI


CIRCUITS 1 p2

Second-Order Transients

The solution can be shown to be an exponential of the form

stKx =

Consider the homogeneous differential equation

0cxdtdx

bdt

xda 2

2

=++

with initial conditions x(0)=X 0 and =X 0 .)0(xtdx

where K and s are constants. Substitution gives

0cKbsKKas ststst2 =++


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CIRCUITS 1 p3

After canceling the exponential term, we get thecharacteristic equation

0cbsas 2 =++Using the quadratic formula, we get the two roots

a24ac-bb-

s,s2

21=

Assuming the roots are real and distinct, thesolution will consist of two exponentials. Thus

ts

2

ts

121

KK)t(x +=K1 and K 2 can be evaluated using x(0) and (0). dtdx


CIRCUITS 1 p4

Series RLC Netw orkConsider the circuitshown. From KVL, weget for t 0

R

iE+

-

t=0v C+

-

L

C

EidtC1Ri

dtdiL =++

Differentiating, we get

0iC1

dtdi

Rdtid

L 22

=++This is a homogeneous second-order differentialequation.


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CIRCUITS 1 p5

The characteristic equation is

0C1RsLs 2 =++

0LC1s

LRs 2 =++

or

From the quadratic formula, we get the two roots

LC1

L2R

L2R

s,s2

21

=

Note: There are three types of root depending onthe value of the term inside the square root sign.


CIRCUITS 1 p6

1. Overdamped Case: The roots are real anddistinct when

LC1

L2R

2

>

ts2

ts1

21 KK)t(x +=The solution is the sum of two exponential terms

2. Critically Dam ped Case: The roots are real butrepeated when

LC1

L2R

2

=


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CIRCUITS 1 p7

st21 )KtK()t(x +=

The solution can be shown to be

LC1

L2R

2


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CIRCUITS 1 p9

Example: The capacitor is initially uncharged. Att=0, the switch is closed. Find

)0(dtdi + R

i12V+

-

t=0vC+

-

1H

C

i(0+) and .

From KVL, we getfor t 0,

EidtC1

Ridtdi

L =++ At t=0 + , i(0 + )=0. Also, we are given that v C(0 + )=0.Substitution gives

A/s12LE)0(

dtdi ==+


CIRCUITS 1 p10

F161

10

i12V+

-

t=0v C+

-

1HExample: The capacitoris initially uncharged. Att=0, the switch isclosed. Find i(t)for t 0.

The characteristic equation is thus

016s10s 2 =++

From KVL, we get for t 0,

12idt16i101 =++ dtdi

Differentiating theequation, we have 0i16dt

di10 =++2

2

dtid


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CIRCUITS 1 p11

F161

10

i12V+

-v C+

-

1HThe characteristicequation is

016s10s 2 =++

whose roots are s 1=-2 and s 2=-8. Thus, we get8t-

22t-

1 KK)t(i +=and

8t-2

2t-1 K8K2dt

di =

From the previous example, weve found that att=0 + ,

A/sdtdi

12LE

)0( ==+i(0 + )=0, v C(0 + )=0 and


CIRCUITS 1 p12

The initial conditions are i(0 + )=0 and =12A/sec. Substitution gives

)0(dtdi +

21 KK0)0(i +==+

21 8K--2K12)0(dtdi ==+

Solving simultaneously, we get K 1=2 and K 2=-2.Thus,

0t Amp 2-2)t(i 8t-2t- =Note: For an over-damped case, the solutionconsists of two distinct exponential terms.


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CIRCUITS 1 p13

F161

8

i12V+

-

t=0v C+

-

1HExample: The capacitor

is initially uncharged. Att=0, the switch isclosed. Find i(t)for t 0.

The characteristic equation can be shown to be

016s8s 2 =++whose roots are s 1=-4 and s 2=-4. Thus, we get

4t-3

4t-2

4t-1 KKK)t(i =+=

A single exponential solution will not work since theoriginal differential equation is second-order.


CIRCUITS 1 p14

Assume . Differentiating twice, we get4t-)t(y)t(i =4t-4t- )t('y)t(y-4

dtdi +=

4t-4t-4t-2

2

)t(''y)t('y8-)t(y16dt

id +=

The original differential equation is

0i16dtdi

8dt

id2

2

=++

-4t-4t-4t )t(''y(t)8y'-)t(y160 +=-4t-4t4t- 16y(t))t('y832y(t)- ++

Substitution gives


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CIRCUITS 1 p15

Integrating twice, we get

1K)t('y =or21 KtK)t(y +=

Simplifying, we get

0)t(''y 4t- =0(t)'y' =

or

4t-

)t(y)t(i =

Finally, the solution is

4t-2

4t-1 KtK)t(i +=

or


CIRCUITS 1 p16

Differentiating the solution, we get

4t-2

4t-1

4t-1 4K-Kt-4Kdt

di +=

We get K 1=12 and K 2=0. Thus

0t Amp t12)t(i 4t- =

2K00)0(i +==+

21 4K-K012)0(dtdi

+==+


)0(dtdi +

4t-2

4t-1 KtK)t(i +=


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CIRCUITS 1 p17

F161

6

i12V+

-

t=0v C+

-

1HExample: The capacitor

is initially uncharged. Att=0, the switch isclosed. Find i(t)for t 0.

The characteristic equation can be shown to be

016s6s 2 =++whose roots are s 1 , s 2=-3 j2.65. Thus, we get

j2.65)t-(-32

j2.65)t(-31 KK)t(i += +

or)KK()t(i t j2.65-2

t j2.651

3t- +=


CIRCUITS 1 p18

Eulers Identities:

xsin jxcos jx +=(1)sin x j-xcos = jx(2)

To prove the first identity, let y=cos x + j sin x.Differentiating, we get

xcos jx-sindxdy +=

x)sin jx(cos j +=xcos jxsin j

dxdy 2 +=

and since , the equation can be re-writtenas

1 j =


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CIRCUITS 1 p19

We get y jdx

dy =

Integrating both sides, we get

Kx jyln +=

dx jdyy1 =

or

Evaluate K. When x=0, y=1.

0KorK j01ln =+=Thus we get ln y = jx, or xsin jxcos jx +=Note: The other Eulers identity can be verifiedfollowing the same analysis.


CIRCUITS 1 p20

Back to the expression for the current

)KK()t(i t j2.65-2t j2.65

13t- +=

tsin jKtcosK[)t(i 11t3 +=

]tsin jKtcosK 22 +

From Eulers identities, we get

where = 2.65. Combining the two cosine termsand the two sine terms, we get

]tsinKtcosK[)t(i 43t3 +=

where K 3 = K 1+K 2 and K 4 = j(K 1-K 2 ).


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CIRCUITS 1 p21

3K0)0(i ==+

.544KorK12)0(dtdi

44 ===+

We get

0t Amp t2.65sin54.4)t(i t3 =


)0(dtdi +

Differentiate to get

]tcosKtsinK[dtdi 43

t3 +=

]tsinKtcosK[3 43t3 +


CIRCUITS 1 p22

R=8

R=10

R=6

P lot of the Currents


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CIRCUITS 1 p23

P arallel RLC Netw ork

Differentiating, we get

0v

L

1

dt

dv

R

1

dt

vdC 22

=++

This is a homogeneous second-order differentialequation.

Consider the circuitshown. From KCL, weget for t 0

IvdtL1v

R1

dtdvC =++

RI v+

-CL


CIRCUITS 1 p24

0LC1s

RC1s 2 =++

or

Note: We get three types of root depending on thevalue of the term inside the square root sign.

The characteristic equation is

0L1s

R1Cs 2 =++

From the quadratic formula, we get the two roots

LC1

RC21

2RC1

-s,s

2

21

=


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CIRCUITS 1 p25

Higher-Order Transients

The solution can be shown to be an exponential of

the form stKx =where K and s are constants.

Consider the homogeneous differential equation

0xadtdxa...

dtxda

dtxda 011n

1n

1nn

n

n =++++

with initial conditions x(0)=X 0 , =X 0)0(xtdx

'''''''01n

1n

X)0(dx

xd =

''02

2

X)0(dt

xd = ,


CIRCUITS 1 p26

Substitution gives

0KasKa...KsaKsa st0st

1st1n

1nstn

n =++++ After canceling the exponential term, we get thecharacteristic equation.

0asa...sasa 011n

1nn

n =++++ This is a polynomial of n th order and there will be nroots. The type of response will depend on the

values of these roots. Assuming all the n roots arereal and distinct, the solution can be shown to be

tsn

ts1n

ts2

ts1

n1n21 KK...KKx ++++=


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CIRCUITS 1 p27

The roots of the characteristic equation can beshown to be s 1=-2, s 2=-4 and s 3 =-8.

Example: Consider the differential equation

0v64dtdv56

dtvd14

dtvd

2

2

3

3 =+++

with initial conditions v(0)=7 volts, (0)=-24 v/sdtdv

and (0)=112 v/s 2 . Find v(t).22

dtvd

064s56s14s 23 =+++The characteristic equation is


CIRCUITS 1 p28

-8t3

-4t2

-2t1 KKK)t(v ++=

Since the roots are real and distinct, the solution is

Differentiating twice, we get

8t-3

4t-2

2t-1 K8K4K2dt

dv =

8t-3

4t-2

2t-12

2

K64K16K4

dt

vd ++=

Evaluate the expressions for v, and at t=0and use the initial conditions. dt

dv2

2

dtvd


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CIRCUITS 1 p29

321 K8K4K224)0(dtdv ==

3212

2

K64K16K4112)0(dt

vd ++==

Solving simultaneously, we get K 1=4, K 2 =2 andK3=1. The final solution is

V 124)t(v -8t-4t-2t ++=

321 KKK7)0(v ++==We get


CIRCUITS 1 p30

Example: Consider the differential equation

0i32dtdi

32dt

id10

dtid

2

2

3

3

=+++

032s32s10s 23 =+++The characteristic equation is

The roots of the characteristic equation can beshown to be s 1=-2, s 2=-4 and s 3 =-4. The solution

is -4t3-4t

2-2t

1 KtKK)t(i ++=The constants K 1 , K 2 and K 3 can be evaluated if the values of i, di/dt and d 2 i/dt 2 are known at t=0.


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CIRCUITS 1 p31

Getting the Differen tial EquationUsing nodal analysis or loop analysis, write theKCL or KVL equations that describe the circuit.

1.

Differentiate the equations, if necessary, toeliminate any integral expressions.

2.

In every equation, replace the derivatives withoperators.

3.

Eliminate all variables, except one, using anyappropriate method.

4.

Simplify as necessary and replace the operatorswith the corresponding derivative terms.

5.


CIRCUITS 1 p32

=++t

1222

0dt)ii(16i10dtdi

Mesh 2:

Mesh 1: =+t

211 )t(vdt)ii(16i8First, write the mesh equations for the circuit.

Then, differentiate the mesh equations to eliminatethe integrals.

Example: Find thedifferential equationsthat describe themesh currents i 1 andi2 in the networkshown.

1H

v(t)+

-

8

i1 i2 10 F161


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CIRCUITS 1 p33

Next, multiply equation (1) by 16 and equation (2)by (8D+16), then add the resulting equations. Thiswill eliminate the current variable i 1 .

0i)16D10D(i16 22

1 =+++ (2)

)t(vdtdi16i16

dtdi8 211 =+

We get

(a)

0i16dtdi

10dt

idi16 222

22

1 =+++ (b)

)t(Dvi16i)16D8( 21 =+ (1)

Using operators, let D= . Substitution givesdtd


CIRCUITS 1 p34

We get

)t(Dv16i)D288D96D8( 223 =++

which simplifies to

)t(v2i)36D12D( 22 =++

The differential equation for the current i 2 is

)t(v2i36dtdi12

dtid

22

22

2

=++

0i)16D10D(i16 22

1 =+++ (2))t(Dvi16i)16D8( 21 =+ (1)16

(8D+16)


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CIRCUITS 1 p35

)t(v)16D10D(Di)D288D96D8( 2123 ++=++

which simplifies to

)t(v)16D10D(i)36D12D( 281

12 ++=++

The differential equation for current i 1 is

2

2

11

21

2

dt

)t(vd125.0i36

dt

di12

dt

id =++

)t(v2dt

)t(dv25.1 ++

Similarly, if we multiply equation (2) by 16 andequation (1) by (D 2 +10D+16), then add theresulting equations, we will eliminate the currentvariable i 2 . We get


CIRCUITS 1 p36

22

85

22

2

161

1 idtdi

dtid

i ++=

Alternative Procedure: First, solve for i 1 from (b)and differentiate the resulting equation. We get

Next, substitute the equations in (a). We get

)t(v2i36dtdi

12dtid

22

22

2

=++A similar procedure, applied on equation (a), willresult in the differential equation for current i 1 .

dtdi

dtid

dtid

dtdi 2

22

2

85

32

3

1611 ++=

and


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CIRCUITS 1 p37

Example: The switch is moved from b to a at t=0.

Find the differential equations that describe thevoltages v 1 and v 2 for t 0.4

E+

-

t=0

a

1H

bv(t)

+

-F

416

REF

+v 1

+v 2

From KCL, we getfor t 0

0dtv6

v-v1

21 =+ Node 1:0

dtdv

41

6vv

4v(t)-v 2122 =++Node 2:

(1)

(2)


CIRCUITS 1 p38

Differentiate (1) and re-write the equations. We get

0dt

dv61v

dtdv

61 2

11 =+

)t(v41

v125

dtdv

41

v61

- 22

1 =++

)t(v3v5dt

dv3v2- 22

1 =++

0dt

dvv6

dtdv 2

11 =+

Simplify into

0Dv6)v(D 21 =+

)t(v3)v5(3D2v- 21 =++


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CIRCUITS 1 p41

For t 0, we get from KVL

)t(vdt)ii(20i4 211 =+ 0dt)ii(20

dtdi

122 =+

or

dt)t(dv

i20i20dtdi

4 211 =+

0i20i20dt

id122

22 =+

3

t=0

1Hv(t)

+

-

F2011

i1 i2

Example: At t=0, the switch is opened. Find the

differential equations that describe the currents i 1and i 2 for t 0.


CIRCUITS 1 p42

)t(Dvi20i)20D4( 21 =+Using operators, we get

(1)

0i)20D(i20 22

1 =++ (2)Multiply equation (1) by (D 2+20) and (2) by 20,then add the resulting equations. This willeliminate the variable i 2 . We get

)t(v)5D(i)20D5D( 241

12 +=++

The differential equation for i 1 is

)t(v5dt

)t(vdi20

dtdi

5dt

id2

2

41

11

21

2

+=++


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CIRCUITS 1 p43

Multiply equation (1) by 20 and (2) by (4D+20),then add the resulting equations. This willeliminate the variable i 1 . We get

)t(v5i)20D5D( 22 =++

The differential equation for i 2 is

)t(v5i20dtdi

5dt

id2

222

2

=++


CIRCUITS 1 p44

Complete Response:1. Steady-state Response2. Transient Response

W hy Get Initial Conditions?

Transient Response: General form is exponentialts

nts

1nts

2ts

1n1n21 KK...KKx ++++=

where K 1 , K 2 , Kn are arbitrary constants.

Answer: The initial conditions are necessary in thedetermination of the numerical values of thearbitrary constants K 1 , K 2 , Kn .


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CIRCUITS 1 p45

Evaluating Initial Conditions

5. Use the KVL and KCL equations for t 0 and

their derivatives, plus the inductor currentsand capacitor voltages at t=0 + to evaluatethe required initial conditions.

1. Assume switching operation at t=0.

2. Evaluate the inductor currents and capacitorvoltages at t=0 - .

3. Find inductor currents and capacitor voltagesat t=0 + .

4. Write the KVL and KCL equations describing thenetwork for t 0.


CIRCUITS 1 p460)0(v

A1.0)0(i

C

ss,L

=

==

1k

100

Example: The circuit has reached steady-statecondition with the switch in position a. At t=0, theswitch is moved to position b. If the capacitor isinitially uncharged, find i(0 + ),

).0(dt

id )0(

dtdi

2

2++ and

1k

i100V+

-

t=0

a

1H0.1 F

b

The circuit is at steadystate for t


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CIRCUITS 1 p47

0idtC1Ri

dtdiL =++ (1)

whose derivative is

0iC1

dtdi

Rdt

idL 2

2

=++(2)

At t=0 + ,

A1.0i)0(iss,L

==+

0)0(v C =+

From KVL, we get for t 0,1k

i 1H0.1 F


CIRCUITS 1 p48

From (2), we get

)]0(iC1)0(

dtdiR[

L1-)0(

dtid2

2 +++ +=

2kA/s-900=

0From (1), we get at t=0 +

0)0(v)0(Ri)0(dtdi

L C =++ +++

which gives

A/s100)0(iLR

)0(dtdi == ++


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CIRCUITS 1 p49

A430

120)0(i -L ==

V80)0(i20)0(v L-

C ==

t=0

i1 i2

10 120V

+

- 1H

20

1 F

20

and .)0(dtdi 2 +

Example: The network is initially at steady-state

condition with the switch open. At t=0, the switchis closed. Find i 1(0 + ), i 2(0 + ),

)0(dtdi1 +

At t=0 - , we get

iL(0 -)

10

120V

+

-

20 +-

vC(0 - )


CIRCUITS 1 p50

For t 0, we get from KVL

i1 i2120V+

- 1H

20

1 F

20

At t=0 + , we get

A4)0(i)0(i -L1 ==+

V80)0(v)0(v-

CC ==+

120i20dtdi

11 =+(1)

120dti10i20t

26

2 =+ (2)

From equation (2), we get

A2)]0(v120[)0(i C201

2 == ++


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CIRCUITS 1 p51


A/s40)0(i20120)0(dtdi

11 == ++

To get an equation involving , differentiate (2).We get dt

di 2

0i10dtdi

20 262 =+

At t=0 + , we get

kA/sec -100)0(dtdi 2 =+

120i20dt

di1

1 =+(1) 120dti10i20t

26

2 =+ (2)


CIRCUITS 1 p52

dtdi1

Example: The network is initially unenergized. Att=0, the switch is closed. Determine i 1(0 + ), i 2 (0 + ),

(0 + ) and (0 + ).dtdi 2

i1E+

-

t=0

i2R1 L

R2C

For t 0, we get fromKVL,

E)ii(RdtiC

12111

=+

(1)

0iRi)RR(dtdi

L 112212 =++(2)


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CIRCUITS 1 p53

Since the circuit is initially unenergized, we knowthat v C(0 + )=0 and i L(0 + )=0. Thus

0)0(i2 =+

From (1), we get

E)0(iR)0(iR)0(v 2111C =+ +++

or

11 R

E)0(i =+

E)ii(RdtiC

12111 =+ (1)

0iRi)RR(dtdiL 11221

2 =++(2)


CIRCUITS 1 p54

0)0(i2 =+1

1 RE

)0(i =+

From (2), we get

0)0(iR)0(i)RR()0(dtdi

L 112212 =++ +++

E)ii(RdtiC1

2111 =+ (1)0iRi)RR(

dtdi

L 112212 =++(2)

LE)0(i

LR)0(

dtdi

112 == ++

which gives


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CIRCUITS 1 p57

For t


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CIRCUITS 1 p59

At t=0 + , we get from (1)

mA9)]0(v)0(v27[R1

)0(i 2C1C2

1 == +++

Since i 1 (0+

) - i 3 (0+

)=i L(0+

), then0)0(i)0(i)0(i L13 == +++

++= dtiC1

dtiC1

iR27 32

21

12(1)

(2) += dtiC1)ii(R0 2

1121

(3) += dtiC1

)ii(dtd

L0 32

13


CIRCUITS 1 p60

For resistor R 1 , we get

)0(v)]0(i)0(i[R)0(v 1C2111R++++ ==

or

mA3R

)0(v)0(i)0(i

1

1C12 ==

+++

++= dtiC1dtiC1iR27 322112(1)(2) += dtiC

1)ii(R0 2

1121

(3) += dtiC1

)ii(dtd

L0 32

13


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CIRCUITS 1 p61

Differentiate equations (1) and (2). We get

32

21

12 iC

1i

C1

dtdi

R0 ++=(4)

(5) 21

11

21 iC

1dtdiR

dtdiR0 +=


s /A3=

)]0(iC1)0(iC1[R1)0(dtdi 322

12

1 +++ +=

++= dtiC1

dtiC1

iR27 32

21

12(1)

(2) += dtiC1)ii(R0 21121


CIRCUITS 1 p62


s /A5.4=

)0(iCR

1)0(dtdi)0(

dtdi

211

12 +++ =


s /A9=

)0(v

L

1)0(

dt

di)0(

dt

di2C

13 +++ =


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CIRCUITS 1 p63

dtdi

Example: The network has reached steady-state

condition with the switch closed. At t=0, the switchis opened. Find i(0 + ), v(0 + ),

(0 + ) and (0 + ).dtdv

t=0

i60V

+

- 1H 10 -3 Fv

+

-

10 20

Equivalent circuit at t=0 -

iL(0 -)60V

+

-+

-

10 20

v C(0 -)

A32060

)0(iL ==

V60)0(v C =


CIRCUITS 1 p64

Equivalent circuit for t 0

i1H 10 -3 Fv

+

-

10 20 vi20dtdi =+(1)

vidt10i10t3 =+ (2)

V30)0(v =+or

At t=0 + , we get

A3)0(i)0(i L == +

V60)0(v)0(v CC == +

)0(v)0(v)0(i10 C+++ =


+vC-


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CIRCUITS 1 p67

dtdv

6vv

4v120 2

41122 +=(2)

Equivalent circuit for t 0 4

1H120V

+

-F

416

REF

+v 1

+v 2

= dtv6vv

112(1)

From KCL, we get

0)0(i6)0(v)0(v L21 == +++From (1), we get

At t=0 + ,V36)0(v)0(v C2 == ++


CIRCUITS 1 p68

From (2), we get

V/s60)0(dt

dv 2 =+

Differentiate equation (1). We get

121 v6

dt

dv

dt

dv =At t=0 + ,

)0(v6)0(dt

dv)0(dt

dv1

21 +++ = V/s60=

dtdv

6vv

4v120 2

41122 +=(2) = dtv6

vv1

12(1)


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CIRCUITS 1 p69

Solving the Differential Equation

Find the steady-state response x ss . This issimilar in form to the forcing function g(t) plusall its unique derivatives.

1.

Consider the n th -order differential equation

)t(gxadtdx

a...dt

xda

dtxd

a 011n1n

1nn

n

n =++++

Find the transient response x t . This is generally

an exponential of the form

2.

tsn

ts2

ts1t

n21 K...KKx +++=


CIRCUITS 1 p70

)0(xtdx )0(

dxxd1n

1n

)0(dt

xd2

2Evaluate the initial conditions. We need the3.

values of x(0), , , .

Find the total response. Add the steady-stateresponse and transient response.

4.

tsn

ts2

ts1ss

n21 K...KKxx(t) ++++=Differentiate the total response (n-1) times.5.

Using the expressions for x(t) and its (n-1)derivatives in step 5, and the initial conditions instep 3, find the arbitrary constants K 1 , K 2 , Kn .

6.


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CIRCUITS 1 p71

First, get the differential equations that describecurrents i 1(t) and i 2 for t 0. The mesh equationsare

24i8i12dtdi

211 =+ (1)

0i8i12dtdi2 12

2 =+ (2)

Example: The network

is initially unenergized.At t=0, the switch isclosed. Find currentsi1(t) and i 2 (t) for t 0.

4

i124V+

-

t=0 1H

i28 2H

4


CIRCUITS 1 p72

24i8i12dtdi

211 =+

0i8i12dtdi

2 122 =+

Using operators we get(D+12)i 1 8i 2 = 24 (a)

-8i 1 + (2D+12)i 2 = 0 (b)

To eliminate i 2 , multiply (a) by (2D+12) and (b)by 8 and add the resulting equations. We get

(D 2 + 18D + 40)i 1 = (D+12)12

144i40dtdi

18dt

id1

121

2

=++or

(1)

(2)


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CIRCUITS 1 p73

(D 2 + 18D + 40)i 2 = 96

Similarly, we eliminate i 1 by multiplying (a) by 8

and (b) by (D+12) and adding the equations. Weget

96i40dtdi

18dt

id2

222

2

=++or

Alternatively, we can solve for i 2 in equation (1)and differentiate the resulting equation. We get

(3)3idtdi

i 1231

81

2 +=

dtdi

dtid

dtdi 1

23

21

2

812 += (4)

and


CIRCUITS 1 p74

Substitute (3) and (4) in equation (2). We get

144i40dtdi

18dt

id1

121

2

=++

This is the required differential equation for i 1 .

2232

41

1 idtdii += (5)

Solve for i 1 in equation (2) and differentiate theresulting equation. We get

dtdi

dtid

dtdi 2

23

22

2

411 += (6)and


96i40dtdi

18dt

id2

222

2

=++This is the required differential equation for i 2 .


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CIRCUITS 1 p75

Next, we determine the steady-state response i 1,ssand i 2,ss which are both constant. We get from

144i40dtdi

18dt

id1

121

2

=++

A6.340

144i ss,1 ==

and from

96i40dtdi

18dt

id2

222

2

=++

A4.24096i ss,2 ==

Since theforcing functionis a constant(24V) thesteady-stateresponse of anycurrent orvoltage shouldalso be aconstant. Thus,

i1,ss = A

di 1,ss / dt = 0d 2 i1,ss / d t 2 = 0


CIRCUITS 1 p76

We can also draw the equivalent circuit at steadystate. We get 4

24V+

-i2,ss8

4

i1,ss24i8i12 ss,2ss,1 =0i12i8 ss,2ss,1 =+

which gives i 1,ss = 3.6 Ampsand i 2,ss = 2.4 Amps.


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CIRCUITS 1 p77

Next, find the transient response. The differential

equations are

Solving for the roots, we get s 1 =-2.6 and s 2 =-15.4.

144i40dtdi

18dt

id1

121

2

=++

040s18s2

=++

Setting the right-hand side of the equation to zeroand changing operators, we get the characteristicequation to be

96i40dtdi

18dt

id2

222

2

=++

Only one characteristicequation defines anycurrent or voltage inthe circuit.


CIRCUITS 1 p78

-15.4t2

-2.6t1t1 KKi +=

Thus, we get the transient response.

-15.4t4

-2.6t3t2 KKi +=

Next, we find the initial conditions. We need i 1 (0 + ),

dtdi 2i2(0 + ), (0 + ) and (0 + ).dt

di1

0)0(i)0(i 21 == ++Since the circuit was initially unenergized, we get

From (1), we get A/s24)0(dtdi 1 =+


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CIRCUITS 1 p79

From (2), we get 0)0(dt

di 2 =+

Next, determine the total response. We get-15.4t

2-2.6t

11 KK6.3)t(i ++=-15.4t

4-2.6t

32 KK4.2)t(i ++=

whose derivative are

15.4t-2

2.6t-1

1 K4.15K6.2dtdi =

15.4t-4

2.6t-3

2 K4.15K6.2dtdi =


CIRCUITS 1 p80

Evaluate the constants K 1 and K 2 . At t=0 + , we get

211 KK6.30)0(i ++==+

211 K4.15K6.224)0(

dtdi ==+

Solving simultaneously, we get K 1=-2.46 andK2=-1.14. The final expression for current i 1 is

A 14.146.26.3)t(i -15.4t-2.6t1 =

Evaluate the constants K 3 and K 4 . We get K 3 =-2.89and K 4=0.49. Thus

A 49.089.24.2)t(i -15.4t-2.6t2 +=


41/53


CIRCUITS 1 p81

I 1

I2

P lot of the Currents

4

i124+

-

t=0 1

i28 2

4


CIRCUITS 1 p82

4

60V+

-

t=0

a

1H

b120V

+

-F

416

REF

+v 1

+v 2

Example: The switch has been in position b for along time. At t=0, the switch is moved to a. Findv 1(t) and v 2(t) for t 0.

In a previous example, we got the differentialequations that describe the voltages v 1 and v 2 . Inanother example, we derived the initial conditions.


42/53


CIRCUITS 1 p83

0v10dt

dv7dt

vd1

1212 =++

720v10dt

dv7dt

vd2

222

2

=++

The differential equations are

0)0(v 1 =+

V36)0(v 2 =+

V/s60)0(dt

dv)0(dt

dv 21 == ++

with initial conditions

(1)

(2)


CIRCUITS 1 p84

Steady-state respo nse:

V7210720

v ss,2 ==

We can also draw the equivalent circuit at steadystate.

v 2,ssv 1,ss

120V+

-

+

-

6

4

+

-

0v ss,1 =

V72)120(106v ss,2 ==

v 1,ss = A

dv 1,ssdt

= 0

d 2 v 1,ssdt 2

= 0

v 2,ss = B

dv 2,ssdt

= 0

d 2 v 1,ssdt 2

= 0

0v10dtdv

7dtvd

1121

2

=++ 720v10dtdv

7dtvd

2222

2

=++

0v ss,1 =From the differential equations (1) and (2), we get


43/53


CIRCUITS 1 p85

Transient response:

whose roots are s 1=-2 and s 2=-5. We get

-5t

2

-2t

1t1 KKv +=-5t

4-2t

3t2 KKv +=

010s7s 2 =++The characteristic equation is

0v10dtdv

7dtvd

11

21

2

=++

720v10dt

dv7

dtvd

22

22

2

=++

The differential equations are

(1)

(2)


CIRCUITS 1 p86

Complete response: We get -5t2

-2t11 KK)t(v +=

-5t4

-2t32 KK72)t(v ++=

The derivatives are

5t-2

2t-1

1 K5K2dt

dv =

5t-4

2t-3

2 K5K2dt

dv =

At t=0 + , we get

211 KK0)0(v +==+ 211 K5K260)0(dtdv ==+


44/53


CIRCUITS 1 p87

Also at t=0 + , we get

432 KK7236)(0v ++==+

432 5K2K60)(0

dtdv ==+

Solving simultaneously, we get K 3=140 andK4=-68. The final expressions are

0t V 2020)t(v -5t-2t1 =

0t V 4 40 72(t)v -5t-2t2 +=

Solving simultaneously, we get K 1=20 and K 2=-20.


CIRCUITS 1 p88

V1

V2

P lot of the Voltages

4

60V+

-

t=0

1120

+

-F

416

REF

+v 1

+v 2


45/53


CIRCUITS 1 p89

Example: The network

is initially unenergized.At t=0, the switch isclosed. Find currenti2(t) for t 0. Assumev(t)=20 cos 4t volts.

8

i1v(t)+

-

t=0 2H

i24 1H

For t 0, the mesh equations are

)t(vi4i12dtdi

2 211 =+ (1)

0i4i4dtdi

122 =+ (2)


CIRCUITS 1 p90

22

41

1 idtdi

i += (3)

Solve for i 1 in equation (2) and differentiate theresulting equation. We get

dtdi

dtid

dtdi 2

22

2

411 += (4)

and


)t(v2i16dtdi10

dtid 2222

2

=++

where v(t)=20 cos 4t volts.

(5)


46/53


CIRCUITS 1 p91

Forced Response: Since the source is sinusoidal,

4tsinB4tcosAi ss,2 +=

4tcosB44tsinA4dt

di ss,2 +=

4tsin16B-4tcosA16dtid

2ss,2

2

=

Substitute in the differential equation (5). We get

4tsin40A-4tsin16B-4tcos16A4tcos404tsin16B4tcos16A4tcos40B =+++

4tcos4016idtdi

10dt

id2

222

2

=++ (5)


CIRCUITS 1 p92

A16B40A1640 ++=B16A40B160 +=

Comparing coefficients, we get

Solving simultaneously, we get A=0 and B=1. Thus

t4sini ss,2 =Transient response: The characteristic equation is

016s10s2

=++The roots are s 1=-2 and s 2 =-8. Thus

t82

t21t,2 KKi

+=


47/53


CIRCUITS 1 p93

Complete response:t8

2t2

12 KK4tsin)t(i

++=t8

2t2

12 K8K2-4tcos4

dtdi =

dtdi 2At t=0 + , i 1 (0 + )=i 2 (0 + )=0. From (2), get (0+)=0.

21 KK0 +=

21 K8K2-40 =Solving simultaneously, we get K 1=-K 2=-2/3. Thus

0t A -4tsin)t(i t832t2322 +=


CIRCUITS 1 p94

P lot of the Current & Voltage

i2

V(t) = 20 cos (4t)


48/53


49/53


CIRCUITS 1 p97


1Hv(t)

+

-F

416

iLv C+

-

KVL:dtdi1i6v LLC +=

KCL: LC

41C i

dtdv

4v)t(v +=

In matrix form, we get

C

L

vi

C

L

v

i

v(t)= +10

1416

with initial conditions i L(0 + )=6 A and v C(0 + )=36 V.


CIRCUITS 1 p98

Example: The circuit hasreached steady-statecondition with the switchclosed. At t=0, the switchis opened. Find v C(0 + ),iL(0 + ) and the stateequations for t 0.

3

t=0

1H12V

+

-

F2011 v C iL

+

-

Equivalent circuit at t=0 -1

12V+

-iL(0 -)

+

-vC(0 -)

)0(v0)0(v C-C +==

)0(iA12)0(i L-

L+==


50/53


CIRCUITS 1 p99


1H12V+

-F

201v C iL

+

-KVL:

dtdi1v LC =

KCL: LC

201C i

dtdv

4v12 +=


C

L

vi

C

L

v

i

12V= +50

52010

with initial conditions i L(0 + )=12 Amps and v C(0 + )=0.


CIRCUITS 1 p100

Solution of the State EquationThe Euler method for integrating a first-orderdifferential equation is of the form

t)t,x(f )t(x)tt(x0t00

++where x=f(x,t). The method can be extended tothe case when x is a vector.

.

Substitution givest)]t(uB)t(xA[)t(x)tt(x 0000 +++)t(utB)t(x)tAI( 00 ++

uBxA)t,x(f x +==

Consider the stateequation

Identity matrix Step-size


51/53


CIRCUITS 1 p101

Example: The network is initially unenergized. At

t=0, the switch is closed. t=0 8

iL1v(t)+

-

2H

iL24 1H

Use the Euler methodwith t=0.02 sec tofind i L1 and i L2 for t 0.Let v(t)=20 cos 4t V.

For t 0, we get from KVL,

)t(vi4i12dtdi

2 211 =+

0i4i4dtdi

122 =+


CIRCUITS 1 p102


2L

1L

ii

2L

1L

i

iv(t)= +

05.0

4426

with initial conditions i L1(0 + ) = i L2(0 + ) = 0.

Recall the Euler method

)t(utB)t(x)tAI()tt(x 000 +++

= +44

26

0.02I+ tA

1001

=92.008.004.088.0


52/53


CIRCUITS 1 p103

We get

= +iL1(t+ t)iL2(t+ t) 92.008.0

04.088.0 iL1(t)

iL2(t)

0.2cos 4t

0

At t=0,

= +iL1(0.02)iL2(0.02) 92.008.0

04.088.0 iL1(0+

)iL2(0 + )

0.20 0

0.2=

where

)t(utB)t(x)tAI()tt(x 000 +++

=I+ tA 92.008.004.088.0

and t=0.2


CIRCUITS 1 p104

At t=0.04 second:

= +iL1(0.04)

iL2(0.04) 92.008.004.088.0 0.3754

0.016 0.0448

0.5284=

0.1974

0

= +iL1(t+ t)iL2(t+ t) 92.008.0

04.088.0 iL1(t)iL2(t)

0.2cos 4t0

and

Weve found

=iL1(0 + )iL2(0 + ) 0

0=

iL1(0.02)iL2(0.02) 0

0.2

At t=0.02 second:

= +iL1(0.04)

iL2(0.04) 92.008.004.088.0 0.2

0 0.016

0.3754=

0.1994

0


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CIRCUITS 1 p105

Comparison o f ResultsFrom a previous example, we got

0t A -4tsin)t(i t832t2

32

2 +=

Time Actual Euler Error

0.02 0.00748 0.0 0.00748

0.04 0.02801 0.016 0.01201

0.06 0.05894 0.04475 0.014190.08 0.09800 0.08344 0.01456

Comparing the actual value with the estimate,

P lot of the Current

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Lecture06 - Higher Order Transient Anaysis

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