Fundamentals of EE 3nts.uni-due.de/downloads/get3-ise/GET3ISE_K3_PO08.pdf · The Heaviside expansion theorem: Conditions: The denominator degree is higher than the nominator degree,

Prof. Dr.-Ing. I. Willms Fundamentals of EE 3 S. 1

FachgebietNachrichtentechnische Systeme

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Fundamentals of EE 3

Chapter 3.1Transient processes -

The Laplace transform



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3.1.1 Introduction

Use of a complex frequency p for time signals. It applies: p j

pt11 û t Re u e

Thus exponential envelopes are representable for sinusoidal functions in addition to those with constant amplitudes:

Example for RLC resonant circuit:

pt22 û t Re u e ptî t Re i e



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3.1.1 Introduction

A network analysis shows here that it applies:

t

1

0 t 0

1 20

di t 1u t i t R L i z dzdt C

di t 1 1 1u t i t R L i z dz i z dz u 0 i z dzdt C C C

Solution can be achived by means of: - solution of the integro-differential equations- the Laplace transform



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3.1.2 Definition of the Laplace transform

An arbitry function is given: 0 0

( )( ) 0

for tf t

g t for t

f(t) exhibits the following characteristics:

1) f(t) has only finite jumps in the interval 1 20 t t t 2

1

( )t

t

f t dt is limited2) 03) ( ) should drop to zero quickly enough: lim ( ) t

tf t f t e

If these conditions are fulfilled

the Laplace transform of the function f(t) exists:

0

( ) ( ) ( ) ( )

ptf t g t G p g t e dtL L



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3.1.2 Definition of the Laplace transform

The relation of original signals and corresponding transform isdescribed by the symbol after DOETSCH:

( )g t ( )G p ( )G p ( )g tRelation with Fourier transform is given for:

- a causal time function and at the same time - a modified time function due to multiplication with

This leads in the integrand of the Fourier Transf. to a term

tet j t pts(t) e e s(t)e

pt0

L s t s t e dt S p

j

1 pt

j

1L S p lim S p e dp s t2 j



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3.1.3 Methods for the determination of theoriginal function from the image function

Direct method:

The original function can be determined according to the relationship:

1( ) lim ( )2

j

pt

j

g t G p e dpj

for t > 0

The use of this formula requireshowever some knowledge of the function theory.



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Use of transformation tables:

Example: In a network it applies to the transform of the current:

0 1( )(1 )

Ui tR p p

LLR

with

The original function i(t) of the current is looked for.

Solution: From a transformation table one receives:

1tae

1(1 )p ap

With a it applies then to the original function: 0( ) 1tUi t e

R



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The method of decomposition into partial fractions:

It accepted that itself the image function ( )G p in a brokenrational function:

( )( )( )

Z pG pN p

Order of the nominator polynomial < Order of the denominator polynomial( )N pThe individual partial fractions have thereby the form:

1, 2,3,...( )

kA

kp p

Now for each partial fraction if the respective original function isintended, then results in itself as sum of these the original function g(t).



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The Heaviside expansion theorem:

Conditions:The denominator degree is higher than the nominator degree, The denominator polynomial has only n simple zeros(k = 1)

is a broken rational function, ( )G p

Thus it applies:1

( )( )( )

n AZ pG pN p p p

( ) ( )( ) ( ) for limiting with ( ) v

p p Z pp p G p A p pN p

1 2

1 2

( ) ( )( ) ( ) ( ) ... ...( )

n

n

p p Z p A AA Ap p G p p pN p p p p p p p p p



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• For further solution the L'Hospital rule must be used (if linear factorform is not given):

0'

'

( ) ( )( ) ( ) ( )lim lim

( )( )p p p p

d p p Z pZ p p p Z pdpA d N pN p

dp

Thus it applies:

'

( )lim( )

p p

Z pAN p

Because of the correspondence

1p p

vp te

the original function gives:1

( ) vn

p tvg t A e



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The modified Heaviside expansion theorem: In case of a pole at the origin the following applies:

1

1

( )( )( )

Z pG p

p N p 10pwith

11'

21 1

( )(0)( ) .(0) ( )

v

np tv

v

Z pZg t eN pN p

For the original function arises then:

Example:

1

( ) .1²(1 ²) 2 ²

Z

pUi tpL p p k

L

In a network is the image function of a branch current is as follows:



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• Solution:

11( ) Z p p

'1 1

1 1( ) ²(1 ²) 2 ; ( ) 2 (1 ²)²

pN p p k N p p k

Thus it applies:

322 3

2 2 3

1 1 1

( ) 1 1 12 (1 ²) 2 (1 ²)

p tp tZ

p pUi t e eL p k p k



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Chapter 3.2Transient processes -

Application of the Laplacetransform



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3.2. Application of the Laplace transform

21

2

I p u 01U p I p R L pI p i 0C p p

u 01I p R pLI p I p i 0 LpC p

t

1

0 t 0

1 20

di t 1u t i t R L i z dzdt C

di t 1 1 1u t i t R L i z dz i z dz u 0 i z dzdt C C C



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• Dissolve for the transform of current and voltage gives forzero-state condition:

1

11

U pI p

R pL 1 pC

U pi t L

R pL 1 pC

1U p I(p) (R pL 1 pC)

11ˆ ˆ ˆû Ri pLi i

pC

For comparison:

NW-Analyses would have given:



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Procedure for general solution of the network varaibles:

1. NW-analysis in the time intervall System of integro-differentialequations

2. Laplace transform of the integro-differential equation systemafter specifying the initial values at t = 0 and determination of theLaplace Transforms

3. Dissolution after Laplace transforms of the unknown variables by meansof algebra rewriting

4. Inverse transform to the desired variables.



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Fundamentals of EE 3Fundamentals of EE 3Chapter 3.3

The classical methodof directly solving

differential equations



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3.3.1 Introduction

( )Ru t

( )Ri t ( ) ( )R Ru t Ri t

( ) ( ) ( )W p t dt u t i t dt

1) Resisitor :

Repetition to network elements concerning current, voltage, energy and/or instantaneous power p(t) :



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L

( )Lu t

( )Li t

( )( )

1( ) ( )

LL

t

L L

di tu t L oderdt

i t u dL

( )

magn0

1W ( ) ( ) ( )²2

Li tt t

Ldip d L i t dt L idi Li tdt

3) Capacitor :

C( )Ci t

( )Cu t

1( ) ( )

t

C Cu t i dC

( )( ) CCdu ti t C

dt

( ) 1( ) ( )²2

Cu tt

el Co

W p d C udu Cu t

2) Coil:

3.3.1 Introduction



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3.3.1 IntroductionIn the network only finite voltages and currents are possible. Thus: No instantaneous change of the energy of these elements is possible

L( )Li t Li cannot change instantaneously( is constant )Li

0( ) ( )L L for ti t t i t t 0 ( 0) ( 0) L Lt i t i t

C

( )Cu t

0( ) ( )C C for tu t t u t t 0 ( 0) ( 0) C Ct u t u t

Cucannot change instantaneouslyCu

( is constant)



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3.3.2 The method of the differential equations

- Kirchhoff‘s equations lead to differential equations- These equations have only constant coefficients- Their solutions provide the desired network varaibles- 1. Step: Solution of the system of the homogeneous differential equations

(which represent the natural oscillations of the network)- 2. Step: Add in each case an individual (particular) solution.

Thus the general solution of the system of the inhomogenousdifferential equations is obtained:

( ) ( ) ( )h pi t i t i t ( ) ( ) ( )h pu t u t u t

For a network with DC or constant sinusoidal excitation, the particulardescribe the steady-state conditions at the time t - > ∞



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Example 1 : Connection of DC voltage to a "RL - circuit".

M

t = 0

i(t)

LR

=0U

At the time t = 0 the network isconnected to a source. The current i(t) iscomputed with the initial condition: ( 0) 0 i t

Solution : For t ³ 0 applies:

0( )( ) di tRi t L U

dt

The differential equation is a usual, linear differential equation of firstorder with constant coefficients, which isinhomogenous.



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A) Solution of the homogeneous differential equation:

( )( ) 0di tRi t Ldt

( ) ( ) 0 di t R i t

dt L

On-set: ( )( )t t

hh

di t Ki t Ke edt

1( ) 0

t t tK R Re Ke Ke

L L

with L

R

is the time constant of the circuit



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( ) 1 ( ) 0di t i tdt

( )R tL

hi t Ke

B) Individual (particular) solution of the inhomogenous differential equation:

After infinitely long time ( )t the current of the coil L shows its

steady-state value. Then the voltage disappears, i.e. for the current i(t) holds:

0( ) ( )pUi t f t for tR

This corresponds to the case: ( ) 0 , . : 0 L Ldi t d h udt



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C) Complete solution:

0( ) ( ) ( )RtL

h pUi t i t i t KeR

Initial condition: The coil currentcannot change instantaneously

( 0) ( 0) 0 i t i t

. 0 0

10

RL UKe

R

0UK

R

For the current i(t) thus results: 0( ) (1 )RtLUi t e

R

for t > 0



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Remarks on the exponential function: ( )t

f t e

0.1353

0.3679

tt

5

( ) 0.0067t

f t e und e

1

t0

1

0

( )/

i tU R

2 3

After 3 to 5 the turn-on transient is practically terminated



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3.3.3 Overview

( ) ( )i t or u tL L

Differential equation in i(t)or u(t)

Individualsolution

Initial conditionsFor determining theunknown constants

Solution of thehomogeneousdifferential equation

Solution for i(t) or u(t)

Time domain

Frequencydomain

Inverse LaplaceTransformation

Laplacetransformation

ClassicalMethod

Method of theLaplacetransform

Transform gives rational, algebraic equations for

Solution to ( ) ( )i t or u tL L

Possiblyapplication of partial fractionmethod



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Chapter 3.4The method of solving

differential equations bythe Laplace transform



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3.4 The method of the Laplace transform• Example 1:

=0U

R L

i(t)

t = 0 Initial condition: ( 0) 0 i t

0 1( ) 1( )Ui tL p p

L

0

0

0

0

( ) 1 ( ) ( )

1 1( ) (0) ( )

1 1( )

Udi t i t tdt L

Up i t i i tL p

Ui t pL p

L L L

L L

L

0( ) 1 ( ) mit Udi t Li tdt L R



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3.4 The method of the Laplace transform

The inverse transform can be performed by means of the modifiedHeaviside` theorem:

11 1'

21 1 1

( )( ) (0)( ) ( )( ) (0) ( )

np t

v

Z pZ p ZI p i t epN p N p N p

-1L with 1 0 und 2 p n

Thus applies: '0

1 1 2 11 1( ) , ( ) , , ( ) 1 UZ p N p p p N p

L

0 01 1( ) 11 1

t tU Ui t e eL L

withLR

This gives for the current i(t):

0( ) (1 )

tUi t e

R for t > 0



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For the inverse transform also the convolution theorem might have been used:

1 11 for 0 1

t

t and ep p

-1 -1L L

0 0 0

0

0

1 ( ) 11

tx

t x tU U Uee dx eL L L

0 1 1( ) 1( )Ui tL p p

-1 -1L LThus it follows:

0 (1 )

tU e

R

2 1 2 1 1 20 0

( ) ( ) ( ) ( ) ( ) ( ) t t

g t g t g x g t x dx g x g t x dx 1 2 1 2( ) ( ) ( ) ( ) G p G p g t g tL



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3.4 The method of the Laplace transformExample 2:

At t = 0 the capacitor ischarged to Q0i(t)

C

t = 0

M

S R

0U

0 00

00

0

( )( ) ( ) ( )

( )( )

t

t

Q tRi t U mit Q t Q i x dxC

Q i x dxi t R U

C

0 0

0

1( ) ( ) t U Qi t i x dx

RC R RC

1) 1) General relationsGeneral relations: :



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3.4 The method of the Laplace transform2) 2) Solution Solution withwith thethe LaplaceLaplace transformtransform:

0 0

0

1( ) ( ) t U Qi t i x dx

RC R RC

0 0

0 0

0 0 0 0

1( ) ( ) ( )

1( ) 1

1 1( ) 11

Integration theorem

Attenuation theorem

U Qi t i t tRC p R RC

U Qi tRCp R RC p

U Q U QRCpi tR RC RCp p R RC pRC

1L L L

1L

L

0 0( )tRCU Qi t e

R RC

for t > 0 also RC

i(t)

t0

0 0QC

0

00Q UC

00

Q UC

0 0Q

C

Diagram of the results:



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3.4 The method of the Laplace transformExample 3 : In this example a series resonant circuit is

connected to a source, with cases a, b and c by an ideal switch S:

a) AC voltage: 0 0ˆ( ) cos( )uu t u t

Desired for all 3 cases: i(t) for t > 0

S '

i(t)M

t = 0 R L C

0 ( )u t b) DC voltage: 0 0( )u t Uc) Mixed case: 0 0 0ˆ( ) cos( )uu t U u t

0) General relations

The loop equation reads:

0( )( ) ( ) ( )C

di tR i t L u t u tdt

and also ( )( ) Cdu ti t Cdt

applies.

0( ) ² ( ) ( ) ( ) C C C

du t d u tRC LC u t u tdt dt



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3.4 The method of the Laplace transform• Solution for case 3 (D < 1) and source case b) (DC source)

S

C

LR( )Cu t

( )i t0U

t

0U0 0 0( ) ( )u t U f t



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The inhomogenous differential equation is subjected to the Laplace Transform:

0 0 0 0² ( ) ( )2 ² ( ) ²

²C C

Cd u t du tD u t U

dt dt for t > 0

With the initial conditions '(0) 0 (0) 0 C Cu und u

' 0 00 0²² ( ) (0) (0) 2 ( ) (0) ² ( ) C C C C C CUp u t pu u D p u t u u t

p L L L

2

2 0 00 0² ( ) 2 ( ) ( )C C C

Up u t D p u t u tp

L L L

it follows:

• Solution by means of the Laplace Transform for case 3 (D < 1) and sourcecase b) (DC source)



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20 0

0 0

( )( ² 2 ²)

C

Uu tp p Dp

L

11

( )( )( )

CZ pu t

pN pL

The original function can be obtained using the modified theorem ofHeaviside:

1 0 0 1 0 0( ) ² , ( ) ² 2 ² Z p U N p p Dp

10 1 2 0

3 0

( ) 2 2 , 0, ( 1 ² )

( 1 ² )

dN p p D p p D j Ddp

p D j D

with

32

3..1 1 0 0 0 0 0 0

'21 1 0 2 2 0 3 3 0

(0) ( ) ² ² ²( )(0) ( ) ² 2( ) 2( )

vp t p tp tvC

v v v

Z Z p U U Uu t e e eN p N p p p D p p D

Thus the original function results to:

( ) :Cu tThus it applies to the Laplace transform of



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Here only the case of conjugated complex poles (i.e. D < 1) is delt with. For the inhomogenous differential equation then holds:

t = 0

S R L

C

0 ( )u t ( )i t( )Cu t0 0 0 0

² ( ) ( )2 ² ( ) ² ( )²

C CC

d u t du tD u t u tdt dt

for t > 0, also 0 0ˆ( ) cos( )uu t u t

• Solution for case 3 (D < 1) and source case a) (AC source)



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Following is the differential equation for ( ) :Cu t

0 0 0 0² ( ) ( )2 ² ( ) ² ( )

²C C

Cd u t du tD u t u t

dt dt

After Laplace transforming of this equation it results:

' 0 0 0 0² ( ) (0) (0) 2 ( ) (0) ² ( ) ² ( ) C C C C C Cp u t pu u D p u t u u t u t L L L L

0 0 0 0² ( ) 2 ( ) ² ( ) ² ( )C C Cp u t D p u t u t u t L L L L

With the initial conditions: '(0) 0 (0) 0 it follows:C Cu und u

• Solution by means of the Laplace transform for case 3 (D < 1) and sourcecase a) (AC source)



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3.4 The method of the Laplace transformThus it applies to the image function of ( ) :Cu t

0 00 0

² ( )( )

² 2 ²

Cu t

u tp Dp

L

L

Concerning the source it holds:

0 0 0cos sinˆ ˆ( ) cos( )

² ²

u uu

pu t u t up

L L

Combining the equations given above leads then to:

0 0 0 0cos sin

ˆ( ) ²² ² ² 2 ²

u uC

pu t u

p p Dp

L



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3.4 The method of the Laplace transformThe image function must then be back-transformed then into theoriginal domain e.g. with help of the method of the decompositioninto partial fractions. Here it applies:

( )( )( )

CZ pu tN p

L

and 0 0( ) ² ² ² 2 ² N p p p Dp

with 0 0ˆ( ) ( cos sin ) u uZ p u p

Due to the periodic case (D < 1) it applies:

1 2, p j p j

3,4 1 0 ( 1 ² ) p j D j D

For the first derivative of the denominator it applies: 3

0 0 0( ) 4 6 ² 2( ² ²) 2 ² dN p p Dp p D

dp



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3.4 The method of the Laplace transformThus finally for the original function holds:

0 0 3 30 0 0 0

cos sin os sinˆ( ) ²4( ) 6 ( )² 2( ² ²) 4( ) 6 ( )² 2( ² ²)( )

j t j tu u u uC

j j cu t u e ej D j j j D j j

1 21 23 2 3 21 0 1 0 1 0 2 0 2 0 2 0

cos sin cos sin4 6 2( ² ²) 2 ² 4 6 2( ² ²) 2 ²

pt p tu u u up pe ep Dp p D p Dp p D

Final problem:

Further simplification to real values is quite complicated !

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Fundamentals of EE 3nts.uni-due.de/downloads/get3-ise/GET3ISE_K3_PO08.pdf · The Heaviside expansion theorem: Conditions: The denominator degree is higher than the nominator degree,