Circuit Network Analysis - [Chapter4] Laplace Transform

Network Analysis

Chapter 4

Laplace Transform and Circuit Analysis

Chien-Jung Li

Department of Electronic Engineering

National Taipei University of Technology

Derivative

2y x=

• Derivative

Isaac Newton (牛頓, 1643-1727)

Method of fluxions

流量: fluent

流數: fluxion

2y x x=i i

2y

xx

=i

iy對應x的導數(變化率)即兩個流數的比

代表對時間的微分

G. Wilhelm Leibniz (萊布尼茲, 1646-1716)dydx

yx

∆ = ∆

J-Louis Lagrange (拉格朗日, 1736-1813)

( )y f x= ( ) dyy f x

dx′ ′= =

2y x= 2y x′ =

( )2

2

d dy d yy f x

dx dx dx ′′ ′′= = =

2/61 Department of Electronic Engineering, NTUT

• Operators do particular mathematical manipulations, such as

Operators

+ − × ÷

( ) ( )dy dy f x y

dx dx′ ′= = = 2 2 32

dx x x

dx=

( ) ( )2 2

2 2

d dy d y dy f x y

dx dx dx dx ′′ ′′= = = =

dD

dx≜

2 2 32x Dx x=

• Louis Arbogast (1759-1803) conceived the calculus as operational

symbols. The formal algebraic manipulation of series investigated by

Lagrange and Laplace.

2 5 320D x x=

• The derivative

nn

n

dD

dx≜

Differential operator


Differential Equations

xy e= x x xdy e De e

dx′ = = = Dy y=

Hence, we can know the solution of the equation should be of the formxy e= xy Ce=

• Oliver Heaviside (1850-1925)

Heaviside was a self-taught English electrical engineer, mathematician, and

physicist who adapted complex numbers to the study of electrical circuits,

invented mathematical techniques to the solution of differential equations

(later found to be equivalent to Laplace transforms. He changed the face of

mathematics and science for years to come.

or

C is a constant

Dy y=

kxy e= kx kx xdy e De ke

dx′ = = = Dy ky=

Similarly, we can know the solution of the equation should be of the

form

Dy ky=kxy Ce=


Solutions of the Differential Equation

• Consider the differential equation: ( )7 8 0y y y′′ ′+ − =2

2 7 8 0d d

y y ydx dx

+ − =

2 7 8 0D y Dy y+ − =

(將D視為代數)

( )2 7 8 0D D y+ − =

( ) ( ) ( )2 7 8 1 8 0D D D D+ − = − + =

Use the differential operator D:

Take this as an algebraic equation:

The solutions:

0y = or 1D = 8D = −

Dy y= 8Dy y= − xy Ae= 6xy Be−=

androots

6x xy Ae Be−= +General Solution:

put y backand

solutionsand(/or)


Fourier Transform

( ) ( ) ω∞ −

−∞= ∫

j tX f x t e dt

The usefulness of the Fourier transform is limited by a series drawback:

if we try to evaluate the Fourier integral, we find that the integral does not

converge for most signals x(t), i.e.,

( ) 0sinx t tω=

( ) 00sin j tX f t e dtωω

∞ −= ⋅∫

( ) 0 0 00 2 20

0 0

sin cossin

j t j tj t j e t e t

X t e dtω ω

ω ω ω ω ωω ωω ω

∞− −∞ − − −= ⋅ =

−∫

( ) 0 02 20

sin cosj jj e eX

ω ω ωωω ω

− ∞ − ∞− ∞ − ∞ +=

−

But what are andsin∞ cos ?∞

• The Fourier Transform


Circumvent the Problem

The evaluation of the Fourier integral would be simpler if the function x(t)

would approach zero for every large values of t. The solution of the Fourier

integral becomes possible if x(t) is multiplied by a damping function ,

where σ is a positive real number.

( ) ( )0

, t j tX f x t e e dtσ ωσ∞ − −′ = ⋅ ∫

( ) 00, sin t j tX f t e e dtσ ωσ ω

∞ − −′ = ⋅ ⋅ ∫

( ) ( ) ( ) ( )

( )0 0 0

220 0

sin cos,

j t j tj e t e tX

j

σ ω σ ωσ ω ω ω ωω σ

ω σ ω

∞− + − +− + −

′ =+ +

• The Laplace transform offers a way to circumvent this problem.

te σ−

( ) ( ) ( ) ( )

( )( )

( )

0 00 02 22 2

0 0

sin cos sin0 cos0,

j jj e e j e eX

j j

σ ω σ ωσ ω ω σ ω ωω σ

ω σ ω ω σ ω

− + ∞ − + ∞− + ∞ − ∞ − + −′ = −

+ + + +

( )( )

022

0

,Xj

ωω σω σ ω

′ =+ +

For ( ) 0sinx t tω=

( ) ( ) 0j t j

te eσ ω σ ω− + − + ∞

=∞= →


The Fourier Transform of the Sinusoid

( )( )

022

0

,Xj

ωω σω σ ω

′ =+ +

• The Fourier integral X(f) is now obtained by letting

( ) 02 20

Xωω

ω ω=

−

Thus the Fourier transform of any function x(t) is obtained by first introducing

a damping function evaluating the integral for and finally

letting .

( )- 00 1te eσσ → =≃

- te σ 0σ >0σ →

• The Fourier integral is obtained with x(t) multiplied by a damping

function:

( ),X ω σ′


Definition of the Laplace Transform

( ) ( )0

, t j tX f x t e e dtσ ωσ∞ − −′ = ⋅ ∫

( ) ( ) ( )0

, j tX f x t e dtσ ωσ∞ − +′ = ⋅∫

• Define s jσ ω= + , where s is called the complex frequency

( ) ( ) ( ),X f X j X sσ σ ω′ = + =

( ) ( )0

stX s x t e dt∞ −= ⋅∫

This is the definition of

the Laplace transform

• As we said, the evaluation of the Fourier integral would be simpler if

the function x(t) is multiplied by a damping function , where σ is a

positive real number.

te σ−

• The Fourier transform can be obtained by letting , i.e.

( ) ( )s j

X X sω

ω=

=

0σ → s jω=


Historical Points of View (I)

Watt 1736-1819Coulomb 1736-1806

Ampere 1775-1836Volta 1745-1827

Ohm 1789-1854

Kirchhoff 1824-1887

1600 19001700

Joule 1818-1889

1609

顯微鏡1752

避雷針1710

溫度計1769

蒸汽機1791

輪船1800

電池

1804

鐵路機車

1807

蒸汽船

1821

電動機

1826

內燃機

1831

發電機

1824

都卜勒效應

1836

縫紉機

1843

冰淇淋

1801

紡織機

1870

汽油引擎

1877

留聲機麥克風

1889

汽車

1893

無線電

Department of Electronic Engineering, NTUT10/61

Lagrange 1736-1813

Newton 1643-1727Leibniz 1646-1716

Fourier 1768-1830Laplace 1749-1827

Arbogast 1759-1803Heaviside 1850-1925

1800

1657

擺鐘1679

壓力鍋1643

晴雨表

1609

克卜勒行星運動定律

1610

伽利略提出太陽自轉

1621

斯耐爾折射定律

1687

牛頓自然哲學的數學原理

(萬有引力, 三大運動定律)

1690

惠更斯以太說

1652

富蘭克林論電與電氣相同

1685

庫侖定律

Historical Points of View (II)

1774-1783

美國獨立戰爭

1774

大陸會議

1776

傑佛遜獨立宣言,

美國成立

1789

美國憲法生效華盛頓第一任總統

1861

南北戰爭爆發

1865

林肯遇刺身亡

1662-1722 康熙 1795-1908 嘉慶, 道光, 咸豐,同治, 光緒1722-35-95 雍正, 乾隆

1754 吳敬梓歿

1763 曹雪芹歿

1796-1804

白蓮教起義1840-42 第一次

鴉片戰爭

1851 洪秀全成立太平天國

1852 曾國藩成立湘軍

1856-60 第二次鴉片戰爭, 英法

聯軍

1861 慈禧垂簾聽政

1865 李鴻章成立江南製造局

1866左宗棠成立福州造船廠

1885劉銘傳任台灣巡撫

1894中日甲午戰爭

1898譚嗣同,

康有為戌戊變法

1900義和團起義

1784 鹿港開港

1863 雞籠開港

1864 打狗開港

1874 牡丹社事件

1884 中法戰爭, 砲轟基隆

1885 台灣脫離福建省, 為台灣省

1887 台灣鐵路

1760 英國工業革命開始

1789-1794 法國大革命

1795 法王路易十六上斷頭台

1799-1814 拿破崙王朝

1871 德意志帝國成立 1889 巴黎艾菲爾鐵塔

1895馬關條約


1600 19001700 1800

-1644明朝末

1644

吳三桂降清

1644-62南明

1650

鄭成功居廈門與金門抗清

1661

鄭成功攻台灣逐荷蘭人

1636 大清國皇太極稱帝

1683

施琅攻台鄭克塽降清

1600

英國東印度公司

1600~

英法殖民者於北美拓殖

1624 荷蘭占台

Definition of The Laplace Transform

Department of Electronic Engineering, NTUT

( ) ( )f t F s = L

( ) ( ) -1 F s f t = L

( ) ( )0

stF s f t e dt∞

−= ∫

• The mathematical definition of the Laplace transform is

One-sided or Unilateral Laplace transform

• The process of transformation is indicated symbolically as

• The process of inverse transformation is indicated symbolically as

12/61

Basic Theorems of Linearity

( ) ( ) ( )Kf t K f t KF s = = L L

( ) ( ) ( ) ( ) ( ) ( )1 2 1 2 1 2f t f t f t f t F s F s + = + = + L L L

( ) ( ) ( ) ( )1 2 1 2f t f t F s F s ⋅ ≠ ⋅ L

• Consider the Laplace transform of a function f(t) is

( ) ( )f t F s = L

Let K represent an arbitrary constant, we have

Let f1(t) and f2(t) represent any arbitrary functions, we have

• It is mentioned here that


Step Function

( ) 0 for 0u t t= <1 for 0t= >

( ) ( )U s u t = L

• The unit step function u(t) can be used to describe the process of

“turning on” a DC level at t=0.

1t

( )u t

( ) ( ) 0 for 0u t Ku t t′ = = < for 0K t= >

K

t

( )u t′

( ) ( )U s u t′ ′ = L

0

1ste dts

∞ −= =∫ 0

st KKe dt

s

∞ −= =∫


Exponential Function

( )0

t t stX s e e e dtα α∞− − − = = ∫L

• The Laplace transform of the exponential function .

for 0α >1

t

( ) tx t e α−=

for 0α <1

t

( ) tx t e α−=

( )0

1s te dts

α

α∞ − += =

+∫

te α−


Sine and Cosine Functions

• The Laplace transform of the sinusoid .0cos tω

( )0 0

0 00

0 0 0 0

1sin

2 2

j t j tj t j tst st st ste e

F s te dt e dt e e dt e e dtj j

ω ωω ωω

∞ ∞ ∞ ∞−−− − − − −= = = −

∫ ∫ ∫ ∫

( ) ( )

( )( )

( )( )0 0 0 0

0 00 0 0 0

1 1 1 12 2

s j t s j t s j t s j te dt e dt e ej j s j s j

ω ω ω ω

ω ω

∞ ∞∞ ∞− − − + − − − +

= − = − − − − +

∫ ∫

( ) ( )( )0 0 0

2 2 2 20 0 0 0

1 1 1 12 2

s j s j

j s j s j j s s

ω ω ωω ω ω ω

+ − −= − = = − + + +

( )0 0

0 00

0 0 0 0

1cos

2 2

j t j tj t j tst st st ste e

F s te dt e dt e e dt e e dtω ω

ω ωω∞ ∞ ∞ ∞−

−− − − − += = = +

∫ ∫ ∫ ∫

( ) ( )

( )( )

( )( )0 0 0 0

0 00 0 0 0

1 1 1 12 2

s j t s j t s j t s j te dt e dt e es j s j

ω ω ω ω

ω ω

∞ ∞∞ ∞− − − + − − − +

= + = + − − − +

∫ ∫

( ) ( )( )0 0

2 2 2 20 0 0 0

1 1 1 12 2

s j s j ss j s j s s

ω ωω ω ω ω

+ + −= + = = − + + +

• The Laplace transform of the sinusoid .0sin tω


Damped Sinusoidal Functions

( ) ( )

0 0

s tt stF s e e dt e dtαα∞ ∞

− +− −= =∫ ∫

( )( )

( )0

1 1s tes s

α

α α∞− += =

− + +

1

( )0cos where 0te tα ω α− >

te α−

( ) ( )

( )0 0 2 20 0 0

cos cos s tt st sF s te e dt te dt

s

αα αω ωα ω

∞ ∞− +− − += = =

+ +∫ ∫

( ) ( )

( )0

0 0 2 20 0 0

sin sin s tt stF s te e dt te dts

αα ωω ωα ω

∞ ∞− +− −= = =

+ +∫ ∫

• The Laplace transform of the damped function .te α−

• The Laplace transform of the damped sine function .0sinte tα ω−

• The Laplace transform of the damped cosine function .0coste tα ω−


Transformation Pairs Encountered in Circuit Analysis

1 ( )u t1s

te α− 1s α+

0sin tω

0cos tω

0sinte tα ω−

( )2 20

s

s

αα ω

++ +

2

1s

t

1

!n

ns +

nt

t ne tα−

( ) 1

!n

n

s α ++

( )tδ 1

( )f t ( ) ( )F s f t = L

or

02 2

0sω

ω+

2 20

ss ω+

( )02 2

0s

ωα ω+ +

0coste tα ω−

( )f t ( ) ( )F s f t = L


Transform Example

( ) 4 210 5 12sin3 4 cos5t tf t e t e t− −= + + +

( ) ( )( )22 2 2

4 210 5 12 34 3 2 5

sF s

s s s s

+⋅= + + ++ + + +

( )2 2

4 210 5 364 9 4 29

s

s s s s s

+= + + +

+ + + +

• Find the Laplace transform of f(t):

<Sol.>

( ) 4 210 5 12sin3 4 cos5t tf t e t e t− −= + + +

constant damped sinusoid damped cosine


OperationOperationOperationOperation f(t)f(t)f(t)f(t) F(s)F(s)F(s)F(s)

Laplace Transform Operations

( )f t′ ( ) ( )0sF s f−

( )0

tf t dt∫

( )F s

s

( )te f tα− ( )F s α+

( ) ( )f t T u t T− − ( )sTe F s−

( )0f ( )lims

sF s→∞

( )limt

f t→∞

( )0

lims

sF s→

poles of sF(s) must be in left-hand

half-plane. (stable)

Differentiation

Integration

Multiplication byte α−

Time shifting

(Frequency shifting)

Initial value theorem

(初值定理)

Final value theorem

(終值定理)


Differentiate Operation

( ) ( )0

stF s f t e dt∞

−= ∫

• If the Laplace transform of f(t) is F(s), prove that Laplace transform of

f’(t) is sF(s)-f(0).

( ) ( )0 0

st stdf tf t e dt e dt

dt

∞ ∞− −′ =∫ ∫

Let ,stu e−= and( )df t

dt dvdt

= ( )v f t=

( ) ( ) ( ) ( ) ( ) ( )0 00

0

st st st st stdf te dt f t e f t de f t e s f t e dt

dt

∞ ∞ ∞∞− − − − − = − = − − ⋅ ∫ ∫ ∫

( ) ( ) ( ) ( ) ( )0

00 0s s stf e f e s f t e dt sF s f

∞− ⋅∞ − ⋅ − = ∞ − + = − ∫


Use Differential Operation to Find the Transform of Sinusoid

( ) 02 2

0

F ss

ωω

=+

( ) 0sinf t tω=

( ) 0 0cosf t tω ω′ =

( ) ( ) ( ) 02 2

0

0 0s

f t sF s fs

ωω

′ = − = − +L

[ ] [ ] 00 0 0 0 2 2

0

cos coss

t ts

ωω ω ω ωω

= =+

L L

[ ] 2 2coss

ts

ωω

=+

L

• Let and its Laplace transform is know as .

Find the Laplace transform of by using the differential

operation.0cos tω

( ) 0sinf t tω=


Use Integral Operation to Find the Transform of Ramp Function

( ) 1f t = ( ) 1F s

s=

( ) ( )0 0

1t t

x t f t dt dt t= = =∫ ∫

( ) ( ) ( )20

1 1 1t F sx t f t dt

s s s s = = = =

∫L L

( ) ( ) [ ] 2

1X s x t t

s = = = L L

( ) 1f t =• Let and its Laplace transform is know as .

Find the Laplace transform of the ramp function by using the

integral operation.

( )x t t=

( ) 1F s

s=


Degree or Order

( ) ( )( )

N sF s

D s=

( ) 11 0

n nn nN s a s a s a−

−= + + +⋯

( ) 11 0

m mm mD s b s b s b−

−= + + +⋯

( ) ( ) ( ) ( )5 4 3 2

2 2

19 160 1086 3896 8919 10440

4 9 4 29

s s s s sF s

s s s s s

+ + + + +=+ + + +

order = m (means m roots)

order = n (means n roots)

• Most transforms of interest in circuit analysis turn out to be expressed

as ratios of polynomials in the Laplace variables s. Define the

transform function F(s) as

Numerator polynomial

Denominator polynomial


Zeros and Poles

( ) 0zN s = ( ) 0zF s =

( ) 0pD s = ( )pF s = ∞

( ) ( )( )

N sF s

D s=• For

ZerosZerosZerosZeros of F(s): roots of the numerator polynomial N(s)

PolesPolesPolesPoles of F(s): roots of the denominator polynomial D(s)

( ) 11 0

m mm mD s b s b s b−

−= + + +⋯ Denominator polynomial

can also be completely specified by its roots except for a constant

multiplier mb in factored form as

( ) ( ) ( ) ( )1 2m mD s b s s s s s s= − − −⋯


Classification of Poles

• The poles can be classified as either real, imaginary, or complex.

• An imaginary or complex pole is always accompanied by its complex

conjugate, i.e., jy is accompanied (-jy) and (x+jy) is with (x-jy).

x jy x+jy

• The poles can also be classified according to their order, which is the

number of times a roots is repeated in the denominator polynomial

(重根).

• The first-order (or simple-order) root is which the root appears only

once. Higher-order roots are referred to as multiple-order roots.

• The roots of second-order equations may be either real or complex. For

third- and higher-degree equations, numerical methods must often be

used.


Example – Inverse Laplace Transform

( ) 2

10 15 203

F ss s s

= + ++

( ) 310 15 20 tf t t e−= + +

( ) 2

8 3025

sF s

s+=+

( ) 2 2 2 2

58 6

5 5s

F ss s

= ++ +

( ) 8cos5 6sin5f t t t= +

( ) ( )( )

( )( ) ( )2 2 22 2 2

2 3 20 32 26 52 4

6 34 3 25 3 5 3 5

s ssF s

s s s s s

+ + ++= = = ++ + + + + + + +

( ) 3 32 cos5 4 sin5t tf t e t e t− −= +

• Find the inverse Laplace transform of F(s):

<Sol.>


<Sol.>


<Sol.>

( ) 2

2 266 34

sF s

s s+=

+ +


Example – Classify the Poles

( ) ( )( ) ( ) ( ) ( )2 2 2 23 2 16 6 34 8 16

N sF s

s s s s s s s s=

+ + + + + + +

9 poles 1 2 9, , ,s s s⋯

1 0s = 2

3 9 81

2s

− + −= = − 3

3 9 82

2s

− − −= = −

4 4s j= + 5 4s j= −

6

6 36 1363 5

2s j

− + −= = − − 7 3 5s j= − +

8

8 64 644

2s

− + −= = − 9

8 64 644

2s

− − −= = −

Real, 1st order (3 poles): s1, s2, s3

Imaginary, 1st order (2 poles): s4, s5

Complex, 1st order (2 poles): s6, s7

Real, 2nd order (2 poles): s8, s9

• Given F(s):

and where N(s) is not specified but known that no roots coincide with

those of D(s). Classify the poles.

<Sol.>


Inverse Laplace Transform Step 1

• For the purpose of inverse transformation, poles will be classified in 4

categories

1. First-order real poles

2. First-order complex poles

3. Multiple-order real poles

4. Multiple-order complex poles

(purely imaginary poles will be considered as a special case of complex poles with zero real part)

• Step 1: Check Poles

( ) ( ) ( )2 2

50 75

3 2 4 20

sF s

s s s s

+=+ + + +

( ) ( ) ( )2

50 75

1 2 4 20

s

s s s s

+=+ + + +

real real complex conjugate


Inverse Laplace Transform Step 2 (I)

• Step 2: Partial Fraction Expansion

( ) ( ) ( ) ( )2

50 75

1 2 4 20

sF s

s s s s

+=+ + + +

( ) ( ) ( )1 2 1 2

21 2 4 20

A A B s Bs s s s

+= + ++ + + +

A1,A2,B1,B2 are constants to be determined.

• Note that the single-pole denominator terms require only a constant in the

numerator, but the quadratic term requires a constant plus a term

proportional to s.

• Various procedures exist for determining the constants, but the results can

always be checked by combining back over a common denominator of

necessary to see if the original function is obtained.


Inverse Laplace Transform Step 2 (II)

( ) ( ) ( ) ( )1 2 1 2

21 2 4 20

A A B s BF s

s s s s

+= + ++ + + +

( ) ( )2 21 2 sin 4t t tf t Ae A e Be t θ− − −= + + +

B and are determined form B1 and B2.

• A first-order real pole corresponds to an exponential time response

term.

• A quadratic factor with complex poles corresponds to a damped

sinusoidal time response term. Said differently, a pair of complex

conjugate poles corresponds to a damped sinusoidal time response

term.

• The inverse transform of :

θ


General Algorithm – Find Coefficients

( ) ( )( ) ( ) ( ) ( ) ( )1

2 2 21 2 1 1 2 2r r rk R R

AF s

s s s s a s b s a s b s a s bα α α=

+ + + + + + + + +⋯ ⋯

1 2, , ,r r rkα α α− − −⋯ ( ) ( ) ( )1 1 2 2, , , R Rj j jα ω α ω α ω− ± − ± − ±⋯

( ) ( ) ( ) ( ) ( ) ( ) ( )1 2 1 2e e ek s s sRf t f t f t f t f t f t f t= + + + + + + +⋯ ⋯

( ) ktek kf t A e α−=

( ) ( )k

k k sA s F s

αα

=−= +

( ) ( )sinRtsR R R Rf t B e tα ω θ−= +

( ) ( )21R

R R

jR R R R rR

R s j

B e B s a s b F sθ

α ω

θω

=− +

= ∠ = + +

• Give F(s) in the factored form:

• The corresponding inverse transform:

coeff. coeff.


Example

( ) 2

6 427 10

sF s

s s+=

+ +

( ) ( )( ) ( )

6 7

2 5

sF s

s s

+=

+ +

( ) ( ) ( ) 2 51 2 1 2

t te ef t f t f t Ae A e− −= + = +

( ) ( ) ( )( )

( )( )1 2

2

6 7 6 2 72 10

5 2 5ss

sA s F s

s=−=−

+ − += + = = =

+ − +

( ) ( ) ( )( )

( )( )2 5

5

6 7 6 5 75 4

2 5 2ss

sA s F s

s=−=−

+ − += + = = = −

+ − +

( ) 2 510 4t tf t e e− −= −


<Sol.>


Example

( )2

3 2

10 42 244 3

s sF s

s s s+ +=

+ +

( ) ( ) ( )210 42 24

1 3s s

F ss s s

+ +=+ +

( ) 31 2 3

t tf t A A e A e− −= + +

( ) ( )2

1

0

10 42 24 248

1 3 1 3s

s sA

s s=

+ += = =+ + ⋅

( )2

2

1

10 42 24 10 42 244

3 1 2s

s sA

s s=−

+ + − += = =+ − ⋅

( ) ( )2

3

3

10 42 24 90 126 242

1 3 2s

s sA

s s=−

+ + − += = = −+ − ⋅ −

( ) 38 4 2t tf t e e− −= + −


<Sol.>


Example

( ) ( )( ) ( )2

20 2

1 2 5

sF s

s s s s

+=

+ + +

( ) ( ) ( ) ( ) ( )1 2 1 2 sin 2t te e sf t f t f t f t A A e Be t θ− −= + + = + + +

( )( )( )1 2

0

20 2 20 28

1 51 2 5s

sA

s s s=

+ ⋅= = =⋅+ + +

( )( )2 2

1

20 2 20 15

1 42 5s

sA

s s s=−

+ ⋅= = = −− ⋅+ +

( ) ( ) ( )( )

( )( ) ( )

2

1 2 1 2

20 2 10 1 21 12 5

2 2 1 1 2 2s j s j

s jB s s F s

s s j jθ

=− + =− +

+ +∠ = + + = =

+ − +

( )( ) ( )

10 2.2361 63.4355 143.13

2.2361 116.565 2 90

⋅ ∠= = ∠ −

∠ ∠

( ) ( )8 5 5 sin 2 143.13t tsf t e e t− −= − + −


<Sol.>


Example

( ) ( ) ( )2 2

100

4 2 10

sF s

s s s=

+ + +

( ) ( ) ( )1 2s sf t f t f t= +

( ) ( )01 1 1sin 2t

sf t B e t θ−= + ( )1 1sin 2B t θ= +

( ) ( )2 2 2sin 3tsf t B e t θ−= +

( ) ( ) ( )( )

21 1 2

2 2

50 21 1 1004

2 2 2 10 4 10 4s j s j

jsB s F s

s s jθ

= =

∠ = + = =+ + − + +

( ) ( ) ( )( )

22 2 22

1 3 1 3

33.33 1 31 1 1002 10 14.6176 127.875

3 3 4 1 3 4s j s j

jsB s s F s

s jθ

=− + =− +

− +∠ = + + = = = ∠ −

+ − + +


<Sol.>

100 9013.8675 56.3099

7.2111 33.6901∠= = ∠

∠

( ) ( ) ( )13.8675sin 2 56.3099 14.6176 sin 3 127.875tsf t t e t−= + + −


Inverse Transform of Multiple-order Poles (I)

• Algorithm for Multiple-order Real Poles

( ) ( )( )i

Q sF s

s α=

+( ) ( ) ( )i

Q s s F sα= +

roots@i s α= −

The time function due to the pole of order i with a value will be

the form:

α−

( ) ( ) ( ) ( )1 2

1 2

1 ! 2 ! !

i i i ktk

m i

C t C t C tf t C e

i i i kα

− − −−

= + + + + + − − − ⋯ ⋯

A give coefficient Ck can be determined from the expression:

( ) ( )1

1

11 !

k

k ks

dC Q s

k ds α

−

−=−

=−


Inverse Transform of Multiple-order Poles (II)

• For a first-order real pole:

( )0 1

110!

t tm

C tf t e C eα α

−− −= =

( ) ( ) ( )1

10! s s

C Q s s F sα α

α=− =−

= = +

• For second-order real poles:

( ) ( ) ( )2Q s s F sα= +

( ) ( )1 2t

mf t C t C e α−= +

( )1 sC Q s

α=−=

( )2

s

dQ sC

dsα=−

=

Department of Electronic Engineering, NTUT

For complex poles, find the complex coefficients Ck with s jα ω= − ±

38/61

s-domain Circuit Analysis

Time domain circuit for which a general solution is desired

Convert circuit to s-domain form

Solve for desired response in s-domain

Determine inverse transform of desired response

Desired time domain response


Transform Impedances (I)

Passive

RLC Circuit

( )i t

( )v t Z(s)

( )I s

( )V ss-domain

( ) ( )I s i t = L

( ) ( )V s v t = L

( ) ( )( )

V sZ s

I s=

( ) ( )( )( )

1 I sY s

Z s V s= =

Transform impedance Z(s)

Transform admittance Y(s)


Transform Impedances (II)

( )i t

( )v t R

( )i t

( )v t

( )i t

( )v t

C

L

( )V s

( )I s

R

1sC

( )V s

( )I s

sL( )V s

( )I s

R

C

L

( )V ω

( )I ω

R

ω1

j C( )V ω

( )I ω

ωj L( )V ω

( )I ω

0σ →

0σ →

0σ →


Example – Transform Impedance

1 kR = Ω

0.5 FC µ=

30 mHL =

1 kΩR =

( )6

6

1 1 2 10

0.5 10sC ss −

⋅= =⋅

( )330 10 0.03sL s s−= ⋅ =


Models for Initially Charged Capacitor

0VC+

− 0Vs

1sC

+−

0CV1sC

0 60 VV =0.2 FC µ=+

−60s

650 10s⋅

+−

612 10−⋅650 10

s⋅

Thevenin’s equivalent circuit Norton’s equivalent circuitInitially charged capacitor

Example:


Models for Initially Fluxed Inductor

0IL0LI

sL

+−

0Is

sL

0 0.4 AI =50 mHL =0.02

0.05s

+−

0.4s0.05s

Thevenin’s equivalent circuit Norton’s equivalent circuitInitially fluxed inductor


Complete Circuit Models

1. Transform the complete circuit from the time-domain to the s-domain

( ) ( )v t V s→ ( ) ( )i t I s→

L sL→1

CsC

→

00

VV

s→ 0

0

II

s→

2. Solve for the desired voltages currents using the s-domain model.

3. Using inverse Laplace transform to determine the corresponding time-

domain forms for the voltages or currents of interest.


Example (I)

20 V+−10cos3t +

−

0t =

2 H1

F4 3 Ω 4 Ω 5 H

+

−8 V

1 F

6( )1i t ( )2i t

20s

+−2

109

ss +

+−

2s4s 3 4 5s

8s

6s( )1I s ( )2I s+

−

Transform from time-domain to s-domain


Example (II)

20s

+−2

109

ss +

+−

2s4s 3 4 5s

8s

6s( )1I s ( )2I s+

−

( ) ( ) ( ) ( ) ( )1 1 1 1 22

10 4 6 82 3 0

9s

sI s I s I s I s I ss s s s−

+ + + + − + = +

( ) ( ) ( ) ( )2 1 2 2

8 6 204 5 10 0I s I s I s sI s

s s s−

+ − + + + + =

Mesh 1:

Mesh 2:

Solve for I1(s) and I2(s)

Apply the circuit laws:


General Forms for Solutions (I)

( ) ( ) ( )n fy t y t y t= +

• Natural and Forced Responses

Let represent some arbitrary general circuit response (either a voltage

or current). When the circuit is excited by one or more sources, a general

response may be represented by the sum of two responses as follows:

( )y t

( )y t

Natural response Forced response

• Natural Response

The form of the natural response is determined by the circuit parameter,

i.e., if the circuit has a time constant of 2 seconds, corresponding to an

exponential , such a term will appear in the response when the circuit

is excited by any type of source.

2te−

• Force Response

The form of the forced response is determined by the excitation source(s),

i.e., if a circuit is excited by a sinusoid having a frequency of 5 kHz, the

general response will always contain a sinusoid with a frequency of 5 kHz.


General Forms for Solutions (II)

( ) 10sin1000sv t t=

( ) ( )22 3 4sin 1000 30t ti t e e t− −= − + +

( ) ( ) ( )4sin 1000 30ss fi t i t t= = + ( ) ( ) 22 3t tt ni t i t e e− −= = −

( ) ( ) ( )t ssy t y t y t= +

• Transient and Steady-state Responses

Transient response Steady-state response

Frequently, transient response and natural response are considered to be

equivalent, and steady-state response and forced response are considered

to be equivalent. The terms transient and steady-state relate to the

common case where the natural response is transient in nature and

eventually vanishes, whereas the forced response continuous as a steady-

state condition indefinitely.

Example:


First-Order Circuits

• First-order Circuit with Arbitrary Input

( ) t tny t Ke Keτ α− −= =

For first-order circuits, the natural response will always be an exponential

term of the form:

( )ny t

K is a constant

is the time constantτis the damping factor1α τ=

Since the exponential term approaches zero as time increases, it is proper to

designate the natural response as a transient response whenever the forced

response continuous indefinitely.


Example of a First-Order Circuit (I)

40sin4t +−

0t =

4 Ω

( )i t1

F12

+

−( )cv t

2

16016s +

+−

4 Ω

( )I s12s

+

−( )cV s

( ) ( )2

160 124 0

16I s I s

s s− + + =

+

( ) ( ) ( )2

40

3 16

sI s

s s=

+ +

( ) ( )34.8 8sin 4 36.87ti t e t−= − + +

( ) ( ) ( ) ( )2

12 480

3 16CV s I ss s s

= =+ +

( ) ( )319.2 24sin 4 53.13tCv t e t−= + −

• Use Laplace transform techniques, determine the current and

voltage for .( )i t

( )Cv t 0t >


Apply KVL:



Time constant = 1/3 = RC

Damping factor = 3


Example of a First-Order Circuit (II)

( ) ( )306 3 0

1I s s I s

s− + + ⋅ =

+

( ) ( ) ( )10

1 2I s

s s=

+ +

( ) 210 10t ti t e e− −= −

• Use Laplace transform techniques, determine the for .( )i t 0t >


Apply KVL:

Forced response Natural response

Time constant = 1/2 = L/R

Damping factor = 2

30 te− +−

0t =

6 Ω

( )i t 3 H

301s +

+−

6

( )I s 3 s


Second-Order Circuits

( ) ( ) ( )10iV

sLI s RI s I ss sC

− + + + =

( )2 1

iVLI sR

s sL LC

=+ +

( ) ( )2

11

i

C

VLCV s I s

RsC s s sL LC

= = + +

• Second-order circuits are of special interest because they are capable

of displaying, on a simple scales, the types of responses that appear

in circuits of arbitrary order. In fact, second-order circuits and systems

occur frequently in practical applications, so their behavior is subject

of considerable interest.

• Series RLC Circuit

iV +−

0t =

R

( )i t

+

−( )cv t

L

C

iVs

+−

R

( )I s

+

−( )CV s

sL

1sC


3 Possible Forms of the Roots

21

22

12 4

s R Rs L L LC

= − ± −

• The roots of the second-order circuit

• Three possibilities for the roots s1 and s2

Overdamped Case (the roots are real and different) :2

2

14RL LC

>

Critically Damped Case (the roots are real and equal) :2

2

14RL LC

=

Underdamped Case (the roots are complex) :2

2

14RL LC

<


Overdamped Case (Series RLC)

( ) ( )( )1 2

iVLI s

s sα α=

+ +

( ) ( ) ( )1 2

i

C

VLCV s

s s sα α=

+ +

( ) 1 20 0

t ti t A e A eα α− −= −

( ) 1 21 2

t tC iv t V Ae A eα α− −= + +

• In the overdamped case, the two poles are real and different. Assume

that the poles are and , the forms for I(s) and VC(s)

can be expressed as1 1s α= − 2 2s α= −

• The inverse transforms are of the forms

The natural response consists of two exponential

terms, each having a different damping factor or

time constant. The forced response for the current

is zero and for the capacitor voltage is the

constant final voltage across the capacitor.


Critically Damped Case (Series RLC)

( )( )2

iVLI s

s α=

+

( )( )2

i

C

VLCV s

s s α=

+

( ) 20

t Rt LiVti t C te e

Lα− −= =

( ) ( ) 21 2

Rt LC iv t V C t C e−= + +

• In the critically damped case, the two poles are real and equal.

Assume that the poles are , the forms for I(s) and VC(s)

can be expressed as1 2s s α= = −

The damping factor:2RL

α =


0 iC V L=

The most significant aspect of the natural

response function for the critically damped case is

the form. Although the t factor increases

with increasing t, the decreases at a faster

rate, so the product eventually approaches zero.

tte α−

tte α−


Underdamped Case (Series RLC)

1 ds jα ω= − + 1 ds jα ω= − −

2RL

α =2

12d

RLC L

ω = −

( )( )2 2

i

d

VLI s

s α ω=

+ +( )

2 1

i

C

VLCV s

Rs s s

L LC

= + +

( ) sintid

d

Vi t e t

Lα ω

ω−=

( ) ( )sintC i dv t V Be tα ω θ−= + +

• In this case, the two poles are complex and denoted as

• The forms for I(s) and VC(s) can be expressed as

and

where and

is called the damped

natural oscillation frequencydω

and


The natural response is oscillatory. Depending on

the value of , this response may damp out very

quickly, or it may continue for a reasonable period

of time.

α


Comparison of Response Forms (I)

Overdamped Case :

Critically Damped Case :

Underdamped Case :

( ) 1 20 0

t ti t A e A eα α− −= −

( ) 20

t Rt LiVti t C te e

Lα− −= =

( ) sintid

d

Vi t e t

Lα ω

ω−=


( )i t

t

Underdamped

Critically damped

Overdamped

Comparison of Response Forms (II)

Overdamped Case :

Critically Damped Case :

Underdamped Case :

( ) 1 21 2

t tC iv t V Ae A eα α− −= + +

( ) ( ) 21 2

Rt LC iv t V C t C e−= + +

( ) ( )sintC i dv t V Be tα ω θ−= + +


overshoot

Underdamped

Critically damped

Overdamped

Final level= iV

( )Cv t

t

Example (I)

40 ViV = +−

0t =

400 Ω

( )i t

+

−( )Cv t

2 H

0.5 Fµ

• Use Laplace transform techniques, determine the current and

voltage for .( )i t

0t >( )Cv t

40s

+−

400

( )I s

+

−( )CV s

2s

62 10s

⋅

( )6 2 62 10 2 400 2 10

2 400s s

Z s ss s

⋅ + + ⋅= + + =

( ) ( )( ) 2 6

40

2 400 2 10

V s sI ss sZ s

s

= =+ + ⋅

( )22 6 2

20 20200 10 100 994.987s s s

= =+ + + +

( ) ( )1000.0201008 sin 994.987ti t e t−= ⋅

( ) ( ) ( ) ( )6

2 6

40 10

200 10V s I s Z s

s s s

⋅= =+ +

( )( )

−= +

× −

10040 40.2015

sin 994.987 95.7392

tCv t e

t


Example (II)

( ) ( )−= + ⋅ − 10040 40.2015 sin 994.987 95.7392

tCv t e t

( ) ( )1000.0201008 sin 994.987ti t e t−= ⋅


( ), mAi t( ), VCv t

20

10

, mst10

20

30

60

40

20

0

−20

Final voltage = 40 V

( )i t

( )Cv t