logics of the laplace transform

DynamicSystem

x(t) input y(t) output

A processor which processes the input signal to produce the output

DynamicSystem

X(f) Y(f )

H(f ) Algebraic equation, not differential equation

LOGICS OF The Laplace Transform BY TARUN GEHLOT

Introduction

(1) System analysis

static system: y(t) = ax(t) => easy (simple processing)

dynamic system:

dy(n )( t )dtn

+a1

dy(n−1)( t )dt n−1

+.. .+an y ( t )=b0

dx(m )( t )dtm

+. . .+bm x (t )

Can we determine y(t) for given u(t) easily?

Easier solution method

Y ( f )=H ( f )X ( f )y ( t )=F−1(Y ( f ))

(1) we have systematic way to obtain H(f) based on the differential equation (2) we can obtain X(f)

Fourier transform: an easier way

(2) Problem Fourier transform of the input signal:

Page 5-1

X ( f )=∫−∞

∞

x ( t )e− j 2π ft dt

|e− j2 π ft|=1if x(t) does not go to zero when t →∞ and t →−∞ X(f) typically does not exist! (the existence of X(f) if not guaranteed)

A very strong condition, can not be satisfied by many signals!

(3) Solution:

Why should we care about t<0 for system analysis? We do not care!

x(t) does not go to zero (when t →∞ ), but x ( t )e−σt may!

(Will be much easier. )

use “single-sided” Fourier transform of x ( t )e−σt, instead of “double-

sided” Fourier transform of x(t).

∫0

∞

( x (t )e−σt )e− jωt dt=∫0

∞

x ( t )e−(σ + jω) t dt

very useful, let’s use a new name for it: Laplace transform.

(4) Laplace Transform

Definition: Laplace transform of x(t)

L[ x ( t ) ]=X ( s )=∫0

∞

x ( t )e−st dt (s=σ+ jω)

What is a Laplace transform of x(t)? A time function? No, t has been eliminated by the integral with respect to t!

A function of s ( s is complex variable)

(5) System analysis using Laplace transform

Page 5-2

DynamicSystem

X(s) Y(s )

G(s )

Inverse Laplace transform

Y ( s )=G( s )X (s )y ( t )=L−1 (Y (s ))

(6) How is the inverse transform defined?

x ( t )= 12 πj

∫σ − j∞

σ + j ∞

X ( s )est ds

(7) Will we often use the definition of the inverse transform to find time function? No! What will we do?

Express X(s) as sum of terms for which we know the inverse transforms!

5-2 Examples of Evaluating Laplace Transforms using the definition

(1) x(t)=1 and step function x(t)=u(t)

L[ x ( t )=u( t ) ]=∫0

∞

x( t )e−st dt=∫0

∞

e−st dt=−1s∫

t=0

t=∞

e−st d (−st )

=−e−st

s|t=0

t=∞=−e−σt

se− jωt|t=0

t=∞=−e−σ ∞

se− jω∞+e−σ 0

se− jω 0

(|e− jω∞|=1 e−σ ∞=0 , (if σ>0 ) e−σ ∞→∞ , ( if σ<0 )

=1s

(cos0− j sin 0 )=1s

⇒L(1 )=L[u ( t )]=1s

( Re(s ))>0

(2) x ( t )=e−αt u (t )

L[ e−αt u( t ) ]=∫0

∞

e−αt e− st dt=∫0

∞

e−(α+ s)t dt

Define a new complex variable s~

=s+α

Page 5-3

∫0

∞

e−s~

t dt

we know ∫0

∞

e−st dt=1s

Re(s )>0

⇒∫0

∞

e−s~

t dt =1

s~ Re(s

~

)>0

⇒∫0

∞

e−(α+s) t dt=1s+α

Re( s+α )>0

⇒L [e−αt u( t ) ]=1s+α

Re (s+α )>0 or Re(s )>−Re(α )

⇒L [e−αt ]=1s+α

(3) x ( t )=δ ( t )

L[ δ( t ) ]=∫0

∞

δ( t )e−st dt

¿e−st|t=0=e−σt e− jωt|t=0

¿e−σt (cos ωt− jω sin ωt )|t=0

¿1 No constraint on s.

5-2B Discussion: Convergence of the Laplace Transform

(1) To assure ∫0

∞

x ( t )e−st dt=∫0

∞

x ( t )e−σt e− jωt dt converge, σ=Re( s ) must be psotive

enough such that x ( t )e−σt goes to zero when t goes to positive infinite

(2) Region of absolute convergence and pole

Page 5-4

(3) How to obtain Fourier transform form Laplace transform:

L[ x ( t ) ]=X ( s ) ⇒s= jω

X ( jω )=F ( x( t ))

Important: why introduce Laplace transform; definition of Laplace transform as a modification of Fourier transform; find the Laplace transforms of the three basic functions based on the (mathematical) definition of Laplace transform.

Chapter structure

Part one: Definition (5.1)Part two: Direct evaluation of Laplace transforms of simple time-

domain function.Easy? Not at all!

complex (not simple) functions: even harder tools for application of Laplace transform in system analysis

Part three: Rest of the chapter What tools: tools for easier Laplace transform evaluation

tools for easy inverse transform

Page 5-5

5-3 Some Laplace Transform theorems (Tools for evaluating Laplace transform based on the Laplace

transforms of the basic functions)

5.3.1 Linearity

Assume x ( t )=a1 x1 ( t )+a2 x2( t ) (a1 and a2 are time independent)

X1 (s )=L[ x1( t ) ] , X2 (s )=L [ x2( t ) ]

then X ( s )=L[ x ( t )]=a1 X1( s )+a2 X 2(s )

HW#2-1: Assume x ( t )=a1( t ) x1 ( t )+a2 ( t )x2( t ) , X1 (s )=L[ x1( t ) ] and X 2(s )=L[ x2( t ) ] .

(a) Is X ( s )=L[ x ( t )] equal to a1( t ) X1( s )+a2 ( t )X 2( s )? (Answer: Yes or No)

(b) If A1 (s )=L [a1( t ) ], is it trueL[ x ( t ) ]=A1( s ) X1( s )+ A2( s ) X2( s ) ?(Answer: Yes or No)

Example 5.1

(1) Find L(cosω0 t )

Key to solution : express (cos ω0 t ) as linear combination of δ ( t ), u( t ) ,

and/or e−αt

:

L[ δ( t ) ]=1

L[u ( t ) ]=1s

L[ e−αt ]=1s+α

let α= jω0

L[ e− jω0 t

]= 1s+ jω0

let α=− jω0

L[ ejω0 t

]=L[ e−(− jω0 )t

]= 1s− jω0

Page 5-6

Can we use e− jω0 t

and ejω0 t

to express cos (ω0 t ) ?

e− jω0 t

=cos(−ω0t )+ jsin (−ω0 t )=cos(ω0 t )− jsin (ω0 t )

ejω0 t

=cos(ω0 t )+ j sin(ω0t )

e− jω0 t

+ejω0 t

=2 cos(ω0t )

⇒ cos (ω0 t )=e− jω0 t

+ejω0 t

2

⇒L [cos (ω0 t )]=12

[ L(e− jω0 t

)+L(ejω

0t) ]

=12 [1s+ jω0

+1s− jω0 ]

=12 [(s− jω0 )+(s+ jω0 )

(s+ jω0 )( s− jω0 ) ] =s

s2+ω02

(2) Find L[ sin ω0 t ]

5-3-2 Transforms of Derivatives

Assume X ( s )=L[ x ( t )]

Then L( dx ( t )

dt )=sX ( s )−x (0−)

Proof:

(1) Definition L[ dx ( t )

d ( t ) ]=∫0

∞ dx ( t )dt

e−st dt=∫0

∞

e−st dx (t )

(2) Integration by parts: General equation:

Page 5-7

∫a

b

u ( t )dv ( t )=u( t )v ( t )|t=at=b−∫

a

b

v ( t )du( t )

(3) Use the above equation

Why? If we assume v ( t )=x (t ) , u( t )=e−st

⇒∫0

∞

v ( t )du( t )=∫0

∞

x ( t )de−st=−s∫0

∞

x ( t )e−st dt

X ( s )=L ( x ( t ))

u( t )=e−st , v ( t )=x ( t )

⇒∫0

∞

u ( t )dv ( t )=∫0

∞

e−st dx ( t )=[ L(dx ( t )dt ) from (1) ]

u( t )v ( t )|t=0t=∞−∫

a

b

v ( t )du( t )

=e−st x ( t )|t=0t=∞+s∫

0

∞

x ( t )e−st dt

=e−st x ( t )|t=0t=∞+sL [ x ( t ) ]

limt →∞

e−st x ( t ) must go to zero. Otherwise,

L[ x ( t ) ]=∫0

∞

x( t )e−st dt does not exist !

⇒ e−st x ( t )|t=0

t=∞=−e−s⋅0 x (0)=−x (0)

use 0−

as lower limit => x (0−)

Page 5-8

⇒u ( t )v ( t )|t=0t=∞−∫

0

∞

v ( t )du( t )

=−x (0−)+sX (s )=sX (s )−x (0−)

HW#2-2: Assume X ( s )=L[ x ( t )] . Prove

L[ d2 x( t )dt ]=s2 X ( s )−sx (0−)−x(1)(0−)

HW#2-3: Express L[ d(n) x ( t )

dtn ] using L[ x ( t ) ]

Example 5-2

Find i(t) using Laplace transform method for t>0

Solution:(1) Before switched from 1 to 2 at t=0

i=42=2 A⇒ i( 0−)=2 A

(2) System equation (t>0)

KVL:

Ldi ( t )dt

+Ri( t )=0 ( L=1 H ) (R=2 ohm)

⇒di ( t )dt

+2i( t )=0

(3) Solve system equation using Laplace transform

Page 5-9

L[di( t )dt

+2 i( t )]=L[di ( t )dt ]+2 L[ i( t ) ]

=sI ( s )−i(0− )+2 I (s )¿ (s+2)I ( s )−2=0

⇒ I ( s )=2s+2

⇒i( t )=2 e−2 t u( t ) A

5-3-3 Laplace Transform of an integral

Assume y ( t )=∫

−∞

t

x ( λ )dλ , X ( s )=L[ x ( t ) ]

Then L[∫

−∞

t

x ( λ )dλ]= X (s )s

+y ( 0−)

s

where y (0−)=∫

−∞

0−

x( λ)dλ

Proof :

L[∫−∞

t

x ( λ )dλ]=∫0−

∞ (∫−∞

0−

x ( λ )dλ)e−st dt

=∫0−

∞

udv =uv|t=0−t=∞ −∫

0−

∞

vdu

(1)

Page 5-10

uv|t=0−t=∞ =(∫

−∞

t

x ( λ)dλ)−e−st

s|

t=0−t=∞

=limt→∞

−e−st

s∫−∞

t

x ( λ )dλ+e−s⋅0

s∫−∞

0−

x ( λ )dλ

=y (0−)s

(2)

∫0−

∞

vdu =−∫0−

∞e−st

sx ( t )dt=−1

s∫0−

∞

x ( t )e−st dt =−1s

X (s )

(3)

L[∫−∞

t

x ( λ )dλ]= y (0−)s

+X (s )

s=

X ( s )s

+y (0−)

s Proved!

Example 5.3:

Find I(s) = L(i(t))

Solution:(1) Differential equationKVL :

x ( t )=vL( t )+vC (t )+vR( t )

0

1

y (0−)

Page 5-11

i( t )=CdvC ( t )dt

¿}dvC ( t )=1C

i( t )dt ¿}∫−∞

t

dvC( λ)=1C ∫

−∞

t

i( λ )dλ ¿}vC (t )=vC (−∞)+1C ∫

−∞

t

i( λ )dλ ¿}¿¿⇒¿ vL( t )=Ldi( t )dt

vC( t )=1C∫−∞

t

i( λ )dλ

vR( t )=Ri ( t )(2) Laplace transform

X ( s )=V L( s )+V C( s )+V R(s )V L(s )=L[ sI (s )−i(0−) ]=LsI ( s ) i(0−)=zero

V C (s )=1C [I ( s )

s+1

s∫−∞

0−

i( λ )dλ ]¿

1Cs

I (s )+1s

1c ∫

−∞

0−

i( λ )dλ

=1Cs

I ( s )+1s

vc (0− )

V R(s )=RI ( s )

Page 5-12

X (s )=LsI ( s )+1Cs

I ( s )+1s

vC (0−)+RI ( s )

⇒ I ( s )=X ( s )−1

svC (0−)

Ls+1Cs

+R

¿sX ( s )−vC (0−)

Ls2+1C

+sR

¿sX ( s )−vC (0−)

L[s2+(RL )s+1

LC ]5-3-4. Complex Frequency shift (s-shift) Theorem

Assume y ( t )=x ( t )e−αt

X ( s )=L[ x ( t )] Y ( s )=L[ y ( t ) ]

Then Y ( s )=X ( s+α )

L[u ( t ) ]=1

s, L [u( t )e−αt ]= 1

s+α

L[ cosω0 t ]= s

s2+ω02

L[ sin ω0 t ]=ω0

s2+ω0

2

=>

L[ cosω0 t⋅e−αt ]= s+α( s+α )2+ω

02

L[sin ω0 t⋅e−αt ]=ω0

(s+α )2+ω0

2

Example 5-4 Find x ( t )=L−1 [ X (s ) ]=L−1 s+8

s2+6 s+13Solution:

X ( s )=s+8s2+6 s+13

=( s+3 )+5

s2+6 s+9+4

¿s+3

( s+3 )2+22+(5 /2)×2

(s+3)2+22

Page 5-13

x ( t )=L−1 [ X (s ) ]=e−3 t cos 2t +52

e−3t sin 2t ( t>0)

5-3-4 Delay Theorem question: How to express delayed function?

Assume L[ x ( t ) ]≡L[ x ( t )u ( t )]=X ( s )

Then L[ x ( t−t0 )u ( t−t0 ) ]=e−st0 X ( s ) ( t0 ¿0 )

(If ( t0<0 ) , it will not be a delay!) Proof :

L[ x ( t−t0 )u ( t−t0 ) ]

¿∫0

∞

x ( t−t0 )u( t−t0 )e−st dt

¿∫0

t0

x ( t−t0 )u( t−t0 )e−st dt +∫t0

∞

x ( t−t0 )u ( t− t0 )e− st dt

=∫t0

∞

x ( t−t 0)e−st dt =∫t

0

∞

x (t−t 0 )e−s (t−t0 )−st0 dt

=e−st0∫

t0

∞

x ( t−t0 )e−s(t−t0 )

d ( t−t 0)

=τ =t−t0

e− st

0∫0

∞

x ( τ )e−sτ dτ =e−st

0∫0

∞

x ( t )e−st dt =e−st

0 X (s )

Page 5-14

Question: will L[ x ( t−t0 )u ( t−t0 ) ]=e−st0 X ( s ) be true if t0<0 ?

No! (it will not be a delay)

Example 5-5: Square wave beginning at t = 0

L[ x sq( t )]=1s−2

1s

e−

T0

2s+2

1s

e−T0s

−21s

e−

3T 0

2s+2

1s

e−2 T 0s

=λ=−e

−12

T 0s

1s+2

1s

λ+21s

λ2 +21s

λ3+21s

λ4+ .. .

=1s+

2s

( λ+λ2+λ3+. .. )

¿1s+2 λ

s(1−λ )=1

s⋅1+ λ1−λ

=1s⋅1−e

−T 0

2s

1+e−

T0

2s

5-3-5 Convolution

Signal 1: x1( t ) Signal 2 : x2( t )

y ( t )=x1 ( t )∗x2( t )=∫−∞

∞

x1( λ )x2 ( t−λ )dλ

if x1( t )=0 ∀ t <0

⇒ y ( t )=∫0

∞

x1( λ )x2 ( t− λ )dλ

Page 5-15

t

if x2( t )=0 ∀ t <0 ( x2( t− λ)=0 ∀ λ>t )

⇒ y ( t )=∫0

t

x1( λ )x2 ( t− λ )dλ

Therefore, if x1( t )=0 , x2( t )=0 ∀ t <0

⇒∫

0

t

x1 ( λ) x2( t−λ )dλ=∫0

∞

x1 ( λ )x2( t− λ)dλ=∫−∞

∞

x1( λ) x2( t−λ )dλ

⇒L [∫0

t

x1( λ )x2 ( t−λ )dλ ]=L[∫0

∞

x1 ( λ )x2( t− λ)dλ ]=L[∫−∞

∞

x1( λ )x2 ( t− λ )dλ ]

∫0

∞

[∫0

∞

x1( λ) x2( t−λ )dλ ]e−st dt=∫0

∞

x1( λ )[∫0

∞

x2 ( t− λ)e−st dt ]dλ

Look at

∫0

∞

x2 ( t− λ)e−st dt=∫0

∞

x2( t−λ )e−s( t−λ )e−sλ dt

=τ=t−λ

dτ=dtt=0→τ=−λ

t=∞→τ=∞

e−sλ∫− λ

∞

x2(τ )e−sτ d τ

=x

2( τ )=0∀ τ∈ [− λ 0 ]

e−sλ∫0

∞

x2(τ )e− sτ d τ=e−sλ X2( s )

Then

Y ( s )=∫0

∞

x1( λ )e−sλ X2( s )d λ

¿ X2 (s )∫0

∞

x1( λ )e−sλ d λ

¿ X1 (s ) X2 (s )

Page 5-16

5-3-7 Product5-3-8 Initial Value Theorem

x ( t )=L−1 [ X (s ) ]⇒ x (0+ )=lim

s →∞sX ( s )

Example: A demonstration where x(0) is obvious

x ( t )=e−αt cos ω0 tu( t )

It is evident: x (0 )=e−0 cos ω0⋅0=1

Using Laplace transform

X ( s )=L[ x ( t )]= s+α

(s+α )2+ω0

2

x (0 )=lims→∞

sX ( s )=lims →∞

s (s+α )(s+α )2+ω

02

¿ lims→∞

s2+αs

s2+2 αs+α 2+ω02

¿ lims→∞

d (s2+αs )/ds

d (s2+2 αs+α 2+ω0

2)/ds(∞∞ )

¿ lims→∞

2 s+α2 s+2 α

=lims→∞

d (2 s+α )/dsd (2 s+2 α )/ds

=lims→∞

22

=1

5-3-9 Final Value Theorem: if x ( t ) and dx ( t )/dt are Laplace transformable, then limt →∞

x ( t )= lims→0

sX ( s )

(condition: sX (s ) has no poles on jω−axis or in the right-half s-plan

or limt →∞ x ( t ) exists)

5-3-10 Scalinga>0: x(at) a times fast (if a>1)

or slow (if a<1) as x(t)

X ( s )=Δ

L [ x ( t ) ] What do we expect on L[ x ( at )]?

Page 5-17

L[ x ( at )]⇒ X ( sa)?

L[ x ( at )]=∫0

∞

x (at )e−st dt =a>01

a∫0

∞

x (at )e−s

a(at )

d( at )

=τ=at

¿

a>0 ¿t=∞ ¿

⇒ τ=∞ ¿¿1a∫0

∞

x (τ )e−( s

a ) τ

dτ=1a

X ( sa ) ¿¿

5-4 Inversion of Rational Functions

(1) Ways to find x ( t ) from X ( s )

(1) x ( t )= 1

2 πj∫

σ −ω

σ +ω

X (s )est dt(Contour Integral)

(2) Transform pair1s

↔u ( t )

1s+α

↔e−αt u ( t )

Therefore X ( s )= 1

s+α⇒ x ( t )=e−αt u( t )

(2) All kinds of Laplace Transform ? No!

We almost only see

b0 sm+b1 sm−1+. ..+bm−1 s+bm

sn+a1 sn−1+. .. an−1 s+an

e−τs

rational function X(s) Delay

Page 5-18

x ( t )=L−1 [ X (s ) ]y ( t )=L−1 [ X (s )e−τs ]=x( t−τ )u( t−τ )

Consider Rational Functions only!

(3) Non proper Rational Function proper Rational Function

Non proper m>=n (b0≠0)proper m<n

Non proper => proper + Polynomial (using long division)

s3+4 s2+6 s+7s2+3 s+2

=s+1+ s+5s2+3 s+2

s+1s2+3 s+2√s3+4 s2+6 s+7

s2+4 s+7s2+3 s+s

s+5 How to find inverse Laplace transform for polynomials?

sn↔ δ(n )( t )L−1( s+1 )=L−1 (s )+L−1(1 )=δ(1)( t )+δ( t )

consider proper rational functions only!

(4) Proper Rational Functions: Partial Fraction Expansion

Page 5-19

sum of

1(s+α )n+1

→ tn e−αt

n!u( t )

s+α(s+α )2+ω

02

→ e−αt cosω0 tu( t )

ω0

(s+α )2+ω02

→ e−αt sin ω0 tu( t )

1→δ( t )1s

→u( t )

1s+α

→e−αt

Let’s look at examples, and then summarize! Techniques:

Common Denominator Factorize first! Specific value of s Expand second! Heaviside’s Expansion Find coefficients third! Matlab

Example 5-9: Simple Factors

Y ( s )=10

s2+10 s+16

Solution:

(1) Factorize and expand Y ( s )=10

( s+8 )(s+2)= A

s+8+ B

s+2(2) Common Denominator Methods

10(s+8 )(s+2)

=A ( s+2 )+B (s+8)( s+8 )( s+2 )

⇒ A( s+2 )+B (s+8)=10⇒¿ { A+B=0⇒ A=−B ¿ ¿¿¿

¿

Page 5-20

Same as real roots!

Y ( s )=−53⋅1s+8

+53⋅1s+2

⇒ y ( t )=(−53

e−8 t+53

e−2 t )u ( t )

specific values of s 10(s+8 )(s+2)

=As+8

+Bs+2

⇒s=0 10

8×2=

A8

+B2

⇒s=2 10

10×4=A

10+B

4 Can you solve for A and B?

Heaviside Expansion10(s+8 )(s+2)

=As+8

+Bs+2

⇒10s+2

=A+B(s+8 )s+2

⇒s=−810

−8+2=A=−5 /3

and ⇒10

s+8=

A( s+2)s+8

+B ⇒s=−2

B=10−2+8

=5/3

Example 5-10 Imaginary Roots

Y ( s )=15 s2+25 s+20( s2+1 )(s+2)( s+8)

Solution: what do we have:

Y ( s )=A1

s+ j+

A2

s− j+

A3

s+2+

A4

s+8

Y ( s )=A1 (s− j)+A2 (s+ j)

s2+1+

A3

s+2+

A4

s+8

¿( A1+ A2 )s+( A2−A1 ) j

s2+1+

A3

s+2+

A4

s+8

A1+A2 must be real number(-A1+A2)j must be real number

Page 5-21

Y ( s )=c1s+c2

s2+1+

A3

s+2+

A4

s+8

Heaviside Expansion => A3=1 and A4= - 2.

Y ( s )=15 s2+25 s+20( s2+1 )(s+2)( s+8)

=c1s+c2

s2+1+ 1

s+2+ −2

s+8

s=1 =>

15+25+202×3×9

=c1+c2

2+ 1

3+−2

9

s=2 =>

15×4+25×2+205×4×10

=2 c1+c2

5+ 1

4+−2

10

Can we solve for c1 and c2?c1=1 c2=1

Y ( s )=s+1

s2+1+

1s+2

−2s+8

¿ s

s2+1+1

s2+1+1

s+2−2

s+8

=>[ cos t+sin t +e−2t−2e−8 t ]u( t )

Too complex: use MATLAB

Example 5-11 Repeated linear Factors

Y ( s )=10 s

( s+2)2( s+8)

¿A1

s+8+

A2

s+2+

A3

( s+2)2

Example 5-12

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Y ( s )=10 s

( s+2)3( s+8)

¿A1

s+8+

A2

s+2+

A3

( s+2)2+

A4

(s+2)3

Example 5-13 Complex - Conjugate Factors

Y ( s )=2 s2+6 s+6( s+2)(s2+2 s+2 )

¿2 s2+6 s+6

( s+2)[(s+1)2+1]=

A1

s+2+

A2

s−1− j+

A3

s−1+ jExample 5-14 Repeated Quadratic Factors

Y ( s )=s4+5 s3+12 s2+7 s+15( s+2)(s2+1 )2

¿A1

s+2+

B1 s+c1

s2+1+

B2s+c2

( s2+1)2

* Summary of Partial–Fraction Expansion(1) Expansion Structure:

Simple Roots (including complex conjugate)

=>

A j

s−α j could be complex.Repeated Roots: m multiplicity

=>

B1

s−β j

+B2

( s−β j )2+. ..+

Bm

( s−β j )m

real number or complex number

(2) Avoid complex number

For complex conjugates: α j=a+ jb

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A j

s−α j

+A

j¿

s−α j¿

⇒cs+D

( s−a )2+b2

Bk

(s−α j )k+

Bk ¿

(s−α j¿ )k

⇒ cs+D

[( s−a )2+b2 ]k

(3) Inverse Laplace transform

A j

s−α j

↔ A j eα jt u( t )

A j

(s−α j )k

↔ Akt k−1 e

α j t

(k−1 )!u( t ) k≥2

A j

(s−α j )k+

B j

(s−αj¿)k

↔t k−1

( k−1 )![ A j e

α jt+B jeα

j¿ t]u( t )

=α

j¿=a− jb

αj=a+ jb

t k−1

(k−1)!eat [ A j(cosbt+ j sin bt )+B j(cosbt− jsin bt ]u( t )

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