The Laplace Transform† National Chiao Tung University Chun-Jen Tsai 10/23/2019 † Chapter 7 in the textbook.

(1)

The Laplace Transform

^†

National Chiao Tung University Chun-Jen Tsai 10/23/2019

† Chapter 7 in the textbook.

(2)

Transform of a Function

 Some operators transform a function into another function:

Differentiation: , or Dx² = 2x Indefinite Integration:

Definite Integration:

 A function may have nicer property in the transformed domain!

x dx x

d ²  2

x c dx

x  



² ₃³

3 9

0

2 



^x ^dx

(3)

Integral Transform

 If f(x, y) is a function of two variables, then a definite

integral of f w.r.t. one of the variable leads to a function of the other variable.

Example:

 Improper integral of a function defines how integration can be calculated over an infinite interval:

2 2 1

2 3

2xy dx  y





  ^b

b K s t f t dt

dt t f t s

K 0

0 ( , ) ( ) lim ( , ) ( )

(4)

Laplace Transform

 Definition: Let f be a function defined for t  0, then the integral

is said to be the Laplace transform of f, provided the integral converges.

The result of the Laplace transform is a function of s, usually referred to as F(s).



^{ }

 0 ( )

)}

(

{ f t e ^st f t dt

(5)

Example: {1}

 By definition:

provided s > 0. The integral diverges for s < 0.

1 , lim 1

lim

lim )

1 ( )

1 (

0

0 0

s s

e s

e

dt e

sb

b st b

b

b st b

st

 

 





















 



(6)

Example: {t}

 By definition:

Using integration by parts and apply l’Hospital’s rule to get lim_t→∞te^–st = 0, s > 0, we have:



^{ }

 0

)

(t e ^sttdt

2 0 0

1 1

} 1 1 1 {

} 1 {

s s

dt s e

s

t te ^st

st

 



 



 



 





^{ }

 

(7)

Example: {e

^at

}

 By definition:

a a s

s

a s

e

dt e

t d e e e

t a s

t a s at

st at

 





 



 



  

 



1 , }

{

0 ) (

0

) ( 0

(8)

Example: {t

ⁿ

}, n  N

 Similarly, let u = tⁿ, dv = e^–stdt,

    ^{ }

. 0

! ,

2 )

1 ... (

) 1 (

} {

1

1 2

2 1

0

1

0





 

 













  



  



s s n

s t n t n

s n t n

s n

dt t

s e e n

s dt t

t e t

n

n n

n

n st n st

n st n



(9)

Example: {sin 2t}

 By definition:

 



 

  







 



0

0 0

0 ,

2 2 cos

2 2 sin

sin 2

sin

s dt t s e

dt t s e

s

t dt e

t e

t

st

st st st

   

^, ⁰

4 2 2

sin 2

4 sin 2

2 2 sin

2 cos 2

2 2

2

0 0

 



















 





^{ }

 

s s t

s t s

dt t s e

s

t e

s

st st

(10)

Linearity of {}

 For a sum of functions, we can write

whenever both integrals converge for s > c, where c is some constant.

Hence,



0^{ }e ^st[f (t) ^ g(t)]dt ^ 0^{ }e ^st f (t)dt ^  0^{ }e ^stg(t)dt

) ( )

(

)}

( { )}

( {

)}

( )

( {

s G s

F

t g t

f

t g t

f























(11)

Example: {3e

^2t

+ 2sin

²

3t}

 {3e^2t + 2sin²3t} = 3 {e^2t} + {2sin²3t}

= 3/(s – 2) + {1 – cos 6t}

= 3/(s – 2) + [1/s – s/(s² + 36)], s > 2.

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Transform of Basic Functions



s } 1 1 { 

, 3 , 2 , 1

! , }

{  _₁ n  s

tⁿ n_n

2

} 2

{sin s k

t k

k  

2

} 2

{sinh

k s t k

k  

a a s

e^a^t s 

 1 , }

{

2

} 2

{cos s k

t s

k  

2

} 2

{cosh

k s t s

k  

(13)

Existence of {f(t)}

 Theorem: If f is piecewise continuous on [0, ), and that f is of exponential order for t > T, where T is a constant, then {f(t)} converges.

 Definition: A function f is said to be of exponential order c if there exists constants c, M > 0, and T > 0 such that |f(t)|  Me^ct for all t > T.

f(t)

a t₁ t₂ t₃ b t

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Examples: Exponential Order

 The functions f(t) = t, f(t) = e^–t, and f(t) = 2 cos t are all of exponential order c = 1 for t > 0, since we have

| t |  e^t, | e^–t |  e^t, | 2 cos t |  2e^t.

t f(t)

T

f(t)

Me^ct, c > 0

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Proof of Existence of {f(t)}

 By the additive interval property of definite integrals,

The integral I₁ exists (finite interval, f piecewise continuous). Now,

 I₂ exists as well  {f(t)} converges.

2

0 ( ) ( ) 1

)}

(

{ f t e f t dt e f t dt I I

T

T st  st  





^



^{ }

. ,

| ) (

|

) ( )

) ( ( 2

c c s

s M e c

s M e dt

e M

dt e e M

dt t

f e I

T c s

T t c s T

t c s

T T

ct st st

 

 









 



   

   



 

(16)

Example: Transform of Piecewise f(t)

 Evaluate {f(t)} for

Solution:

 

. 0 2 ,

2

2 )

0 (

) (

) ( )

(

3

3 3

0 0

0















 



 

  

 



s s e s

e

dt e

dt t f e

dt t f e t

f

s st

st st









 

. 3 ,

2

3 0

, ) 0

( t

t t f

t y

2

3

(17)

Behavior of F(s) as s  

 If f is piecewise continuous on [0, ) and of exponential order for t > T, then lim_s {f(t)} = 0.

Proof:

Since f(t) is piecewise continuous on 0 ≤ t ≤ T, it is necessarily bounded on the interval. That is

| f(t)| ≤ M₁e^0t. Also, | f(t)| ≤ M₂e^^t for t > T. If M denotes the maximum of {M₁, M₂} and c denotes the maximum of {0,



}, then for s > c:

. as

0

| ) (

| )}

( {

) (

0 0





 

 









 



 



c s s

M c

s M e

dt e e

M dt

t f e

t f

t c s

ct st

st

(18)

Inverse Laplace Transform

 If F(s) is the Laplace transform of a function f(t), namely, {f(t)} = F(s), then we say that f(t) is the inverse

Laplace transform of F(s), that is, f(t) = ^–1{F(s)}.

 Example:

1 = ^–1{1/s}, t = ^–1{1/s²}, and e^–3t = ^–1{1/(s + 3)}.

(19)

Examples: Inverse Transforms

 Evaluate ^–1{1/s⁵} Solution:

 Evaluate ^–1{1/(s² + 7)}

Solution:

4 5

1 5

1

24 1

! 4

1

1 t

s s

-

- 







 









s t s

-

- sin 7

7 1 7

7 7

1 7

1

2 1

2

1 









 











(20)

Linearity of

^–1

{}

 The inverse Laplace transform is also a linear transform; that is, for constant



^and



^,

 Example: Evaluate ^–1{(–2s + 6) / (s² + 4)}

)}

( { )}

( )

(

{ ¹ ¹

1 F s G s ^- F s ^- G s

-















t t

s s

s s s

s s

s

2 sin 3 2

cos 2

4 2 2

6 2 4

4 6 4

2 4

6 2

2 1

2 2

1 2

1















 









 











 



 













(21)

Example: Partial Fractions (1/2)

 Evaluate

Solution:

There exists unique constants A, B, C such that:

By comparing terms, we have















 

) 4 )(

2 )(

1 (

9

2 6

1

s s

s

s s

) 4 )(

2 )(

1 (

) 2 )(

1 (

) 4 )(

1 (

) 4 )(

2 (

) 4 (

) 2 (

) 1 (

) 4 )(

2 )(

1 (

9

2 6











 

 

 

 







s s

s

s s

C s

s B s

s A

s C s

B s

A s

s s

(22)

Example: Partial Fractions (2/2)

Partial fractions:

Therefore

) 4 (

) 2 (

) 1 (

) 4 )(

2 )(

1 (

9

6 ¹⁶₅ ²⁵₆ ₃₀¹

2

 

 

 







s s

t t

t e e

e

s s

s

s s

s

s s

4 2

1 1

1 2 1

30 1 6

25 5

16

4 1 30

1 2

1 6

25 1

1 5

16

) 4 )(

2 )(

1 (

9 6

















 









 









 

















(23)

Partial Fraction Decompositions

 Inverse Laplace transform usually involves partial fractions decomposition, let P(s) be a polynomial function with degree less than n:

 Linear factor decomposition:

where A₁, A₂, …, A_n are constants.

 Quadratic factor decomposition

where A₁, …, A_n and B₁, …, B_n are constants.

) , ... (

) (

2 2 1

n n

n s a

A a

s A a

s s P

 

 

 

] , )

... [(

] )

[(

) (

] )

[(

) (

2 2 2

2 2

1 1

2

2 n

n n

n s a b

B s A b

a s

B s A b

a s

B s A b

a s

s P





 



 





 





(24)

Transforming a Derivative

 What is the Laplace transform of f '(t)?

Therefore

 Note that this derivation only works if f (t) is a continuous function

 

⁽ ⁾

) 0 (

) ( )

( )

( 0 0 0

t f s

f

dt t f e s

t f e dt

t f e t

f ^st ^st ^st









 



^{ } ^ ^



^{ }



^f ^⁽^t⁾



^ ^sF⁽^s⁾ ^ ^f ⁽⁰⁾

(25)

Derivative Transform Theorem

 Theorem: If the function f(t) is continuous and

piecewise smooth for t  0 and is of exponential order as t  +, so that there exist nonnegative constants M, c, and T such that

|f(t)|  Me^ct for t  T.

Then {f (t)} exists for s > c, and

Proof:

Perform (finite) piece-by-piece integration of



^f ^⁽^t⁾



^ ^sF⁽^s⁾ ^ ^f ⁽⁰^).



^{ }^e ^st ^f ^^{( dt}^t⁾

(26)

General Derivative Transform

 Theorem: If f, f ', …, f ^(n–1) are continuous on [0, ) and are of exponential order and if f ⁽ⁿ⁾(t) is piecewise

continuous on [0, ), then

where F(s) = {f(t)}.

), 0 ( )

0 ( )

( )}

(

{ f ⁽ⁿ⁾ t  sⁿF s  sⁿ^¹ f  sⁿ^² f   f ⁽ⁿ^¹⁾

(27)

Solving Linear IVPs (1/2)

 The Laplace transform of a linear DE with constant coefficients becomes an algebraic equation in X(s).

That is,

becomes

or

a_n[sⁿX(s) – s^n–1x(0) – s^n–2x(0) – … – x^(n–1)(0)]

+ a_n–1[s^n–1X(s) – s^n–2x(0) – … – x^(n–2)(0)]+…+ a₀X(s) = F(s)

 

⁽ ⁾

1 0 1

1 a x f t

dt x a d

dt x

a d _n

n n n

n

n 







  _ ^_ 

   

⁽ ⁾ ^,

1 0 1

1 a x f t

dt x a d

dt x

a d _n

n n n

n

n   







 











 

(28)

Solving Linear IVPs (2/2)

 Given initial conditions x(0) = x₀, x'(0) = x₁, …, x^(n–1)(0) = x_n–1, we have Z(s)X(s) = I(s) + F(s), or

where Z(s) = a_nsⁿ + a_n–1s^n–1 + … + a₀ and I(s) = (a_ns^n–1 + a_n–1s^n–2+ … + a₁)x(0)

+(a_ns^n–2 + a_n–1s^n–3+ … + a₂)x'(0) + … + a_nx^(n–1)(0).

steady state behavior

transient behavior ( )

) ( )

( ) ) (

( Z s

s F s

Z s s I

X  

(29)

Example: (1/2)

 Since

{dy/dt} = sY(s) － y(0) = sY(s)－6, and {sin2t} = 2/(s² + 4), we have

or

6 )

0 ( ,

2 sin 13

3  

 y t y

dt dy

}, 2 {sin 13

} {

3 y t

dt

dy  









4 , ) 26

( 3 6 )

( ₂

 



 Y s s

s sY

4 , 6 26

) ( ) 3

( ₂

 



 Y s s

s

) . 4 )(

3 (

50 6

) 4 )(

3 (

26 )

3 (

) 6

( ₂

2

2  

 



 

 

 s s

s s

s s s

Y

(30)

Example: (2/2)

Assume that

we have A = 8, B = –2, C = 6. Therefore 4 ,

3 )

4 )(

3 (

50 6

2 2

2



 

 



s

C Bs

s A s

s s









 









 









 ^  ^ ^

4 3 2

2 4 3

8 1 )

( ¹ ¹ ₂ ¹ ₂

s s

s t s

y

t t

e t

y( )  8 ³^t  2cos2 3sin 2

 ^

6 )

0 ( ,

2 sin 13

3  

 y t y

dt

dy

(31)

Example:

 Solution:

5 )

0 ( ,

1 )

0 ( ,

2 3   

⁴

  

  y y e

^

y y

y

^t

} {

} { 2

3 ⁴

2 2

e t

dt y dy dt

y

d _









 









4 ) 1

( 2 )]

0 ( )

( [

3 ) 0 ( )

0 ( )

2 (

 





 



 sy y sY s y Y s s

s Y s

) 4 )(

2 )(

1 (

9 6

) 4 )(

2 3

(

1 2

3 ) 2

(

2

2 2







 



 





 

s s

s

s s

s s s

Y

t t

t e e

e s

Y t

y ¹ 16 25 ² 1 ⁴

)}

( { )

(  ^     ^



(32)

s-axis Translation Theorem

 Theorem: If {f(t)} = F(s) and a is any real number, then

{e^atf(t)} = F(s – a).

Proof:

. ),

(

, ) (

) ( )}

( {

0

) ( 0

a s

F

a s

dt t f e

dt t f e e t

f e

t a s

at st at













  

 

(33)

Example: {e

^5t

t

³

} and {e

^–2t

cos 4t}

 Solution:

4 5

5 4 3

3 5

) 5 (

6

! } 3

{ }

{    







 s s

t t

e

s s s

s t

16 )

2 (

2 16

} 4 {cos }

4 cos {

2 2

2

) 2 ( 2



 









 

s s s

s

t t

e

s s

s s t

(34)

Inverse of s-axis Translation

 The inverse Laplace transform of F(s – a), can be computed multiplying f(t) = ^–1{F(s)} by e^at:

 Example: Compute ^–1{(2s+5)/(s–3)²}.

Since

) ( }

) ( { )}

(

{ ¹

1 F s  a  ^ F s _s__s__a  e^at f t



11 , 2

) 3 (

5 2

3 2

2





 





s

s s

s

. 11 2

11 1 2 1

11 2

3 3

3 2

1 3

2 1

t e e

s s

t t

s s s

s s

s











 







 







  















(35)

Example: y" – 6y' + 9y = t

²

e

^3t

 Solve the DE with initial conditions y(0) = 2, y'(0) = 17.

3 5

3 2

5 2

2 11

2

) 3 (

2 )

3 (

5 ) 2

(









 



 



 

s s s

s s

s s Y

12 . 11 1

2

! 4

2 11 1

2 1 ) (

3 4 3

3

3 5

1 3

2 1 3

1

t t

t

s s s

s s

s

e t te

e

s s

t s y











 







 







 















3 2

) 3 (

! ) 2

( 9 )) 0 ( )

( ( 6 ) 0 ( )

0 ( )

(        

s s Y y

s sY y

sy s

Y s

(36)

Unit Step Function

 The unit step function u(t – a) is defined to be

u(t – a) is often denoted as u_a(t). Note that u_a(t) is only defined on the non-negative axis since the Laplace transform is only defined on this domain.

, . 1

0 , ) 0

( 





 

 t a

a a t

t u

t 8

1

a

(37)

Rewrite of a Piecewise Function

 A piecewise defined function can be rewritten in a compact form using u(t – a).

For example,

is the same as f(t) = g(t) – g(t)u(t – a) + h(t)u(t – a).









 

a t

t h

a t

t t g

f ( ),

0 , ) ) (

(

y(t) y(t – a)

(38)

Laplace Transform of u(t – a), a > 0

 By definition,

Therefore,

 

. lim

) (

)

( 0

b

a t st

b

a

st st

s e

dt e

dt a t

u e a

t u









 



 













 



⁽ ^ ⁾



^ ^ ⁽^s ^ ⁰^, ^a ^ ⁰ ^.)

s a e

t u

as

(39)

t-axis Translation Theorem

 Theorem: If F(s) = {f(t)} and a > 0, then {f(t – a)u(t – a)} = e^–asF(s).

Proof:

Let v = t – a, dv = dt, dt a t

f e

dt a t

u a t

f e dt

a t

u a t

f e

dt a t

u a t

f e

a

st

a a st st

st



 



 















) (

0 0



⁽ ⁾ ⁽ ⁾



⁽ ⁾

 

⁽ ⁾

0

)

( f v dv e f t

e a

t u a t

f   



^ ^^s ^v^^a  ^^as

(40)

Inverse of t-axis Translation

 If f(t) = ^–1{F(s)} and a > 0, the inverse form of the t- axis translation theorem is:

 Example:

).

( ) (

)}

(

1{e^^asF s  f t  a u t  a





⁰^,



^, ^if^if ^.

) 2 (

) 1

( ₂

12 2

3 1









 



 





 ^



a t

a a t

t a

t s u

e ^as

(41)

Alternative Form of t-axis Translation

 For g(t)u(t – a), we can derive an alternative form:

 Example: Since g(t＋



) = cos (t＋



) = –cos t, )}.

( { )}

( ) ( {

)}

( { )

(

) (

) ( )}

( ) ( {

0

) (

a t

g e

a t

u t g

a t

g e

dv a v

g e

e

dv a v

g e

dt t g e

a t

u t g

as

as sv

sa

a v s a

st























  



  

 



1 . }

cos {

)}

(

{cos ^s ₂ e ^s

s t s

e t

u

t



^^ ^^

 









(42)

Example:

 Note that f(t) = 3cos t u(t –



), we have {y'} + {y} = 3 {cos t u(t –



)},









 







 t

t t t

f y

t f y

y 3 cos ,

0 ,

) 0 ( , 5 ) 0 ( ), (





 

 

 

 ^ ^s ^ ^s e^ ^s

s e s

e s s

s s

Y ^ ^ ^

1 1

1 2

3 1 ) 5

( ₂ ₂

).

( ) 2 cos(

) 3 2 sin(

3 2

5 3

) (

) 2 cos(

) 3 2 sin(

3 2

5 3 )

(

) (



 

 

  





 

 

    







t u t t

e e

t u t

t e

e t

y

t t

(43)

Derivatives of Transforms

 Theorem: If f(t) is piecewise continuous and f(t) is of exponential order, then

Proof:

Note:

 



⁽ ⁾



) (

) ( )

( )

(

0

0 0

t tf dt

t tf e

dt t

f s e

dt t f ds e

s d ds F

d

st

st st











 





 

 

⁽ ⁾

 

^f ⁽^t⁾

ds t d

tf  



⁽ ⁾

  

^f ⁽^t⁾ ^F ⁽^s^).

ds t d

tf   



#



⁽ ⁾



) 1

( ¹ F s

t t

f   ^- 



(44)

nth-Order Derivatives of Transforms

 Theorem: If F(s) = {f(t)} and n = 1,2,3…, then

Proof:

The proof can be done by mathematical induction.

Here, we only check the 2^nd-order case.

 Example: Compute {t sin kt}.

) ( )

1 ( )}

(

{ F s

ds t d

f

t _n

n n

n  

2 2 2

2

2 ( )

} 2 {sin

} sin

{ s k

s k k

s k ds

t d ds k

t d k

t   



 





 









² ⁽ ⁾

 ^

⁽ ⁾

^ ^

⁽ ⁾

^

²₂

^{ }

^f ⁽^t⁾

ds t d

ds tf t d

tf t t

f

t     

(45)

Convolution of Two Functions

 If f and g are piecewise continuous on [0, ), then a special product, denoted by f  g, is defined by the integral

and is called the convolution of f and g. The

convolution is a function of t. Note that f  g = g ^ f.

 Example:



^



 g ^t f g t d f 0 (



) (



)



).

cos sin

2 ( ) 1

sin(

sin 0 t t

t t e t d t t e

e  



^ 

^ ^

   

(46)

Convolution Theorem

 Theorem: If f(t) and g(t) are piecewise continuous on [0, ) and of exponential order, then

Proof:

Let t =



+



, dt = d



, so that



^f ^ ^g



^

   

^f ⁽^t⁾ ^g⁽^t⁾ ^ ^F⁽^s⁾^G⁽^s^).

. )

( )

(

) ( )

( )

( ) (

0 0

) (

0 0

 



   

 



 







 





 























d d

g e

f

d g

e d

f e

s G s F

s

s s

 

^.

) (

) ( )

( )

(s G s 0 e 0 f g t d dt f g

F ^st   





 





^ ^



^

^ ^ ^

(47)

Example: Compute

 Solution:

 _

₀^t ^e^ ^sin(^t ^

^

⁾^d

^ 

^.

  ^{ } ^{ }

) 1 )(

1 (

1 1

1

sin )

sin(

2 2

0



 

 

 









s s

t e

d t

e ^t

t ^

 

(48)

Inverse Form of Convolution

 Theorem:

 Example:

Let F(s) = G(s) = 1/(s² + k²),



⁽ ⁾ ⁽ ⁾



^.

1 F s G s f g

-  









 ² ²

2 1

) (

1 k s

-

 

^.

2

cos cos sin

) 2

( 2 cos

1

, ) (

sin 1 sin

) (

1

0 3 2

2 0 2

2 2

1

k

kt kt

d kt kt t

k k

d t

k k k

k s

t - t

 







 













(49)

Transforms of Integrals

 Theorem: The Laplace transform of the integral of a piecewise continuous function f(t) of exponential order is

The inverse form is:

(Recall that: {f'(t)} = sF(s) – f(0)).

 _

₀^t ^f ⁽

^

⁾^d

^ 

^ ^F⁽_s^s⁾ ^.

) . ) (

( ¹

0 



 



^t ^f

^

^d

^

^- ^F_s^s

(50)

Proof of Transform of Integrals

 Since f(t) is piecewise continuous, by fundamental theorem of calculus, if g(t) = ^t f()d, g(t) is continuous and g'(t) = f(t) where f(t) is continuous.

Because f(t) is of exponential order, there exists constants M and c such that

 g(t) is of exponential order as t  +.

Thus, {f(t)} = {g'(t)} = s {g(t)} – g(0).

But g(0) = 0, therefore,

. )

1 (

) ( )

( 0 0

ct t c ct

t e

c e M

c d M

e M

d f

t

g 

 ^ ^





^

  

 _

₀^t ^f ⁽

^

⁾^d

^ 

^

^{ }

^g⁽^t⁾ ^ ^F_s⁽^s⁾^.

(51)

Example: Inverse by Integration

 Starting with f(t) = sin t, F(s) = 1/(s²+1) we have:

. cos 1

) sin ) (

1 (

1

, sin )

cos 1

) ( 1 (

1

, cos 1

) sin 1 (

1

2 12 2 0

3 1

2 0 2 1

2 0 1

t t

s d s

t t

s d s

t s d

s

- t - t - t









 















 













 













(52)

Integral Equations

 We can use convolution theorem to solve differential equations as well as “integral equations”.

For example, the Volterra integral equation:

where g(t) and h(t) are known.

, )

( ) ( )

( )

(^t ^ ^g ^t ^



0^t ^f ^h ^t ^ ^d

f

  

(53)

Example:

 Solution: notice that h(t) = e^t. Take the Laplace transform of each term:

The inverse transform then gives:

f(t) = 3t² – t³ + 1 – 2e^–t. 1.

2 1

6 6

1 ) 1

1 ( 1 3 2

) (

4 3

3

 







 

 







s s

s s s F

s s F

. )

( 3

)

(

0

2

^

^ 

^

 t e

^t ^t

f e

^t

d t

f 

^



(54)

Series Circuits

 The current in a circuit is governed by the integrodifferential equation

.) ( )

1 ( )

) ( (



0 ^



 ^ti d E t

t C dt Ri

t

L di  

C

L R

E

(55)

Example: Single-loop LRC Circuit

 Given L = 0.1h, R = 2, C = 0.1f, i(0) = 0, and E(t) = 120t－120t (t－1), find i(t).

Solution:

Since

and , we have

, ) 1 (

120 120

) ( 10

2 1

.

0 _dt^di ^ ⁱ ^



0^tⁱ

^

^d

^

^ ^t ^ ^t ^t ^

 _

₀^tⁱ⁽

^

⁾ ^d

^ 

^ ^I⁽^s⁾^/^s

1 . 1

120 1 )

10 ( )

( 2 ) ( 1 .

0 ₂ ₂



  





 ^^s e^^s

e s s s

s s s I

I s

sI

) . 10 (

1 )

10 (

1 )

10 (

1200 1 )

( ₂ ₂ ₂ 



 





 

 

 ^^s e^^s

e s s

s s

s s I

(56)

Example: continued



















  _ ^ _ _^ _ _ _

1 ,

) 1 (

1080 120

12 12

1 0

, 120

12 12

) 1 ( 10 10

10 10

t e

t te

e e

t te

e

t t

20 i

10 0 -10 -20 -30

0 0.5 1 1.5 2 2.5

t

(57)

Transform of a Periodic Function

 If a periodic function f has period T, T > 0, then

f(t + T) = f(t). The Laplace transform of a periodic

function can be obtained by integration over one period.

 Theorem: If f(t) is piecewise continuous on [0, ), of exponential order, and periodic with period T, then

. ) 1 (

)} 1 (

{ ^f ^t ^ _ _e^^sT



0^T ^e^^st ^f ^t ^dt

(58)

Proof of Periodic Transform Theorem

 Proof:

let t = u + T, then the 2^nd term becomes

Therefore

, ) ( )

( )}

(

{ ^



0^T ^ ^



T^{ } st

st f t dt e f t dt

e t

f

 

⁽ ⁾

) (

0 0

) (

t f e

du u

f e

e

du T

u f e

dt t

f e

sT su

sT T

T u s st

  



    









 

   

 

⁽ ⁾ ¹ ⁽ ⁾ ^.

) ( )

( )

( 0













T st

T st sT

dt t

f e t

f

t f e

dt t

f e t

f

(59)

Example: Square-Wave Transform

 Find the transform of a square-wave.

Solution:

One period of E(t) can be defined as:











 

2 1

, 0

1 0

, ) 1

( t

t t E

 

) . 1

( 1 1

1 1

0 1 1

1

) 1 (

2

1 0

2 2 1

2 2 0

s s

s

st st

s

st s

e s

s e e

dt e

dt e e

dt t E e e

t E



 

 

 



   

 

 



^t

E(t)

1 3

1

2 4

(60)

Example: Periodic Input Voltage (1/3)

 The DE for i(t) in a single-loop LR series circuit is

Determine i(t) when i(0) = 0 and E(t) is the square-wave as in the previous example.

Solution:

Since

).

(t E dt Ri

L di  

) 1

(

1 )

/ (

/ ) 1

) ( 1

( ) 1

( )

( _s _s

e L

R s

s s L

e I s s

RI s

LsI _ _

 

 

 













 

3

1 2

1

1 x x x

x     

  __s e^^s e^ ^s e^ ^s e

3

1 2

1 1