Laplace Transform Exercises

LAPLACE TRANSFORM EXERCISES

Quiz

Quiz

Exercise No.1 Exercise No.1

Calculate the Inverse Laplace Transform of the function:

) 8 ( 3 ) ( − = s s s F Solution: Solution:

By using the partial fractions technique we get:

) 8 ( 3 ) ( − = s s s F 8 − + = s B s A ( ) ( ) s B s A − + = 8 3 When s₌8， ( ) ( )0 8 3= A +B ， 8 3 = B When s=0，3= A( ) ( )−8 +B0 ∴ 8 3 − = A Then, ( ) ₍ 3 ₈₎ − = s s s F ( 8) 8 3 8 3 − + − = s s

Now computing the Inverse Laplace Transform:

8 3 ) 8 ( 3 1 ₌₋       − − s s L 8 3 1 1 ₊       − s L −_ ₋ _ 8 1 1 s L

From the table of Laplace Transforms we know that: { } a s at

e

L

= ₋1

{ } s a a L = Then,       − − ) 8 ( 3 1 s s L ( )

( )

_e8t 8 3 1 8 3 + − = Conclusion: Conclusion:       − − ) 8 ( 3 1 s s L

₌

8

₍

1

₎

3 8 − t e Quiz

Calculate the Laplace Transform of f(t) = sin(8t).

Solution: Solution: ) 8 sin( ) (t t f =

1) The function is piecewise continuous for each t. 2) The function is of exponential order for C≥_0.

-1.5 -1 -0.5 0 0.5 1 1.5 0 1 2 3 4 5

∫ ∞ _⋅ − = t e dt s F _sin(8) st 0 ) (

We can convert the sine function using Euler’s formula. We also consider the function as it approaches infinity:

      ⋅ − ∞ → = ∫ − − dt e j e e B B s F st t j t j 2 0 lim ) ( 8 8       − ∞ → = _{∫ ∫} − − − dt j e B dt j e B B s F t s j t s j 2 0 2 0 lim ) ( ) 8 ( ) 8 (

Evaluating the integral gives us:

                ⋅ − − −       ⋅ − ∞ → = − − − B t s j B t s j e s j e s j j B s F 0 ) 8 ( 0 ) 8 ( 8 1 8 1 2 1 lim ) (       _{⋅ −} + + − ⋅ − ∞ → = − ₍ − − ₁₎ 8 1 ) 1 ( 8 1 2 1 lim ) ( ( j8 s) B ( j8 s)B e s j e s j j B s F

Further analysis takes us at the final result:

       + − − ⋅ = 8 1 8 1 2 1 ) ( j s j s j s F 64 8 ) ( ₂ + = s s F Homework Homework

t e t f ( )=5 4 t t e t f () = 4 sin Solution: Solution:

In order to evaluate the Laplace transform, the following two sufficient conditions must be satisfied.

1) The function must be of exponential order

ct Ke t f () ≤

{ f f (t t ) must not grow faster than exponential }

2) The function must be piecewise continuous

A t

≤

≤

0

A

>

>

0

-100 -50 0 50 100 150 200 250 300 -1.5 -1 -0.5 0 0.5 1 1.5

∫

∞

₋

=

=

0 ) ( )] ( [ ) ( s L f t f t e st dt F

∫

∫

∫

∞

−

⇒

∞

−

⇒

∞

−

−

=

0 0 0 ) 4 ( 5 ) 4 ( 5 4 5 ) ( s e t e s t d t et sd t e t s d t F t e t f ( )=5 4 t e4 5

≤

k ec t

_k

_{≥ 5 and c ≥ 4 so}

_s

_>

_c

Piecewise continuous

B s t e s B dt s t e B s F B 0 ) 4 ( ) 4 ( 1 lim 5 ) 4 ( lim 5 ) ( 0 _        _{− −} − − ∞ → ⇒ − − ∞ → =

_∫

         _{− −} − − − ∞ → − − = lim ( 4) 0( 4) ) 4 ( 5 ) ( e B s e s B s s F

(

)

) 4 ( 5 1 0 ) 4 ( 5 ) (

−

⇒

−

−

−

=

s s s F t t e t f () = 4 sin

f(t) is a function of exponential order and piecewise continuous

∫

∞

−

=

=

0 ( ) )] ( [ ) ( s L f t f t e st dt F

∫

∫

∞

−

⇒

∞

−

−

=

0 0 )) 4 ( ( sin 4 sin ) ( s te t e s t d t tet s d t F Using ∫

(

a b x b b x

)

b a e b x d x ea xs i n a x s i n c o s . 2 2

−

+

=

⇒

−

∞

→

=

_∫

−

B d t s t t B s F

e

0

))

4

(

(

s in

lim

)

(

; s

>

c

t t e4 sin

≤

k ec t

c

≤

4

k 1

≥

s

≥

c

( )

−

−

⇒

+−

∞→

=

























−

−

−−

B

t

t

s

s

t

s

B

s F

e

0

)c o

s i n

) )4

((

1) )4

((

l i m

)(

22

)4(

( )             − − − + − − − − ∞ → = ( ( 4))sin cos ) 2 1 2 )) 4 ( ( ) 4 ( lim ) ( s B B s B s e B s F ( )           − − − + − − − − ) 0 cos 0 sin )) 4 ( ( 2 1 2 )) 4 ( ( 0 ) 4 ( s s s e

when the integral is evaluated at

∞

, the answer is

0

B s e

−

(

−

4

)

=e− ∞=

0

because

s

> c ,

s

> 4

(

)

[

]

(

    

+

−

−

−

−

−

=

−

0 sin 2 1 2 )) 4 ( ( 0 ) cos sin )) 4 ( (( 0 ) ( s e B B s s F

=

) ( s F

₀

1 ) 4 ( 1 2 1 2 )) 4 ( ( 1 0 2 ₊ − ⇒ + − − −          

−

s s 1 ) 4 ( 1 ) ( 2

₊

−

=

s s F Homework Homework

Solving a Differential Equation Using the Laplace Transform Method

• A second order differential equation:

( ) ( ) ( ) ( ) L t E t q t q L R t q′′ + ′ + 2 = ω

Calculate its Laplace Transform

Assuming:

( )

( ) ( )

[

]

( )

( )

( )











=

⇒

=

′

=

=

=

=

−

−

−

=

0 0 0 0 ; 0 ; 0 5 ; 2 i q q q R s b s a b t H a t H E t E o o

Then the equation becomes:

( ) ( ) ( 2 2) 5 2 2 2 1 ω ω + − + + = − − s s e e L E q s s s Q o s s o

Find q( )t . We use partial fractions expansion to calculate the inverse Laplace

Transform.

1. - For the first term, we have:

    + − 2 2 1 ω s sq L o =       − + − − 2 2 1 1 1 s s R s s R L ₍₁₎ ( ) ( )( ) 2 2 lim 1 o o o j s q j q j j s j s sq j s R _= =      + − − = → ω ω ω ω ω ω 2 R = _; s =1 jω , s2=− jω

substituting the values of R1, R2,s1 and s2 into (1): ( )t q e q e q j s q j s q L o t j o t j o o o _ω ω ω ω ω cos 2 2 2 2 1 = + =       + + − − −

2. - For the second term, we have:

      + − − L E s s e L o s ) ( 2 2 2 1 ω and ( ) _     + − − − L E s s e L o s 2 2 5 1 ω

the first inverse Laplace transform gives:

      + − − L E s s e L o s ) ( 2 2 2 1 ω _            + + − + = o − _e− s j s R j s R s R L L E 1 1 2 2 2 ω ω (2) ( ) ( ) 2 3 2 2 2 2 2 0 1 2 1 1 lim 21 1 lim 1 1 lim ω ω ω ω ω ω ω ω =       − = − =       + = =     + = − → → → j s s R j s s R s R j s j s s

substituting R1, R2 and R3 into (2) and calculating its inverse Laplace

transform, we have: ( ) ( ) _{( )} ( ) _{( ) [ ( ) ] ( )} 2 2 cos 1 2 2 1 2 2 1 2 1 2 2 2 2 2 2 = + − −    _{− −} − _{− +} − − _{− t H t} L E t H e t H e t H L E o j t j t o _ω ω ω ω ω ω ω

By analogy, the second inverse Laplace transform is:

( 2 2) 2[ 1 cos ( 5)] ( 5) 5 1 _{=− + − −}       + − − − t H t L E L E s s e L s o o ω ω ω

Now, let qo= E o=L=1. Arranging terms, we obtain: ( )t = q ( ) ( ) ( ) ( ) ( ) 2 2 5 cos 1 5 2 cos 1 2 cos ω ω ω ω ω _t + _{H t} − − t − − _{H t} − − t − Graphs ( )ω t cos ( ) ( ) 2 2 cos 1 2 ω ω − − − t t H q1(t) q1(t) -1 0 1 tt q2(t) q2(t) -1 0 1 tt ( ) 2( ) 5 cos 1 5 ω ω − − − t t H q( )t q3(t) q3(t) -1 0 1 tt q(t) q(t) -2 -1 0 1 2 tt

HOMEWORK : find the Laplace transform of the following functions: 1) sin(at)

2) cos(at) 3) tan(at)

and also include a discussion of the “a” parameter.

SOLUTION:The Laplace transform of the cosine function is the integral of the function multiplied by the kernel of t

1

he Laplace transform:

L

[sin(at)]

= ∫

= ∫

sin(at) e

-st

dt

In order to evaluate if such transform exists, we have to find if there are sufficient conditions for the Laplace transform to exist. If either one of the necessary conditions for the Laplace transform are not meet the Laplace transform doesn’t exist:

1) sin(at) is piecewise continuous for ∀t :

-7 -5 -3 -1 1 3 5 7

2) sin(at) is of exponential order…

K = a

in general:in general:

│sin(at)│ ≤ Ke

– c t

_{c = 0}

_{K = a}

t = 0

c > 0

t > 0

Since we now know that the sin function has sufficient conditions for the Laplace transform to exist, we can proceed to calculate the transform.

1

00

∞

L

[sin(at)]

= sin(at) e – s t_dt

u = sin(at) dv =e-st_dt

du = a cos(at) dt v = (1/-s) e-st

L

[sin(at)]

= (1/-s) e– s t_{sin(at) – (1/-s) a e} – s t _{cos(at) dt}

B

lim (1/-s) e – s t _{sin(at) = 0}

B

L

_[sin(at)]

= (a/s) cos(at) e – s t _dt

u = cos (at) dv = e – s t_dt

du = -a sin(at) dt v = (1/-s) e – s t

L

[sin(at)]

= (a/s) [ (1/-s) e – s t _{cos(at)– (-a/-s) sin(at) e} – s t _{dt ]} B

lim (1/-s) e – s t _{cos(at) = 1/s}

B

L

_[sin(at)]

= (a/s2_{) – (a}2_/s2_{) sin(at) e} – s t _dt

∞ 0 0 0 ∞ ∞

∫

∫

∫

∫

∞ ∞ ∞ ∞ 0 0 0 0 0 ∞ ∞

∫

0

*

butL

[sin(at)]

= sin(at) e – s t_{dt so …}

sin(at) e – s t_{dt = (a/s}2_{) – (a}2_/s2_{) sin(at) e} – s t _dt

[1 + (a2_/s2_{)] sin(at) e} – s t_{dt = (a/s}2₎

sin(at) e – s t_{dt = (a/s}2_{) [ 1 / (1 + {a}2_/s2_{} ) ]}

The Laplace transform of the cosine function is the integral of the function multiplied by the kernel of the Laplace transform:

L

[cos(at)]

= ∫

= ∫

cos(at) e

-st

dt

In order to evaluate if such transform exists, we have to find if there are sufficient conditions for the Laplace transform to exist. If either one of the necessary conditions for the Laplace transform are not meet the Laplace transform doesn’t exist: 0

∫

∞

∫

_∫

∞

∫

∞ ∞ ∞ 0 0 0 0

∫

L

[sin(at)]

= a . s2_{+ a}2 00 ∞

3) The cosine function is piecewise continuous∀t :

-7 -5 -3 -1 1 3 5 7

4) cos(at) is of exponential order…

K = a

in general:in general:

│cos(at)│ ≤ Ke

– c t

_{c = 0}

_{K = a}

t = 0

c > 0

t > 0

Since we now know that the sin function has sufficient conditions for the Laplace transform to exist, we can proceed to calculate the transform.

L

[cos(at)]

== cos(at) e-st_dt

u = cos(at) dv = e – s t_dt

du = -asin(at)dt v = (1/-s) e – s t

L

[cos(at)]

==(1/-s) e – s t _{cos(at) – (-a/-s) sin(at) e} – s t _dt

lim (1/-s) e – s t _cos(at)  _{= 1/s} B

{

∫

0 ∞ ∞ ∞ ∞

∫

0∞ 0 0

L

[cos(at)]

==(1/s) – (a/s) sin(at) e – s t _dt

u = sin(at) dv = e – s t_dt

du = acos(at)dt v = (1/-s) e – s t

L

_[cos(at)]

==(1/s) – (a/s) [ (1/-s) e

– s t

sin(at) – (a/-s) cos(at) e

– s t

dt ]

B

lim (1/-s) e – s t _{sin(at) = 0}

B

L

[cos(at)]

==(1/s) – (a/s) [ (a/s) cos(at) e – s t _{dt ]}

L

_[cos(at)]

==(1/s) – (a2_/s2_{) cos(at) e} – s t _dt

*butL

[cos(at)]

== cos(at) e-st_{dt so…}

cos(at) e-st_{dt = (1/s) – (a}2_/s2_{) cos(at) e} – s t _dt [ 1 + (a2_{/ s}2_{) ] cos(at) e}-st_{dt = 1/s}

∫

0∞ ∞ ∞

∫

∫

∫

∞ ∞ ∞ 0 0 0 0 0 0

∫

∫

∫

∫

∞ ∞ ∞ ∞ 0 0 0

cos(at) e-st_{dt = (1/s) [ 1 + (a}2_{/ s}2_{) ]}

The Laplace transform of the cosine function is the integral of the function multiplied by the kernel of the Laplace transform:

L

[tan(at)]

= ∫

= ∫

tan(at) e

-st

dt

In order to evaluate if such transform exists, we have to find if there are sufficient conditions for the Laplace transform to exist. If either one of the necessary conditions for the Laplace transform are not meet the Laplace transform doesn’t exist:

5) The tangent function is not piecewise continuous∀t :

-4 -3 - 2 - 1 0 1 2 3 4

/2 is the discontinuity point (+ /2 & - /2) 0 L

[cos(at)]

== s . s2_{+ a}2 00 ∞

and since the mean value at the discontinuity does not converge…

f (x+_{) + f (x}--_{) never converges as xπ/2}

2

because as xπ/2 the function has a very large value, not a finite limit

the Laplace transform of

the Laplace transform of tan(at) doesn’t exist.tan(at) doesn’t exist.

Discussion of the “a” parameter

f

f (t) (t) = = sin(at) sin(at) ; F(s) ; F(s) = = aa ..

ss22_{+ a+ a}22

The “a” parameter is a constant in both the t domain and the s domain. Moreover, as “a” changes the amplitude of the graph of F(s) changes; therefore a∝_amplitude: F(s) F(s) 0 1 s s f f (t) (t) = = cos(at) cos(at) ; ; F(s) F(s) = = ss .. ss22_{+ a+ a}22

The “a” is also a constant in both t & s domains. Likewise, a change in “a” changes the amplitude of F(s); therefore a∝amplitude in the s domain.

F(s) F(s) -0.6 0 0.6 s s

Laplace Transform

Table of Transform Pairs

It is assumed tat all f(t) exist for t≥0 and f(t)=0 for t<0. Each of the functions from (*) to the end can be considered as being multiplied by u(t).

ff((tt)) CCoommmmeennttss FF((ss)) dt t df ( ) ) 0 ( ) ( s − f + sF 2 2 ) ( dt t f d ) 0 ( ) 0 ( ) ( 2 ₋ + ₋ + dt df sf s F s n n dt t f d () ( ) 0 ( ) 0 ( ) ( ₁ 1 2 2 2 1 2 − − + − + − _{− − −} − _n n n n dt f d dt f d s dt df s s F s 

∫

∞ =0 ( ) ) (_{t f} τ _d g _s g s s F ( ) (0+) + ) ( ) (t or u t H Heviside funct. s 1 ) (t δ _{Dirac funct. 1} t (*) 2 1 s )! 1 ( 1 − − n t n I ∈ n n s 1 at e− a s+ 1 at te− _{( )}2 1 a s+ at n e t −1 − n_∈I ( ) n a s+ 1 t ω sin _{2 2} ω ω + s t ω cos _{2 2} ω + s s ( ω t +θ )

sin sin₂ cos₂

ω θ ω θ + + s s ( ω _t +θ ) cos 2 2 sin cos ω θ ω θ + − s s t e−at _sinω ( )

[

2 _ω 2 2 _ω 2

]

ω + + s t e−at _cosω ( )

[

2_+ω 2 2_+ω 2

]

+ s a s t te−at _{sinω ( )} ( )

[

22 2 2 2

]

ω ω ω + + + s s a t te at ω cos − ( ) ( )

[

2

]

2 2 ω − +a s