Laplace Transform Exercises

8/14/2019 Laplace Transform Exercises

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LAPLACE TRANSFORM EXERCISES

Quiz

Exercise No.1

Calculate the Inverse Laplace Transform of the function:

)8(

3)(

−=

s s s F

Solution:

By using the partial fractions technique we get:

)8(

3)(

−=

s s s F

8−+= s

B

s

A

( ) ( ) s B s A +−= 83

When 8= s ，( ) ( )803 B A += ，

8

3= B

When 0= s ， ( ) ( )083 B A +−= ∴83−= A

Then,)8(

3)(

−=

s s s F ( )88

3

8

3

−+−=

s s

Now computing the Inverse Laplace Transform:

8

3

)8(

31 −=

−

−

s s L

8

311 +

−

s L

−−

8

11

s L

From the table of Laplace Transforms we know that:

{ }a s

at e L−

=1



{ } s

aa L =

Then,

−−

)8(

31

s s L ( ) ( )t e8

8

31

8

3+−=

Conclusion:

−−

)8(

31

s s L = ( )1

8

3 8 −t e

Quiz

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

Solution:

)8sin()( t t f =

1) The function is piecewise continuous for each t.2) The function is of exponential order for C ≥0.

We can therefore proceed with the Laplace transformation:

2

-1.5

-1

-0.5

0

0.5

1

1.5

0 1 2 3 4 5



∫ −⋅

∞= dt et s F st )8sin(

0)(

We can convert the sine function using Euler’s formula. We also consider

the function as it approaches infinity:

⋅

−∞→

= ∫ −−

dt e j

ee B

B s F

st t jt j

20

lim)(

88

−

∞→= ∫ ∫

−−−

dt j

e Bdt

j

e B

B s F

t s jt s j

2020

lim)(

)8()8(

Evaluating the integral gives us:

⋅

−−−

⋅

−∞→= −−−

B

t s j

B

t s je

s je

s j j B s F

0

)8(

0

)8(

8

1

8

1

2

1lim)(

−⋅

+

+−⋅

−∞→

= −−− )1(

8

1)1(

8

1

2

1lim)( )8()8( B s j B s j

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

Evaluate the Laplace transform of the given functions.

3



t et f 45)( =

t t et f sin4)( =

Solution:

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

1) The function must be of exponential order

ct Ket f ≤)(

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

2) The function must be piecewise continuous

At ≤≤0 A > 0

-100

-50

0

50

100

150

200

250

300

-1.5 -1 -0.5 0 0.5 1 1.5

∫ ∞ −== 0 )()]([)( dt st et f t f L s F

∫ ∫ ∫ ∞∞∞ −−

⇒−

⇒−=000

)4(5

)4(545)( d t

st ed t

st ed t

s t e

t e s F

4

t et f 45)( =

t e4

5 c t k e

≤

5≥k and 4≥c so s > c

Piecewisecontinuous



B st

e s B

dt st

e B

s F B

0

)4(

)4(

1lim5)4(lim5)(

0

−−−−

∞→⇒−−

∞→= ∫

−−

−

−−

∞→−−=

)4(0)4(lim)4(

5)(

s

e

s B

e B s s F

( ))4(

510)4(

5)(−

⇒−−

−= s s

s F

t t et f sin4)( =

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

∫ ∞ −

== 0)()]([)( dt

st et f t f L s F

∫ ∫ ∞∞ −−

⇒−=00

))4((sin

4sin)( d t

st ted t

s t et

te s F

Using ∫ ( )b xbb xaba

eb x d xe

a xa x c o ss i ns i n

.22

−+

=

⇒−∞→= ∫ − B d t st t B

s F e0

) )4((s inl im)(

5

; s > c

t t e sin4 c t k e≤ c 4≤ k 1≥ s c≥



( ) ⇒−−+−∞→=

−−

−− B

tt s

s

t s

B s F

e

0

)c os i n) )4((1) )4((

l i m)( 22

)4(

( )

−−−+−−

−−

∞→= )cossin))4((

212))4((

)4(lim)( B B s

s

B se

B s F

( )

−−−+−−

−−)0cos0sin))4((

212))4((

0)4( s

s

se

when the integral is evaluated at ∞ , the answer is 0

B se

)4( −− = ∞−e = 0 because s c> , s > 4

( )[ ] (

+−−−−−= − sin

21

2))4((

0)cossin))4(((0)(

s

e B B s s F

6



=)( s F 0 1)4(

1212))4((

10

2 +−⇒

+−−−

− s s

1)4(

1)(2 +−

= s

s F

Homework

7



Solving a Differential Equation Using the Laplace Transform Method

•A second order differential equation:

( ) ( ) ( ) ( ) L

t E t qt q

L

Rt q =+′+′′ 2

ω

Calculate its Laplace Transform

Assuming:

( ) ( ) ( )[ ]

( ) ( ) ( )

=⇒=′====

−−−=

0000;0;0

5;2

iqqq R

sb sa

bt H at H E t E

o

o

Then the equation becomes:

( ) ( )( )22

52

22

1

ω ω +−+

+= −−

s see

L

E q

s

s sQ s so

o

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

1. - For the first term, we have:

+

−22

1

ω s

sq L o

=

−+

−

−

2

2

1

11

s s

R

s s

R L

(1)

( )( )( ) 22

lim1

ooo

j s

q

j

q j

j s j s

sq j s R ==

+−

−=→ ω

ω

ω ω

ω ω

2 R= ; ω j s =1 , ω j s −=2

8



substituting the values of 121,, s R R and 2 s into (1):

( )t qe

qe

q j s

q

j s

q L o

t j

o

t j

o

ooω

ω ω

ω ω

cos22

221 =+=

+

+

−

−−

2. - For the second term, we have:

+

−−

L

E

s s

e L o

s

)( 22

21

ω and ( )

+

−−

−

L

E

s s

e L o

s

22

51

ω

the first inverse Laplace transform gives:

+

−−

L

E

s s

e L o

s

)( 22

21

ω

+

+−

+= −− so e j s

R

j s

R

s

R L

L

E 22211

ω ω (2)

( )

( ) 23

22

22201

2

11lim

2

11lim

11lim

ω ω

ω ω

ω ω

ω

ω

=

−=

−=

+=

=

+=

−→

→

→

j s s R

j s s R

s R

j s

j s

s

substituting 21 , R R and 3 R into (2) and calculating its inverse Laplace

transform, we have:

( ) ( ) ( ) ( ) ( ) ( )[ ] ( )22cos122

12

2

12

12

2

2

2

22−−+=

−+−−− −−− t H t

L

E t H et H et H

L

E ot jt joω

ω ω ω ω

ω ω

By analogy, the second inverse Laplace transform is:

( )( )[ ] ( )55cos1

222

51 −−+−=

+

−−

− t H t L

E

L

E

s s

e L oo

s

ω

ω ω

9



Now, let 1=== L E q oo . Arranging terms, we obtain:

( ) =t q ( ) ( )( )

( )( )

22

5cos15

2cos12cos

ω

ω

ω

ω ω

−−−−

−−−+

t t H

t t H t

Graphs

( )t ω cos ( ) ( )22cos12

ω

ω −−− t t H

q1(t)

-1

0

1

t

q2(t)

-1

0

1

t

( ) ( )2

5cos15

ω

ω −−−

t t H ( )t q

q3(t)

-1

0

1

t

q(t)

-2

-1

0

1

2

t

10



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 thefunction multiplied by the kernel of t1he Laplace transform:

L[sin(at)] = ∫ sin(at) e-st dtIn 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 thenecessary conditions for the Laplace transform are not meet the Laplace transformdoesn’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:

│sin(at)│ ≤ Ke – c t c = 0 K = at = 0 c > 0

t > 0

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

1

11

0

∞

{



L[sin(at)] = sin(at) e – s t dt

u = sin(at) dv =e-st dtdu = 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

Blim (1/-s) e – s t sin(at) = 0B

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 ]

Blim (1/-s) e – s t cos(at) = 1/sB

L[sin(at)] = (a/s2) – (a2/s2) sin(at) e – s t dt

12

∞

0

0

0∞

∞

∫

∫∫

∫

∞

∞

∞

∞

0

0

0

0

0∞

∞

∫0



* but L[sin(at)] = sin(at) e – s t dt so …

sin(at) e – s t dt = (a/s2) – (a2/s2) sin(at) e – s t dt

[1 + (a2/s2)] sin(at) e – s t dt = (a/s2)

sin(at) e – s t dt = (a/s2) [ 1 / (1 + {a2/s2} ) ]

The Laplace transform of the cosine function is the integral of the functionmultiplied 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 aresufficient conditions for the Laplace transform to exist. If either one of thenecessary conditions for the Laplace transform are not meet the Laplace transformdoesn’t exist:

13

0∫

∞

∫ ∫∫

∞

∞∞

∞0

00

0∫L[sin(at)] = a .

s2 + a2

0

∞



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:

│cos(at)│ ≤ Ke – c t c = 0 K = at = 0 c > 0

t > 0

Since we now know that the sin function has sufficient conditions for theLaplace 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 dtdu = -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/sB

14

{

∫0

∞

∞

∞

∞ ∫

∫

∞

∞

0

0

0



L[cos(at)] = (1/s) – (a/s) sin(at) e – s t dt

u = sin(at) dv = e – s t dtdu = 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) = 0B

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

*but L[cos(at)] = cos(at) e-st dt so…

cos(at) e-st dt = (1/s) – (a2/s2) cos(at) e – s t dt

[ 1 + (a2

/ s2

) ] cos(at) e-st

dt = 1/s

15

∫∞

0

∞

∞

∫

∫∫

∞

∞

∞

0

0

0

0

0

0

∫ ∫

∫

∫

∞ ∞

∞

∞

0 0

0



cos(at) e-st dt = (1/s) [ 1 + (a2 / s2) ]

The Laplace transform of the cosine function is the integral of the functionmultiplied 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 aresufficient conditions for the Laplace transform to exist. If either one of thenecessary conditions for the Laplace transform are not meet the Laplace transformdoesn’t exist:

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

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

π/2 is the discontinuity point(+ π/2 & - π/2)

tan(π / 2) = sin( π / 2) = 1 =cos(π / 2) 0

16

0

L[cos(at)] = s . s2 + a2

0

∞

∞



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

f (x + ) + f (x - ) never converges as x π/22

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

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

Discussion of the “a” parameter

f (t) = sin(at) ; F(s) = a . s2 + a2

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; thereforea ∝ amplitude:

F(s)

0

1

s

----------------------------------------------------------------------------------------------------

f (t) = cos(at) ; F(s) = s .

s2 + a2

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.

17



F(s)

-0.6

0

0.6

s

In both cases, “a” changes the frequency of the function in the t domain.

18



Laplace TransformTable 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 consideredas being multiplied by u(t).

19



f(t) Comments F(s)

dt

t df )()0()(

+− f s 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

2212

−

−+−+− −−−−

n

nnn

dt

f d

dt

f d s

dt

df s s F s

∫ ∞

=0

)()( τ d f t g s

g

s

s F )0()( +

+

)()( t uor t H Heviside funct. s

1

)(t δ Dirac funct. 1

t (*)2

1

s

)!1(

1

−

−

n

t nI∈n

n s

1

at e−a s +

1

at te −

( ) 21

a s +

at n et −−1 I∈n( ) na s +

1

t ω sin22

ω

ω

+ s

t ω cos22

ω + s

s

( )θ ω +t sin22

cossin

ω

θ ω θ

+

+

s

s

( )θ ω +t cos22

sincos

ω

θ ω θ

+

−

s

s

t eat

ω sin−

( )[ ]2222ω ω

ω

++ s

t eat

ω cos−

( )[ ]2

2

22 ω ω ++

+

s

a s

t teat

ω sin−

( )

( )[ ]2222

2

ω ω

ω

++

+

s

a s

t teat

ω cos−

( )

( )[ ]2222

22

ω ω

ω

++

−+

s

a s

20



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Laplace Transform Exercises

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