Laplace Transform Exercises

LAPLACE TRANSFORM EXERCISES Quiz Exercise No.1

Calculate the Inverse Laplace Transform of the function: F ( s) =

3 s ( s −8)

Solution:

By using the partial fractions technique we get: F (s) =

3 s ( s −8)

(

=

A s

+

B s −8

) ()

3 = A s −8 + B s

3

When

s

When

s =0

Then,

F ( s) =

=8

，

，

3 = A 0 +B 8

()

()

，

( ) ()

3 = A −8 + B 0

3 s ( s −8)

=−

3 8s

B

=8 3

∴ A = −8 +

3

(

8 s −8

)

Now computing the Inverse Laplace Transform: L−

1

 3  3  s( s − 8)  = − 8  

L

−1

1  3 +  s   8

L

−1

 1   s −8  

From the table of Laplace Transforms we know that:

L{e } = s −1 a at

L{a} = a s

Then, L

−1

  3 s ( s −8)   

=−

3 8

(1) + 3 (e t ) 8

8

Conclusion: −1

L





3

 s ( s −8)  

3

=

8

8t

(e −1) Quiz

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

1.5

1

0.5

0 012345 -0.5

-1

-1.5

f (t ) = sin(8t )

1) continuous forCeach t. ≥0. 2) The The function function is is piecewise of exponential order for We can therefore proceed with the Laplace transformation: 2

F (s) = ∫

∞

sin(8t ) ⋅ e −st dt

0

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

 B e j8t − e − j 8t −st  ⋅ e dt   B → ∞ 0 j2  ∫ lim

=

F (s) =

B e ( − j 8−s )t  B e ( j 8−s )t  dt − ∫ dt  ∫ B →∞ 0 0 j2 j2  lim

Evaluating the integral gives us: F ( s)

=

F ( s)

=

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

lim

1

1  1 ⋅ (e ( j 8−s ) B −1) + →∞ j2   j8 − s

lim

B

1

j8 + s

 ⋅ (e ( − j 8−s ) B −1) 

Further analysis takes us at the final result: F (s)

=

F ( s) =

j2

 1 1   ⋅  −  s − j8 s + j8 

s2

+ 64

1

8

Homework

Evaluate the Laplace transform of the given functions. 3

f (t ) =5e 4t f (t ) = e4t sin t

Solution:

In order to evaluate the Laplace transform, the following two sufficient conditions must be satisfied. 1) The function must be of exponential order f (t ) ≤ Ke ct { f(t) must not grow faster than exponential }

2) The function must be piecewise continuous

0≤ t

≤

A >0

A

300

250

200

150

f (t ) =5e 4t

100

50

Piecewise continuous

0 -1.5

-1

-0.5

0

0.5

1

1.5

-50

-100

ct 4t ∞ k −≥ st F (s5)e= L≤[ fk e(t )] = f (t )e 5 dtand c ≥ 4 so

F ( s)

=

∞ 4t − s t 5e e d t ⇒

∫0

∫0∞ ∫0

5e

t (4 − s )

dt⇒

∞ − t (s − 4)

∫0 5e

s>c

dt

4

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

∫

5

F (s ) = −

(s − 4) B



lim

→∞

5

F (s)

− (s

=

(0

 − e  

B(s − 4)

B

0

− e − 0(s − 4)   

;s>c

5

1)

− ⇒ (s − 4)

− 4)

f (t ) = e4t sin t

f(t) is a function of exponential order and piecewise continuous e 4t sin t

F (s)

F (s)

=

Using

≤

c≤ 4

k ec t

= L[ f (t)] = ∫0∞

c

4t e− s td t ⇒ ∞ tet(− (s − 4))d t sin

∞

∫0

ea xs i nb x d x=

F s

()=

s≥

f (t )e− st dt

∫0 sinte ∫

k≥ 1

B

ea x 2

a +b

.

( a s i nb x− b c o bs x)

t(− (s − te

lim

B

→∞

0 s in

∫

2

4))

dt

⇒

5

B  − (s − 4)t   em  l i F s)( =  − (( s − 4) s t)− ci t)n o⇒ B ∞→  2 2   − (( s − 4) + 1)  0

(

F (s) =

 

B

)

 − ( s − 4) B  e  →∞(−( s − 4)) 2 +12  

lim

e

((−(s − 4)) sin

−( s − 4)0

  (−( s 

−4)) 2 +12

 

B − cos B) )   

 

((−(s −4)) sin

0 −cos 0)  )  

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

e

F (s)

− (s − 4)B = e −

0

because s > c , s > 4

B

    (−(s 

∞=

= [0(((−(s − 4)) sin

− cos

B))]

−

e0

(sin 0

− 4)) 2 + 12

6



 F (s) = 0 − 0 −  

F (s)

=

1 (−(s − 4)) 2

   2 + 1 

⇒

1 (s − 4) 2 + 1

1 (s − 4)2

+1

Homework 7

Solving a Differential Equation Using the Laplace Transform Method • A second order differential equation: + R()q ′ t +() ω 2 q t = E ( t )

q ′′()t

L

L

Calculate its Laplace Transform

 E)( t = E(o [) H( ) t − a − H t − b ]  Assuming:  a = 2 s; b = 5s  R = 0; q() 0 = q (); q ′ 0 =() 0 ⇒ i 0 = 0 o  Then the equation becomes: Q( s )

=

s s2

+ω 2

qo

+

E o −2s (e L

− e −5 s )

1

s(s 2

+ω2 )

Find q (t ) . We use partial fractions expansion to calculate the inverse Laplace Transform. 1. - For the first term, we have:  sqo  =  s 2 + ω 2 

L−1 

R1

 sq o = slim ( s − jω ) →j  ( s − j)(ω s +) jω  ω

 R1 R  + 2  − − s2  s s s  1

L−1 

 jωq o q o  = 2 jω = 2 = R 2 

(1) ;

s1 = jω

,

s 2 = − jω

8

substituting the values of  qo s −

L−1 

2

jω

R1 , R2 , s1

+

and

s2

into (1):

 ej t e− j t = qo + qo = qo cos(ωt )  s + jω  2 2 ω

qo 2

ω

2. - For the second term, we have: e −2 s



L−1 

 s(s + ω 2

2

Eo   ) L 

 

L−1 −

and

e −5 s s( s

2

+ω2 )

Eo  L

 

the first inverse Laplace transform gives: 

e −2 s

Eo   2 L  s ( s + ω ) 

L−1 

2

Eo

=

L

 R1  R R  + 2 + 2 e −2 s  s s − j ω s + j ω   

L−1 

(2)

1  1  = 2 2 2  s + ω  ω   1 1 R2 = slim → jω  s ( s + jω )  = − 2ω 2     1 1 R3 = lim   = 2ω 2 s →− jω s ( s − jω )  

R1 = lim  s →0

substituting R , R and transform, we have: 1

Eo L

2

 1 H ( t − 2) − 1 e j ω 2 2ω 2

ω

R3

( t −2)

into (2) and calculating its inverse Laplace

H ( t − 2) +

1 2ω

2

e− j

ω

( t − 2)

 E o [1( )+ cos ω t − 2 ] H ( t − 2)  Lω 2

H ( t) − 2  =

By analogy, the second inverse Laplace transform is: L−1 −



e −5 s Eo  = − Eo [1 + cosω t − 5 ] H t − 5 ( ) ( )  s( s 2 + ω 2 ) L  Lω 2

9

Now, let

qo = Eo = L = 1

. Arranging terms, we obtain:

q (t ) = cos( ω) t

+( H ) t − 2

1 − cosω ( t − 2) ω

2

− H ( t − 5)

1 − cosω ( t − 5) ω

2

Graphs cos

H ( t − 2)

(ωt )

(

1 − cos ω t − 2 ω

)

2

q2(t)

q1(t) 1

1

0

0

-1

-1

t

H ( t − 5)

(

1 − cos ω t − 5 ω

2

t

)

q (t )

q3(t) 1

q(t) 2 1

0

0 -1

-1

-2

t

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 the 1

function multiplied by the kernel of t∞he Laplace transform: -st

L[sin(at)] = ∫0

sin(at) e 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 │sin(at)│ ≤ Ke

–ct

t=0

in general:

c=0

K=a

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

11

L[sin(at)] =

∫

∞

sin(at) e – s t dt

0

u = sin(at) dv =e -st dt du = a cos(at) dt v = (1/-s) e -st ∞ ∞

L[sin(at)] = (1/-s) e

–st

sin(at) 0 –

∫

(1/-s) a e – s t cos(at) dt

0

B

lim (1/-s) e – s t sin(at)  0 = 0 B ∞ ∞

L[sin(at)] = (a/s) cos(at) e – s t dt 0

∫

–st

u = cos (at) dv = e dt du = -a sin(at) dt v = (1/-s) e – s t ∞ ∞

∫

L[sin(at)] = (a/s) [ (1/-s) e – s t cos(at) 0 – (-a/-s) sin(at) e – s t dt ] 0

B

lim (1/-s) e – s t cos(at) = 1/s 0 B ∞ ∞

L[sin(at)] = (a/s2) – (a2/s2)

∫

sin(at) e – s t dt

0

12

*

but L[sin(at)] =

∫

∞

sin(at) e – s t dt so … 0

∞

∞

sin(at) e – s t dt = (a/s2) – (a2/s2) 0

∫ ∫

∫

[1 + (a2/s2)] ∞

∞

sin(at) e – s t dt 0

∫

sin(at) e – s t dt = (a/s2)

0

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

0

L[sin(at)] = a 2

.

2

s +a

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

L[cos(at)] = ∫0

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:

13

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 │cos(at)│ ≤ Ke – c t

t=0

in general:

c=0

K=a

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

0 u = cos(at) dv = e – s t dt du = -asin(at)dt v = (1/-s) e

–st

∞

L[cos(at)] = (1/-s) e – s t cos(at) 0 –

∫

∞

(-a/-s) sin(at) e – s t dt

0

∞ –st lim (1/-s) ∞ e cos(at) B



0 = 1/s

∫

∞

14

L[cos(at)] = (1/s) – (a/s) sin(at) e – s t dt 0 u = sin(at) dv = e – s t dt du = acos(at)dt v = (1/-s) e

–st

∞ ∞

–st

L[cos(at)] = (1/s) – (a/s) [ (1/-s) e

∫

B

–st lim (1/-s) ∞ e sin(at)  0= 0 B

∫

L[cos(at)] = (1/s) – (a/s) [ (a/s)

–st

sin(at)0 – 0(a/-s) cos(at) e

dt ]

∞

cos(at) e – s t dt ]

0

∞

L[cos(at)] = (1/s) – (a2/s2) cos(at) e – s t dt

*but L[cos(at)] =

∫

0

∫ ∫ ∞

cos(at) e-st dt so…

0

∞

∫

cos(at) e-st dt = (1/s) – (a2/s2)

0

∞

cos(at) e – s t dt

0

∞

[ 1 + (a2 / s2) ]

cos(at) e-st dt = 1/s 0

∫

∞

∫

15

cos(at) e-st dt = (1/s) [ 1 + (a2 / s2) ] 0

L[cos(at)] =

s2 + a2

s

.

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

L[tan(at)] = ∫0

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) tan(

/ 2) = sin( / 2) = ∞ 1 = cos( / 2) 0 16

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 tan(at) doesn’t exist. Discussion of the “a” parameter f (t) = sin(at) ;

F(s) = s2 + a2

a

.

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) 1

0

s

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

s 2 + 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

s

-0.6

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

18

Laplace Transform Table of Transform Pairs It is assumed tat all f(t) exist for t≥ 0 and f(t)=0 for t