6-1

Chapter 6. Laplace Transforms 6.4 Short Impulses. Dirac Delta Function.

Partial Fraction

6.5 Convolution. Integral Equations 6.6 Differentiation and Integration of Transforms 6.7 Systems of ODEs

6-2 6.4 Short Impulses. Dirac Delta Function. Partial Fraction

Dirac delta function or unit impulse function is defined as

( ) if 0 otherwise

t a t aδ − = ∞ =

=

( ) 1a

a

t a dt+ε

−ε

δ − =∫

The delta function can be obtained by taking the limit of kf

( ) ( )0

lim kkt a f t a

→δ − = −

Sifting property of delta function

( ) ( ) ( )a

a

g t t a dt g a+ε

−ε

δ − =∫

The Laplace transform of delta function. Start from kf

( ) ( ) ( ) 1kf t a u t a u t a k

k − = − − − +

→ Take the limit 0k → and apply l’Hopital’s rule to the quotient.

0 0

1lim lim

ks ksas as

k k

e see e

ks s

− −− −

→ →

−⇒

→ ( ) ast a e−δ − = L

6-3 Example 1 Mass-Spring system under a square wave Input is of the form of a rectangular function

The subsidiary equation

Use the partial fraction expansion : The inverse transform : Using t-shifting

( ) ( ) ( ) ( ) ( ) ( ) ( )1 2 1 2 2 21 1 1 11 2

2 2 2 2t t t ty t e e u t e e u t− − − − − − − − = − + − − − + −

Example 2 Hammer blow response of a mass-spring system The input is given by a delta function

Solving algebraically

The solution ( ) ( ) ( )( 1) 2( 1)1 1t ty t e u t e u t− − − −= − − −

6-4 Example 3 Four-Terminal RLC-Network Find the output voltage if 420 , 1 , 10R L H C F−= Ω = = . The input is a delta function and current and charge are zero at t=0.

The voltage drops on R, L, C should be equal to the input. Using 'i q=

The subsidiary equation

Using s-shifting and 29900 99.50≈

The solution

More on Partial Fractions

The solution of a subsidiary equation is of the form ( )( )

F sY

G s=

Partial fraction representation may be needed.

(1) Unrepeated factor (s-a) in G(s) → Partial fraction should be ( )

As a−

(2) Repeated factor ( )2s a− in G(s) → Partial fractions ( ) ( )2

A Bs as a

+−−

Repeated factor ( )3s a− in G(s) → Partial fractions ( ) ( ) ( )3 2

A B Cs as a s a

+ +−− −

(3) Unrepeated complex factors ( )2 2s −α +β → Partial fraction ( )2 2

As B

s

+ −α +β

6-5 Example 4 Unrepeated Complex Factors. A damped mass-spring system under a sinusoidal force.

( )" 2 ' 2y y y r t+ + = , ( ) 10sin2 for 0

0 for r t t t

t = < < π

= > π, ( ) ( )0 1, ' 0 5y y= = −

The subsidiary equation

The solution

(6) • The partial fraction of the first term

Multiplying the common denominator

Terms of like powers of s should be equal on the right and left sides

→ A=-2, B=-2, M=2, N=6 Therefore the first term becomes

The inverse transform

(8)

• The inverse of the second term of (6) is obtained from (8) using t-shifting ( )u t − π (11) • Rewrite the third term of (6)

( )2 2

3 ( 1) 42 2 1 1

s ss s s

− + −⇒

+ + + +

The inverse using s-shifting ( )cos 4sinte t t− − (7) • The final solution For 0 t< < π y(t)= Eq. (8) + Eq. (7) For t > π y(t)= Eq. (8) + Eq. (7) + Eq. (11)

6-6 6.5 Convolution. Integral Equations

The convolution of two functions f and g is defined as

( ) ( ) ( )0

*t

f g f g t d≡ τ − τ τ∫ : Note the integration interval

Theorem 1 Convolution theorem

If F and G are Laplace transforms of f and g, respectively, the multiplication FG is the Laplace transform of the convolution (f*g)

Proof:

Set p t= − τ , then

Calculate the multiplication

↑ ↑ ↑ G can be inside of F For fixed τ , integrate from τ to ∞ . because and t τ are independent. ( The integration over blue region ) The integration can be changed as

• Some properties of convolution

( )0

*ste f g dt∞

−⇒ ∫

6-7 Example 1 Convolution

Let ( ) ( )1

H ss a s

=−

. Find h(t).

Rearrange : ( ) ( )1 1

H ss a s

= −

↑ ↑ F(s) G(s) Inverse transforms : ( ) ( ), 1atf t e g t= =

Using convolution theorem : ( ) ( ) ( ) ( )0

1* 1 1

ta ath t f t g t e d e

aτ= ⇒ ⋅ τ⇒ −∫

Example 2 Convolution

Let ( )( )22 2

1H s

s=

+ω. Find h(t).

Rearrange : ( )( ) ( ) ( )2 2 2 2 22 2

1 1 1H s

s ss= ⇒

+ω +ω+ω

Inverse of ( )2 2

1s +ω

: sin tωω

Using convolution theorem : ( ) ( )2 20

sin sin 1 1 sin* sin sin cos

2

tt t th t t d t t

ω ω ω = ⇒ ωτ ω − τ τ⇒ − ω + ω ω ωω ω ∫

)]cos()cos([2/1sinsin yxyxyx −++−= Example 3 Unusual Properties of Convolution *1f f≠ in general → ( )* 0f f ≥ may not hold → Applications to Nonhomogeneous Linear ODEs Nonhomogeneous linear ODE in standard form ( )" 'y ay by r t+ + = : a and b, constant The solution

( ) ( ) ( ) ( ) ( ) ( ) ( )0 ' 0Y s s a y y Q s R s Q s= + + + : ( ) 2

1Q s

s as b=

+ +, transfer function

The inverse of the first right term can be easily obtained. The inverse of the second term, assuming ( ) ( )0 ' 0 0y y= =

( ) ( ) ( )0

t

y t q t r d= − τ τ τ∫

The output is given by the convolution of the impulse response q(t) and the driving force r(t). Example 5 Mass-spring system

6-8 Solve

( )" 3 ' 2y y y r t+ + = , ( ) 1 for 1 2

0 otherwiser t t = < <

= ( ) ( )0 ' 0 0y y= =

The transfer function

Its inverse

Since ( ) ( )0 ' 0 0y y= = , the solution is given by the convolution of q and r.

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 212

10 0 1

1 2t t t t

t t t ty t q t r d q t u u d e e d e eτ=

− −τ − −τ − −τ − −τ

τ= = − τ τ τ⇒ − τ τ − − τ − τ⇒ − τ⇒ − ∫ ∫ ∫

↑ ↑ r(t)=1 only for 1<t<2 Note the change in the lower limit. t should be less than 2. For t<1 : y(t) = 0

For 1<t<2 : The upper limit is t, ( ) ( ) ( ) ( ) ( )2 1 2 112

1

1 12 2

tt t t ty t e e e e

τ=− −τ − −τ − − − −

τ= = − ⇒ − +

For t>2 : The upper limit is 2, ( ) ( ) ( ) ( ) ( ) ( ) ( )22 2 2 2 1 2 11

21

1 12 2

t t t t t ty t e e e e e eτ=

− −τ − −τ − − − − − − − −

τ=

= − ⇒ − − −

Integral Equations Convolutions can be used to solve certain integral equations Example 6 A volterra Integrals Equation of the Second Kind Solve

↑ Convolution of ( ) ( ) and siny t t Using the convolution theorem we obtain the subsidiary equation

( ) ( ) ( )2

2 2 2

1 11 1

sY s Y s Y s

s s s− = ⇒

+ + → ( )

2

4 2 4

1 1 1sY s

s s s+

= ⇒ +

The answer is

6-9 6.6 Differentiation and Integration of Transforms. Differentiation of Transforms If F(s) is the transform of f(t), then its derivative is

( ) ( )0

stF s f t e dt∞

−= ∫ → ( ) ( ) ( )0

' stdF sF s t f t e dt

ds

∞−= = −∫

Consequently ( ) ( )'t f t F s= − L and ( ) ( )1 'F s t f t− = − L

Example 1 Differentiation of Transforms The table can be proved using differentiation of F(s).

The second one

[ ]( )2 2 22 2

2sin

sdt t

ds s s

β ββ = − ⇒ +β +β

L

Integration of Transforms If f(t) has a transform and ( )

0lim /

tf t t

→ + exists,

( ) ( )

s

f tF s ds

t

∞ =

∫L and ( ) ( )1

s

f tF s ds

t

∞−

= ∫L

Proof: From the definition

( ) ( ) ( ) ( )0 0

st st

s s s

f tF s ds e f t dt ds f t e ds dt

t

∞ ∞ ∞ ∞ ∞− −

= ⇒ ⇒

∫ ∫ ∫ ∫ ∫ L

↑ ↑ Reverse the order of integration. = /ste t−

6-10 Example 2 Differentiation and Integration of Transforms Find the inverse transform of F(s) = Its derivative

Take the inverse transform

( ) ( )1 12 2

2 2' 2cos 2

sF s t t f t

ss− − ⇒ − ⇒ ω − = − + ω

L L

→ ( ) 2cos 2tf t

tω −

= −

• Using integration of transforms Let → Then

( ) ( ) ( ) ( )' 0s s

F s F F s ds G s ds∞ ∞

= ∞ − ⇒ −∫ ∫

Take the inverse transform of both sides

( ) ( )g tf t

t= − ( ) ( )2 cos 1

t

f ttω −

→ = −

Special Linear ODEs with Variable Coefficients Use differentiation of transform to solve ODEs. Let [ ]y Y=L → [ ] ( )' 0y sY y= −L . Using differentiation of transform

[ ] ( )' 0d dY

ty sY y Y sds ds

= − − ⇒ − − L

Similarly, using [ ] ( ) ( )2" 0 ' 0y s Y sy y= − −L

[ ] ( ) ( ) ( )2 2" 0 ' 0 2 0d dY

ty s Y sy y sY s yds ds

= − − − ⇒ − − + L

6-11 Example 3 Laguerre’s Equation Laguerre’s ODE is ( )" 1 ' 0ty t y ny+ − + = n=0, 1, 2, … The subsidiary equation

→ Separating variables, using partial fractions

2

1 11

dY n s n nds ds

Y s ss s+ − + = − ⇒ − −−

→ ( ) ( ) ( )1

1ln ln 1 1 ln ln

n

n

sY n s n s

s +

−= − − + ⇒ →

( )1

1 n

n

sY

s +

−=

The inverse transform is given by Rodrigues’s formula [ ]1

nl Y−= L → n=1, 2, … • Prove Rodrigues’s formula Using s-shifting

Using the n-th derivative of f,

After another s-shifting

6-12 6.7 Systems of ODEs

The Laplace transform can be used to solve systems of ODEs. Consider a first-order linear system with constant coefficients

The subsidiary equations

Rearrange

( ) ( ) ( )

( ) ( ) ( )11 1 12 2 1 1

21 1 22 2 2 2

0

0

a s Y a Y y G s

a Y a s Y y G s

− + = − −

+ − = − −

Solve this system algebraically for ( ) ( )1 2 and Y s Y s and take the inverse transform for ( ) ( )1 2 and y t y t Example 2 Electrical Network Find the currents ( ) ( )1 2 and i t i t .

( ) 100 only for 0 0.5v t volts t= ≤ ≤ and ( ) ( )0 = ' 0 0i i = From Kirchhoff’s voltage law in the lower and the upper circuits,

Rearrange

The subsidiary equations using ( ) ( )1 20 = 0 0i i =

Solve algebraically for 1 2 and I I

( )( )( ) ( ) ( ) ( ) ( )

( )( ) ( ) ( ) ( ) ( )

/2 /21 1 7 1 7

2 2 2 2

/2 /21 1 7 1 7

2 2 2 2

125 1 500 125 6251 1

7 3 21

125 500 250 2501 1

7 3 21

s s

s s

sI e e

s s s s s s

I e es s s s s s

− −

− −

+= − ⇒ − − −

+ + + +

= − ⇒ − + − + + + +

6-13 The inverse transform of the square bracket terms using s-shifting.

/2 7 /2

/2 7 /2

500 125 6257 3 21

500 250 2507 3 21

t t

t t

e e

e e

− −

− −

− −

− +

Using t-shifting

( ) ( )

( ) ( )

/2 7 /2 ( 0.5)/2 7( 0.5)/21

/2 7 /2 ( 0.5)/2 7( 0.5)/22

500 125 625 500 125 6250.5

7 3 21 7 3 21500 250 250 500 250 250

0.57 3 21 7 3 21

t t t t

t t t t

i t e e e e u t

i t e e e e u t

− − − − − −

− − − − − −

= − − − − − − = − + − − + −

Note that the solution for 1

2t ≥ is different from that for 120 t≤ ≤ due to the unit step function.

Example 3 Two masses on Springs Ignoring the mass of the springs and the damping

↑ ↑ ↑ Newton’s second law(mass X acceleration) ↑ ↑ Hooke’s law (restoring force) Initial conditions

( ) ( )( ) ( )

1 2

1 2

0 0 1

' 0 3 , ' 0 3

y y

y k y k

= =

= = −

The subsidiary equations

The algebraic solution using Cramer’s rule

The final solution

→