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
6-1
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
→