Laplace Transform
National Chiao Tung University
Chun-Jen Tsai
10/19/2011
Transform of a Function
Some operators transform a function into another
function:
d 2
x = 2 x , or Dx2 = 2x
Differentiation:
dx
x3
2
Indefinite Integration: ∫ x dx = + c
3
Definite Integration:
∫
3
0
x 2 dx = 9
→ A function may have nicer property in the
transformed domain!
2/73
Integral Transform
If f(x, y) is a function of two variables, then a definite
integral of f w.r.t. one of the variable leads to a
function of the other variable.
Example:
∫
2
1
2 xy 2 dx = 3 y 2
Improper integral of a function defines how
integration can be calculated over an infinite interval:
∫
∞
0
b
K ( s, t ) f (t )dt ≡ lim ∫ K ( s, t ) f (t )dt
b →∞ 0
3/73
Laplace Transform
Definition: Let f be a function defined for t ≥ 0, then
the integral
∞
L { f (t )} = ∫ e − st f (t ) dt
0
is said to be the Laplace transform of f, provided the
integral converges.
The result of the Laplace transform is a function of s,
usually referred to as F(s).
4/73
Example: L{1}
By definition:
∞
b
L (1) = ∫ e (1)dt = lim ∫ e − st dt
− st
b →∞ 0
0
− st b
−e
= lim
b →∞
s
0
− e − sb + 1 1
= lim
= ,
b→∞
s
s
provided s > 0. The integral diverges for s < 0.
5/73
Example: L{t}
By definition:
∞
L (t ) = ∫ e − st tdt
0
Using integration by parts and limt→∞te–st = 0, s > 0,
we have:
− te
L {t} =
s
− st ∞
0
1 ∞ − st
+ ∫ e dt
s 0
1
11 1
= L {1} =   = 2
s
ss s
6/73
Example: L{eat}
By definition:
∞
∞
L {e } = ∫ e e d t = ∫ e −( s − a ) t dt
at
− st at
0
0
−( s − a )t ∞
−e
=
s−a
0
1
=
,s>a
s−a
7/73
Gamma Function Γ(x)
The gamma function is defined for x > 0 as
∞
Γ( x) = ∫ e −t t x −1dt.
0
Since
∞
−t x ∞
Γ( x + 1) = ∫ e t dt = − e t
0
−t x
0
∞
+ x ∫ e −t t x −1dt.
0
For x > 0, we have Γ(x+1) = x Γ(x), therefore, gamma
function is called the generalized factorial function.
Note: Γ(1) = 1 and Γ(x+1) = x! if x is a positive integer.
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Example: L{ta}
If we let u = st, t = u/s, and dt = du/s, then
by definition, for a > –1,
∞
L {t } = ∫ e t d t =
a
− st a
0
1
∞
s a +1 ∫0
Γ(a + 1)
,
e u du =
a +1
s
−u
a
for all s > 0 (so that st > 0).
If n is a nonnegative integer, Γ(n+1) = n!, and we have
n!
L {t } = n +1 , s > 0.
s
n
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Example: L{sin 2t}
By definition:
∞
L {sin 2t} = ∫ e
0
− st
sin 2t dt =
−e
− st
sin 2t
s
∞
0
2 ∞ − st
+ ∫ e cos 2t dt
s 0
2 ∞ − st
= ∫ e cos 2t dt , s > 0
s 0
∞
− st


2 − e cos 2t
2 ∞ − st
= 
− ∫ e sin 2t dt 
0
s
s
s

0

2 4
2
= 2 − 2 L {sin 2t} → L {sin 2t } = 2
,s>0
s
s
s +4
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Linearity of L{}
For a sum of functions, we can write
∫
∞
0
∞
e [αf (t ) + β g (t )]dt = α ∫ e
− st
0
− st
∞
f (t )dt + β ∫ e − st g (t )dt
0
whenever both integrals converge for s > c.
Hence,
L {α f (t ) + β g (t )}
= α L { f (t )} + β L {g (t )}
= α F ( s) + βG ( s)
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Example: L{3e2t + 2sin23t}
L{3e2t + 2sin23t} = 3L{e2t} + L{2sin23t}
= 3/(s – 2) + L{1 – cos 6t}
= 3/(s – 2) + [1/s – s/(s2 + 36)], s > 2.
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Unit Step Function
The unit step function u(t – a) is defined to be
0 , 0 ≤ t < a
.
u (t − a ) = 
t≥a
1 ,
u(t – a) is often denoted as ua(t). Note that ua(t) is only
defined on the non-negative axis since the Laplace
transform is only defined on this domain.
8
1
t
a
13/73
Rewrite of a Piecewise Function
A piecewise defined function can be rewritten in a
compact form using u(t – a).
For example,
 g (t ) , 0 ≤ t < a
f (t ) = 
t≥a
h(t ),
is the same as f(t) = g(t) – g(t)u(t – a) + h(t)u(t – a).
y(t)
y(t – a)
t
a
t
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Laplace Transform of ua(t), a > 0
By definition,
∞
∞
L {ua (t )} = ∫ e ua (t ) dt = ∫ e − st dt
− st
0
a
b
 e 
= lim −
 .
b→∞
 s  t =a
− st
Therefore,
e − as
L {ua (t )} = −
( s > 0, a > 0).
s
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Transform of Basic Functions
L {1} =
1
s
n
L {t } =
L {e a t } =
n!
, n = 1, 2, 3, L
n +1
s
L {sin k t} =
k
s2 + k 2
k
L {sinh k t} = 2
s − k2
1
s−a
L {cos k t} =
s
s2 + k 2
L {cosh k t} =
s
s2 − k 2
e − as
L {u (t − a )} =
s
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Existence of L{f(t)}
Theorem: If f is piecewise continuous on [0, ∞), and
that f is of exponential order for t > T, where T is a
constant, then L{f(t)} converges.
f(t)
t
a t1
t2
t3 b
Definition: A function f is said to be of exponential
order c if there exists constants c, M > 0, and T > 0
such that |f(t)| ≤ Mect for all t > T.
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Examples: Exponential Order
The functions f(t) = t, f(t) = e–t, and f(t) = 2 cos t are all
of exponential order c = 1 for t > 0, since we have
| t | ≤ et, | e–t | ≤ et, | 2 cos t | ≤ 2et.
Mect, c > 0
f(t)
f(t)
t
T
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Proof of Existence of L{f(t)}
By the additive interval property of definite integrals,
T
L { f (t )} = ∫ e
0
− st
∞
f (t ) dt + ∫ e − st f (t ) dt = I1 + I 2
T
The integral I1 exists (finite interval, f piecewise
continuous). Now,
∞
| I2 | ≤ ∫ | e
− st
T
=M∫
∞
T
∞
f (t ) | dt ≤ M ∫ e − st e ct dt
T
−( s −c )t ∞
e
e −( s −c ) t dt = − M
s−c
T
e − ( s − c )T
=M
, s > c.
s−c
→ I2 exists as well → L{f(t)} converges.
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Example: Transform of Piecewise f(t)
Evaluate L{f(t)} for
y
0, 0 ≤ t < 3
f (t ) = 
 2, t ≥ 3.
2
Solution:
t
3
∞
L { f (t )} = ∫ e − st f (t )dt
0
∞
=∫ e
− st
0
3
f (t )dt = ∫ (e
0
− st ∞
2e
=−
s
3
− st
∞
⋅ 0)dt + ∫ 2e − st dt
3
2e − 3 s
=
, s > 0.
s
20/73
Inverse Laplace Transform
If F(s) is the Laplace transform of a function f(t),
namely, L{f(t)} = F(s), then we say that f(t) is the
inverse Laplace transform of F(s), that is,
f(t) = L–1{F(s)}.
Example:
1 = L–1{1/s}, t = L–1{1/s2}, and e–3t = L–1{1/(s + 3)}.
21/73
Examples: Inverse Transforms
Evaluate L–1{1/s5}
Solution:
1 1
 4!  1
L -1  5  = L -1  5  = t 4
 s  4!
 s  24
Evaluate L–1{1/(s2 + 7)}
Solution:
 1  1 -1  7  1
L  2
L  2
sin 7t
=
=
7
7
s + 7
s + 7
-1
22/73
Linearity of L–1{}
The inverse Laplace transform is also a linear
transform; that is, for constant α and β,
L -1{α F ( s ) + β G ( s )} = α L -1{F ( s )} + β L -1{G ( s )}
Example: Evaluate L–1{(–2s + 6) / (s2 + 4)}
6 
 − 2s + 6 
−1  − 2 s
L  2
+ 2
=L  2

s
4
s
4
s
4
+
+
+




s  6 −1  2 
−1 
= −2L  2

+ L  2
s
4
2
s
4
+
+




= −2 cos 2t + 3 sin 2t
−1
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Example: Partial Fractions (1/2)
Evaluate
2


s
+ 6s + 9
−1
L 

 ( s − 1)( s − 2)( s + 4) 
Solution:
There exists unique constants A, B, C such that:
s 2 + 6s + 9
A
B
C
=
+
+
( s − 1)( s − 2)( s + 4) ( s − 1) ( s − 2) ( s + 4)
A( s − 2)( s + 4) + B ( s − 1)( s + 4) + C ( s − 1)( s − 2)
=
( s − 1)( s − 2)( s + 4)
By comparing terms, we have
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Example: Partial Fractions (2/2)
Partial fractions:
16
25
1
s 2 + 6s + 9
= 5 + 6 + 30
( s − 1)( s − 2)( s + 4) ( s − 1) ( s − 2) ( s + 4)
Therefore
2


s
+ 6s + 9
−1
L 

(
1
)(
2
)(
4
)
s
−
s
−
s
+


16 −1  1  25 −1  1  1 −1  1 
=− L 
+ L 
+ L 

5
 s − 1 6
 s − 2  30
s + 4
16
25
1
= − et + e 2t + e − 4t
5
6
30
25/73
Transforming a Derivative
What is the Laplace transform of f'(t)?
∞
L { f ′(t )} = ∫ e
0
− st
f ′(t )dt = e
− st
∞
∞
0
0
f (t ) + s ∫ e − st f (t )dt
= − f (0) + sL { f (t )}
Therefore
L { f ′(t )} = sF ( s ) − f (0)
→ Note that this derivation only works if f′(t) is a
continuous function
26/73
Derivative Transform Theorem
Theorem: If the function f(t) is continuous and
piecewise smooth for t ≥ 0 and is of exponential order
as t → +∞, so that there exist nonnegative constants
M, c, and T such that
|f(t)| ≤ Mect for t ≥ T.
Then L{f′(t)} exists for s > c, and
L { f ′(t )} = sF ( s ) − f (0).
Proof:
Perform (finite) piece-by-piece
integration of
∞
− st
e
∫ f ′(t )dt
0
27/73
General Derivative Transform
Theorem: If f, f', …, f(n–1) are continuous on [0, ∞) and
are of exponential order and if f(n–1)(t) is piecewise
continuous on [0, ∞), then
L { f ( n ) (t )} = s n F ( s ) − s n −1 f (0) − s n − 2 f ′(0) − L − f ( n −1) (0),
where F(s) = L{f(t)}.
28/73
Solving Linear IVPs (1/2)
The Laplace transform of a linear DE with constant
coefficients becomes an algebraic equation in X(s).
That is,
 d nx

d n −1 x
L an n + an −1 n −1 + L + a0 x  = L { f (t )}
dt
 dt

becomes
d nx 
 d n −1 x 
anL  n  + an −1L  n −1  + L + a0L {x} = L { f (t )},
 dt 
 dt 
or
an[snX(s) – sn–1x(0) – sn–2x′(0) – … – x(n–1)(0)]
+ an–1[sn–1X(s) – sn–2x(0) – … – x(n–2)(0)]+…+ a0X(s) = F(s)
29/73
Solving Linear IVPs (2/2)
Given initial conditions,
x(0) = x0, x'(0) = x1, …, x(n–1)(0) = xn–1,
we have Z(s)X(s) = I(s) + F(s), or
F ( s) I ( s)
X (s) =
+
,
Z ( s) Z (s)
steady state behavior
transient state behavior
where Z(s) = ansn + an–1sn–1 + … + a0 and
I(s) = –(ansn–1 + an–1sn–2+ …
+ a1)x(0)
–(ansn–2 + an–1sn–3+ … + a2)x'(0)
– … – anx(n–1)(0).
30/73
dy
+ 3 y = 13 sin 2t , y (0) = 6 (1/2)
Example:
dt
Since
 dy 
L   + 3L { y} = 13L {sin 2t},
 dt 
L{dy/dx} = sY(s) － y(0) = sY(s)－6, and
L{sin2t} = 2/(s2 + 4), we have
26
sY ( s) − 6 + 3Y ( s) = 2
,
s +4
or
26
( s + 3)Y ( s ) = 6 + 2
,
s +4
6
26
6s 2 + 50
→ Y (s) =
+
=
.
2
2
( s + 3) ( s + 3)( s + 4) ( s + 3)( s + 4)
31/73
dy
+ 3 y = 13 sin 2t , y (0) = 6 (2/2)
Example:
dt
Assume that
6s 2 + 50
A
Bs + C
=
+ 2
,
2
( s + 3)( s + 4) s + 3 s + 4
we have A = 8, B = –2, C = 6. Therefore
s 
2 
 1 
−1
−1
y (t ) = 8L 
 − 2L  2
 + 3L  2

 s + 3
s + 4
s + 4
−1
→ y (t ) = 8e −3t − 2 cos 2t + 3 sin 2t
32/73
−4 t
′
′
′
y
−
3
y
+
2
y
=
e
, y (0) = 1, y′(0) = 5
Example:
Solution:
d 2 y 
 dy 
L  2  − 3L   + 2L { y} = L {e − 4t }
 dt 
 dt 
1
s 2Y ( s ) − sy (0) − y′(0) − 3[ sY ( s ) − y (0)] + 2Y ( s ) =
s+4
1
s+2
+ 2
Y ( s) = 2
s − 3s + 2 ( s − 3s + 2)( s + 4)
s 2 + 6s + 9
=
( s − 1)( s − 2)( s + 4)
16 t 25 2t 1 − 4t
→ y (t ) = L {Y ( s)} = − e + e + e
5
6
30
−1
33/73
Transform of Integrals
Theorem: The Laplace transform of the integral of a
piecewise continuous function f(t) of exponential
order is
t
F (s)
L ∫ f (τ )dτ =
.
0
s
The inverse form is:
t
-1  F ( s ) 
f
(
τ
)
d
τ
=
L

.
∫0
 s 
{
}
(Recall that: L{f'(t)} = sF(s) – f(0)).
34/73
Proof of Transform of Integrals
Since f(t) is piecewise continuous, by fundamental
theorem of calculus, if g(t) = ∫0t f(τ)dτ, g(t) is
continuous and g'(t) = f(t) where f(t) is continuous.
Because f(t) is of exponential order, there exists
constants M and c such that
t
t
M ct
M ct
g (t ) ≤ ∫ f (τ ) dτ ≤ M ∫ e cτ dτ =
(e − 1) <
e .
0
0
c
c
→ g(t) is of exponential order as t → +∞.
Thus, L{f(t)} = L{g'(t)} = sL{g(t)} – g(0).
But g(0) = 0, therefore,
L
{∫
t
0
}
F ( s)
f (τ )dτ = L {g (t )} =
.
s
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Example: Inverse by Integration
Starting with f(t) = sin t, F(s) = 1/(s2+1) we have:
 1  t
L  2
 = ∫0 sin τ dτ = 1 − cos t ,
 s ( s + 1) 
-1

 t
1
L  2 2
 = ∫0 (1 − cos τ ) dτ = t − sin t ,
 s ( s + 1) 
-1

 t
1
1 2
L  3 2
=
(
τ
−
sin
τ
)
d
τ
=
 ∫0
2 t − 1 + cos t .
 s ( s + 1) 
-1
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Behavior of F(s) as s → ∞
If f is piecewise continuous on [0, ∞) and of
exponential order for t > T, then lims→∞L{f(t)} = 0.
Proof:
Since f(t) is piecewise continuous on 0 ≤ t ≤ T, it is
necessarily bounded on the interval. That is
| f(t)| ≤ M1e0t. Also, | f(t)| ≤ M2eγt for t > T. If M denotes
the maximum of {M1, M2} and c denotes the
maximum of {0, γ}, then for s > c:
∞
L { f (t )} ≤ ∫ e
0
− st
∞
| f (t ) | dt ≤ M ∫ e − st ⋅ e ct dt
0
−( s −c )t ∞
e
= −M
s−c
0
M
=
→ 0 as s → ∞.
s−c
37/73
Partial Fraction Decompositions
Finding inverse Laplace transform usually involves
partial fractions decomposition, let P(s) be a
polynomial function with degree less than n:
Linear factor decomposition:
An
P( s )
A1
A2
=
+
+
...
+
,
n
2
n
(s − a)
s − a (s − a)
( s − a)
where A1, A2, …, An are constants.
Quadratic factor decomposition
An s + Bn
P( s )
A1s + B1
A2 s + B2
=
+
+
...
+
,
2
2 n
2
2
2
2 2
2
2 n
[( s − a ) + b ]
( s − a ) + b [( s − a ) + b ]
[( s − a ) + b ]
where A1, …, An and B1, …, Bn are constants.
38/73
s-axis Translation Theorem
Theorem: If L{f(t)} = F(s) and a is any real number,
then
L{eatf(t)} = F(s – a).
Proof:
∞
L {e f (t )} = ∫ e − st e at f (t )dt
at
0
∞
= ∫ e −( s − a ) t f (t )dt
0
= F ( s − a ).
39/73
Example: L{e5tt3} and L{e–2tcos 4t}
Solution:
5t 3
3
L {e t } = L {t }s → s −5
3!
= 4
s
s → s −5
6
=
( s − 5) 4
L {e −2t cos 4t} = L {cos 4t}s → s −( −2 )
s
s+2
= 2
=
s + 16 s → s + 2 ( s + 2) 2 + 16
40/73
Inverse of s-axis Translation
The inverse Laplace transform of F(s – a), can be
computed by multiply f(t) = L–1{F(s)} by eat, as follows.
L −1{F ( s − a )} = L −1{F ( s ) s → s − a } = e at f (t )
Example: Compute L–1{(2s+5)/(s–3)2}.
2 s + 5 2s + 11
Since
=
,
2
2
( s − 3)
s
s → s −3


 2 s + 11
−1 1
−1 1
→ L −1 2
2
11
=
L
+
L



 2
s → s −3 
 s
 s s → s −3 
s
= 2e 3t + 11e 3t t.


s → s −3 
41/73
Example: y" – 6y' + 9y = t2e3t
Solve the DE with initial conditions y(0) = 2, y'(0) = 17.
2s + 5
2
Y ( s) =
+
2
( s − 3) ( s − 3) 5
2s + 11
2
=
+ 5
2
s
s
s → s −3
s → s −3
1

−1 1
y (t ) = 2L 
 + 11L  2
 s s → s −3 
s
1 4 3t
3t
3t
= 2e + 11te + t e .
12
−1
 2 −1 4!

+ L  5

4
!
s
s → s −3 
 s → s −3 
42/73
Convolution of Two Functions
If f and g are piecewise continuous on [0, ∞), then a
special product, denoted by f ∗ g, is defined by the
integral
t
f ∗ g = ∫ f (τ ) g (t − τ )dτ
0
and is called the convolution of f and g. The
convolution is a function of t. Note that f ∗ g = g ∗ f.
Example:
1
e ∗ sin t = ∫ e sin(t − τ )dτ = (− sin t − cos t + et ).
0
2
t
t
τ
43/73
Convolution Theorem
Theorem: If f(t) and g(t) are piecewise continuous on
[0, ∞) and of exponential order, then
L { f ∗ g } = L { f (t )}L {g (t )} = F ( s )G ( s ).
Proof:
∞
∞
− sτ



F ( s)G ( s) =  ∫ e f (τ )dτ  ∫ e − sβ g ( β )dβ 
 0
 0

∞

= ∫ f (τ ) ∫ e − s (τ + β ) g ( β )dβ dτ .
0
 0

Let t = τ + β, dt = dβ, so that
∞
F ( s)G ( s) = ∫
∞
0
∞

e  ∫ f (τ ) g (t − τ )dτ  dt = L { f ∗ g }.
 0

− st
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Example:
L
L
{∫ e sin(t −τ )dτ }.
t
τ
0
{∫ e sin(t −τ )dτ }= L {e }⋅L {sin t}
t
τ
t
0
1
1
1
=
⋅ 2
=
s − 1 s + 1 ( s − 1)( s 2 + 1)
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Inverse Form of Convolution
Theorem:
L -1{F ( s)G ( s)} = f ∗ g.


1
Example: L  2
2 2
 (s + k ) 
Let F(s) = G(s) = 1/(s2 + k2),
-1

 1 t
1
L  2
= 2 ∫ sin kτ sin k (t − τ )dτ ,
2 2
 (s + k )  k 0
1 t
sin kt − kt cos kt
= 2 ∫ [cos k (2τ − t ) − cos kt ]dτ =
.
3
0
2k
2k
-1
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Solving Integral Equations
We can use convolution theorem to solve differential
equations as well as “integral equations”.
For example, the Volterra integral equation:
t
f (t ) = g (t ) + ∫ f (τ )h(t − τ ) dτ ,
0
where g(t) and h(t) are known.
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Example:
t
f (t ) = 3t − e − ∫ f (τ )e t −τ dτ .
2
−t
0
Solution: notice that h(t) = et. Take the Laplace
transform of each term:
2
1
1
F ( s) = 3 ⋅ 3 −
− F ( s) ⋅
s s +1
s −1
6 6 1
2
= 3− 4+ −
.
s s
s s +1
The inverse transform then gives:
f(t) = 3t2 – t3 + 1 – 2e–t.
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Differentiation of Transform Theorem
Theorem: If f(t) is piecewise continuous and f(t) is of
exponential order, then
d
L {− tf (t )} = L { f (t )} = F ′( s ).
ds
Proof:
∞ ∂
d
d ∞ − st
F ( s) = ∫ e f (t )dt = ∫
e − st f (t ) dt
0 ∂s
ds
ds 0
[
]
∞
= ∫ − e − st tf (t )dt = L {− tf (t )}
0
d
→ L {tf (t )} = − L { f (t )}
ds
1 -1
→ f (t ) = − L {F ′( s )}
t
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nth-Order Transform Differentiation
Theorem: If F(s) = L{f(t)} and n = 1,2,3…, then
dn
L {t f (t )} = (−1)
F ( s)
n
ds
n
n
Proof:
The proof can be done by mathematical induction.
Here, we only check the 2nd-order case.
2
d
d
L t 2 f (t ) = L {t ⋅ tf (t )} = − L {tf (t )} = 2 L { f (t )}
ds
ds
{
}
Example: Compute L{t sin kt}.
d
d  k 
2k s
= 2
L {t sin k t} = − L {sin k t} = −  2
2 
ds
ds  s + k  ( s + k 2 ) 2
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Example: x"+ 16x = cos4t, x(0) = 0, x'(0) = 1
Solution:
L{x"} + L{16x} = L{cos 4t},
1
s
→ X (s) = 2
+ 2
.
2
s + 16 ( s + 16)
Since, from previous example,
 2k s 
L  2
= t sin k t
2 2
 (s + k ) 
-1

1 -1  4  1 -1  8s
→ x(t ) = L  2
+ L  2
2
4
 s + 16  8
 ( s + 16) 
1
1
= sin 4t + t sin 4t.
4
8
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Integration of Transform Theorem
Theorem: If f(t) is piecewise continuous with
exponential order, and that limt→0+f(t)/t exists and is
finite. Then,
 f (t )  ∞
L
 = ∫s F (σ )dσ .
 t 
In addition, we have
f (t ) = L
-1
{
∞
{F ( s)} = tL ∫s
-1
}
F (σ ) dσ .
Proof:
∫
∞
s
F (σ )dσ = ∫
∞
∞
[
s
 ∞ f (t )e −σt dt dσ = ∞  ∞ f (t )e −σt dσ dt
∫0  ∫s
 ∫0


=∫ e
0
−σt
]
∞
/(−t ) σ = s f (t )dt = L { f (t ) / t}.
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Example: L{(sinh t)/t}
We first verify that the function is bounded when t→0+:
sinh t
et − e −t
et + e −t
lim+
= lim
= lim
= 1.
t
→
0
t
→
0
t →0
t
2t
2
Now, we have
∞ dσ
 sinh t  ∞
L
 = ∫s L {sinh t}dσ = ∫s 2
σ −1
 t 
∞
1 ∞ 1
1 
1  σ −1 
= ∫ 
−
dσ = ln
s
2  σ −1 σ + 1 
2  σ + 1  s
1 s −1
.
= ln
2 s +1
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t-axis Translation Theorem
Theorem: If F(s) = L{f(t)} and a > 0, then
L{f(t – a)u(t – a)} = e–asF(s).
Proof:
∫
∞
0
e − st f (t − a )u (t − a ) dt
a
=∫ e
0
− st
∞
f (t − a )u (t − a ) dt + ∫ e − st f (t − a )u (t − a ) dt
a
∞
= ∫ e − st f (t − a ) dt
a
Let v = t – a, dv = dt,
∞
L { f (t − a )u (t − a )} = ∫ e − s ( v + a ) f (v) dv = e − asL { f (t )}
0
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Inverse of t-axis Translation
If f(t) = L –1{F(s)} and a > 0, the inverse form of the t-
axis translation theorem is:
L −1{e − as F ( s )} = f (t − a )u (t − a ).
Example:
if t < a
 0,
 e − as 
1
2
L  3  = u (t − a ) (t − a ) =  1
.
2
2
 s 
 2 (t − a ) , if t ≥ a
−1
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Alternative Form of t-axis Translation
For g(t) that lacks the precise shifted form g(t – a), we
can derive an alternative form:
∞
∞
L {g (t )u (t − a )} = ∫ e g (t )dt = ∫ e − s ( v + a ) g (v + a )dv
− st
0
a
→ L {g (t )u (t − a )} = e − asL {g (t + a )}.
Example: Since g(t＋π) = cos (t＋π) = –cos t,
s
L {cos t u (t − π )} = −e L {cos t} = − 2
e −πs .
s +1
−πs
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Example:
0≤t <π
0 ,
′
y + y = f (t ), y (0) = 5, f (t ) = 
3 cos t , t ≥ π
Note that f(t) = 3cos t u(t – π), we have
L{y'} + L{y} = 3 L{cos t u(t – π)},
5
3  1 −πs
1 −πs
s −πs 
Y (s) =
− −
e + 2 e + 2 e 
s +1 2  s +1
s +1
s +1

3
 3 −(t −π ) 3

y (t ) = 5e +  e
− sin(t − π ) − cos(t − π )u (t − π )
2
2
2

3
 3 −( t −π ) 3

−t
= 5e +  e
+ sin(t ) + cos(t )u (t − π ).
2
2
2

−t
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Series Circuits
The current in a circuit is governed by the
integrodifferential equation
di (t )
1 t
L
+ Ri (t ) + ∫ i (τ ) dτ = E (t ).
dt
C 0
E
L
R
C
58/73
Example: Single-loop LRC Circuit
Given L = 0.1h, R = 2Ω, C = 0.1f, i(0) = 0, and
E(t) = 120t－120tU(t－1), find i(t).
Solution:
t
di
Since 0.1 + 2i + 10∫ i (τ ) dτ = 120t − 120tU (t − 1),
0
dt
and L
{∫ i(τ ) dτ }= I (s)/ s , we have
t
0
I ( s)
 1 1 −s 1 −s 
0.1sI ( s) + 2 I ( s ) + 10
= 120 2 − 2 e − e .
s
s
s 
s

1
1
1
−s
−s 
→ I ( s ) = 1200 
−
e −
e .
2
2
2
s ( s + 10)
( s + 10)
 s ( s + 10)

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Example: continued
 12 − 12e −10t − 120te −10t ,
0 ≤ t <1
→
−10 t
−10 ( t −1)
−10 t
−10 ( t −1)
−
12
e
+
12
e
−
120
te
−
1080
(
t
−
1
)
e
,
t ≥1

20
i
10
0
t
-10
-20
-30
0
0.5
1
1.5
2
2.5
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Transform of a Periodic Function
If a periodic function f has period T, T > 0, then
f(t + T) = f(t). The Laplace transform of a periodic
function can be obtained by integration over one
period.
Theorem: If f(t) is piecewise continuous on [0, ∞), of
exponential order, and periodic with period T, then
1
L { f (t )} =
1 − e − sT
∫
T
0
e − st f (t ) dt.
61/73
Proof of Periodic Transform Theorem
Proof:
T
L { f (t )} = ∫ e
∞
f (t ) dt + ∫ e − st f (t ) dt ,
− st
0
T
let t = u + T, then the 2nd term becomes
∫
∞
T
e
− st
∞
f (t ) dt = ∫ e − s ( u +T ) f (u + T ) du
0
=e
− sT
∫
∞
0
e − su f (u ) du = e − sT L { f (t )}
Therefore
T
L { f (t )} = ∫ e − st f (t ) dt + e − sT L { f (t )}
0
1
→ L { f (t )} =
1 − e − sT
∫
T
0
e − st f (t ) dt.
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Example: Square-Wave Transform
Find the transform of a square-wave.
Solution:
One period of E(t) can be defined as:
E(t)
 1, 0 ≤ t < 1
E (t ) = 
1
 0, 1 ≤ t < 2
1
2
1
− st
e E (t ) dt
L {E (t )} =
−2 s ∫0
1− e
2
1  1 − st
− st

=
e
⋅
1
dt
+
e
⋅
0
dt
∫1

1 − e − 2 s  ∫ 0
1
1 − e−s
1
=
⋅
=
.
−2 s
−s
1− e
s
s (1 + e )
t
2
3
4
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Example: Periodic Input Voltage (1/3)
The DE for i(t) in a single-loop LR series circuit is
di
L + Ri = E (t ).
dt
Determine i(t) when i(0) = 0 and E(t) is the squarewave as in the previous example.
Solution:
1
1/ L
1
LsI ( s ) + RI ( s ) =
→ I (s) =
⋅
−s
s (1 + e )
s ( s + R / L) (1 + e − s )
Since
1
1
2
3
−s
−2 s
−3 s
=
1
−
e
+
e
−
e
+K
= 1− x + x − x +K →
−s
1+ e
1+ x
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Example: Periodic Input Voltage (2/3)
Since
1
L/R
L/R
=
−
,
s ( s + R / L)
s
s+R/L
we have
1 1
1

−s
−2 s
−3 s
I ( s) =  −
 (1 − e + e − e + L).
Rs s+R/L
By applying the t-axis translation theorem:
1
i (t ) = (1 − u (t − 1) + u (t − 2) − u (t − 3) + L)
R
1 − Rt / L − R (t −1) / L
− (e
−e
u (t − 1) + e − R ( t − 2 ) / L u (t − 2) − L)
R
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Example: Periodic Input Voltage (3/3)
Therefore
∞
1
1
i (t ) = (1 − e − Rt / L ) + ∑ (−1) n (1 − e − R (t − n ) / L )u (t − n).
R
R n =1
For example, if R = 1, L = 1, and 0 ≤ t < 4, we have
2
i
1.5
1
0.5
0
t
0
1
2
3
4
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Unit Impulse
Quite often, the input to a physical system is a short
period, large magnitude function. This type of
function can be described by

0 ≤ t < t0 − a
0,
 1
δ a (t − t0 ) =  , t0 − a ≤ t < t0 + a .
 2a
0,
t ≥ t0 + a

y
2a
1/2a
t
t0 – a
t0
t0 + a
The function δa(t – t0) is called unit impulse because
∫
∞
0
δ a (t − t0 ) dt = 1.
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Dirac Delta Function
Define δ (t − t0 ) = lim δ a (t − t0 ). The function δ(t – t0)
a →0
is called Dirac delta function. δ(t – t0) is characterized
by:
∞, t = t0
(i ) δ (t − t0 ) = 
,
 0, t ≠ t0
∞
(ii ) ∫ δ (t − t 0 ) dt = 1.
0
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Transform of δ(t – t0)
Theorem: For t > 0, L{δ(t – t0)} = e–st .
0
Proof:
1
δ a (t − t0 ) = [u (t − (t0 − a)) − u (t − (t0 + a))],
2a
sa
− sa


1  e − s ( t0 − a ) e − s ( t0 + a ) 
e
−
e
− st0
,
L {δ a (t − t0 )} =
−

 = e 
2a  s
s 
 2sa 
sa
− sa

 − st0
e
−
e
− st0
 = e .
L {δ (t − t0 )} = lim L {δ a (t − t0 )} = e lim
a →0
a →0
 2sa 
Note that L{δ(t)} = 1. δ(t) is not a “normal” function
since L{δ(t)} → 1 as s → ∞.
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Example: Two IVPs (1/2)
Solve y" + y = 4δ(t – 2π), with initial conditions
(a) y(0) = 1, y'(0) = 0, and (b) y(0) = 0, y'(0) = 0.
Solution (a):
The Laplace transform is: s2Y(s) – s + Y(s) = 4e–2πs,
s
4e −2πs
Y (s) = 2
+ 2
.
s +1 s +1
→ y (t ) = cos t + 4 sin(t − 2π )u (t − 2π ).
y
1
0 ≤ t < 2π
cos t ,
→ y (t ) = 
.
t ≥ 2π
cos t + 4 sin t ,
t
-1
2π
4π
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Example: Two IVPs (2/2)
Solution (b)
4e −2πs
.
The Laplace transform is Y ( s ) = 2
s +1
Therefore,
y
y (t ) = 4 sin(t − 2π )u (t − 2π )
0 ≤ t < 2π
 0,
=
.
t ≥ 2π
 4 sin t ,
1
t
-1
2π
4π
71/73
Impulse Response
Consider a 2nd-order linear system with unit impulse
input at t = 0:
a2x" + a1 x' + a0x = δ(t), x(0) = 0, x' (0) = 0.
Applying Laplace transform to the system:
X ( s) =
1
1
−1  1 
=
=
W
(
s
)
→
x
(
t
)
=
L

 = w(t ).
2
a2 s + a1s + a0 Z ( s)
 Z (s) 
w(t) is the zero-state response of the system to a unit
impulse, therefore, w(t) is called the impulse
response of the system.
72/73
Linear Dynamic Systems
Recall that for a general linear dynamic system, we
have
X (s) =
F (s) I (s)
+
.
Z ( s) Z ( s)
W(s) = 1/Z(s) is called the transfer function of the
system. Note that
x(t ) = L −1{W ( s ) F ( s)}+ L −1{W ( s ) I ( s )} .
zero-state response
zero-input response
73/73