M. J. Roberts - 2/18/07
Web Appendix N - Derivations of the
Properties of the LaplaceTransform
N.1 Linearity
()
()
()
Let z t = x t + y t where and are constants. Then
( ) x (t ) + y (t ) e
Z s =
0
st
()
()
()
0
0
and the linearity property is
()
()
()
()
L
x t + y t X s + Y s .
N.2 Time Shifting
() (
Let z t = g t t0
)
, t0 0 . Then
( ) z (t ) e
Z s =
0
st
dt =
g (t t ) e
st
0
0
dt .
Let = t t0 d = dt . Then
()
Z s =
()
s ( t )
st
g e 0 d = e 0
t0
g ( ) e
t0
s
d = e
st0
N.3 Complex-Frequency Shifting
Let s0 be a complex constant. Then
()
()
L
e 0 g t e 0 g t e st dt
st
()
()
dt = x t e dt + y t e st dt = X s + Y s
st
st
0
()
(
s s t
st
L
e 0 g t g t e ( 0 ) dt = G s s0
0
N-1
)
()
G s .
M. J. Roberts - 2/18/07
The complex-frequency-shifting property of the Laplace transform is
()
(
L
e 0 g t G s s0
st
)
(N.1)
N.4 Time Scaling
( )
Let a be any positive real constant . Then the Laplace transform of g at is
( )
( )
L
g at g at e st dt
0
Let = at and d = adt
( )
()
g at g e
L
0
s / a
( )
1 s
d 1
= g e s / a d = G , a > 0
a 0
a a
a
()
(
) (
)
L
g at 1/ a G s / a , a > 0
(N.2)
N.5 Frequency Scaling
Let a be any positive real constant. Then, using the time-scaling property of the
( )
( ) ( )
or (1 / b) g ( t / b) G ( bs )
L
1/ a G s / a , a > 0.
Laplace transform g at ( )
( )
L
g t / b bG bs , b > 0
scaling property of the Laplace transform is
L
(1 / a ) g (t / a ) G ( as)
L
Let
b = 1/ a .
Then
, b > 0 and the frequency-
, a>0
(N.3)
N.6 First Time Derivative
The Laplace transform is defined by
( ) g (t ) e
G s =
0
Evaluate the integral by parts using
Then
st
dt .
udv = uv vdu and let u = g (t )
N-2
and dv = e st dt .
M. J. Roberts - 2/18/07
( ( ))
d
1
g t dt and v = e st
dt
s
du =
and
1 st 1 d
st
g t e dt = g t s e + s dt g t e dt
0
0
0
()
()
st
( )
()
( ( ))
( ( ))
1
1 d
G s = g 0 + g t e st dt
s
s 0 dt
()
()
(where it is understood that Re s = is chosen to make G s exist). Then
d
d
L g t = g t e st dt = sG s g 0
dt
0 dt
( ( ))
( ( ))
() ( )
and the first-time-derivative property of the Laplace transform is
( ( ))
() ( )
d
L
g t sG s g 0
dt
.
(N.4)
N.7 Nth Time Derivative
This property can be proven using the previous property for the first time
derivative and applying it to a first time derivative to form a second time derivative and
then generalizing the result to the Nth time derivative. The second time derivative of a
()
function g t is
dd
d2
g t = g t 2
dt dt
dt
( ( ))
( ( ))
Therefore, using
( ( ))
() ( )
d
L
g t sG s g 0
dt
we get
d 2
d
d
L 2 g t = sL g t g t
dt
dt
dt
( ( ))
( ( ))
d 2
L 2 g t = s sG s g 0
dt
( ( ))
{ ( ) ( )} dtd (g (t ))
( ( ))
()
( )
= s2 G s s g 0 t =0
The second-time-derivative property of the Laplace transform is
N-3
t =0
( ( ))
d
g t
dt
t =0
M. J. Roberts - 2/18/07
( ( ))
()
( ( ))
( )
d
d2
L
g t s2 G s s g 0 g t
2
dt
dt
t =0
.
After seeing the derivation of this property from the previous differentiation
property we can inductively generalize to the Nth derivative.
N
d n1
dN
L
N
N n
g
t
s
G
s
s
n1 g t N
dt
n=1
dt
t =0
( ( ))
( ( ))
()
(N.5)
N.8 Complex-Frequency Differentiation
Start with the definition of the Laplace transform
( ) g (t ) e
G s =
0
st
dt .
Differentiating with respect to s
( ( ))
()
( () )
()
( ( ))
d
d
d
G s =
g t e st dt = g t e st dt = t g t e st dt = L t g t
ds 0
ds
ds
0
0
()
L
t g t ( ( ))
d
G s
ds
N.9 Multiplication-Convolution Duality
()
()
The convolution of g t with h t is
( ) ( ) g ( ) h (t ) d .
g t h t =
()
Since g t is zero for time t < 0 ,
( ) ( ) g ( ) h (t ) d .
g t h t =
0
From the definition of the Laplace transform,
N-4
(N.6)
M. J. Roberts - 2/18/07
() ()
st
h
t
d
g
e dt
0
0 () ()
st
g
e h t dt d .
0
0
L g t h t =
L g t h t =
() (
)
()
(
)
()
Since h t is zero for time t < 0 ,
() ()
L g t h t =
st
g e h t dt d
()
0
(
)
Let = t and d = dt . Then
L g t h t =
() ()
s( + )
e
h
d
g
d
0
0
() ()
s
s
g
e
h
d
e
d
0
0
L g t h t =
()
( )
()
( )
()
H s
() ()
( )e
L g t h t = H s
s
()
() ()
g d = G s H s
0
()
=G s
The time-domain convolution property of the Laplace transform is
() ()
() ()
L
g t h t G s H s
(N.7)
The Laplace transform of a product of time-domain functions is
( ) ( ) g (t ) h (t ) e
L g t h t =
0
st
dt
1 + j
wt
st
L g t h t = G
w
e
dw
h t e dt
j2
j
0 () ()
()
( )
()
()
where is chosen to make G s and H s exist. Doing the t integration first,
N-5
M. J. Roberts - 2/18/07
+ j
1
( s w) t
L g t h t =
G
w
h
t
e
dt
dw .
j2 j
0
() ()
()
( )
()
If H s exists then
h (t ) e
0
(
)
s w t
(
dt = H s w
)
and
+ j
() ()
( ) (
)
1
L g t h t =
G w H s w dw .
j2 j
Therefore
() ()
+ j
( ) (
)
1
g t h t G w H s w dw
j2 j
L
(N.8)
N.10Integration
The integration property is easy to prove, using the convolution property just
proven in the last section and the fact that
t
0
( ) ( ) g ( ) u (t ) d = g ( ) d
g t u t =
() ()
() ()
()
L
g t u t G s U s = G s / s
Therefore
t
g ( ) d s G ( s) .
L
0
1
(N.9)
N.11Initial Value Theorem
Start with the first-time-derivative property of the Laplace transform
d
d
L g t = g t e st dt = sG s g 0 .
dt
dt
0
( ( ))
Let s . Then
lim s
0
( ( ))
( ( ))
() ( )
() ( )
d
g t e st dt = lim sG s g 0 s
dt
N-6
M. J. Roberts - 2/18/07
d
st sG s g 0 .
lim
g
t
e
dt = lim
s dt
s 0
( ( ))
() ( )
()
Case I. g t is continuous at t = 0
()
()
()
If the Laplace transform of g t , which is G s , exists for Re s = > 0 , the
quantity
( ( ))
d
g t e st approaches zero as s approaches infinity and
dt
() ( )
( )
()
0 = lim sG s g 0 g 0 = lim sG s
s
s
( ) ( )
()
and, since g t is continuous at t = 0 , g 0 = g 0+ and
( )
()
g 0+ = lim sG s .
s
()
Case II. g t is discontinuous at t = 0
()
()
In this case, the discontinuity of g t at t = 0 means that the derivative of g t
( ) ( )
has an impulse at t = 0 and the strength of the impulse is g 0+ g 0 . Now the
integral lim s
0
( ( ))
d
g t e st dt becomes
dt
0+
( ( ))
( ) ( ) ()
( ( ))
d
d
lim g t e st dt = lim g 0+ g 0 t e st dt + lim g t e st dt .
s
s
s
dt
dt
0
0
0+
=0
and, using the sampling property of the impulse in the first integral ,
lim s
Therefore,
0
( ( ))
( ) ( )
( ) ( )
d
g t e st dt = lim g 0+ g 0 = g 0+ g 0
s
dt
( ) ( )
() ( )
() ( )
g 0+ g 0 = lim sG s g 0 = lim sG s g 0
s
s
or
( )
()
g 0+ = lim sG s
and the result is the same as in Case I.
N-7
s
(N.10)
M. J. Roberts - 2/18/07
N.12Final Value Theorem
From the first-time-derivative property of the Laplace transform,
lim s0
0
( ( ))
d
lim dt ( g (t )) e
0
() ( )
d
g t e st dt = lim sG s g 0 s0
dt
st
s0
sG s g 0 dt = lim
s0 () ( )
dt ( g (t )) dt = lim sG ( s) g (0
d
s0
0
() ( )
)
() ( )
lim g t g 0 = lim sG s g 0 t
s0
()
Then, if the limit lim t g t exists, the final value theorem of the Laplace transform is
()
()
limg t = lim sG s .
t
s0
N-8
(N.11)