Laplace Transform Properties
A being constant
Z∞
L [Af (t)] = A
f (t) e−st dt = AF (s)
0
Similarly A and B being constant
L [Af1 (t) + Bf2 (t)] = AF1 (s) + BF2 (s)
Laplace transform of differential equations:
df (t)
f˙ (t) =
dt
Z∞ df (t)
df (t) −st
L
=
e dt
dt
dt
L [f (t)]
and
0
u = e−st dt
v = f (t)
du = −se−st dt
dv = df (t)
Z∞
−
udv = uv
0
Z∞
0
e−st df (t) =
Z∞
Z∞
∞
0
vdu
0
df (t) −st
e dt = e−st f (t)
dt
0
∞

Z∞
− −s
0
0
As t → ∞, f (t)rises slower than e−st decreases to zero. Hence
= −f (0) + sF (s)
h
i
df (t)
⇒L
= L f˙ (t) = sF (s) − f (0)
dt

f (t) e−st dt
f (0) is the initial condition of f (t) function at t = 0. If f (0) = 0 at initial condition
then,
h
i
L f˙ (t) = sF (s)
If the initial conditions are zero, then it is much easier to obtain Laplace transform
d
of differential equations. Then s ↔
dt
A
dy (t)
+ y (t) = x (t)
dt
Taking A as constant and initial condition y (0) = 0
AsY (s) + Y (s) = X (s)
Similarly
2
(2) d f (t)
L
=
L
f (t) = s2 F (s) − sf (0) − f˙ (0)
dt2
..
..
..
.
.
.
n
d f (t)
(n) (t) = sn F (s) − sn−1 f (0) − sn−2 f˙ (0) . . . . . . − f (n−1) (0)
=
L
f
L
dtn
If all initial conditions are zero
d2 f (t)
L
= s2 F (s)
dt2
..
..
.
.
n
d f (t)
= sn F (s)
L
dtn
Laplace transform of integral of a function:
Z
L
Z∞ Z
f (t) dt =
f (t) dt e−st dt
0
Using the partial integration rule:
Z
L
F (s) f (−1) (0)
−
f (t) dt =
s
s
R
Here f (−1) (t) = f (t) dt and f (−1) (0) is the integral initial condition at t = 0. If
the initial condition is zero ⇒ f (−1) (0) = 0
Z
F (s)
f (t) dt =
L
s
2
Hence when the initial condition is zero
tion operator ⇒
Rt
1
↔ Similarly:
s
0
1
becomes the replacement for the integras
h
i F (s) f (−1) (0) f (−2) (0)
L f (−2) (t) dt =
−
−
s2
s2
s
Final-value theorem
This theorem is used to determine the value of f (t) function at t = ∞ through
Laplace transform. Final-value theorem can be applied for functions that have a
limit at t = ∞
Let’s again rewrite the derivative transformation
i Z∞
˙
L f (t) = f˙ (t) e−st dt = sF (s) − f (0)
h
0
replacing s with zero:
Z∞
f˙ (t) dt = [sF (s)]s=0 − f (0)
0
in other way:
Z∞
∞
f˙ (t) dt = f (t)
= f (∞) − f (0)
0
0
using these two equalities:
f (∞) = [sF (s)]s=0
This is the mathematical representation of final-value theorem. It can be also represented as:
lim f (t) = lim sF (s)
t→∞
s→0
Initial-value theorem
This theorem is used to determine the value of f (t) function at t = 0+ directly from
Laplace transform.
Let’s consider the derivative transform:
+
Z0
Z∞
h
i Z∞
−st
L f˙ (t) = f˙ (t) e dt = f˙ (t) dt + f˙ (t) e−st dt
0
0
3
0+
Let’s consider the limit s → ∞
+
Z∞
h
i Z0
˙
˙
lim L f (t) = f (t) dt + lim
f˙ (t) e−st dt
s→∞
s→∞
0+
0
for s → ∞ ⇒ e−∞ ≈ 0
h
i
lim L f˙ (t) = f (t)
s→∞
Other direction:
0+
= f 0+ − f (0)
0
h
i
L f˙ (t) = sF (s) − f (0)
Let’s consider the limit for both when s → ∞
h
i
lim L f˙ (t) = lim sF (s) − f (0)
s→∞
s→∞
⇒ f 0+ = lim sF (s)
s→∞
This is the mathematical representation of initial-value theorem. It is valid at t = 0+
and can be re-written as
lim f (t) = lim sF (s)
s→∞
t→0+
4