Laplace transform of a periodic function1
16 November 2001
If f (t) is a periodic function with period a we have f (t + a) = f (t). The Laplace transfrom
of f (t) can be found by integrating over only one period:
Z a
1
f (t)e−st dt
(1)
L(f ) =
1 − e−as 0
This equation can be derived as follows:
Z ∞
f (t)e−st dt
L(f ) =
0
Z
Z a
−st
f (t)e dt +
=
0
=
∞ Z
X
n=0
2a
f (t)e−st dt + . . .
a
(n+1)a
f (t)e−st dt
(2)
na
Next, we integrate one term of the sum using the change of variable t0 = t − na:
Z
(n+1)a
f (t)e
−st
dt =
na
Z
a
0
f (t + na)e
−st0 −nas
0
dt =
0
Z
a
0
f (t0 )e−st −nas dt0
(3)
0
Hence, dropping the prime on t we get
L(f ) =
∞
X
e−san
n=0
!Z
a
f (t)e−st dt
(4)
0
and, finally, we use
∞
X
xn =
n=0
to make the replacement
∞
X
e−san =
n=0
and this proves the formula.
1
1
1−x
(5)
1
1 − e−sa
(6)
Conor Houghton, houghton@maths.tcd.ie please send me any corrections.
1