2E2 Tutorial sheet 1 Solutions
[Wednesday October 25th, 2000]
1. Find the Laplace transform of the step function
0, t < a
Ha (t) =
1, t ≥ a
where a > 0.
Solution: From the definition of the Laplace transform, we have
Z ∞
L [Ha ] (s) =
Ha (t)e−st dt
0
and since Ha (t) = 0 for t < a, this is
Z ∞
L [Ha ] (s) =
Ha (t)e−st dt
Za ∞
=
1e−st dt
a
Z b
= lim
e−st dt
b→∞
a
(using the definition of an improper integral of this type)
t=b
−1 −st
e
= lim
b→∞
s
t=a
−1 −sb
−1 −sa
= lim
e −
e
b→∞
s
s
1
= 0 + e−sa (if s > 0)
s
2. Find the Laplace transform of 1 − Ha (t).
Solution: We could do this in the same way as the first question, or we could use linearity
of the Laplace transform plus the earlier calculation we did that L[1](s) = 1/s (s > 0) and
the result of the previous question:
L [1 − Ha ] (s) = L[1](s) − L [Ha ] (s) =
1 1 −sa 1 − e−sa
− e
=
s s
s
(for s > 0).
In fact, if we did it directly we could show that
Z a
−st
L [1 − Ha ] (s) =
e dt =
0
1−e−sa
s
a
s 6= 0
s = 0.
3. Find the Laplace transform of the function

 0, t < a
1, a ≤ t < b
f (t) =

0, t ≥ b
where 0 < a < b.
Solution: Again we could use previous results and linearity, because f (t) = Ha (t) − Hb (t)
and so
1
1
L[f ](s) = L [Ha − Hb ] (s) = e−sa − e−sb
s
s
(for s > 0).
Or, we could apply the definition of the Laplace transform to show directly that
Z b
−1 −sb
−1 −sa
−st
L[f ](s) =
1e dt =
e −
e
s
s
a
(as calculated in question 1) without any restriction on s except s 6= 0. For s = 0 we get
the value b − a.
4. Find the Laplace transform of both sides of the differential equation
d2 x
dx
+ 2 + 2x = 1
2
dt
dt
dx
(0) = 0.
dt
Solution: Using linearity of L, plus the property of Laplace transforms of derivatives (simpler with zero initial conditions), we get
2
dx
dx
L
+ 2 + 2x (s) = L[1](s)
dt2
dt
2 dx
dx
1
(s) + 2L
(s) + 2L[x](s) =
L
2
dt
dt
s
1
s2 L[x](s) + 2sL[x](s) + 2L[x](s) =
s
1
(s2 + 2s + 2)L[x](s) =
s
with initial conditions x(0) =
Richard M. Timoney
2