Step Functions
To deal effectively with functions having jump discontinuities, it is very helpful to
introduce a function known as the unit step function or Heaviside function.
This function is denoted by uc where c ≥ 0, and it is defined by:
!0!if!t < c
u c (t) = "
#1!if!t > c
When c = 0, we have u0(t) = 1
Notice
"1!if!0 < t < c
u 0 (t) ! u c (t) = #
$ 0!if!t > c
A variety of so-called step functions can be expressed as a linear combination of unit step
functions.
Example:
1)
!0!if!0 < t < 3
!0!if!0 < t < 3
f(t) = "
= 5"
= 5u 3 (t)
# 1!if!t > 3
# 5!if!t > 3
2)
!0!if!0 < t < 7
! 3!if!0 < t < 7
!0!if!0 < t < 7
g(t) = "
= 3+ "
=!3 + 6 "
= 3u 0 (t) + 6u 7 (t)
# 9!if!t > 7
# 1!if!t > 7
# 6!if!t > 7
3)
"0!if!0 < t < 1
"0!if!0 < t < 1 "0!if!0 < t < 3
"0!if!0 < t < 1 "0!if!0 < t < 3
%
h(t) = # 2!if!1 < t < 3 = #
+#
=2#
!#
=
2!if!t
>
1
!1!if!t
>
3
1!if!t
>
1
1!if!t
>
3
$
$
$
$
% 1!if!t > 3
$
2u1 (t) ! u 3 (t)
4)
" 2!if!0 < t < 3
" 0!if!0 < t < 3
"0!if!0 < t < 3 "0!if!0 < t < 5
$
$
p(t) = #!3!if!3 < t < 5 !!= 2 + #!5!if!3 < t < 5 !!!=!2 + #
+#
=
% !5!if!t > 3 % 9!if!t > 5
$ 6!if!t > 5
$ 4!if!t > 5
%
%
"0!if!0 < t < 3
"0!if!0 < t < 5
2 ! 5#
+ !9 #
= !2u 0 (t) ! 5u 3 (t) + 9u 5(t)
% 1!if!t > 3
% 1!if!t > 5
5)
" 1!if!0 < t < 1
" 0!if!0 < t < 1
$ 2!if!1 < t < 4
$ 1!if!1 < t < 4
"0!if!0 < t < 1 "0!if!0 < t < 4 "0!if!0 < t < 8
$
$
g(t) = #
= 1+#
= 1+#
+#
+#
% 1!if!t > 4
% !5!if!t > 8
% 1!if!t > 1
$ 3!if!4 < t < 8
$2!if!4 < t < 8
$% !2!if!t > 8
$% !3!if!t > 8
"0!if!0 < t < 1 "0!if!0 < t < 4
"0!if!0 < t < 8
1+ #
+#
! 5#
= u 0 (t) + u 4 (t) ! 5u 8 (t)
% 1!if!t > 4
% 1!if!t > 8
% 1!if!t > 1
Since the function uc(t) is bounded and piecewise continuous, then it has a Laplace
transform.
L{uc(t)} =
# !st
"0 e
u c (t)!dt =
c !st
"0 e
c
A #st
e dt
A!" c
lim
*
#
0!dt + " e !st 1!dt =
$ #e #st
= lim &
A!"
% s
A'
c )
(
=
e #sc
, for s > 0
s
Example:
" 3!if!0 < t < 2
" 0!if!0 < t < 2
"0!if!0 < t < 2 "0!if!0 < t < 6
$
$
f(t) = #!4!if!2 < t < 6 !!= 3 + #!7!if!2 < t < 6 !!!=!3 + #
+#
=
% !7!if!t > 2 % 11!if!t > 6
$ 7!if!t > 6
$ 4!if!t > 6
%
%
"0!if!0 < t < 2
"0!if!0 < t < 6
3! 7#
+ !11 #
= !3u 0 (t) ! 7u 2 (t) + 11!u 6 (t)
% 1!if!t > 2
% 1!if!t > 6
L{f(t)} = L{3u0(t) – 7u2(t) + 11 u6(t)} = 3
e0
e !2s
e !6s
!7
+ 11
.
s
s
s
The Second Translation Property or Shifting Property
Consider the function f(t) for t > 0, we graph
Consider a new function g(t) that results from translating f(t) a distance of a units into the
positive direction
The new function is defined by:
" 0!if!0 < t < a
g(t) = #
$f(t ! a)!if!t > a
so
g(t) = ua(t) f(t –a)
The Laplace transform of the new function satisfies the following property:
Theorem: Shifting Property or Second Translation Property
Assume f(t) has a Laplace Transform
" !st
#0 e
L{f(t)} = F(s) =
f(t)!dt
Let g(t) = ua(t) f(t –a), then
L{g(t)} = L{ua(t) f(t –a)}= e-as L{f(t)} = e-as F(s)
Proof:
L{ua(t) f(t –a)} =
" !st
e u a (t)f(t
0
#
! a)!dt =
Taking τ = t – a, since a < t < ∞, then τ > 0 and
# !s( "+a )
$0 e
" !st
e f(t
a
#
#
f(")!d" = e !sa $ e !s" f(")!d" = e !as F(s)
0
Example:
1)
"0!if!0 < t < 5
g(t) = #
$ t ! 3!if!t > 5
We express t – 3 = (t – 5) + 2, so
" 0!if!0 < t < 5
g(t) = #
$( t ! 5 ) + 2!if!t > 5
so f(t) = t + 2, t > 0 and
g(t) = u5(t) f(t -5)
1 2
+
s2 s
" 1 2%
L{u5(t)f(t – 5)} = e-5s F(s) = e !5s $ 2 + '
#s
s&
F(s) = L{f(t)} = L{t + 2} = L{t} + L{2} =
! a)!dt =
2)
"$0!if!0 < t < !2
g(t) = #
!
$% sin t!if!t > 2
since sin t = cos(t – π/2)
#% 0!if!0 < t < !2
g(t) = $
!
!
%&cos(t " 2 )!if!t > 2
then g(t) = uπ/2(t) cos(t – π/2) where f(t) = cos t, t > 0
s
Since, F(s) = L{f(t)} = L{cos t} = 2
s +1
then
L{uπ/2(t)f(t – π/2)} = e-π/2s F(s) = e ! 2 s
"
s
s +1
2
3)
#%
sin t!if!0 < t < !4
g(t) = $
!
&%sin t + cos(t " 4 )!if!t >
#%
sin t!if!0 < t < !4
g(t) = $
!
&%sin t + cos(t " 4 )!if!t >
where f(t) = cos t, t > 0
!
4
!
4
#% 0!if!0 < t < !4
= sin t + $
!
&%cos(t " 4 )!if!t >
!
4
= sin t + u !4 (t)f(t " !4 )
1
s
1 + se ! 4
! "4 s
+e
= 2
L{g(t)} = L{sin t} + L{uπ/4(t) f(t – 5)} = 2
s +1
s1 + 1
s +1
"
4)
" 3!if!0 < t < 4
$
h(t) = #!5!if!4 < t < 6
$ 7! t
% e !if!t > 6
Since
" 3!if!0 < t < 4
" 0!if!0 < t < 4
"$0!if!0 < t < 4 "$ 0!if!0 < t < 6
$
$
h(t) = #!5!if!4 < t < 6 = 3 + # !8!if!4 < t < 6 = 3 + #
+ # 7! t
=
%$e + 5!if!t > 6
$ 7! t
$ 7! t
%$ !8!if!t > 4
% e !if!t > 6
%e ! 3!if!t > 6
"0!if!0 < t < 4 "$ 0!if!0 < t < 4
3! 8#
+ # !(t!6)
= 3u 0 (t) ! 8u 4 (t) + u 6 (t)f(t ! 6)
e + 5!if!t > 6
$%e
% 1!if!t > 4
where!f(t) = e ! t e + 5,!t > 0.
F(s) = L {f(t)} = eL{e-t} + L{5} = e/(s+1) + 5/s.
L{h(t)} = 3 L{u0(t)} - 8 L{u4(t)} + L{u6(t) f(t-6)} = 3/s - 8e-4s/s + e-6s(e/(s+1) + 5/s) =
3/s - 8e-4s/s + e-6s+1/(s+1) + 5e-6s/s