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MATH 2270
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Matthew Demers
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Mathematics

MATH 2270

Matthew Demers

Fall

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1
4.5 Creating other Functions from the Heaviside Function
The Heaviside function is really interesting, because it can be used to create
many other functions. To do this, we add or subtract two or more di▯erent Heaviside
functions to or from one another. Here are some examples:
Rectangular Impulse Function
If a < b and t ▯ 0, then a rectangular impulse r(t) can be created by
r(t) = H(t ▯ a) ▯ H(t ▯ b) (4.97)
8
>
< 0; 0 ▯ t < a
= > 1; a ▯ t < b (4.98)
>
: 0; t ▯ b
Figure 4.1: On the left, a graph of the rectangular impulse function r(t), with a = 1
and b = 2. This sort of function is sometimes depicted as appears on the right.
The Laplace Transform is then given by
e▯as▯ ebs
Lfra;b)g = ; s > 0 (4.99)
s 2
Staircase Function
Let a > 0 and t ▯ 0. Then, let a staircase function be given by
stair(t) =8H(t ▯ a) + H(t ▯ 2a) + H(t ▯ 3a) (4.100)
>
> 0; 0 ▯ t < a
>
< 1; a ▯ t < 2a
= : (4.101)
> 2; 2a ▯ t < 3a
>
: 3; t ▯ 3a
Figure 4.2: On the left, a graph of stair(t) when a = 1. On the right, a di▯erent
depiction.
The Laplace Transform is then given by
e▯a+ e▯2a+ e▯3as
Lfstair(t)g = ; s > 0: (4.102)
s
This staircase function has three steps. We could easily create a staircase function
with n steps by adding together n Heaviside functions.
\Simple Switch O▯" Function
Let a > 0 be the point of time at which we want a switch from 1 to 0, for t ▯ 0.
Then, let a switching-o▯ function be given by 3
off(t) = 1 8 H(t ▯ a) (4.103)
<
1; 0 ▯ t < a
= : : (4.104)
0; t ▯ a
Figure 4.3: Two depictions of the \Simple Switch O▯" Function, with a = 2.
The Laplace Transform is
▯as
Lfoff(t)g = 1 ▯ e ; s > 0: (4.105)
s
\Switch F on" Function
Let a > 0 be the point of time at which we want a behaviour given by a function
F(t) to switch on, for t ▯ 0. Then, we de▯ne the important \Switch F on" function
by
SwitchFOn(t) = H(t ▯ a)F(t ▯ a) (4.106)
8
< 0; 0 ▯ t < a
= : (4.107)
: F(t ▯ a); t ▯ a 4
Figure 4.4: The \Switch F on" Function, where a = 3 and F(t) = sin(t).
The Laplace Transform for this special function is
▯as
LfH(t ▯ a)F(t ▯ a)g = e f(s); (4.108)
where f(s) = LfF(t)g. (We could derive this from ▯rst principles!)
Going backwards, we also have
▯1 ▯as
L fe f(s)g = H(t ▯ a)F(t ▯ a); (4.109)
▯1
where F(t) = L ff(s)g. 5
Example:
1 ▯2s
Evaluate LfH(t ▯ 2)g = se .
This is done using the basic Laplace transform of the Heaviside function.
Example 2:
Evaluate Lf(t ▯ 2)H(t ▯ 2)g = e ▯2s1 .
s2
This is done using the identity (4.109).
Example 3:
Evaluate LftH(t ▯ 2)g.
This is not quite as straightforward. We need to get this into a form we can use!
It would help if the \t" multiplying in front was instead a \t ▯ 2". But, observe that
LftH(t ▯ 2)g = Lf(t ▯ 2)H(t ▯ 2) + 2H(t ▯ 2)g (4.110)
▯2s 1 1 ▯2s
= e 2 + 2 e : (4.111)
s s
Example 4:
e▯6s
Evaluate L ▯1f g.
s + 9
We can rewrite this as
▯1 ▯6s 1
L fe s + 9 g (4.112)
= 1 L▯1fe▯6s 3 g (4.113)
3 s + 9
1
= H(t ▯ 6)sin(3(t ▯ 6)) (4.114)
3
A rule for writing piecewise functions in terms of Heaviside functions:
Given a piecewise function:
▯ Write down the ▯rst branch 6
▯ For each value \a" that separates the branches, add
H(t ▯ a)[The right branch’s function ▯ The left branch’s function] (4.115)
A rule for writing combinations of Heaviside functions as a piecewise func-
tion:
Given a combination of Heaviside functions:
▯ Arrange the terms by the values a in ihe Heaviside functions H(t ▯ a ) that i
appear, in order of increasing a . Put any terms not multiplied by a Heaviside
i
function before all of the others.
▯ Locate the ▯rst terms, those that are NOT multiplied by a Heaviside function.
This is your ▯rst branch:
Those terms, if 0 ▯ t < (the lowest value of a apiearing in any term):
(4.116)
▯ Now, add a branch for each new term. Each branch should be of the form
The function for this term PLUS all previous functions, if a previous▯ t < a next biggest
(4.117)
▯ In this way, continue until the last branch, which should be the sum of ALL of
the functions appearing in each term, for t ▯ a
biggest
4.6 Using Laplace Transforms to Solve DEs with Discontin-
uous Forcing Terms
Some DEs have forcing terms f(t) with discontinuities/jumps at particular val-
ues of the independent variable. Such DEs in general have solutions that are not
di▯erentiable at those points, so we might be hesitant to talk about such di▯erential
equations. We require that solutions for these DEs are: 7
▯ di▯erentiable on each open interval where the forcing term is continuous; and
▯ continuous at each point at which there is a discontinuity in the forcing term.
It is important that discontinuous forcing terms are written in terms of Heaviside
functions if they are not already, so that we can take the Laplace transforms easily.
Example:
Find the solution to the IVP
00
y + 4y = H(t ▯ ▯) ▯ H(t ▯ 2▯) (4.118)
y(0) = 1 (4.119)
y (0) = 0: (4.120)
We take the Laplace transforms of each part.
▯▯s ▯2▯s
2 0 e ▯ e
s Lfyg ▯ sf(0) ▯ f (0) + 4Lfyg = s (4.121)
▯▯s ▯2▯s
2 e ▯ e
(s + 4)Lfyg = s + s (4.122)
▯▯s ▯2▯s
Therefore, Lfyg = e ▯ e + s : (4.123)
s(s + 4) s(s + 4) s + 4
We can use partial fraction decomposition on the ▯rst two terms. This yields
▯ ▯ ▯ ▯
1 e ▯▯s se▯▯s 1 e ▯2▯s se▯2▯s s
Lfyg = ▯ ▯ ▯ + : (4.124)
4 s s + 4 4 s s + 4 s + 4
Taking inverse transforms,
1 1 1
y(t) = 4H(t ▯ ▯) ▯ 4(t ▯ ▯)cos(2(t ▯ ▯)) ▯ H4t ▯ 2▯) + (4.125)
1H(t ▯ 2▯)cos(2(t ▯ 2▯)) + cos(2t) (4.126)
4
Or, grouping by Heaviside function,
▯ ▯ ▯ ▯
1 1 1 1
y(t) = cos(2t) + H(t ▯ ▯) ▯ cos(2t) + H(t ▯ 2▯) ▯ + cos(2t) ; (4.127)
4 4 4 4 8
noting that cos(2t) has a period of ▯ and so cos(2(t ▯ ▯)) = cos(2t). How can we
make sense of this? Sort the di▯erent terms by the values of a in each of the Heaviside
functions, and construct the piecewise function of t from left to right. Analyzing it
this way, we see that:
▯ We have cos(2t) alone to begin with, until we get to t = ▯.
▯ At t = ▯, we add 1 ▯ 1cos(2t).
4 4
▯ At t = 2▯, we add ▯ + 1 1cos(2t).
4 4
So, at t = 2▯, the changes from before are undone, and the function once again
becomes cos(2t). A graph of this solution function can be seen in Figure 4.5.
Figure 4.5: A graph of the solution y(t) of the IVP described in Equations (4.118 -
4.120).
4.7 The Dirac Delta Function
Another function that is often seen in applications is an impulse function, repre-
senting a sudden behaviour at a particular moment of time. To create this function, 9
we ▯rst de▯ne:
8
1
< 2▯; if ▯▯ < t < ▯
▯▯(t) = :
: 0; if jtj ▯ ▯
Note that the area under ▯ ▯t) is always equal to 1, no matter the value of ▯, since
Z Z
1 ▯ 1 1
▯▯(t) dt = dt = (2▯) = 1: (4.128)
▯1 ▯▯ 2▯ 2▯
As ▯ decreases, the height of the step increases, while the width decreases, but the
area underneath maintains a constant area of 1. We may decrease ▯ arbitrarily, and
even speak of a limit. Let
▯(t) = lim▯▯(t); (4.129)
▯!0
Z 1
with the property that ▯(t) dt = 1. Then, we call this function the Dirac Delta,
▯1
or unit impulse function at t = 0. The function ▯(t▯t )0is therefore a unit impulse
at time t = t0instead. Let’s ▯nd the Laplace Transform of this function!
Lf▯(t ▯ t0)g = limLf▯ (t ▯ t )g0 (4.130)
▯!0 Z
1
= lim e▯st▯▯(t ▯ 0 ) dt (4.131)
▯!0 0
Z 0 +▯
▯st1
= li▯!0 e 2▯ dt (4.132)
0 ▯▯ ▯
1 ▯ e▯st▯▯t0+▯
= lim ▯ (4.133)
▯!0 2▯ ▯s ▯
t0▯▯
1 ▯ ▯s(t0+▯) ▯s(0 ▯▯)
= lim ▯e + e (4.134)
▯!

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