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Week 11- Piecewise Functions; Periodic Functions
The following graphs depicts a periodic function. For each
function,(a) write the function in piecewise form,
(b) write the function using step functions, and
(c) find the Laplace transform of the function
1. Repeated step function with amplitude 5, period 4.
0 1 2 3 4 5 6 7 8
10
5
5
10
(a) In piecewise form,
f(t) =
5, 0< t
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the period T= 4, so
L(f) = 1
1 e4sL(5u0 5u2)
= 1
1 e4s
5
e0s
s 5
e2s
s
= 1
1 e4s
5
1
s5
e2s
s
2. Square wave, amplitude 2, period 10.
0 5 10 15 20 25 30 35 40
3
2
1
1
2
3
(a) In piecewise form,
f(t) =
2, 0< t
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3. Ramp function with max value 8, period 4.
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
4
2
2
4
6
8
10
(a) In piecewise form,
f(t) =
2t, 0< t
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(b) In step form,f(t) = 100sin(t)(u0 u) + 100sin(t )(u u2) + . .
.
(c) Identifying the first p eriod functionf1,
f1(t) = 100 sin(t)(u0 u)
At and after t = , the +/
combinations of step functions will have have a value ofzero,
sof1 is zero for all t > , as needed.
L(f) = 1
1 esL(100 sin(t)u0 100 sin(t)u)
= 100 1
1 es
e0sL(sin(t+ 0)) esL(sin(t+pi))
Noting that sin(t+ ) = sin(t),
= 100 1
1 es
1
s2 + 1
es
1
s2 + 1
For each of the following functions,(a) find the inverse Laplace
transform,
(b) write thet-domain version in piecewise form, and
(c) sketch the graph off(t) = L1 (F(s)))
5. F(s) = 1
1 e5s
3
s
2e2s
s
e5s
s
(a) Looking at the leading factor 11e5s , the function in the
time domain is periodic withperiod T= 5. To find its shape on one
cycle, we look at the inverse transform of theremainder of the
function.
f1(t) = L
13
s 2e2s
s e5s
s
= 32u2 u5
(b) In piecewise form, this would be
f1(t) =
3, 0< t
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(c) The graph is shown below.
0 5 10 15
1
1
2
3
6. F(s) = 1
1 e8s
1
s2(12e4s + e8s)
(a) To compute the inverse transform, we need to express this
function in a form entrieslike the table, then each term multiplied
by exponentials as necessary.
F(s) = 1
1 e8s
1
s2(1 2e4s + e8s)
= 1
1 e8s
1
s22e4s
1
s2+ e8s
1
s2
From the leading factor, we know we are looking for a periodic
function with period
T= 8. We focus on the second factor, which gives us the
transform of the first periodof the function:
f1(t) = L1
1
s22e4s
1
s2+ e8s
1
s2
=t 2u4 (t 4) + u8 (t 8)
(b) In piecewise form,
f1(t) =
t, 0< t
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(c) The graph of the periodic functionf(t) is shown below.
0 4 8 12 16 20 24
2
2
4
Solve the following initial value problems using Laplace
transforms.
7. x + 4x= f(t), x(0) = 1, where f(t) =
0, 0< t
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Writing the non-homogeneous part using step functions, f(t) =
1(u1u2)+1(u4u6) + . . ..
This function is defined by its values over the first period,
f1(t) = 1(u1 u2). The periodof the function is T = 4. We use the
formula for the transform of periodic functions,
L(f) = 1
1 eTsL(f1), to help us take the Laplace transform of both sides
of the DE.
L(x + 9x) = 1
1 e4sL(u0 u2)
(s2X(s) s 0 0) + 9X(s) = 1
1 e4s
1
s e2s
1
s
(s2 + 9)X(s) = 1
1 e4s
1
s e2s
1
s
Solving for X(s), X(s) = 1
1 e4s
1
s(s2 + 9) e2s
1
s(s2 + 9)
Note that we need to find the partial fraction decomposition of
1s(s2 + 9) to proceed:
1
s(s2 + 9)=
A
s +
B s + C
s2 + 9 = (some work) =
1/9
s
s/9
s2 + 9
We can then rewrite our form ofX(s) to make it easier to find
the inverse transform:
X(s) = 1
1 e4s
1
s(s2 + 9) e2s
1
s(s2 + 9)
= 1
1 e4s1/9
s 1
9
s
s2 + 9 e
2s 1/9
s 1
9
s
s2 + 9
The first factor indicates that x(t) is periodic with period T =
4. We look at the inversetransform of the second factor to obtain
the definition of the function on its first period:call this
x1(t).
x1(t) = L1
1/9
s
1
9
s
s2 + 9 e2s
1/9
s
1
9
s
s2 + 9
=1
9[1 cos(3t) u2(1cos(3(t 2))]
The solutionx(t) will be this function repeated periodically
every T= 4 seconds.
