Download - Laplace Periodic Solutions
The Laplace Transform of a Periodic Function
Periodic Functions
Definition. A function f is periodic (with period T ) if f(t + T ) = f(t) for all t in thedomain of f .
A periodic function has regular repetitive behavior.
Example. f(t) = sin(πt) has period T = 2 since sin(π(t + 2)) = sin(πt). The graph of frepeats itself every 2 units.
Example. The following function f also repeats every two units. So, f has period T = 2.
We can describe f on one of its periods, [0, 2], with the piecewise function:
{2t 0 ≤ t < 1−1 1 ≤ t < 2
.
Exercise 1. What is the period of the following function? T = 3
Describe f on one of its periods, [0, 3], with the piecewise function:
{23t 0 ≤ t < 3/2−2
3t+ 2 3/2 ≤ t < 3
.
Exercise 2. What is the period of the following function? T = 6
Describe f on one of its periods, [0, 6], with the piecewise function:
{3 0 ≤ t < 3−3 3 ≤ t < 6
.
Exercise 3. What is the period of the following function? T = 2π(It is a half-wave rectification of cos t.)
Describe f on one of its periods, [0, 2π], with the piecewise function:
cos t 0 ≤ t < π
2
0 π2≤ t < 3π
2
cos t 3π2≤ t < 2π
.
Laplace Transform of a Periodic Function
You already know the Laplace Transform of some periodic functions. For example,
L {sin(πt)} =π
s2 + π2
L {4 cos(3t)} =4s
s2 + 9
Now we will derive a formula for computing the Laplace Transform for periodic functionssuch as the ones in the first section.
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Derivation of Formula (look at page 311 for help)
Suppose f is a periodic function with period T . Then, we know for any value of t thatf(t + T ) = f(t). So, for example, these are all true: f(6 + T ) = f(6), f(u + T ) = f(u),f(, + T ) = f(,). Let’s evaluate L {f(t)}.
L {f(t)} =
∫ ∞0
e−stf(t)dt =
∫ T
0
e−stf(t)dt+
∫ ∞T
e−stf(t)dt
We have written the Laplace Transform of f(t) as the sum of two integrals. Apply thesubstitution t = u + T, dt = du to the second integral. (Don’t forget to change the limitsof integration! When t=T what is u?) Then, factor out e−sT and replace f(u+T ) with f(u).∫ ∞
T
e−stf(t)dt =
∫ ∞0
e−s(u+T )f(u+ T )du = e−sT∫ ∞0
e−suf(u)du
After applying the substitution and simplifying you have e−sT∫ ∞0
e−suf(u)du. The name
of the variable of integration is not important; notice the integral is L {f(t)}. So, we now
know that
∫ ∞T
e−stf(t)dt = e−sTL {f(t)}. Replace
∫ ∞T
e−stf(t)dt with e−sTL {f(t)} in the
boxed equation above and then solve the equation for L {f(t)}.
L {f(t)} =
∫ T
0
e−stf(t)dt+ e−sTL {f(t)}
L {f(t)} =1
1− e−sT
∫ T
0
e−stf(t)dt
This is the formula for computing the Laplace Transform of a periodic function with periodT . Don’t forget it! To apply this formula, all you need to know is the period T of thefunction and a description of f on the period [0, T ]. Be careful, T is a number and t is thevariable of integration. Don’t confuse their roles in the formula.
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Use example 6 on page 312 as a guide to complete the following exercises.
Exercise 4. Use the formula you derived to find the Laplace Transform of the function inExercise 1.
L {f(t)} =1
1− e−3s
∫ 3
0
e−stf(t)dt =1
1− e−3s
(∫ 3/2
0
e−st(
2
3t
)dt+
∫ 3
3/2
e−st(−2
3t+ 2
)dt
)
=1
1− e−3s
(−2te−st
3s
∣∣∣3/20
+2
3s
∫ 3/2
0
e−stdt+
(2te−st
3s− 2e−st
s
) ∣∣∣33/2− 2
3s
∫ 3
3/2
e−stdt
)=
1
1− e−3s
(−e−3/2s
s− 2
3s2e−st
∣∣∣3/20
+
(6e−s3
3s− 2e−3s
s
)−(e−3s/2
s− 2e−3s/2
s
)+
2
3s2e−st
∣∣∣33/2
)=
1
1− e−3s
(−4e−3s/2
3s2+
2
3s2+
2e−3s
3s2
)
Exercise 5. Use the formula you derived to find the Laplace Transform of the function inExercise 2.
L {f(t)} =1
1− e−6s
∫ 6
0
e−stf(t)dt =1
1− e−6s
(∫ 3
0
3e−stdt+
∫ 6
3
(−3)e−stdt
)=
1
1− e−6s
(−3
se−st
∣∣∣30
+3
se−st
∣∣∣63
)=
1
1− e−6s
(−3e−3s
s+
3
s+
3e−6s
s− 3e−3s
s
)=
1
1− e−6s
(3
s− 6e−3s
s+
3e−6s
s
)
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