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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