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Chapter 4 Laplace Transforms

4 Introduction

Reading assignment: In this chapter we will cover Sections 4.1 – 4.5.

4.1 Definition and the Laplace transform of simple functions

Given f , a function of time, with value f(t) at time t, the Laplace transform of f which is

denoted by L(f) (or F ) is defined by

L(f)(s) = F (s) = ∫ ∞ 0

e−stf(t) dt s > 0. (1)

Example 4.1. 1. The Laplace transform is linear: L(αf(t)+βg(t)) = αL(f(t))+βL(g(t))

∫ ∞ 0

e−st (αf(t) + βg(t)) dt = α

∫ ∞ 0

e−st f(t) dt+ β

∫ ∞ 0

e−st g(t) dt.

2. f(t) = 1 ⇒

F (s) =

∫ ∞ 0

e−st 1 dt = −1 s e−st

∣∣∣∣∞ 0

= 1

s

3. f(t) = eat ⇒ for s > a

F (s) =

∫ ∞ 0

e−st eat dt =

∫ ∞ 0

e−(s−a)t dt = − 1 (s− a)

e−(s−a)t ∣∣∣∣∞ 0

= 1

(s− a)

4. For a positive integer n, f(t) = tn ⇒

F (s) = L(tn) = ∫ ∞ 0

e−st tn dt =

∫ ∞ 0

( −1 s e−st

)′ tn dt

=

( −1 s e−st

) tn ∣∣∣∣∞ 0

− ( −n s

)∫ ∞ 0

e−sttn−1 dt

= (n s

)∫ ∞ 0

e−st tn−1 dt = (n s

) L(tn−1)

1

So we find that L(tn) = (n s

) L(tn−1). We can now use this formula over and over to

successively reduce the power of t to obtain

L(tn) = (n s

) L(tn−1) =

( n(n− 1)

s2

) L(tn−2) = · · · =

( n!

sn

) L(1) =

( n!

sn+1

) .

5. f(t) = cos(at) ⇒ To compute the Laplace transform we will use the Euler formula

described in the notes for Chapter 3.

eiθ = cos(θ) + i sin(θ) (2)

which implies that

cos(θ) = eiθ + eiθ

2 .

Also, using i2 = −1 we can write

(s+ ib)(s− ib) = s2 − (ib)2 = s2 + b2.

Combing the above we can write

L(cos(bt)) =L ( eibt + eibt

2

)

= 1

2

( 1

s− ib +

1

s+ ib

)

= 1

2

( (s+ ib)

(s2 + b2) +

(s− ib) (s2 + b2)

)

= 1

2

( (s+ ib) + (s− ib)

(s2 + b2)

) =

s

(s2 + b2) .

So we arrive at

L(cos(bt) = s (s2 + b2)

.

2

6. Given a function f(t) we can find L(f ′(t)) by applying integration by parts as follows

L(f ′(t)) = ∫ ∞ 0

e−st f ′(t) dt = f(t)e−st ∣∣∞ 0 − (−s)

∫ ∞ 0

e−st f(t) dt = sL(f(t))− f(0)

or

L(f ′(t)) = sL(f(t))− f(0)

7. Given a function f(t) find L(f ′′(t)) can be easily computed by using the previous

formula

L(f ′′(t)) = sL(f ′(t))− f ′(0) = s ( sL(f(t))− f(0)

) − f ′(0) = s2L(f(t))− sf(0)− f ′(0).

So we have

L(f ′′(t)) = s2L(f(t))− sf(0)− f ′(0).

8. To compute the Laplace transform of f(t) = sin(bt) we will use two of the previous

formulas.

L(sin(bt)) =−1 b L(cos(bt)′)

= −1 b

[sL(cos(bt))− cos(0)]

= −1 b

[ s

( s

(s2 + b2)

) − 1 ]

= −1 b

[ s2

(s2 + b2) − 1 ]

= −1 b

[ s2 − (s2 + b2)

(s2 + b2)

] =

b

(s2 + b2)

Therefore

L(sin(bt) = b (s2 + b2)

.

Let us consider a few examples of finding Laplace transforms.

Example 4.2. 1. L(2t4) = 2× 4! s5

= 48

s5 .

3

2. L(t2 + 6t− 3) = L(t2) + 6L(t)− 3L(1) = 2 s3

+ 6

s2 − 3 s

.

3. L(2 cos(3t) + 3 sin(2t) − 3e−7t) = 2L(cos(3t)) + 3L(sin(2t)) − 6L(e−7t) = 2s s2 + 9

+

6

s2 + 4 − 6

(s+ 7) .

4. L(2e−t + 6e3t) = 2 (s+ 1)

+ 6

(s− 3) .

Example 4.3. Find the Laplace transform of f(t) = (1 + e2t)2. To do this we first note that

f(t) = 1 + 2e2t + e4t so we have

L(f(t)) = L(1 + 2e2t + e4t) = 1 s

+ 2

(s− 2) +

1

(s− 4) .

Example 4.4. Find the Laplace transform of f(t) = (cos(t) + sin(t))2. To do this we first

note that

f(t) = 1 + 2 sin(t) cos(t) = 1 + sin(2t)

so we have

L(f(t)) = L(1 + sin(2t)) = 1 s

+ 2

(s2 + 4) .

In order to do the next example we need one of the addition formulas from trig

sin(α± β) = sin(α) cos(β)± sin(β) cos(α)

cos(α± β) = cos(α) cos(β)∓ sin(α) sin(β)

Example 4.5. Find the Laplace transform of f(t) = sin(t+ π/2).

f(t) = sin(t) cos(π/2) + sin(π/2) cos(t) = cos(t).

Therefore

L(f(t)) = L(cos(t)) = s (s2 + 1)

.

4.2 The Inverse Laplace Transform

Given a function f(t) the operation of taking the Laplace transform is denoted by L(f(t)) =

F (s) and the inverse process is denoted by L−1(F (s)) = f(t). The process of computing

4

the Laplace transform of a function turns out to be more challenging than most students

like. It involves lots of algebra and using a table of Laplace transforms backwards. For

example, if we were asked to find L−1(3/s3) we would write

L−1(3/s3) = 3 2 L−1(2/s3) = 3

2 t2

since we know that L(t2) = 2/s3 and we can adjust the constants to work out. Most

generally this process will require the use of the method of partial fractions.

Partial Fractions These notes are concerned with decomposing rational functions

P (s)

Q(s) = aMs

M + aM−1s M−1 + · · ·+ a1s+ a0

sN + bN−1sN−1 + · · ·+ b1s+ b0

Note: We can (without loss of generality) assume that the coefficient of sN in the denom-

inator is 1. Also in our intended applications we will always have M < N .

By the fundamental theorem of algebra we know that the denominator factors into a

product of powers of linear and quadratic terms where the quadratic terms correspond to

complex roots. Namely, it can be written in the form

(s− r1)m1 · · · (s− rk)mk (s2 − 2α1s+ α21 + β21)p1 · · · (s2 − 2α`s+ α2` + β2` )p` ,

where k∑ j=1

mj + 2 ∑̀ j=1

pj = n.

The process referred to as Partial Fractions is a method to reduce a complex rational

function into a sum of much simpler terms of the form

c

(s− r)j or

cs+ d

(s2 − 2αs+ α2 + β2)j

The most important point is to learn how to deal with certain types of terms that can

appear. But first there is a special case that arises and is worth a special attention. This

is the case of non-repeated linear terms.

5

I. Nonrepeated Linear Factors

If Q(s) = (s− r1)(s− r2) · · · (s− rn) and ri 6= rj for i 6= j P (s)

Q(s) =

A1 (s− r1)

+ A2

(s− r2) + · · ·+ An

(s− rn)

II. Repeated Linear Factors

If Q(s) contains a factor of the form (s−r)m then you must have the following terms A1

(s− r) +

A2 (s− r)2

+ · · ·+ Am (s− r)m

III. A Nonrepeated Quadratic Factor

If Q(s) contains a factor of the form (s2 − 2αs + α2 + β2) = (s − α)2 + β2 then you

must have the following term

A1s+B1 (s2 − 2αs+ α2 + β2)

IV. Repeated Quadratic Factors

If Q(s) contains a factor of the form (s2 − 2αs + α2 + β2)m then you must have the

following terms

A1s+B1 (s2 − 2αs+ α2 + β2)

+ A2s+B2

(s2 − 2αs+ α2 + β2)2 + · · ·+ Ams+Bm

(s2 − 2αs+ α2 + β2)m

Lets consider some examples of computing inverse Laplace transforms:

Example 4.6. 1. Find L−1 (

1

s3

) =

1

2! L−1

( 2!

s3

) =

1

2 t2.

2. Find L−1 (

1− s s2

) = L−1

( 1

s2 − 1 s

) = t− 1.

3. Find L−1 (

(3s+ 7)

s2 + 16

) = 3L−1

( s

s2 + 42 +

7

4

4

s2 + 42

) = 3 cos(4t) +

7

4 sin(4t).

6

Example 4.7. Find L−1 (

(s+ 1)3

s4

) . We need to first expand the numerator to get

(s+ 1)3

s4 = s3 + 3s2 + 3s+ 1

s4 =

1

s +

3

s2 +

3

s3 +

1

s4 .

So we have

L−1 (

(s+ 1)3

s4

) = L−1

( 1

s +

3

s2 +

3

s3 +

1

s4

) = 1 + 3t+

3

2 t2 +

1

6 t3.

Example 4.8. Find L−1 (

1

(4s+ 1)

) . We have

L−1 (

1

(4s+ 1)

) =

1

4 L−1

( 1

(s+ 1/4)

) =

1

4 e−t/4.

Example 4.9. Find L−1 (

2s− 6 (s2 + 9)

) . For this example we have

L−1 (

2s− 6 (s2 + 9)

) = 2L−1

( s

(s2 + 9)

) − 2L−1

( 3

(s2 + 9)

) = 2 cos(3t)− 2 sin(3t).

Example 4.10. Find L−1 (

4s

(s2 + 2s− 3)

) . For this example we first note that (s2+2s−3) =

(s+ 3)(s− 1) so we have

L−1 (

4s

(s2 + 2s− 3)

)