# Chapter 4 Laplace Transforms 4 gilliam/ttu/s16/chapter_4_lecture_notes.pdfChapter 4 Laplace Transforms

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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

estf(t) dt s > 0. (1)

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

0

est (f(t) + g(t)) dt =

0

est f(t) dt+

0

est g(t) dt.

2. f(t) = 1

F (s) =

0

est 1 dt = 1sest

0

=1

s

3. f(t) = eat for s > a

F (s) =

0

est eat dt =

0

e(sa)t dt = 1(s a)

e(sa)t0

=1

(s a)

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

F (s) = L(tn) = 0

est tn dt =

0

(1sest

)tn dt

=

(1sest

)tn0

(ns

) 0

esttn1 dt

=(ns

) 0

est tn1 dt =(ns

)L(tn1)

1

• So we find that L(tn) =(ns

)L(tn1). We can now use this formula over and over to

successively reduce the power of t to obtain

L(tn) =(ns

)L(tn1) =

(n(n 1)

s2

)L(tn2) = =

(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.

ei = cos() + i sin() (2)

which implies that

cos() =ei + ei

2.

Therefore, using i2 = 1 and

(s+ ib)(s ib) = s2 (ib)2 = s2 + b2,

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) find L(f (t)) we apply integration by parts as follows

L(f (t)) = 0

est f (t) dt = f(t)est0 (s)

0

est 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)) . This can be easily done 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. f(t) = sin(at) To compute this Laplace transform we will use two previous

formulas.

L(sin(bt)) =1bL(cos(bt))

=1b

[sL(cos(bt)) cos(0)]

=1b

[s

(s

(s2 + b2)

) 1]

=1b

[s2

(s2 + b2) 1]

=1b

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

3

• f(t) = 1 + 2e2t + e4t so we have

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

+2

(s 2)+

1

(s 4).

Example 4.3. 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)) = 1s

+2

(s2 + 4).

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

sin( ) = sin() cos() sin() cos()

cos( ) = cos() cos() sin() sin()

Example 4.4. Find the Laplace transform of f(t) = sin(t+ /2).

f(t) = sin(t) cos(/2) + sin(/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 L1(F (s)) = f(t). The process of computing

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 L1(3/s3) we would write

L1(3/s3) = 32L1(2/s3) = 3

2t2

4

• 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 FractionsThese notes are concerned with decomposing rational functions

P (s)

Q(s)=aMs

M + aM1sM1 + + a1s+ a0

sN + bN1sN1 + + 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 21s+ 21 + 21)p1 (s2 2`s+ 2` + 2` )p` ,

wherekj=1

mj + 2j=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)jor

cs+ d

(s2 2s+ 2 + 2)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.

I. Nonrepeated Linear Factors

If Q(s) = (s r1)(s r2) (s rn) and ri 6= rj for i 6= jP (s)

Q(s)=

A1(s r1)

+A2

(s r2)+ + An

(s rn)

5

• II. Repeated Linear Factors

If Q(s) contains a factor of the form (sr)m then you must have the following termsA1

(s r)+

A2(s r)2

+ + Am(s r)m

If Q(s) contains a factor of the form (s2 2s + 2 + 2) = (s )2 + 2 then you

must have the following term

A1s+B1(s2 2s+ 2 + 2)

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

following terms

A1s+B1(s2 2s+ 2 + 2)

+A2s+B2

(s2 2s+ 2 + 2)2+ + Ams+Bm

(s2 2s+ 2 + 2)m

Lets consider some examples of computing inverse Laplace transforms:

Example 4.5. Find

L1(

1

s3

)=

1

2!L1

(2!

s3

)=

1

2t2.

Example 4.6. Find L1(

(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

L1(

(s+ 1)3

s4

)= L1

(1

s+

3

s2+

3

s3+

1

s4

)= 1 + 3t+

3

2t2 +

1

6t3.

6

• Example 4.7. Find L1(

1

(4s+ 1)

). We have

L1(

1

(4s+ 1)

)=

1

4L1

(1

(s+ 1/4)

)=

1

4et/4.

Example 4.8. Find L1(

2s 6(s2 + 9)

). For this example we have

L1(

2s 6(s2 + 9)

)= 2L1

(s

(s2 + 9)

) 2L1

(3

(s2 + 9)

)= 2 cos(3t) 2 sin(3t).

Example 4.9. Find L1(

4s

(s2 + 2s 3)

). For this example we first note that (s2+2s3) =

(s+ 3)(s 1) so we have

L1(

4s

(s2 + 2s 3)

)= L1

(4s

(s+ 3)(s 1)

).

In order to carry out this inverse Laplace transform we must use partial fractions. Since

the denominator consists of non-repeated terms we can use the first box (in the formulas

for partial fractions) to write

4s

(s2 + 2s 3)=

A

(s+ 3)+

B

(s 1).

To find A and B we can use the cover-up method to find that A = 3 and B = 1 so that

L1(

4s

(s2 + 2s 3)

)= L1

(3

(s+ 3)

)+ L1

(1

(s 1)

)= 3e3t + et.

Now lets solve an initial value problem using the formula for the Laplace transform of

a derivitive.

Example 4.10. Find the solution of y y = 1 with y(0) = 0. First we take the Laplace

transform of both sides (using Y = L(y)) to get

L(y) L(y) = L(1) (sY 0) Y = 1s Y = 1

s(s 1)

7

• So to find y we need to take the inverse Laplace transform which requires we do partial

fractions.

y = L1(Y ) = L1(

1

s(s 1)

)= L1

(1s

)+ L1

(1

(s 1)

)= 1 + et.

Example 4.11. Find the solution of y+ 5y+ 4y = 0 with y(0) = 3 and y(0) = 0. First we

take the Laplace transform of both sides (using Y = L(y)) to get

L(y) + 5L(y) + 4L(y) = 0 (s2Y 3s) + 5(sY 3) + 4Y = 0

Solving for Y we get

Y =3(s+ 5)

(s2 + 5s+ 4) Y = 3(s+ 5)

(s+ 4)(s+ 1)

So to find y we need to take the inverse Laplace transform which requires we do partial

fractions.

y = L1(Y ) = L1(

3(s+ 5)

(s+ 4)(s 1)

)= L1

(1

(s+ 4)

)+ L1

(4

(s+ 1)

)= e4t + 4et.

4.3 The Shift Theorems

Two of the most important and useful results we need to discuss are the first and second

shift (or Translation) theorems.

Theorem 4.1 (First Shift Theorem). If L(f(t)) = F (s) then L(eat f(t)) = F (s a) for any

real number a.

Proof.

L(eat f(t)) = 0

est eat f(t) dt =

0

e(sa)tf(t) dt = F (s a).

8

• Lets consider some examples.

Example 4.12.

L(t2e3t) = L(t2)s3 =

2

(s 3)3

Example 4.13.

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

=2

(s+ 1)2 + 4

Example 4.14.

L(t(et

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