Chapter 5

Laplace Transform Method

In this chapter, we introduce one of the crucial tools in engineering mathematics called

Laplace transform. This tool enables us to solve diﬀerential equations with discontin-

uous terms in an eﬃcient and straightforward way. Besides solving ordinary diﬀerential

equations, with the aid of the Laplace tr a nsform, we are able to deﬁne the concept of the

transfer function of complex systems in engineering and the idea of the fundamental solu-

tion.

5.1 Deﬁni tion of the transformation

The unilateral Laplace or L-transform of a function f (t) is deﬁned by the following inte-

gral

L(f)(s) =

¡st

f(t) dt; (5.1)

as long as the integral exists for some values of s. Note that (5.1) transforms a function of

t (that we can interpret as time) to a function of s (can be c o nsidered just as a para-

meter). In this chapter, we use notations f

(s), L(f) for the transform. It is called unilat-

eral because f(t) for t < 0 (the history of f ) does not aﬀect the transformation.

Definition 5.1. A function f(t) is called L-admissible if there is an interval of s for

which the integral ( 5.1) converges.

Remark 5.1. The class of L-admissible functions is wide, however, the sub-exponential

functi ons are well-known to be

-adm is sible (see the problem set). A fun ction f (t) is

called sub-exponential if there a re a > 0; b > 0 such that jf(t)j< ae

for all t 2[0; 1).

At the ﬁrst glance, formula (5.1) looks peculiar, nevertheless, the transform possesses

nice properties that makes it desirable in engin eering mathematics. For example, it trans-

forms the derivative operation to an algebraic multiplication, that is, if f

(t) is the deriva-

tive of a function f (t), then

L(f

) = s L(f ) ¡ f(0):

The above property ﬁnds its importance to solve ordinary diﬀerential equations.

5.1.1 L-Transform of basic functions

1. The Heaviside or unit step function u(t) is deﬁned by the following two valued

function

u(t) =



0 t < 0

1 t > 0

: (5.2)

In some texts, the value u(0) is considered

, or even 1, but u(0) does not aﬀect its

transformation. We have

u^(s) =

¡st

u(t) dt =

¡st

dt =

lim

t!1

¡st

¡1

: (5.3)

Clearly, the above integral converges to

if s > 0. Note that L(1) =

since two

function u(t) and 1 are equal for t ≥0.

2. Consider the unit pulse

p(t) =



1 0 < t < 1

0 otherwise

: (5.4)

Its L-transform is

p^(s) =

¡st

dt =

1 ¡e

¡s

: (5.5)

Note that the value of p(t) at the end points does not aﬀect the transformation.

3. The transform of the ramp function r(t) = t is

L(t) =

¡st

dt = ¡lim

t!1

¡st

dt: (5.6)

If s > 0, we have

lim

t!1

¡st

= 0;

and thus L(t ) =

4. The transform of the exponential function f (t) = e

L(e

) =

¡st

dt =

¡(s¡a)t

dt: (5.7)

Clearly the integral converges for s > a and then L(e

¡at

) =

s ¡a

5. The L-transform of the function f(t) = sin(!t) is simply derived by the aid of the

Euler formula

sin(!t) =

i!t

¡e

¡i!t

A simple algebraic simpliﬁcation g ives

L(sin(!t)) =

fL(e

i!t

) ¡L(e

¡i!t

)g=

+ !

: (5.8)

A similar argument applies for the transform of f (t) = cos(!t)

L(cos(!t)) =

+ !

2 Laplace Transform Method

6. Remember the formula sinh(!t) =

¡e

¡!t

. For s > j!j, we have

L(sinh(!t)) =

fL(e

) ¡L(e

¡!t

)g=

¡!

: (5.9)

A similar argument applies for the transform of cosh(!t)

L(cosh(!t)) =

¡!

: (5.10)

5.1.2 The inverse of Laplace transform

We saw that L(u(t)) =

for s > 0, and it implies that functions u(t) in t-domain has the

representation

in s-domain through the Laplace transform; see the ﬁgure (5.1)

1 2 3

(

)

L-Transfo rm

1 2 3

(

)

Figure 5.1.

An inverse transformation from s-domain to t-domain is also can be deﬁned through

the following relation

L(f(t)) = f

(s) , f(t) = L

¡1

(s)):

However, sinc e the transform is un ilateral, we just write

¡1

(s)) = f (t) for t > 0.

Accordingly, since L(t) =

, we write L

¡1



= t; t > 0. There is still a minor problem

here. Let f

(t) be the function

(t) =



t t =/ 1

0 t = 1

It is simply seen that L(t) = L(f

) =

, and therefore it is not clear that L

¡1



how

should be deﬁned. In fact the transformation L is not one-to-one. Here we make the fol-

lowing convention.

Convention . The function L

¡1

(s)) refers to the most possibly continuous function f (t)

for t > 0 such that L(f (t)) = f

(s).

Accordi ng to the above conventi on, we define

¡1



= 1,

¡1

s ¡ a



= e

¡1

+ !



= cos(t), and L

¡1

+ !



= sin(t) all for t > 0.

5.1 Definition of the transformation 3

Proposition 5.1. Assume that f (t) is sub-exponential, then

lim

s!1

(s) = 0: (5.11)

Proof. We have

lim

s!1

(s) = lim

s!1

¡st

f(t) dt: (5.12)

Since the function f (t) is sub-exponential, the function je

¡st

f(t)j is bounded. By the

dominant convergence theorem, we can pass the limit inside the integral and write it as

lim

s!1

(s) =

lim

s!1

¡st

f(t) dt = 0; (5.13)

and this completes the proof. 

We have also the following fact.

Proposition 5.2. Assume th at f (t) is sub-exponential, then f

(s) is continuous in its

domain of deﬁnition.

Proof. Since the domain of f

(s) is open, let us consider the fraction

(s + h) ¡ f

(s) =

¡st

f(t)(e

¡ht

¡1) dt:

Now, for h !0, we have

lim

h!0

(s + h) ¡f

(s)) = lim

h!0

¡st

f(t)(e

¡h t

¡1) dt:

Now, if jf(t)j < ae

for some a; b > 0 holds, then the condition of dominant convergence

theorem (see the appendix to this book) holds for the above integral since

¡st

f(t)(e

¡ht

¡1)j≤ae

¡(s¡b)t

¡ht

¡1j≤ae

¡(s¡b+h)t

< a

if s is suﬃciently large. Therefore

lim

h!0

¡st

f(t)(e

¡ht

¡1) dt =

¡st

f(t)lim

h!0

¡h t

¡1) dt = 0 ;

and this completes the proof. 

Problems

Problem 5.1. Assume that the function f (t) is sub-exponential, that is, there are constants a; b such

that jf(t)j< ae

for all t 2[0; 1). Show that f(t) is L-admissible.

Problem 5.2. Assume that f(t) is a sub-exponential function and f

(t) is L-admissible.

a) Use the deﬁnition and show

L(f

) =

¡st

(t) dt + s

¡st

f(t) dt:

4 Laplace Transform Method

b) Show the relation

lim

t!1

¡st

f(t) = 0;

and conclude

L(f

) = s L(f ) ¡ f (0):

Problem 5.3. Use the deﬁnition and ﬁnd the L-transform of the following functions

f(t) =



1 0 < t < 1

¡1 1 < t < 2

ii.

f(t) =

(

¡1 0 < t < 1

0 t > 1

iii. f(t) = e

¡e

¡2t

Problem 5.4. Use the deﬁnition and calculate the transform of the following functions. For each

function determine the domain of s for which the transform exists

f(t) =

(

0 ≤t < 1

t t > 1

b) f(t) = cos

(t)

c) f (t) = sinh

(t)

f(t) =



1 1 < t < 2

¡1 2 < t < 3

Problem 5.5. If f

(s) is given we can recover the function f (t) such that L(f (t)) = f

(s). For the fol-

lowing functions, ﬁnd f (t) such that L(f ) = f

i. f

(s) =

+ 1

ii. f

(s) =

2 ¡ 3s

+ 4

iii. f

(s) =

s ¡ 2

iv. f

(s) =

(s + 1)(s + 2)

Problem 5.6. Is there any L-transform of the function f(t) = e

? Is it possible the function f

(s) = 1

to be the transform of a sub-exponential function?

Problem 5.7. Determine which one of the following functions are sub-exponential in [0; 1)

i. f (x) = t sin(t)

ii. f (x) = t sin(1/t)

iii. f (x) = 1 + e

+ e

+ ···+ e

iv. f (x) = 1 + e

+ e

+ ···

v. f (x) = u(t) + u (t ¡1) + u(t ¡2) + ···

Problem 5.8. Show L( t

) =

¡3/2

. Hint: Remember the integral

¡u

du =

5.1 Definition of the transformation 5

5.2 Properties of L aplace transform

Here we study only important properties of L-transform that we need for solving initial

value problems.

Linearity.

L-transform is linear in the sense that for any pair of admissible functions f(t); g(t) and

any arbitrary constants c

; c

, it satisﬁes the relation

L(c

f + c

g) = c

L(f) + c

L(g) (5.14)

The property is directly proved by the deﬁnition (5.1):

L(c

f + c

g) =

¡st

f(t) + c

g(t))dt = c

¡st

f(t) dt +

¡st

g(t) dt = c

L(f) + c

L(g):

The linearity of L-transform implies the linearity of L

¡1

as well. In fact, by the relation

f + c

g = L

¡1

L(c

f + c

g) = L

¡1

+ c

g^);

we obtain

¡1

(s) + c

g^(s)) = c

¡1

(s)) + c

¡1

(g^(s)):

Example 5.1. Let us ﬁnd L

¡1



s(s + 1)



. According to the relation

s(s + 1)

s + 1

;

we can write

¡1



s(s + 1)



= L

¡1





¡L

¡1



s + 1



= (1 ¡e

¡t

); t > 0:

Multiplication by t.

If f (t) is an sub-exponential functions then

(s) =

¡st

f(t) dt = ¡

¡st

tf (t) dt (5.15)

In other word we can write

(s) = ¡L(tf (t)); (5.16)

or equivalently

L(tf (t)) = ¡

(s): (5.17)

It is left as an exercise to the reader to verify the following formula by the mathematical

induction

L(t

f(t)) = (¡1)

(s): (5.18)

6 Laplace Transform Method

Example 5.2. We calculate the inverse transform o f the fo llowing function

(s) = ln

s ¡1

: (5.19)

Taking derivative of f

(s) gives

(s) =

s(s ¡1)

s ¡1

: (5.20)

According to the linearity of

¡1

, we have

¡1

(

) = e

¡ 1. Now use the property

¡1

(

) = ¡tf(t) and write

f(t) = L

¡1

(s)) =

1 ¡e

; t > 0: (5.21)

Note that f(t) is a sub-exponential function.

Division by t.

Let f (t) be a sub-exponential function that satisﬁes the relation

lim

t#0



f(t)



<1: (5.22)

Consider the function g(t) =

f(t)

. By (5.16), we have

g^( s) = ¡L (tg(t)) = ¡L(f (t)) = ¡f

(s): (5.23)

Integrating both sides of the above formula in [s; 1) gives

¡g^(s) + lim

s!1

g^(s) = ¡

(v) dv: (5.24)

By (5.11) we have

lim

s!1

g^(s) = 0;

and thus



f(t)



(v) dv: (5.25)

Example 5.3. Some deﬁnite integrals can be calculated by the aid of the above property.

For example, let us ﬁnd the fo llowing integral

I =

sin(t)

dt: (5.26)

We can write

sin(t)

dt =

lim

s!0

¡st

sin(t)

dt: (5.27)

5.2 Propert ies of L aplace transfo rm 7

Since

sin(t)

is sub-exponential, we can take the limit out of the integral and write

sin(t)

dt = lim

s!0

¡st

sin(t)

dt = lim

s!0



sin(t)



But, we have



sin(t)



+ 1

¡tan

¡1

(s): (5.28)

Finally for s !0, we obtain

sin(t)

dt =

¡lim

s!0

tan

¡1

(s) =

: (5.29)

Shift in t-domain.

Suppose that f (t) is an admissible function and g(t) = f(t ¡t

)u(t ¡t

) is the time shift of

f by t

≥0; see the ﬁgure (5.2).

f (t ¡a)u(t ¡a)

f (t)

Figure 5.2. f (t) and its shifted graph f (t ¡t

We have

L(g) =

¡st

f(t ¡t

)u(t ¡t

) dt =

¡st

f(t ¡t

) dt: (5.30)

For τ = t ¡t

, we obtain

¡st

f(t ¡t

) dt =

¡s(τ +t

)

f(τ ) dτ = e

¡t

¡sτ

f(τ ) dτ ; (5.31)

and then

L(f(t ¡t

) u(t ¡t

)) = e

¡t

(s): (5.32)

Example 5.4. The above property is used to calculate the inverse transform of some s-

functions of the form g^(s) = e

¡t

(s). For example,

¡1



¡2s

¡1



= L

¡1



¡1



t!t¡2

u(t ¡2): (5.33)

But,

¡1



¡1



= cosh(t): (5.34)

and hence

¡1



¡2s

¡1



= cosh(t ¡2) u(t ¡2): (5.35)

8 Laplace Transform Method

Example 5.5. Let g(t) be the following function:

g(t) =

k=0

f(t ¡kt

)u(t ¡kt

); (5.36)

and assume that f (t) is sub-exponential. We have

jg(t)j≤

k=0

jf (t ¡kt

)ju(t ¡kt

) ≤a

k=0

b(t¡kt

)

u(t ¡kt

) ≤

≤ae

k=0

¡bkt

u(t ¡kt

Since

k=0

¡bkt

u(t ¡kt

) ≤

1 ¡e

¡bt

; (5.37)

we conclude

jg(t)j≤

1 ¡e

¡bt

: (5.38)

The above argument guarantees that g(t) is a sub-exponential function. Let us deﬁne

(t) =

k=0

f(t ¡kt

)u(t ¡kt

); (5.39)

and then

g^(s) = lim

n!1

(s) = lim

n!1

¡st

(t) dt: (5.40)

But, we have

¡st

(t) dt =

k=0

L(f(t ¡kt

) u(t ¡kt

)) =

k=0

¡kt

(s); (5.41)

and ﬁnally

g^(s) = f

(s) lim

n!1

k=0

¡kt

1 ¡e

¡t

(s); (5.42)

or equivalently

¡1



1 ¡e

¡t

(s)



k=0

f(t ¡kt

) u(t ¡kt

): (5.43)

In particular, if f(t) is a T -periodic function, that is, f (t + T ) = f (t), then

L(f(t)) =

1 ¡e

¡sT

¡st

f(t) dt:

Shift in s-domain.

There is a beautiful duality between the shift in t dom a in and the shift in s domain some-

times called phase shift. For g(t) = e

f(t), we have

L(g) =

¡st

f(t) dt =

¡(s¡a)t

f(t) dt = f

(s ¡a); (5.44)

5.2 Propert ies of L aplace transfo rm 9

or equivalently

¡1

(s ¡a)) = e

¡1

(s)): (5.45)

Example 5.6. Let us use the above property to ﬁnd

¡1



¡2s

s ¡1

(s ¡1)

¡4



: (5.46)

We have

¡1



¡2s

s ¡1

(s ¡1)

¡4



= L

¡1



s ¡1

(s ¡1)

¡4



t!t¡2

u(t ¡2): (5.47)

Since

¡1



s ¡1

(s ¡1)

¡4



= e

cosh(2t); (5.48)

we obtain

¡1



¡2s

s ¡1

(s ¡1)

¡4



= e

(t¡2)

cosh2(t ¡2)u(t ¡2): (5.49)

Derivative.

If f (t) is a sub-exponential diﬀerentiable function, then

¡st

(t) dt = lim

t!1

¡st

f(t) ¡ lim

t!0

f(t) + s

¡st

f(t) dt: (5.50)

Since f(t) is sub-exponential, we have lim

t!1

¡st

f(t) = 0. If f(t) is continuous at t = 0,

then lim

t!0

f(t) = f (0), and then

L(f

) = s f

(s) ¡f (0): (5.51)

Example 5.7. Consider the following problem



+ y = u(t ¡1)

y(0) = y

Taking the L-transform to the both sides of the equati on gives

s y^(s) ¡ y

+ y^(s) =

¡s

;

or equivalently

(s + 1)y^(s) =

¡s

+ y

This yields y^(s) as

y^(s) =

s + 1

¡s

s(s + 1)

To obtain y(t), we should take inverse transform of y^(s). Note that

¡1



¡s

s(s + 1)



= (1 ¡e

¡(t¡1)

) u(t ¡1);

and

¡1



s + 1



= y

¡t

; t > 0:

10 Laplace Transform Method

Here we should take a notice. The term L

¡1

s + 1



is associated to the initial condition

of the equation at time t = 0 and L

¡1



¡s

s(s + 1)



is assoc iated to the external term that is

exercised for t > 1. For this the homogeneous solution y

¡t

is not only deﬁned for t > 0

but also for all t 2(¡1; 1). Therefore, we can write the s o luti o n to the equation a s

y(t) = y

¡t

+ (1 ¡e

¡(t¡1)

) u(t ¡1); (5.52)

for all t 2 (¡1; 1). Note also that the solution is continuous everywhere and diﬀeren-

tiable, but it is not diﬀerentiabl e at t = 1 due t the term u(t ¡1).

The following formula is simply proved by the mathematical induction:

L(f

(n)

(t)) = s

(s) ¡s

n¡1

f(0) ¡···¡ f

(n¡1)

(0): (5.53)

Integration.

Let us d erive the L-transform of the integral function

g(t) =

f(τ) dτ ; t > 0: (5.54)

Since g

(t) = f (t) and lim

t!0

g(t) = 0, we obtain

(s) = L(g

) = s g^(s) (5. 55)

and thus



f(τ ) dτ



(s); (5.56)

or equivalently

f(τ ) dτ = L

¡1



(s)



; (5.57)

for t > 0.

Example 5.8. By the aid of the integral property, we are able to calculate the inverse of

some s-functions of the form g^(s) =

(s). Consider the function

(s) =

s(s

+ 1)

: (5.58)

We have

¡1

(s)) = L

¡1



+ 1)



¡1



+ 1



sin (v) dv = (1 ¡cos(t))u(t): (5.59)

Example 5.9. Let us calculate L

¡1



(s)

s + a



. According to the phase shift property, we

have



¡at

f(τ )



= L



f(τ )



s!s+a

(s)

s!s+a

(s + a)

s + a

5.2 Propert ies of L aplace transfo rm 11

Since we need f

(s) in the top, we write



¡at

aτ

f(τ )



= L



aτ

f(τ )



s!s+a

But



aτ

f(τ )



L(e

f(t)) =

(s ¡a)

;

and therefore



¡at

aτ

f(τ )



(s ¡a)

s!s+a

(s)

s + a

;

or equivalently

¡1

(s)

s + a



¡at

aτ

f(τ ) dτ



u(t)

5.2.1 Dirac delta function and its Laplace tr ansform

The Dirac delta functio n δ, i ntroduced by the E nglish physicist Paul Dirac (1902-

1984) is a speciﬁc type of singularity used in advanced physics. There are diﬀerent ways

to deﬁne the Dirac delta function. One way that we adopt here is to deﬁne it as the limit

of a sequence of functions called the delta-sequence functions. Consider the following

sequence

(t) =

(

< t <

0 otherwise

; (5.60)

for n = 1; 2; 3; ···. The graph is shown below.

(t)

¡1

Figure 5.3. An example of δ

-sequence functions.

Assume that f (t) is a continu o us function. We can write

min

¡1

;

f(t) ≤

¡1

f(t)δ

(t) dt ≤ max

¡1

;

f(t): (5.61)

12 Laplace Transform Method

Since

lim

n!1

min

¡1

;

f(t) = lim

n!1

max

¡1

;

f(t) = f(0);

then by the squeeze theorem, we can write

lim

n!1

¡1

f(t)δ

(t) dt = f(0):

As it is seen, δ

(t) converges to a spike at t = 0 with inﬁnite height and zero width with

the property

¡1

(t) dt = 0; 8n ≥1:

This limiting function w hich is a singularity at t = 0 is called the Dirac delta function.

The above deﬁned sequence is not unique and there are other sequences with the same

property. For example, consider the sequence of function G

(t) =

¡n

for n = 1; 2; ···

shown below. Note that all G

(x) are derived from the Gaussian function G (t) =

¡t

by the relation G

(t) = nG(nt).

−

2 2

= 1

= 8

Figure 5.4.

When n approaches inﬁnity, the function approaches to a spike with the width 0 a nd

the height inﬁnity at t = 0. Let f(t) be a continuous and bounded function. We have

¡1

f(t) G

(t) dt =

¡1

f(t)

¡n

dt;

and by taking τ = nt, we obtain

¡1

f(t)G

(t) dt =

¡1





¡τ

dτ :

5.2 Propert ies of L aplace transfo rm 13

For n !1, we can write

lim

n!1

¡1

f(t)G

(t) dt =

¡1

lim

n!1





¡τ

dτ =

¡1

f(0)e

¡τ

dτ = f (0):

Note that in the above calculations, we took the limit inside the integral which is j ustiﬁ-

able by the assumption on f . Note also the relation

¡1

(t) dt = 1; 8n ≥1:

Deﬁnition 5.2. A sequence of function (δ

(x)) is called a δ-sequence function if the fol-

lowing relation holds for any bounded and continuous function f

lim

n!1

¡1

f(t) δ

(t)dt = f (0):

The limit function of a δ-sequence function is called the Dirac delta function and is

denoted by δ(t).

An alternative way for deﬁning δ is by the following equality

¡a

f(t) δ(t)dt = f (0); (5.62)

for any a > 0 and any continuous function f. The formula (5.62) gives immediately the

following relation

t¡a

t+a

f(τ )δ(t ¡τ )dτ = f (t); (5.63)

or equivalently

¡1

f(τ ) δ(t ¡τ ) dτ = f(t):

That δ(t) i s not a function in the usual sense is seen from the following observation. If

δ was a classical function then it would had to satisfy the following condition

lim

a!0

¡a

f(t) δ(t) d t = 0;

which contradicts the relation (5.62). In engineering literature, δ(t) is called an impulse

and is usually denoted by a unit arrow; see Fig.5.5. For this rea son, δ(t) is som etime

deﬁned (and it is not technically correct) by the following relation

δ(t) =



1 t = 0

0 otherwise

: (5.64)

14 Laplace Transform Method

δ(t)

Figure 5.5. The graph of δ function.

For any τ > 0, the relation (5.63) implies

L(δ(t ¡a)) =

¡st

δ(t ¡a) dt = e

¡as

We accept this deﬁnition for a = 0 as well and write

L(δ) = 1; and L

¡1

(1) = δ(t): (5.65)

By this de ﬁnition, the following relation is immediately followe d

L(f(t)δ(t ¡a)) = f(a)e

¡as

: (5.66)

Problems

Problem 5.9. Find the transform of the following functions

i. f (t) = e

u(t ¡1)

ii. f (t) = te

¡t

u(t ¡1)

iii. f (t) = u(t) ¡e

u(t¡1)

iv. f (t) = e

sin(πt)

δ(t ¡2)

v. f (t) = (1 ¡u(t ¡2)) u(t ¡1).

Problem 5.10. Compare the transformation of the function

f(t) =



1 0 < t < 1

0 otherwise

;

and the function

g(t) =



t 0 < t < 1

0 otherwise

;

and conclude g^(s) = ¡

(s).

Problem 5.11. Use the deﬁnition and derive the Laplace transform of the function

f(t) =



1 0 < t < 1

0 otherwise

; g(t) =



1 1 < t < 2

0 otherwise

;

and conclude g^(s) = e

¡s

(s).

Problem 5.12. Use the Laplace transform properties to calculate

i. L(te

)

ii. L(t sin(!t))

5.2 Propert ies of L aplace transfo rm 15

iii. L(t cos(!t))

iv. L(t sinh(!t))

v. L(t cosh(!t))

Problem 5.13. Find the transformation of the following functions

i. L(t

3/2

ii. L(t

¡1/2

)

iii. L(e

¡t

¡1/2

)

Problem 5.14. Use mathematical induction to prove the identities

L(t

f(t)) = (¡1)

(s);

and conclude

i. L(t

) =

n+1

ii. L(t

¡at

) =

(s + a)

n+1

Problem 5.15. Use partial fraction to ﬁnd the inverse transformation o f the following functions:

i. f

(s) =

¡3s + 2

ii. f

(s) =

s(s

+ 4)

iii. f

(s) =

s + 2

¡s

+ s ¡ 1

Problem 5.16. Use the properties of Laplace transform to obtain the following integrals:

1 ¡e

¡t

dt,

ii.

1 ¡cos(t)

¡t

dt,

iii.

sin(t)

¡t

iv.

¡1

¡jtj

1 ¡ cos(t)

ln(2) jtj

Problem 5.17. Find the inverse transformation of the following functions:

i. f

(s) =

¡4s + 5

ii. f

(s) =

s + 3

+ 2s + 5

iii. f

(s) =

(s ¡ 1)

+ 3

iv. f

(s) = e

¡s

log

s + 1



v. f

(s) = e

¡2s

+ 3)

vi. f

(s) = e

¡s

+ 3s + 2

Problem 5.18. Find the inverse transformation of following functions:

i. f

(s) =

1 ¡ e

¡s

¡1

ii. f

(s) =

1 + e

¡s

1 ¡ e

¡2s

+ 1

Problem 5.19. If we deﬁne

g(t) =

k=0

(¡1)

f(t ¡ka)u(t ¡ka)

16 Laplace Transform Method

show that

g^(s) =

1 + e

¡as

(s):

Using the above argument calculate the following inverse transformation

¡1



1 + e

¡2s

(s ¡3)



Problem 5.20. Use mathematical induction to prove

L(f

(n)

(t)) = s

(s) ¡s

n¡1

f(0) ¡···¡ f

(n¡1)

(0):

Problem 5.21. Use Laplace tra nsform of tsin(!t) and tcos(!t) to ﬁnd the inverse transformation of

the following functions:

i. f

(s) =

+ 1)

ii. f

(s) =

((s ¡ 2)

+ 1)

Problem 5.22. Find the transformation of the following periodic functions:

i. f (t) = t; 0 < t < 1; f (t + 1) = f(t),

ii. f (t) =



1 0 < t < 1

¡1 1 < t < 2

, f(t + 2) = f (t).

Problem 5.23. Show the following relation

¡1

(s)

(s + a)

= e

¡at



f(s) ds



dτ :

Use integration by part formula and rewrite the right hand side of the above formula as

¡at



f(s) ds



dτ =

(t ¡τ )e

¡a(t¡τ )

f(τ) dτ :

Problem 5.24. Even though the derivative of the unit step function u(t ¡ a) is not deﬁned at t = a

(it is n ot even continuous at th is point), show that δ(t ¡ a) and u(t ¡ a) are related in the following

way

L(δ(t ¡a)) = s L(u(t ¡a));

that is δ(t ¡a) can be considered as the generalized derivative of u(t ¡a).

Problem 5.25. Consider the sequence of functions

(t) =

nt 0 ≤t ≤

1 t ≥

i. Find f

(s) and then obtain lim

n!1

(s).

ii. What is the limit function lim

n!1

(t)?

iii. Compare lim

n!1

(s) with the the L-transform of the limiting function you obtained in part

(ii).

Problem 5.26. In the proposition (5.2) we proved that f

(s) is continuous in its domain of deﬁnition.

i. For f (t) = e

¡t

sin(t) ﬁnd

f(t)dt.

5.2 Propert ies of L aplace transfo rm 17

ii. Find f

(s) and let lim

s!0

(s) and explain why it gives the same result in part (i).

iii. Now for f (t) = sin(t) we know f

(s) =

+ 1

and lim

s!0

(s) = 1. Does it imply that

sin(t) dt =

1? why?

Problem 5.27. It is not always allowed to pas s the limit inside the integral.

i. Let f

(t) =

1 + n

. Find

(t).

ii. Compare lim

n!1

(t)dt and

lim

n!1

(t)dt and explain why they are not equal.

Problem 5.28. Let f

(t) = sin(nt).

i. Find lim

n!1

(s).

ii. Can we pass the limit in side the integral of L-transform?

5.3 In itial value probl ems and L-tr ans form

In this section, we solve initial value problems by the aid of the Laplace transform. As

we will see, this method offers considerable advantag e over the classical methods for

solving initial value problem, specially when the fo rcing terms are discontinuous. Consider

the following initial value problem



+ a y

+ by = f(t)

y(0) = y

; y

(0) = y

; (5.67)

where a; b are constants. The L-transform of the equation transforms the equation into the

following algebraic one

L(y

) + a L(y

) + b L(y) = f

(s): (5.68)

Using the relations

L(y

) = s

y^(s) ¡sy

¡y

; and L(y

) = sy^(s) ¡ y

;

we obtain

y^(s) ¡sy

¡y

+ as y^(s) ¡ay

+ b y^(s) = f

(s); (5.69)

and thus we reach

+ as + b) y^(s) = (s + a)y

+ y

+ f

(s): (5.70)

Note that the coeﬃcient of y^(s) i s the characteristic polynomial of the diﬀerential equa-

tion. Therefore,

y^(s) =

(s)

+ as + b

s + a

+ as + b

; (5.71)

and ﬁnally

y(t) = y

¡1



s + a

+ as + b



+ y

¡1



+ as + b



+ L

¡1

(s)

+ as + b

(5.72)

18 Laplace Transform Method

Example 5.10. Consider the following problem



+ y = u(t ¡2) ¡u( t ¡8)

y(0) = y

(0) = 0

: (5.73)

Here a pulse in the period (2; 8) is applied to a harmonic oscillator. By the method of

Laplace transform, the problem reduces to the following algebraic one

+ 1)y^(s) =

¡2s

¡e

¡8s

;

and thus

y(t) = L

¡1



¡2s

¡e

¡8s

s(s

+ 1)



= L

¡1



¡2s

s(s

+ 1)



¡L

¡1



¡8s

s(s

+ 1)



By the property

¡1

¡as

(s)) = f (t ¡a) u(t ¡a);

we have

¡1



¡2s

s(s

+ 1)



= L

¡1



s(s

+ 1)



t=t¡2

u(t ¡2):

Since

¡1



s(s

+ 1)



¡1



+ 1



sinvdv = 1 ¡cost;

we conclude

y(t) = [1 ¡cos(t ¡2)] u(t ¡2) ¡[1 ¡cos(t ¡8)] u(t ¡8): (5.74)

The ﬁgure (5.6) shows the graph o f this solution.

2 8 16

(

)

Figure 5.6. The graph of the solution y(t).

Observe from the ﬁgure that y(t) ≡ 0 for t < 2 th a t conﬁrms our expectation because

the system is at rest for t < 2. The solution y(t) is smooth of order 1 at t = 2 since y

(t)

has a ﬁnite jump at that time. A similar argument holds at t = 8 where f (t) jumps from 1

to 0. By the classical method of previous chapters, to solve the problem, we had to split

the problem into the three dub-domains t < 2, 2 < t < 8 and t > 8 as follow s. For t < 0 the

problem is



+ y = 0;

y(0) = y

(0) = 0

; (5.75)

5.3 Initial value pro blems and L-transform 19

and thus the unique solution is y(t) = 0. In the interval (2; 8), the problem reads



+ y = 1;

y(2) = y

(2) = 0

; (5.76)

with the solution y(t) = 1 ¡cos(t ¡2). In the interval (8; 1), we have



+ y = 0; for t > 8

y(8) = 1 ¡cos(6); y

(8) = sin(6)

; (5.77)

with the solution y = cos(t ¡ 2) ¡ cos(t ¡ 8). Note the initial conditions in the above

problem.

Example 5.11. Let us solve the following problem



+ y = u(t ¡1)

y(1) = 1; y

(1) = 0

Here the initial data is given at t = 1 instead of t = 0 . To solve the problem, we pro cess as

if conditions y(0) = y

and y

(0) = y

are known. Taking L-transform we obtain

y^(s) =

¡s

s(s

+ 1)

+ y

+ 1

+ y

+ 1

This implies

y(t) = (1 ¡cos(t ¡1))u(t ¡1) + y

cos(t) + y

sin(t):

Now, we apply the data y(1) and y

(1) and get y

= cos(1) and y

= sin(1) and thus

y(t) = (1 ¡cos(t ¡1))u(t ¡1) + cos(t ¡1) =



cos(t ¡1) t ≤1

1 t ≥1

The graph o f the solution for t > 1 is shown in the ﬁgure (5.7).

5 1

0 1

5 2

Figure 5.7.

Note that two branches of the solution for t ≤ 1 and for t ≥ 1 connect together

smoothly of ﬁrst order. This is because y

has a ﬁnite jump at t = 1 due to the term u(t ¡

1) and therefore y

is c o ntinuous at this point. Note that the initial condition at t = 1

determines y(t) for t < 1.

20 Laplace Transform Method

Example 5.12. Let f (t) be the function

f(t) = u(t) ¡2u



t ¡



+ u(t ¡π);

and let g(t) be the π-periodic extension of f in (0; 1); see th e ﬁgure (5.8).

¡1

f(t)

2π

3π

Figure 5.8.

Note that g has the representation

g(t) =

k=0

f(t ¡kπ) u(t ¡kπ):

Now consider the problem



+ y = f (t)

y(0) = y

(0) = 0

: (5.78)

We have

L(f ) =

k=0

L(g(t ¡kπ)u(t ¡kπ)) = g^(s)

k=0

¡kπs

1 ¡e

¡πs

g^(s):

By the L-transform method, we have

y^(s) =

1 ¡e

¡πs

g^(s)

+ 1

;

and thus

y(t) = L

¡1



1 ¡e

¡πs

g^(s)

+ 1



: (5.79)

We use the formula (5.42) to write y(t) as

y(t) =

k=0

¡1



g^(s)

+ 1



t!t¡kπ

u(t ¡kπ): (5.80)

But

g^(s) =

1 ¡2e

¡πs/2

+ e

¡πs

;

and therefore

¡1



g^(s)

+ 1



= (1 ¡cos(t)) u (t) ¡2 (1 ¡sin(t)) u



t ¡



+ (1 + co s(t)) u(t ¡π):

The ﬁgure (5.9) shows the solution to the problem.

5.3 Initial value pro blems and L-transform 21

-1

4 8 12 16 20

Figure 5.9. The graph of the solution y(t).

Example 5.13. Consider the initial value problem

(

+ y = e

sin(log(t/π))

δ(t ¡π)

y(0) = y

(0) = 0

: (5.81)

The transform reduces the equation to the following algebraic one

+ 1)y^(s) = L(e

sin(log(t/π))

δ(t ¡π)): (5.82)

It is simply seen that

L(e

sin(log(t/π))

δ(t ¡π)) = e

sin(log(π/π))

¡πs

= e

¡πs

; (5.83)

and thus y^(s) =

¡πs

+ 1

which gives

y = sin(t ¡π) u(t ¡π): (5.84)

Observe that the solution starts at t = π, the time when the external source is applied.

Problems

Problem 5.29. Use the Laplace transform method to solve the following initial value problems



+ y = sin(t ¡1) u(t ¡1)

y(0) = 0:

ii.

(

+ 4y

+ 3y = e

y(1) = 0; y

(1) = 1

iii.



+ y = sin(t)

y(0) = 0; y

(0) = 1

iv.



+ y = u(t ¡1)

y(0) = 0; y

(1) = 1

(

+ 3y

+ 2y = 2u(t ¡1) + e

δ(t ¡2)

y(0) = y

(0) = 0

22 Laplace Transform Method

vi.

(

+ y = u(t ¡1) ¡u(t ¡2);

y(0) = y(

); y

(0) = 1:

vii.



+ y = sin(t) u(t ¡π)

y(0) = y

(0) = 0

;

viii.

(

000

¡3y

+ 3y

+ y = e

(1 + t)

y(0) = y

(0) = y

(0) = 0

Problem 5.30. Let us solve the prob lem (5.11) as follows. Let τ = t ¡ 1 and deﬁned z(τ ) = y(t).

Then we have



(τ) + z(τ ) = u(τ )

z(0) = 1; z

(0) = 0

The L-transform of the problem gives

z^(s) =

s(s

+ 1)

+ 1

Therefore z(τ) = u(τ) and y(t) = u(t ¡1) (!) What is wrong here?

Problem 5.31. Solve the f ollowing equations and use a computer software to draw the solutions

i. y

+ 2y

+ y = f (t); y(0) = y

(0) = 0 and f(t) is a period ic function as follows

f(t) = u(t) ¡u(t ¡1), f(t + 2) = f (t)

ii. y

+ y = f (t), f (t) = t for 0 < t < 1 and f(t + 1) = f(t)

iii. y

+ y = f (t) where f (t) is given in the following ﬁgure

2 3

Problem 5.32. Assume that s(t) is a continuous function such that

¡1

s(t) dt = 1: (5.85)

Prove that the sequence s

(t) = ns(n t) is a δ-sequence functions for the class of continuous and

bo unded functions f (t). Use this property to conclude that

¡1

n cos(t)

π(1 + n

)

n!1

Problem 5.33. If L

¡1

(F (s)) = f (t), show that

¡1

¡2s

sF (s)) = f

(t ¡2)u(t ¡2) + f (0) δ(t ¡2):

Problem 5.34. Solve the f ollowing problem and draw the solution y(t) in the interval [0; 10]

(

+ y =

k=1

δ(t ¡k)

y(0) = y

(0) = 0

5.3 Initial value pro blems and L-transform 23

Problem 5.35. Solve the f ollowing problem and draw the solution y(t) in the interval [0; 10]

(

(4)

+ 2y

+ y =

k=1

¡k

δ(t ¡k)

(0) = 0; k = 0; 1; 2; 3

5.4 Impulse respo nse and convolu tion

5.4.1 Convolution

Deﬁnition 5.3. Assume that f(t) and g(t) are two integrable functions deﬁned on (¡1;

1). The convolution f ∗g is deﬁned by the following relation integral relation

(f ∗ g) (t) =

¡1

f(τ ) g(t ¡τ ) dτ : (5.86)

Although it looks odd at ﬁrst glance, we will see the importance of convolu tion in real

applications. Before that, let us solve a few e xamples.

Example 5.14. If f (t) = δ(t), then f o r arbitrary integrable function g(t), we have

(δ ∗ g) (t) =

¡1

δ(τ) g(t ¡τ ) dτ = g(t): (5.87 )

If f (t) = u(t) then for arbitrary integrable function g(t) we have

(u ∗ g)(t) =

¡1

u(τ) g(t ¡τ ) dτ =

g(t ¡τ ) dτ =

¡1

g(τ) dτ ; (5.88)

and in particular, if g(t) = 0 for t < 0, then

(u ∗ g)(t) =

g(τ )dτ:

Assume f (t) = e

¡t

u(t), then

(f ∗f )(t) =

¡1

¡τ

u(τ ) e

¡(t¡τ)

u(t ¡τ ) dτ =

¡t

dτ = te

¡t

Remark 5.2. If f ; g are zero for t < 0 (as we usually assume for the Laplace transform)

the integral reads

(f ∗ g )(t) =

f(τ) g(t ¡τ ) d τ : (5.89)

Remark 5.3. It is simply veriﬁed that the convolution relation is commutative, that is,

f ∗ g = g ∗ f

(f ∗ g)(t) =

¡1

f(τ )g(t ¡τ) dτ =

¡1

f(t ¡τ ) g(τ ) dτ = (g ∗ f )(t):

24 Laplace Transform Method

The equality is veriﬁed simply by the change of variable v = t ¡τ .

Theorem 5.1. For any admissible function f (t) and g(t) we have

L(f ∗ g) = f

(s)g^(s): (5.90)

Proof. By de ﬁnition, we can write

L(f ∗g) =

¡st

f(τ ) g(t ¡τ ) dτdt: (5.91)

Remember the follow ing change of order of the integration from calculus

h(t; τ ) dτdt =

h(t; τ ) dtdτ : (5.92)

The following ﬁgure justiﬁes the above relation

τ = t

Figure 5.10.

Using the above formula for b = 1, we can write

¡st

f(τ ) g(t ¡τ ) dτdt =

¡st

f(τ ) g(t ¡τ ) dtdτ =

f(τ )



¡st

g(t ¡τ ) dt



dτ:

By the change of variable t ¡τ = v, we reach

¡st

f(τ ) g(t ¡τ ) dτdt =



¡sτ

f(τ ) dτ



¡sv

g(v) dv



= f

(s) g^(s);

and this completes the proof. 

As it is seen, the Laplace transform, transforms the convolution relation to a simple

algebraic multiplication. As we will see below, this relation oﬀers signiﬁcant a simpliﬁca-

tion in the calculations of linear systems. But before that, let us solve a few examples.

Example 5.15. Let us start with the following inverse transformation

¡1



(s)



5.4 Impulse response and convolution 25

As we know, it is equal to

¡1



(s)



f(t) dτ :

By the convolution relation, we can write

¡1



(s)



= L

¡1





∗L

¡1

(s)g= u(t) ∗ f(t) =

f(τ )u(t ¡τ ) dτ =

f(τ )dτ ;

for t > 0. Similarly, we can write

¡1



s + a

(s)



= L

¡1



s + a



∗L

¡1

(s)g= e

¡at

∗f (t) =

f(τ) e

¡a(t¡τ)

dτ = e

¡at

f(τ ) e

aτ

dτ :

The above relation is the solution of the familiar initial value problem



+ ay = f(t )

y(0) = 0

;

which is by the Laplace transform is equivalent to

y^(s) =

(s)

s + a

Example 5.16. Let us solve the following integral equation

y ¡

t¡τ

y(τ) dτ = δ(t ¡π): (5.93)

Since the integral term in the equation is equal to e

∗ y, the Laplace transform of the

equation is

y^(s) ¡L(e

∗y) = e

¡πs

; (5.94)

and therefore

y^(s) ¡

y^(s)

s ¡1

= e

¡πs

: (5.95 )

The above equation is solved for y^ as

y^(s) =

s ¡1

s ¡2

¡πs

; (5.96)

and ﬁnally

y(t) = L

¡1



s ¡1

s ¡2

¡πs



= δ(t ¡π) + e

2(t¡π)

u(t ¡π): (5.97)

5.4.2 System approach and transfer function

The word system is a prevalent term in the whole of applied sciences. Almost all physical

systems, incl uding man-machine and natural systems, are formulated in te rms of three

fundamental terms: input, output, and transaction.

26 Laplace Transform Method

Transaction

input

output

For example, the movement of a mass connected to a spring of stiﬀness k under the inﬂu-

ence o f an external force f(t), or the change of inﬂation rate in a n economy, or the reac-

tion of a body to some virus all of them can be interpreted in terms o f the above simple

block-diagram. The advantage of block-diagram representation is find out the exis tent

functional similarities between seemingly diﬀerent systems. For ex a mple, the dynamic of

a mass-spring system is describ ed by the following diﬀerential equation

+ kx = f (t); (5.98)

where k is the stiﬀness of the spring, m is the mass of the body, and f(t) is th e external

force exerting to the mass. In the block-diagram representation, we can show the above

system as follows

+ k

x^(s)

(s)

In the above block-diagram, the transaction is represented by the Laplace transform of

the diﬀerential equation

x^(s) =

+ k

(s): (5.99)

The expression h

(s) =

+ k

is als o called the transfer function of the mass-spring system.

Now, let us consider the following LC circuit and assume that the switch S connects at

t = 0.

According to the voltage-current relationships of the capacitor C and the inductor L,

we can write

+ V

= v

(t) u(t): (5.100)

Assuming that the system is at rest, the Laplace transform of the equation is

(LCs

+1)V

(s) = v^

(s); (5.101)

or equivalently

(s) =

LCs

+ 1

(s); (5.102)

which has the following block-diagram.

5.4 Impulse response and convolution 27

LCs

+ 1

(s)

^(s)

As it is seen, two systems have the completely similar transfer function and thus they

behave completely similar from the functional point of view.

Note that if the input in the above examples is δ(t), then the Laplace transform of

the output will be h

(s). In fact, fo r the mass-spring functional equation (5.99), if f (t) =

δ(t), then f

(s) = 1 and thus x^(s) =

+ k

:= h

(s).

Deﬁnition 5.4. The response of a system to the input δ(t) is called the impulse response

of the system and is usually denoted by h(t). The Laplace transform of the impulse

response is called the transfer function of a system and is denoted by h

(s).

For example, for the impulse response of the mass-spring system is

h(t) = L

¡1

(s)) =

sin





u(t); (5.103)

and the impulse response of the above LC circuit is

h(t) =

sin





u(t): (5.104)

Example 5.17. Consider the following initial value prob lem



+ y = δ(t ¡τ ); τ > 0

y(0) = 0

By Laplace transform method, the response of the system to the input δ(t ¡τ ) is

h(t ¡τ ) = e

¡(t¡τ)

u(t ¡τ ):

Note that since y

(t) is a Dirac singularity at t = τ , the response y(t) is of a ﬁnite jump

at that time. For τ !0, we take

h(t) =



¡t

t > 0

0 t ≤0

On the other hand, the response of the system to input r(t) is

y^(s) =

r^(s)

s + 1

;

and hence,

y(t) = h(t) ∗r(t) = e

¡t

r(τ) dτ :

5.4.3 LTI systems

General systems are usually classiﬁed into the fo llowing classes:

a) Causal system: The output y at time t depends only on the input x(τ) for τ ≤ t

and is independent of τ > t.

28 Laplace Transform Method

b) Memory-less system: The output y at time t depends on ly on the in put x at time t,

not on the previous or future values of t.

c) Time Invariant: The output y is not sensitive to the time of applying input x . In

other word, if y(t) is the response to the input x(t), then y(t ¡ τ ) will be the

response of x(t ¡τ ).

d) Linear system: If y

; y

are respectively the response to inputs x

; x

, then the

response of the system to αx

+ βx

for arbi trary constants α; β is αy

+ βy

Example 5.18. Suppose that the transaction T is a derivative operator, that is,

y(t) = T fx(t)g= x

(t):

The system is not causal and memory-less b ecause of the deﬁnition of the derivative:

(t) = lim

h!0

x(t + h) ¡x(t)

The system is linear due to the relation T fax

+ bx

g= ax

+ bx

= ay

+ by

. In addition,

this system is time invariant y(t ¡t

) = x

(t ¡t

): S imilarly, an integrator

y(t) =

¡1

x(τ ) dτ ;

is a linear, time invariant and causal but not memory-less.

Deﬁnition 5.5. Consider a system (S) with input x(t) and output y(t). The system is

called linear if it satisﬁes the following two conditions:

1. for arbitrary λ 2R, the response to input λx(t) is λy(t), ( homogeneity property)

2. if y

, y

are responses t o inputs x

and x

respectively, the response to the input

(t) + x

(t) is y

(t) + y

(t) ( additive property)

This concept is shown in the ﬁgure (5.11) schematically.

λyλx

Figure 5.11. A linear system.

Deﬁnition 5.6. A system (S) with input x(t) and output y(t) is called Time Invariant

(TI) if response to x(t ¡τ ) is y(t ¡τ ) for arbitrary τ ≥0.

5.4 Impulse response and convolution 29

Example 5.19. Consider the mass-spring system (5.99). It is almost straightforward to

verify that the system is linear. In fact, if f(t) is replaced by λf (t) (for arbitrary constant

λ), the response changes to λx(t). Furthermore, if f(t) is replaced by f

(t) + f

(t), then

by superposition principle, we can write

x(t) = x

(t) + x

(t): (5.105)

In addition, if f(t)u(t) is replaced by f(t ¡τ)u(t ¡τ ) for s o me τ ≥0, then the response of

the system changes to

¡1

¡sτ

+ k

= L

¡1

+ k

t!t¡τ

u(t ¡τ ) = x(t ¡τ ) u(t ¡τ ): (5.106)

Therefore, the mass spring system is an LTI sy stem. The reader can check that the LC

circuit in the previo us example is LTI as well.

Example 5.20. Consider the following initial value prob lem

(

(n)

+ a

(n¡1)

+ ···+ a

y = x(t)

(k)

(0) = 0; for k = 0; :::; n ¡1

; (5.107)

where a

are some constants. We write the equation as L[y] = x(t), for L the op erator

L := D

+ a

n¡1

+ ···+ a

: (5.108)

It is simply veriﬁed that the equations represents an LTI system with x(t); y(t) as its

input and output respectively. In fact, for arbitrary λ

; λ

2 R, the solution to the input

x(t) = λ

(t) + λ

(t), is y(t) = λ

(t) + λ

(t), where L[y

] = x

(t) and L[y

] = x

(t).

Furthermore, according to the relation

(n)

(t ¡τ ) + a

(n¡1)

(t ¡τ ) + ···+ a

y(t ¡τ ) = x(t ¡τ ); (5.109)

we have L[y](t ¡τ ) = x(t ¡τ ). The transfer function of this system is:

(s) =

+ a

n¡1

+ ···+ a

: (5.110)

In this section, we study causal linear time invariant (LTI) systems described by linear

diﬀerential equations with constant coeﬃcients



+ ay + by = r(t)

y(t

) = 0; y

) = 0

: (5.111)

Or in the operator form as

L[y](t) = r(t);

where L stands for the diﬀerential operator

+ a

+ b. Accordingly, we can write

y(t) = L

¡1

[r(t)];

where L is the inverse of the diﬀerential operator L.

r(t)

y(t)

y^(s)

+ as + b

r^(s)

¡1

30 Laplace Transform Method

5.4.4 Convolution and LTI control systems

The importance of convolution lies in the f o llowing theorem.

Theorem 5.2. Assume that h(t) is the impulse response of an LTI system. Then the

response of the sy stem to an arbitrary input x(t) is determined by the following convolu-

tion

y(t) =

h(t ¡τ ) x(τ ) dτ : (5.112)

Proof. Note that we can write x(t) as the convolution as

x(t) =

x(τ) δ(t ¡τ ) dτ : (5.113)

Let us write the above integral as the following Riemann sum

x(τ) δ(t ¡τ ) dτ = lim

n!1

k=1

x(τ

) δ(t ¡τ

) ∆τ

; (5.114)

and deﬁne x

(t) by the following partial sum

(t) =

k=1

x(τ

)δ(t ¡τ

) ∆τ

: (5.115)

Since the system is linear time-invariant, y

(t) is determined by the following summation

(t) =

k=1

x(τ

) h(t ¡τ

) ∆τ

; (5.116)

where h(t ¡τ

) is the response to the impulse δ(t ¡τ

). Now, we can write

y(t) = lim

n!1

(t) = lim

n!1

k=1

x(τ

) h(t ¡τ

) ∆τ

x(τ) h(t ¡τ ) dτ ; (5.117)

and this completes the proof. 

Example 5.21. For a LTI system, we know that the impulse response function h(t) is

h(t) = e

¡t

sin(t) u(t):

We would like to ﬁnd the response of the system to the input function r(t) = e

¡t

u(t).

According to the above theorem, we can write

y(t) = r(t) ∗h(t) =

¡(t¡τ)

¡τ

sin(τ) dτ = e

¡t

(1 ¡cos(t)) u(t):

Problem 5.36. Following problems:

a) If we know that the impulse response of an LTI system is sin(t), ﬁnd the response to the input

x(t) = cos(t).

5.4 Impulse response and convolution 31

b) If we know that the response of an LTI system to input u(t) is sin(t), ﬁnd the response to the

input x(t) = cos(t).

Working with convolution integrals is not always as simple as above, specially for con-

trol systems. For example consider a simple feedback control system shown below.

c(t)

y(t)

x(t)

h(t)

Figure 5.12.

By a a straightforward calculation, we can write y(t) as the following convolution:

y(t) = (x(t) ¡c(t) ∗ y(t)) ∗h(t); (5.118)

or equivalently in the integral form as

y(t) =



x(τ) ¡

c(τ ¡v)y(v) dv



h(t ¡τ ) dτ :

As it is seen, the formula looks very complicated even for such a simple feedback system.

Here we s ee how the Laplace transform simpliﬁes the calculations. Remember the re la-

tion

Lff ∗ gg= f

(s) g^(s);

and thus fo r (5.118), we can write

y^(s) = h

(s)[x^(s) ¡c^(s) y^(s)];

which is solved simply for y^(s) as

y^(s) =

(s)

1 + c^(s) h

(s)

x^(s):

Therefore, the transfer function of the above control system is

(s) =

(s)

1 + c^(s) h

(s)

Problem 5.37. Find the transfer function of the control system shown in the following diagram

(t)

x(t)

(t)

y(t)

32 Laplace Transform Method

Problems

Problem 5.38. Show that an integrator

y(t) =

¡1

x(τ ) dτ ;

is time invariant system.

Problem 5.39. Consider the discrete system y[n] = x[2n] for n 2Z. Give and example that shows the

system is not time invariant.

Problem 5.40. Consider the equation

+ ay = r(t);

where r(t) = 0 for t < 0. Sh ow that the transaction L

¡1

deﬁned by

¡1

[r](t) =

¡a(t¡τ)

r(τ)dτ ;

is a system representation of the given equation. Show that the system L

¡1

is linear t ime invariant.

Problem 5.41. Consider a system that its response is described by the following diﬀerential equation

+ ay = r(t):

i. Find the im pulse response of the system.

ii. Write the solution in t he convolution form.

The solution is the same as we obtained through solving a linear ﬁrst order equation in previous chap-

ters.

Problem 5.42. Consider a system that its response is described by the following diﬀerential equation

+y = r(t):

i. Find the im pulse response of the system.

ii. Write the solution in t he convolution form.

iii. Verify that the obtained formula is a solution to the problem



+ y = r(t)

y(0) = y

(0) = 0

Problem 5.43. Consider the following initial value problem



+ a

y = r(t)

y(0) = y

(0) = 0

i. If the characteristic polynomial of the equation has two distinct roots λ

=/ λ

, use convolution

theorem to s how that the solution to the equation is

φ(t) =

¡λ

∗r(t) ¡e

∗r(t)):

ii. If the characteristic polynomial has a repeated root λ, use convolution theorem to show that

φ(t) = te

λt

∗r(t):

Problem 5.44. For the RC circuit shown in the ﬁgure (5.13), ﬁnd the impulse response h(t) and the

transfer function h

(s) if v

(t) is considered as the response of the s ystem. Find v

(t) if v(t) = 1 and

the switch S connect s at time t = 1.

5.4 Impulse response and convolution 33

Figure 5.13.

Problem 5.45. Consider the circuit shown in the ﬁgure (5.14) where v

(t) and v

(t) are the input

and output of the system respectively. Find the transfer function and the impulse response of the

system. If R = L = C = 1, ﬁnd the response v

(t) provided that v

(t) = sin(t).

Figure 5.14.

Problem 5.46. Use the deﬁnition of convolution t o calculate the co nvolution of following function.

Use Laplace transform method for convolution and compare the results.

i. f (t) = tu(t) , g(t) = u(t)

ii. f (t) = e

¡t

, g(t) = u(t) ¡u(t ¡1)

iii. f (t) = u(t), g (t) = t[u(t) ¡u(t ¡1)]

iv. f (t) = u(t) ¡u(t ¡1), g(t) = u(t ¡1) ¡u(t ¡2)

Problem 5.47. Use convolution property to calculate the inverse transformation of the following

function

¡1



s(s

+ 1)



ii.

¡1



+ 1)



iii.

¡1



¡s

(s ¡1)(s

+ 2s + 2)



Problem 5.48. Show that convolution is commutative, that is, f ∗ g = g ∗ f .

Problem 5.49. Us Laplace transform method to solve the following integro-diﬀerential equations.

The initial condition(s) for all equations is assumed zero.

i. y + 3

y(v) sin(t ¡v) dv = u(t ¡1),

ii. y

+ e

¡τ

y(τ) dτ = cos(t) δ(t ¡π)

iii. y +

(τ) e

t¡τ

dτ = u(t ¡1).

iv. y

y(v) u(t ¡v) dv = δ(t ¡1) + u(t ¡1)

5.5 Systems of diﬀerential equations

In this section we see how the method is employed to solve the linear systems of diﬀeren-

34 Laplace Transform Method

tial equations with constant coeﬃcients. Consider the following system

= a

+ a

+ b

(t)

= a

+ a

+ b

(t)

(0) = y

(0) = 0

: (5.119)

The Laplace transform of the system reduces it to the following algebraic system

(

(s ¡a

)y^

(s) ¡a

(s) = b

(s)

(s ¡a

)y^

(s) ¡a

(s) = b

(s)

; (5.120)

that can be put in the following matrix form in turn



s ¡a

¡a

s ¡a



(s)



(s)

: (5.121)

The system (5.121) is solvable if the coeﬃcient matrix is invertible. If so, we can write



(s)



p(s)



s ¡a



(s)

; (5.122)

where p(s) is the characteri sti c polynomial of the matrix

p(s) = s

¡(a

+ a

)s + a

¡a

: (5.123)

Now, y

(t) and y

(t) can be restored by the inverse transform of y^

(s) and y^

(s) respec-

tively.

Example 5.22. Consider the following system

= y

¡y

= y

+ δ(t ¡1)

(0) = y

(0) = 0

: (5.124)

By Laplace transform, we write the above system as the following algebraic one



(s ¡1)y^

(s) + y^

(s) = 0

sy^

(s) ¡ y^

(s) = e

¡s

;

or in the matrix form



s ¡1 1

¡1 s



(s)





¡s



The above system is solved for the vector (y^

; y^

) by the formula



(s)



¡s + 1



s ¡1

1 s ¡1



¡s



¡s + 1



¡e

¡s

(s ¡1) e

¡s



Thus y^

(s) =

¡e

¡s

¡s + 1

and therefore

(t) =

¡2

t¡ 1

sin



(t ¡1)



u(t ¡1):

5.5 Systems of differential equations 35

Likewise, we have y^

(s) =

(s ¡1)e

¡s

¡s + 1

and therefore

(t) = e

(t¡1)/ 2

cos



(t ¡1)



u(t ¡1) ¡

t¡ 1

sin



(t ¡1)



u(t ¡1):

The method is applied to higher order systems as well. The f o llowing example presents

the method for a second order system.

Example 5.23. Consider the following system

= 2 y

+ u(t)

= 8y

(0) = y

(0) = 0

The system reduces to the algebraic one

(

^ (s) = 2y^

(s) +

(s) = 8y

(s)

Solving the system for y^

(s) and y^

(s) gives

(s) =

¡16

, y^

(s) =

s(s

¡16)

The inverse transform yields y

(t) and y

(t) as follows

(t) =

cosh(2t) ¡

cos(2t) and y

(t) =

cosh(2t) +

cos(2t) ¡

Problems

In the following problems, use Laplace transform method to solve the following systems.

Problem 5.50.

(

= y

+ δ(t ¡1)

= ¡y

+ u(t)

; y

(0) = y

(0) = 0

Problem 5.51.

(

= ¡y

¡ y

+ δ(t ¡1)

= ¡2y

; y

(0) = 0; y

(0) = 1

Problem 5.52.

(

= 2y

+ y

= 12y

¡2y

; y

(0) = 1; y

(0) = y

(0) = 0

Problem 5.53.

(

= 2 y

¡ y

+ y

¡y

= ¡y

; y

(0) = y

(0) = 1; y

(0) = y

(0) = 0

Problem 5.54.

(

+ 3y

= y

+ δ(t ¡1)

+ 3y

= 2y

; y

(0) = y

(0) = 0

Problem 5.55. Use the Laplace transform method to prove tha t the following two systems have

the same solution

36 Laplace Transform Method

1. y~

= A

2×2

y~ ; y~(0) = y~

2. y~

= A

2×2

y~ + δ(t)y~

; y~(0) = 0.

Here y~ stands for the vector





and the matrix A

2×2

is a constant matrix

2×2





5.5 Systems of differential equations 37