2. Conditions for the existence of Laplace transforms

(1)

CHAPTER 1 Laplace Transforms

1. Introduction and Definition

The dynamics of physical phenomena are represented mathematically by differential equations. Laplace transform is an efficient tool for solving both ordinary and partial differential equations. Ordinary differential equations are transformed into algebraic equations which simplifies the problem. Similarly Laplace transform converts partial differential equations into ordinary differential equations which are easier to solve. Laplace transform can be used to analyze and study the responses of electrical and mechanical systems subjected to discontinuous forcing terms. This property facilitates studying circuits where the input may be a voltage which remains on for a while, switches off and switches on periodically. Thus Laplace transform finds applications in signal analysis and is also a tool for investigating the effect of sudden impulsive forces on dynamical systems like the spring mass coupled system .

Mathematically Laplace transform is a mapping which assigns to a function ^f ^(t⁾ another function ^F⁽^s) ,called the Laplace transform of function ^f⁽^t^). It is defined for all ^{t ≥}⁰

F(s)=L{f(t) }=∫

0

∞

e^−stf(t)dt (1)

for all values of ^s for which this integral converges. This is an improper integral over an infinite interval , it can be represented as a sum of improper integrals within intervals which are bounded. As for example,

∫0

∞

h(t)dt=lim

a →∞∫

0 a

h(t)dt

If this limit exists , then the given improper integral is said to converge. In eq.(1) ,the integral will converge to a function of ^s which may converge for some range or values of ^s and diverge for

(2)

others. While deriving expressions for Laplace transform of functions it is therefore essential to state the condition on ^s for the transform to exist.

2. Conditions for the existence of Laplace transforms

Definition 2.1

A function ^f^(t⁾ is piecewise continuous over an interval if it can be divided into several sub intervals such that ^f^(t) is continuous within these subintervals and has finite limits at the ends of every subinterval except possibly at ^{t → ± ∞} .

Fig 1 piecewise continuous function

Definition 2.2

A function ^f^(t⁾ with ^t^>⁰ is said to be of exponential order if

|f(t)|≤ k e^pt

as ^{t → ∞} with ^{k , p ,T} all nonnegative for all ^{t ≥ T} .

This implies that the given function ^f ^(t⁾ , for ^{t → ∞ ,} grows less slowly than a positive multiple of some exponential function.

Existence Theorem: The Laplace transform of ^f⁽^t⁾ defined as

L{f(t)=F(s) exists for ^s^>^p if

a) ^f^(t⁾ is defined and is piecewise continuous on [ ⁰^{, ∞¿}

b) ^f^(t⁾ is of exponential order with respect to ^e^pt as ^{t → ∞.}

(3)

Given a function ^h^(t^), ∫

c d

h(t)dt exists if ^h⁽^t⁾ is piecewise continuous on the bounded interval [c,d].

Thus ∫

0 d

f(t)e^−stdt exists for ^d^{<∞ .} To limit the growth of ^f ^(t⁾ as

t → ∞ , we put a constraint on ^f⁽^t⁾ . Proof:

e^pt ≥1 if ^{t ≥}⁰ , ^|^f^(t⁾^|^{≤ k e}^pt^; Consider ∫

0 a

|e^−stf(t)|dt ; We let ^{a → ∞,} and check for the boundedness of this integral

∫0 a

|e^−stf(t)|dt ≤∫

0 a

|e^−stk e^pt|dt=k∫

0 a

e⁻⁽^s−p⁾^tdt

¿ k

(s−p)(1−e^−a⁽^s−p⁾)

If ^s^>^{p ,∧a→ ∞ ,} then the second term within the bracket ^→⁰ and the integral becomes _s−^k_p .

k∫

0 a

e⁻⁽^s−p⁾^tdt ≤ k∫

0

∞

e^−(s−p)tdt

F(s)=∫

0

∞

f(t)e⁻^stdt

|^F^(s)|⁼

|

^∫^∞⁰ ^f^(t^)e^−st^dt

|

^≤^∫^∞⁰^|^e^−st^f⁽^t⁾^|^{dt ≤}^s−^k^p ^.

For a function ^F(^s) to be a Laplace transform of some function

f(t) the condition is

lim

s → ∞, F(s)=0 (2) Example 2.1

^F⁽^s)=_s+5^s , _{s → ∞}^lim _s+^s₅⁼¹

lim

s → ∞F(s)≠0

(4)

∴ F(s) cannot be the Laplace transform of any function.

Initial Value Theorem

lim

t →0 f(t)=lim

s → ∞sF(s)

L{f^'(t) } = ∫

0

∞

e^−stf^'(t)dt=sF(s)−f(0)

Assuming the continuity of ^f^(t) at ^t⁼⁰ , and assuming that the

L{f^'(t) } exists and therefore is piecewise continuous and of exponential order,

s → ∞lim∫

0

∞

e^−stf^'(t)dt=0=¿ lim

s → ∞sF(s)−f(0) or

lim

s → ∞sF(s)=f(0) (3) Final Value Theorem

lim

t → ∞f (t)=lim

s →0 sF(s)

L{f^'(t) } = ∫

0

∞

e^−stf^'(t)dt ¿sF(s)−f(0)

lims →0∫

0

∞

e^−stf^'(t)dt=lim

k → ∞[f(t)]₀^k =

f(k) [¿−f(0)]

lim

k →∞

¿ RHS ^lim_{s →0}^[^sF⁽^s^)−f^(0)]

comparing both ,we have

f(t) [¿] lim

s →0[sF(s)]=lim

t →∞¿ (4)

3. Laplace Transform of functions

Example 3. 1:

Determine the Laplace transform of the following functions

(5)

a) ^f⁽^t^)=tⁿ Definition 3.1

Gamma function ^Γ⁽ⁿ⁾ is defined by the following integral

Γ(n)=∫

0

∞

e^−xxⁿ⁻¹dx

F(s)=∫

0

∞

tⁿe^−stdt = ^Γ^(n+1)/^sⁿ⁺¹ b) ^f⁽^t⁾^=cos^ωt

F(s)=∫

0

∞

e^−stcosωt dt = ∫

0

∞ e^−t(s+iω) 2 dt+∫

0

∞ e^−t^(s−iω)

2 dt = _s²_+ω^s ² c) ^f⁽^t⁾^=sin^ωt

F(s)=∫

0

∞

sinωt e^−stdt = _s²^ω

+ω²

d) ^f⁽^t⁾^=cosh^ωt

F(s)=∫

0

∞

coshωt e^−stdt = _s²_−ω^s ² f ) ^f^(t⁾ = ^sinh^ωt ; ^F⁽^s)=∫

0

∞

sinhωt e^−stdt= ω s²−ω²

g) ^f⁽^t^)=e^at ; ^F⁽^s)=∫

0

∞

e^−t^(s−a)dt= 1 s−a

4. Properties of Laplace transform

a) Linearity property

L{^{c f} ^(t^{)+d g}⁽^t⁾}^{=c L}{^f⁽^t⁾}⁺^{d L}^{g^(t^{) }} with ^c and ^d as scalars.

L{^{c f}⁽^t^{)+d g}⁽^t⁾}^=cF^(s⁾⁺^dG(^s) (5) Example 4.1:

Determine the Laplace transform of the following function

(6)

f(t)=e^2t+sin 4t−cos 2t

L{e²^t+sin 4t−cos 2t}=L{e²^t}+L{sin 4t}−L{cos 2t}

¿ 1

s−2+ 4

s²+16− s s²+4

b) Shifting theorem: First Translational Property

L{e^atf(t)}=F(s−a) (6) Proof: Applying the definition of Laplace transform, we have

L{e^atf (t)}=∫

0

∞

e^−ste^atf (t)dt=∫

0

∞

e^−(s^−a)tf(t)dt=F(s−a)

Multiplying a function by êât results in the shifting of the Laplace transform along the ^s axis by â .

Example 4.1:

Evaluate ^L{e^3t^{sin 4}^t^}

Since ^{L{sin 4}^t^}=_s²₊₁₆⁴ , Thus applying the shifting theorem , we obtain, ^L^{e^3t^{sin 4}^t^} = ₍_s−3⁴₎²₊₁₆

Example 4.2:

Evaluate ^L^{cos²^t^}

cos²t=(1+cos 2t)/2 ; using the linearity property

L{cos²t}=s 2 +

s (¿¿2+4²)2

s

¿

c) Second translation property

In various problems in engineering ,the dependent variables may have jumps which results in the functions having discontinuities at these points. These discontinuities can be represented mathematically

(7)

by a unit step function ^U^(t^−a) , ^t^=a being a point of discontinuity for the given function.

A unit step function is defined as

U(t−a)=1for t ≥ a (7)

U(t−a)=0for t<a

L{f(t−a)U(t−a)}=e^−asF(s)

Fig 2 unit step function

Proof: Using integration by parts and the basic definition of LaplaceTransform

L{f(t−a)U(t−a)}=∫

0 a

f(t−a)0e^−stdt+∫

a

∞

f(t−a)e^−stdt

substitute ^t^−a=^{x ,} ^dt=dx ;thelimits transform¿ (0 ^{, ∞¿} and the expression becomes ∫

0

∞

f(x)e^−(x+a)^sdx=e^−asF(s)

The unit step function is also expressed as ^U^(t^−a^)=U^a^(t⁾

L{^Ua}⁼^e

−as

s (8) Within a given interval if there are several discontinuities which can be expressed as step functions, a linear combination of unit step functions can be invoked to express these discontinuities. For example

f(t)=a , for t∈(0,t₁)

f(t)=b , for t∈(t₁, t₂)

(8)

f(t)=c for t∈(t₂,t₃) and so on. The complete function ^f⁽^t⁾ is expressed as

f(t)=a+(b−a)U_t₁+(c−b)U_t₂+…

Example 4.3:

Evaluate ^L^{⁽^t⁻¹⁾^U⁽^t⁻¹⁾^}

comparing the expression, we obtain ^a=1,f ^(t)=t , thus F(s)=1

s² and on substitution the result is ^e^−s^/^s²

Example 4.4:

Find the LT of ^{cost U}^(t⁻²^π⁾

since the periodicity of ^cos function is ²^π , thus ^cos^t^=cos(t⁻²^π⁾ thus ^L^{cost ^U^(t⁻²^{π) }=L}^{cos (t-2π) ^U^(t−2^{π) }}

using the second translation property, and linearity property we obtain

L { ^cos^⁡^(t⁻²^π⁾^U^(t−2^π⁾ } ^¿^s ^e_s2⁻²^πs

+1 Example 4.5:

Evaluate ^L^{^f⁽^t⁾^}^,

f(t)=0 for ⁰^{≤t ≤}³^; ^f⁽^t⁾⁼p , for t ≥3

This function has a discontinuity at ^t⁼³ .

using the first principles we obtain the LT of the given function,

∫0

∞

f(t)dt e^−st = ∫

0 3

0dt+∫

3

∞

e^−stp dt=pe⁻³^s s ;s>0

Example 4.6

A continuous electric pulse which is switched on only for 2 s say between ^t⁼³ and ^t=5 can be modelled using step functions.

Assuming that the pulse is of 4 volts.

f(t)=0∈(0,3) and ⁽⁵^{, ∞)}

(9)

f(t)=4∈(3,5)

f(t)=0+(4−0)U₃(t)+(0−5)U₅(t)

L{^f(t)}⁼⁴Û3(t)−5U₅(t) ⁴ê^−3s_s ⁻⁵ê⁻⁵_s ^s 5. Laplace transform of derivatives

L[f^'(t)]⁼^¿ ∫

0

∞

f^'(t)e^−stdt=[f(t)e^−st]₀^∞ - ∫

0

∞

f(t)(−s)e^−stdt

with ^f⁽^t⁾ bounded , the upper limit ^→⁰ in the first term.

L[f^'(t)]=s F(s)−f(0) (9)

L[f^{' '}(t)]⁼sG(s)−g(0)=s[s F(s)−f(0)]−g(0) with

g(t)=f^'(t) ; and ^G⁽^s⁾⁼^L[^g⁽^t⁾^] . Thus

L[f^{' '}(t)]⁼s²F(s)−s f(0)−(f^'(t))t=0 (10)

This property is particularly useful for converting ordinary differential equations into algebraic equations.

6. Laplace transforms of integrals

If L{f(t)}=F(s),

L

{

^∫⁰^x ^f⁽^t⁾^dt

}

⁼^F^(s)^s (11) The result of the integration will be some function of ^{x ,} say ^p(^x) Let ∫

0 x

f(t)dt=p(x) ; then ^p^(0)=0∧^p^'⁽^x^)=f^(x⁾

Considering the Laplace transform ^L^{^p^'⁽^x⁾^}⁼^L^{^f⁽^x⁾^} or ^sP⁽^s⁾^−¿ ^p^(0)=F⁽^s) or

s L

{

^∫⁰^x ^f⁽^t⁾^dt

}

^=F^(s⁾ Delhi Univ.(D.U.)

(10)

Example 6.1:

Evaluate ^L

{

^∫⁰^x ^e⁻²^t^dt

}

Applying the above theorem , ^f⁽^t^)=e⁻²^t , ^L{e⁻²^t^}=_S+2¹ Thus ^L

{

^∫⁰^x ^e⁻²^t^dt

}

⁼^s^(s¹⁺²⁾