ADVANCED ENGINEERING MATHEMATICS

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Advanced Engineering Mathematics - The Laplace Transforms

Preprint · May 2020

DOI: 10.13140/RG.2.2.24340.17285

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Moh Kamalul Wafi (May 2020), ”TheL-Transforms,”

The Kamal Institute, Vol.1(5), 2020, Article ID TF181301 https://orcid.org/0000-0002-2016-5721

ADVANCED ENGINEERING MATHEMATICS

The Laplace (

L

)−Transforms Moh. Kamalul Wafi^a,b,c

aLaboratory of Embedded Systems, Institut Teknologi Sepuluh Nopember;

bControl Systems - Imperial College London;

cMember, IEEE Control Systems Society (CSS)

ARTICLE HISTORY Compiled March 27, 2022

ABSTRACT

This material covers the Laplace (L) transform constituting the definition, the important formulas along with the properties. The followings are the inverse with some applications in the end, from the linear systems of differential equations, oscillator harmonic, electrical to mechanical structure

KEYWORDS

The LaplaceL−transform

1. Introduction

The physical phenomena are mostly interpreted with the ordinary or partial differential equations depending on the boundaries. One of the trivial solutions to cope with those cases is to apply Laplace transform converting the differential equations into an algebraic expressions which could be switched into original problem. The discontinu- ity functions could also be handled by the transformation whereas other methods fail.

Moreover, various algorithms such as Fourier transform derive from the Laplace transform. Dealing with either ODE or PDE, even control theories, with certain boundaries without solving the arbitrary constants are the instances of the Laplace benefits.

2. Definition

Function of timef(t),∀t >0 comprises a real- or complex-valued and its Laplace (L) transform is denoted as the semi-infinite integral, therefore

L[f(t)] =F(s) = Z ∞

0

e^−stf(t)dt or lim

τ→∞

Z τ 0

e^−stf(t)dt (1) where sis defined as a real or complex parameter and supposed to be huge positive number to ensure the convergence of such integral. Indeed, the variable of s places in the real and imaginary axes. The limit should exist to guarantee the convergence, otherwise it diverges and the Laplace transform is no transformation off(t).

Contact: [email protected]

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3. Transform of Important Sequences

The whole operators could be transformed into Laplace, but there are certain important sequences as a basis of some complex mixed operators. Therefore, this chapter comprehensively delivers seven (7) theorems with proofs to those key sequences.

Theorem 3.1. Suppose f(t) is a constant value of λ,∀t >0. The Laplace transform is then written as,

L[λ] = Z ∞

0

e^−stλ dt=−λ s

e^−st∞ 0

=−λ

s(0−1) = λ

s −→s >0 (2)

Theorem 3.2. Suppose the function of (t) is defined as f(t) =e^λt with certain constant λ. The Laplace transform of such function is written as follows,

Lh e^λti

= Z ∞

0

e^−ste^λtdt=− 1 s−λ

h

e^−(s−λ)ti∞ 0

=− 1

s−λ(0−1) = 1

s−λ−→s > λ (3)

Keep in mind that as λis negative such that e^−λt, the Laplace transform is positive.

Theorem 3.3. Suppose f(t) =tⁿ where (n) as a positive integer. Transforming into Laplace is initiated by partial integral resulting a general formula of sequences,

L[tⁿ] = Z ∞

0

e^−sttⁿdt=−1 s

tⁿe^−st∞ 0 +n

s Z ∞

0

tⁿ⁻¹e^−stdt (integral by part)

= 0 +n s

Z ∞ 0

tⁿ⁻¹e^−stdt ϕn= n

sϕn−1 −→ϕn−1 = n−1

s ϕn−2 (general formulae)

ϕn= n

s ·n−1

s ·n−2

s ·n−3

s · · · n−(n−1)

s ϕ0 (ϕ0 =t⁰)

L[tⁿ] = n!

sⁿ⁺¹ (4)

Theorem 3.4. Supposef(t) = sin βt, its Laplace transform requires the alteration of (sin)into exponential (e) term, therefore

L[sinβt] = Z ∞

0

e^−st

e^iβt−e^−iβt 2i

dt

= 1 2i

Z ∞ 0

e^−ste^iβtdt− Z ∞

0

e^−ste^−iβtdt

= 1 2i

1

s−iβ − 1 s+iβ

(Theorem3.2)

= 1

2i· 2iβ

s²+β² = β

s²+β² −→s > λ (5)

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Theorem 3.5. Likewise, the transformation of function f(t) = cos βtresults in L[cos βt] =

Z ∞ 0

e^−st

e^iβt+e^−iβt 2

dt

= 1 2

Z ∞ 0

e^−ste^iβtdt+ Z ∞

0

e^−ste^−iβtdt

= 1 2

1

s−iβ + 1 s+iβ

(Theorem3.2)

= 1 2· 2s

s²+β² = s

s²+β² −→s > λ (6) The only different between(cos)and(sin)lies in the numerator of(s) and(β) in turn.

Theorem 3.6. The hyperbolic sine function of f(t) = sinh βtneeds also to change to exponential(e) term without imaginary(i) axis, such that

L[sinhβt] = Z ∞

0

e^−st

e^βt−e^−βt 2

dt

= 1 2

Z ∞ 0

e^−ste^βtdt− Z ∞

0

e^−ste^−βtdt

= 1 2

1

s−β − 1 s+β

(Theorem3.2)

= 1 2· 2β

s²−β² = β

s²−β² −→s > λ (7) Theorem 3.7. The counterpart hyperbolic cosine function off(t) = cosh βt uses the same procedure as the preceding, yielding

L[coshβt] = Z ∞

0

e^−st

e^βt+e^−βt 2

dt

= 1 2

Z ∞ 0

e^−ste^βtdt+ Z ∞

0

e^−ste^−βtdt

= 1 2

1

s−β + 1 s+β

(Theorem3.2)

= 1 2 · 2s

s²−β² = s

s²−β² −→s > λ (8) Example 1. Find the Laplace transform of the following functions according to the Theorem (3.1) - (3.7)!

(1) L[3]

(2) L[−7]

(3) L e^−πt (4) L

e^20t (5) L

t³

(6) L t⁵ (7) L[sin 3t]

(8) L[cos 4t]

(9) L[sinh (−5t)]

(10) L[cosh (−6t)]

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4. Properties of L−Transforms

Having discussed the important formulas, the properties of Laplace transform are divided into eleven theorems so as to ease the transformation.

4.1. Linearity or Superposition

If there are two functions of time (t), x_t and y_t, multiplied by arbitrary constant of (a) and (b) respectively, the Laplace transform is defined as follows,

L(a[xt] +b[yt]) =aX(s) +bY(s) (9) Proof.

L(a[x_t] +b[y_t]) = Z ∞

0

e^−st(ax_t+by_t) dt

=a Z ∞

0

e^−stxtdt+b Z ∞

0

e^−stytdt →aX(s) +bY(s) Example 2. Find the Laplace transform of the following functions!

(1) L

3e^−5t+ 2t (2) L[2π+ cos 3t]

(3) L[3 sin 3t+ 4 cos 2t]

(4) L[2 sin 2t−5 sinh 3t]

(5) L

3e^9πt+ cosh 3πt (6) L

2t³+ 3t²+ 5t+ 1 (7) L[xt] ifxt=

(t

k ,0< t≤k 1 , t > k 4.2. The First-Shifting Theorem

if the Laplace transform ofxtequals toX(s) withαcomprises any real numbers, then L

e^αtxt

=X(s−α) (10) Proof. The±α in function of (e^±α) generatesX(s∓α) in Laplace transform

L e^αtxt

= Z ∞

0

e^−st e^αtxt

dt

= Z ∞

0

e^−(s−α)txtdt →X(s−α) Corollary 4.2.1. From Eq.(10) further formulae could be then assigned, such that

L

e^αtsin t

= β

(s−α)²+β² L

e^−αtsinh t

= β

(s+α)²−β² L

e^−αtcos t

= s+α

(s+α)²+β² L

e^αtcosh t

= s−α

(s−α)²−β² L

e^αttⁿ

= n!

(s−α)ⁿ⁺¹ L

e^−αttⁿ

= n!

(s+α)ⁿ⁺¹

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Example 3. Find the Laplace transform of the following functions!

(1) L

e^−5tsin πt (2) L

e^7tcos 2t (3) L

e^−2tsinh 3t (4) L

e^−4tcosh 5t (5) L

3e^−5tt²

(6) L

5e^−3tt³ (7) L

cos²t (8) L

e^tsin²t (9) L[sin 2tcos 3t]

(10) L sin³t

4.3. Derivative of x_t

Suppose the Laplace ofxt equalsX(s), then the Laplace transform of its derivativex_t is denoted as in the following with (x₀) represents the function of (x_t) at t= 0

L x⁰_t

=sL[x_t]−x₀ (11)

Proof.

L x⁰_t

= Z ∞

0

e^−stx⁰_tdt

= e^−stxt

∞

0 −

Z ∞ 0

−se^−st xtdt

=−x₀+sL[x_t] →sL[x_t]−x₀ Corollary 4.3.1. The term ϕⁿ in s−domain and ϕ⁽ⁿ⁾ in (x₀) indicate the power of nand the n−th derivative of xt in turn.

L h

x⁽ⁿ⁾_t i

=sⁿL[xt]−sⁿ⁻¹L[x0]−sⁿ⁻²L x⁰₀

−sⁿ⁻³L x⁰⁰₀

− · · · −x⁽ⁿ⁻¹⁾₀ (12) Proof.

L x⁰⁰_t

=sL x⁰_t

−x⁰₀

=s²L[xt]−sx0−x⁰₀ (second derivative) L

h x⁽³⁾_t

i

=sL x⁰⁰_t

−x⁰⁰₀

=s³L[xt]−s²x0−sx⁰₀−x⁰⁰₀ (third derivative) Lh

x⁽⁴⁾_t i

=sLh x⁽³⁾_t i

−x⁽³⁾₀

=s⁴L[xt]−s³x0−s²x⁰₀−sx⁰⁰₀ −x⁽³⁾₀ (fourth derivative) ...

L h

x⁽ⁿ⁾_t i

=sⁿL[xt]−sⁿ⁻¹L[x0]−sⁿ⁻²L x⁰₀

−sⁿ⁻³L x⁰⁰₀

− · · · −x⁽ⁿ⁻¹⁾₀ As the physical problems such as mechanical and electrical structure along with oscillator harmonic are formulated into differential equations, this term is absolutely a key formula and one of the most trivial to deal with those cases.

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4.4. Integration of x_t

ifL[x_t] =X(s), then its Laplace transform of integralx_tfrom 0→tis shown below L

Z t 0

x_tdt

= 1

sX(s) (13)

Proof. Letϕt= Z t

0

xtdt andϕ0 = 0, thereforeϕ⁰_t=xt, such that L

ϕ⁰_t

=sL[ϕt]−ϕ0

L[ϕ_t] = 1 sL

ϕ⁰_t L

Z t 0

x_tdt

= 1

sL[x_t] → 1

sX(s)

4.5. Multiplication by tⁿ

Supposen= 1 and L[x_t] =X(s), then the Laplace transform ofx_t being multiplied by t is depicted as follows. As the number of (n) is becoming higher, the power of derivative would parallel with that (n).

L[t x_t] =

−d ds

X(s) (14)

Proof.

L[t xt] = Z ∞

0

e^−st(t xt)dt

= Z ∞

0

− d dse^−st

xtdt

=−d ds

Z ∞ 0

e^−stx_tdt → −d

dsX(s)

Corollary 4.5.1. If the positive integer number ofn >1, the (n) refers to the power of its derivative in the Laplace domain, therefore

L[tⁿxt] = Z ∞

0

e^−st(tⁿxt) dt= Z ∞

0

− d ds

n

e^−st

xtdt

=

− d ds

nZ ∞ 0

e^−stxtdt →

− d ds

n

X(s) (15)

Supposext= sinh tand n= 2, the Laplace transform is written as L

t² sinh αt

=

−d

ds −d ds

α s²+α²

=

−d ds

2αs

(s²+α²)² → 6as²−2a³ (s²+α²)³

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Example 4. Find the Laplace transform of the following functions!

(1) L

t²sin 3t (2) L

t²cos πt (3) L

t²sinh 2t (4) L

t²cosh 4t (5) L

Z t 0

t³e^−3tdt

(6) L

tsin²3t (7) L

tcos²2t (8) L

t e^αt sin βt (9) L

t e^−t cosh πt (10) L

Z t 0

t²e^−2t cost dt

4.6. Division by ¹_t

ifL[x_t] =X(s), then its Laplace transform ofx_tdivided by (t) is the integral ofX(s) with respect tosfrom s→ ∞, such that

L 1

tx_t

= Z ∞

s

X(s)ds−→ provided lim

t→0

x_t

t exists (16)

where the followings are the important derivative of angle (sin⁻¹,cos⁻¹,tan⁻¹) and their counterpart (csc⁻¹,sec⁻¹,cot⁻¹)

d

dxsin⁻¹ x= x⁰

√1−x²

d

dxtan⁻¹ x= x⁰ 1 +x²

d

dxsec⁻¹ x= x⁰ x√

x²−1 d

dxcos⁻¹ x= −x⁰

√ 1−x²

d

dxcot⁻¹ x= −x⁰ 1 +x²

d

dxcsc⁻¹ x= −x⁰ x√

x²−1

Proof.

Z ∞ s

X(s)ds= Z ∞

s

Z ∞ 0

e^−stx_tdt

ds

= Z ∞

0

xt

Z ∞ s

e^−stds

dt= Z ∞

0

xt

e^−st

−t ∞

s

dt

= Z ∞

0

−xt

t 0−e^−st dt

= Z ∞

0

e^−st 1

txt

dt →L 1

txt

Corollary 4.6.1. Let t power (n),∀n >1, the integral ofX(s) would be n−th times according to (n), known as Cauchy’s integral formula (sequences). The general result is written in Eq. (17) whereas the steps to obtain are explained in the followings as Eq.(18) with further definition of σn(x) in Eq. (19). Note that the difference between ϕ⁽⁻ⁿ⁾ and ϕ⁽ⁿ⁾ lies in integral (−) and derivative (+) respectively.

L 1

tⁿx_t

= Z n

X(s)dsⁿ=X⁽⁻ⁿ⁾(s) +σ_n(s),∀n≥1 (17)

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Z

ϕ(x)dx=ϕ⁽⁻¹⁾(x) +C₁ (first)

Z Z

ϕ(x)dx dx=ϕ⁽⁻²⁾(x) +xC1+C2 (second) Z Z Z

ϕ(x)dx dx dx=ϕ⁽⁻³⁾(x) +1

2x²C₁+xC₂+C₃ (third) ...

Z n

ϕ(x)dxⁿ=ϕ⁽⁻ⁿ⁾(x) +σn(x),∀n≥1 (18) The proposed variable of σn(x) is to cover the function of (x) in Cn and the general formula is written as follows with the example ofσ₅(x)

σn(x) = 1

(n−1)!xⁿ⁻¹C1+ 1

(n−2)!xⁿ⁻²C2+· · ·+ 1

1!x¹Cn−1+ 1

0!x⁰Cn (19) such that,

σ₅(x) = 1

4!x⁴C₁+ 1

3!x³C₂+ 1

2!x²C₃+ 1

1!x¹C₄+ 1 0!x⁰C₅ Example 5. Find the Laplace transform of the following functions!

(1) L

sin βt t

(2) L

1−cosh βt t

(3) L

e^−4tsinh 3t t

(4) L Z t

0

e^αt−1 t dt

(5) L

e^−αt−e^−βt t

(6) L

e^−αt−cosπt t

(7) L

1−cost t²

(8) L Z t

0

cos βt−cosγt

t dt

4.7. Unit-Impulse or Dirac Delta Function

This impulse function is the outlier of any properties of Laplace transform because it requires a very limited time of (k). Moreover, it is denoted from other functions while the common transformations are affected by its inputs and corresponding outputs. if xt= 1 and kplaced in between of p andq, such that p < k≤q, then

Z ∞

−∞

δ(t−k) dt=

Z p :0

−∞

δ(t−k) dt+ Z q

p

δ(t−k) dt+

Z ∞:0

q

δ(t−k) dt

= 0 + 1 + 0 →1

This means that the dirac delta function describes the value of certain time (k) as shown in Fig. (??) in a function of x_t, therefore the multiplication of those two result

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Figure 1.:Dirac-Delta Function

inx_k. Note that the boundary is not always ∓∞ meaning that the closest boundary tokis satisfied, such that

Z ∞

−∞

x_tδ(t−k) =x_k (20)

Suppose the dirac-delta function is differentiated and placed anywhere apart from k= 0, it then yields,

Z ∞

−∞

xtδ⁰(t−k) = [xtδ(t−k)]^∞_−∞− Z ∞

−∞

x⁰_tδ(t−k) dt

= (0−0)−x⁰_k → −x⁰_k

Proof. Letδ(k), k→0, then the Laplace transform of the shifting comprises, L[δ(t−k)] =

Z ∞ 0

e^−stδ(t−k) dt−→x_t=e^−st

=e^−ks

Example 6. Evaluate these dirac-delta functions!

(1) L[5δ(t−3)]

(2) L

t³δ(t−5) (3) L

sin 5t δ t− ^π₂ (4) L[cosh 4t δ(t)]

(5) L[cos 3t ln(t)δ(t−π)]

(6) L

e^−4tδ(t−3) (7) L

e^−2tδ⁰(t−2) (8) L

sin²t δ⁰ t−^π₄ 4.8. Heaviside Unit-Step Function

Suppose a function of step (λ) in Theorem (3.1), yet starting from certain (k),u(t−k) instead of zero,u(t), occurring atk= 0 as shown in Fig.2. This function is denoted as the Heaviside unit-step function withk≥0. The further explanation of this function is described in Fig.3that the function ofx_t=t² is then cut withu(t−k) so that the graph ofxt constitutes xt=t²u(t−k).

u(t−k) =

(0, t < k 1, t≥k

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Figure 2.:(Lef t): The function step of 1·u(t) whereas (Right): The function ofxt= 1 with the effect of Heaviside unit-step fromu(t−k)

Figure 3.:Function ofx_t=t²due to the effect of Heaviside unit-step at certaink The idea is that the effect of u(t−k) suppresses the t²,∀ − ∞ ≤ t < k and then turn on the function of t²,∀t≥ k. Indeed, the Laplace transform of such Heaviside unit-step is written in the following. If k = 0, that would be the same as Theorem (3.1), otherwise it is on par with Eq. (21). Note that Fig. 4 is the questions to deal with Heaviside unit-step function.

L[u(t−k)] = e^−ks

s (21)

Proof.

L[u(t−k)] = Z ∞

0

e^−stu(t−k) dt

= Z k

0

e^−st0dt+ Z ∞

k

e^−st1dt

= 0 + e^−st

−s ∞

k

→ e^−ks

s

Figure 4.:Two function ofx_twith different Heaviside unit-step function

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Suppose a function x_t in Fig 4 (left) which could be solved with two different ways, conventional and Laplace, yet it is focused on the properties of Laplace:

x_t=

(7, t <2

5, t≥2 −→7u(t)−2u(t−2) such that,

L[7u(t)−2u(t−2)] = 7

s −2e^−2s s Regarding the (right) question in Fig. 4, the functionxt,

x_t=

(ϕ, α≤t < β

0, t≥β −→ϕ(u(t−α)−u(t−β)) therefore,

L[ϕ(u(t−α)−u(t−β))] =ϕ e^−αs

s − e^−βs s

To comprehensively understand the Heaviside unit-step, Fig.5shows the effects of the properties with four different conditions. The first graph (top-left) is the function of xt= sin(t)u(t) while the (top-right) constitutes xt= sin(t)u(t−90^◦). The Heaviside cuts the function before 90^◦ and turns on after it. Furthermore, the (bottom-left) is shifting the original function from sin(t) to sin(t−90^◦), meaning that the starting point oft= 0 is 90^◦. The last is (bottom-right) which is then suppressed at 90^◦, such that the function of x_t = sin(t−90^◦)u(t−90^◦). This is also the same as x_t = sin(t) being shifted alongt−axis via an interval of k^◦ unit. Another terse-instance is Fig.6 with exponential function (e) in which the schemes are on par with those of Fig. 5.

Figure 5.: Two function ofx_t= sin(t) with different Heaviside unit-step function

(13)

Figure 6.: Two function ofxt=e^−twith different Heaviside unit-step function Similarly, the functions in Fig.6 are accounted for the following formulas. While the ordinary graph e^−t in (top-left) is then suppressed prior to t = k in (top-right), the shifted graph of the ordinarye^−(t−k) is started att=kin (bottom-left) and it is also switched on att=α before -off at t=β in (bottom-right), such that

xt=e^−tu(t) −→e^−tu(t−k) (top)

x_t=e^−(t−k)u(t−k) −→e^−t[u(t−α)−u(t−β]) (bottom) 4.9. The Second-Shifting Theorem

IfL[x_t] =F(s), then the Laplace transform ofxt−ku(t−k) as in Fig.5(bottom-right) and Fig.6(bottom-left) is written by

L[xt−ku(t−k)] =e^−ksX(s) (22) Proof.

L[xt−ku(t−k)] = Z ∞

0

e^−stxt−ku(t−k) dt

=

Z k:0

0

e^−stx_t−kdt+ Z ∞

k

e^−stx_t−kdt

= Z ∞

0

e^−s(v+k)x_vdv

Note that v =t−k so that dt = dv ask → constant and ift =k then the value of v= 0 whereas if t→ ∞ thenv → ∞ which means that the equation above could be also written as in the following

=e^−ks Z ∞

0

e^−svx_vdv =e^−ksX(s)

(14)

Corollary 4.9.1. Suppose L[x_t] =F(s) and a function of x_t as defined in top-right of Fig. 5 and Fig. 6, the Laplace transform is then written as

L[xtu(t−k)] =e^−ksL[xt+k] (23) Proof.

L[x_tu(t−k)] = Z ∞

0

e^−stx_tu(t−k)dt

=

*0 Z _k

0

e^−stx_tdt+ Z ∞

k

e^−stx_tdt

=e^−ks Z ∞

0

e^−svx_v+kdv=e^−ks Z ∞

0

e^−stx_t+kdt →e^−ksL[x_t+k] Example 7. Evaluate these second-shifting theorem functions!

(1) L h

(t−2)³ u(t−2) i

(2) L[sin 3(t−1)u(t−1)]

(3) L

e^(t−5)u(t−5)

(4) L[cosh 2(t−90^◦)u(t−90^◦)]

(5) L

t²u(t−3) (6) L

e^−2tu(t−π)

(7) L[sinh 2t] at 2π≤t <4π (8) L[cos 3t] at 3π≤t <6π

4.10. Periodic Function

If xt is a periodic function at nT,∀n ≥1 so that xt+nT = xt as portrayed in Fig. 7, the Laplace transform of such periodic functionx_t is

Figure 7.:Periodic Function off(t) atnT,∀n≥1

L[xt] = X^†(s)

1−e^−sT −→X^†(s) = Z T

0

e^−stxtdt (24) Suppose x^[i]_t as the i−th periodic function with the period of T. The first cycle and the second would be the same apart from the boundaries of the function of intervalT. The general formula ofn−th periodic function with interval T is defined in Eq. (25), therefore

x^[1]_t =

(x_t, 0≤t < T

0, otherwise −→x^[2]_t =xt−T u(t−T) =

(x_t, T ≤t <2T 0, otherwise

(15)

x^[i+1]_t =xt−iTu(t−iT) = (

xt, iT ≤t <(i+ 1)T

0, otherwise −→ ∀i≥0 (25)

According the information of Eq. (25), the sequence is denoted in Eq. (26), meaning that each sequence ofx^[i]_t has the boundary to turn on and -off the function

xt=x^[1]_t +x^[2]_t +x^[3]_t +. . .

=x_tu(t) +xt−T u(t−T) +xt−2Tu(t−2T) +. . . (26) Proof.

L[x_t] =Lh

x⁽¹⁾_t +x⁽²⁾_t +x⁽³⁾_t +. . .i

L h

x⁽¹⁾_t i

= Z ∞

0

e^−stx⁽¹⁾_t dt= Z T

0

e^−stx_tdt=F^†(s) Lh

x⁽²⁾_t i

= Z ∞

0

e^−stx⁽²⁾_t dt=e^−sT Z T

0

e^−stx_tdt=e^−sTF^†(s) L

h x⁽³⁾_t

i

= Z ∞

0

e^−stx⁽³⁾_t dt=e^−2sT Z T

0

e^−stxtdt=e^−2sTF^†(s) ...

L h

x⁽ⁱ⁾_t i

= Z ∞

0

e^−stx⁽ⁱ⁾_t dt=e^−(i−1)sT Z T

0

e^−stx_tdt=e^−(i−1)sTF^†(s)

L[x_t] = 1 +e^−sT +e^−2sT +. . .

F^†(s) → F^†(s)

1−e^−sT

Figure 8.:Two different periodic function ofxt

f(t) =

(3, 0< t≤2

0, 2< t≤4 −→f(t+ 4) =f(t), T = 4 and then the Laplace transform is,

L[f(t)] = 1 1−e^−4s

Z 4 0

e^−stf(t)dt

= 1

1−e^−4s Z 2

0

e^−st3dt+ 0

→ 3

s(1 +e^−2s)

(16)

While the right-side periodic function in Fig. (26) comprises, f(t) = t

2, 0< t≤3−→f(t+ 3) =f(t) L[f(t)] = 1

1−e^−3s Z ₃

0

e^−stf(t)dt

= 1

1−e^−3s 1

2 Z 3

0

e^−stt dt+ 0

−→

Z

p dq=pq− Z

q dp

= 1 2s²

1− 3s e^3s−1

Example 8. Evaluate these periodic functions!

(1) x_t=e^tfor 0< t≤5 (2) xt=

(

8 sin t, 0< t≤π 0, π < t≤2π (3) xt= 2t

3 for 0< t≤3

(4) Square wave function x_t in Eq. (27) (5) xt=

(

sin ωt, 0< t≤ ^π_ω 0, ^π_ω < t≤ ^2π_ω (6) xt= Kt

T for 0< t≤T

xt=k[u0(t)−2ua(t)−2u2a(t)−2u3a(t)−. . .] (27) 4.11. Convolution

SupposeL[xt] =X(s) andL[yt] =Y(s), the convolution of the two functions is written in Eq. (28) whereas its Laplace transform is made of in Eq. (29). The convolution w.r.t certain variable (t) should be altered first into any other term (τ), such that

xt∗yt= Z t

0

x(τ)y(t−τ)dτ (28)

L[x_t∗y_t] =X(s)·Y(s) (29) Proof.

L[x_t∗y_t] = Z ∞

0

e^−st Z _t

0

x(τ)y(t−τ)dτ dt

= Z ∞

0

Z ∞ τ

e^−stx(τ)y(t−τ)dt dτ

= Z ∞

0

e^−sτx(τ)dτ Z ∞

τ

e^−s(t−τ)y(t−τ)dt (v=t−τ) Note that v =t−τ so that dt = dv ask → constant and ift =τ then the value of v= 0 whereas ift→ ∞ thenv→ ∞ meaning the equation above could be written as

= Z ∞

0

e^−sτx(τ)dτ Z ∞

0

e^−svy(v)dv →X(s)·Y(s)

(17)

5. Inverse of Laplace Transform

There are various methods to solve the inverse Laplace transform based on the structure of suchX(s) and the followings are the ways to deal with certain X(s).

5.1. Inspection

This kind of method is just the exact picture of Laplace transform. Without any further scheme, the inverse should be able to be obtained.

L⁻¹[X(s)] =x_t (30)

Example 9. Find the inverse Laplace transform of the following functions of X(s)!

(1) L⁻¹ 1

s

(2) L⁻¹ 1

s−a

(3) L⁻¹ 1

s²−9

(4) L⁻¹

1 (s−2)²+ 16

(5) L⁻¹ 1

2s−7

(6) L⁻¹

s s⁴+ 2s²a²+a⁴

(7) L⁻¹ 1

(s²+ 4)²

(8) L⁻¹

s²−a² (s²+a²)²

(9) L⁻¹

s² (s²+ 4)²

(10) L⁻¹

2s−5 (9s²−25)²

5.2. Multiplication by s

This method occurs when there is a (s) multiplication of X(s). Keep in mind that certain method sometimes overlaps another so that this (s) multiplication constitutes the most. To solve the inverse is due to the preference way of the test-taker, therefore

L⁻¹[sX(s)] = d

dtx_t+x₀ (31)

(1) L⁻¹ s

s²+ 1

(2) L⁻¹ s

4s²−25

(3) L⁻¹ 3s

2s+ 9

(4) L⁻¹ s

s+ 5

(5) L⁻¹ 2s

3s+ 6

(6) L⁻¹ s

2s²−1

(7) L⁻¹ s²

s²+β²

(8) L⁻¹

s²+ 4 s²+ 9

(18)

(9) L⁻¹

s² (s²+ 4)²

(10) L⁻¹ s

(s−3)²

5.3. Division by ¹_s

By contrast, ifX(s) is divided by sⁿ with n= 1, the inverse refers to the concept of integrator whereas ∀n≥2, it follows the Cauchy’s integral explained in Eq. (18),

L⁻¹ X(s)

s

= Z t

0

xtdt (32)

(1) L⁻¹ 1

s(s+ 2)

(2) L⁻¹

1 2s(s−3)

(3) L⁻¹

1 s(s²+ 1)

(4) L⁻¹

1 s(s²−16)

(5) L⁻¹

1 s(s²+α²)

(6) L⁻¹

1 s(s²−β²)

(7) L⁻¹

s²+ 3 s(s²+ 9)

(8) L⁻¹

s²+ 2 s(s²+ 4)

(9) L⁻¹

1 s²(s+ 1)

(10) L⁻¹

1 s³(s²+ 1)

5.4. First-Shifting Theorem

The first-shifting always emphasizes on the power in exponential (e^±ϕ), yielding the X(s∓ϕ) in its Laplace transform and that also happens in the inverse, such that

L⁻¹[X(s+a)] =e^−atxt (33) Example 12. Find the inverse Laplace transform of the following functions of X(s)!

(1) L⁻¹ 1

(s+ 2)⁵

(2) L⁻¹

s s²+ 4s+ 13

(3) L⁻¹

1 9s²+ 6s+ 1

(4) L⁻¹

s+ 8 s²+ 4s+ 5

(5) L⁻¹

s (s+ 3)²+ 4

(6) L⁻¹ s

(s+ 7)⁴

(7) L⁻¹

s+ 2 s²−2s−8

(8) L⁻¹

s s²+ 6s+ 25

(9) L⁻¹

1 2 (s−1)²+ 32

(10) L⁻¹

s−4 4 (s−3)²+ 16

(19)

5.5. Second-Shifting Theorem

In this method, to simplify it is important to note that the common way to write the shifted functionu(t−ϕ) is uϕ(t)

L⁻¹

e^−asF(s)

=xt−au(t−a) (34)

Example 13. Find the inverse Laplace transform of the following functions ofX(s) and from 6 to 10, sketch the function ofxt!

(1) L⁻¹ e^−πs

s+ 3

(2) L⁻¹

e^−s (s+ 1)³

(3) L⁻¹

6e^−2s s²+ 4

(4) L⁻¹

se^−s s²+ 9

(5) L⁻¹

se^−0.5s+πe^−s s²+π²

(6) L⁻¹ 3

s− 4e^−s

s² +5e^−2s s²

(7) L⁻¹ 2

s+ 3e^−s

s² −3e^−3s s²

(8) L⁻¹

"

1−e^−2s

1 +e^−4s s²

#

(9) L⁻¹ 4

s− 3e^−2s

s +2e^−2s s²

(10) L⁻¹ 6

s− 2e^−s

s² +2e^−3s

s² +2e^−3s s

5.6. Derivative of F(s)

To cope with this method, the six properties in (4.6) saying the derivative of trigonom- etry are the key. Keep in mind that the method being used makes of in the Eq. (35).

L⁻¹ d

dsF(s)

=−tL⁻¹[F(s)] =−t x_t L⁻¹[F(s)] =−1

tL⁻¹ d

dsF(s)

(35) Example 14. Find the inverse Laplace transform of the following functions of X(s)!

(1) L⁻¹

tan⁻¹1 s

(2) L⁻¹

lns²−1 s²

(3) L⁻¹

cot⁻¹(s+ 1) (4) L⁻¹

ln

1 +ω²

s²

(5) L⁻¹

s (s²+a²)²

(6) L⁻¹

s s⁴+s²+ 1

(7) L⁻¹

s+ 11 (s²+ 6s+ 13)²

(8) L⁻¹

tan⁻¹(s+ 1) (9) L⁻¹

sln s

√

s²+ 1+ cot⁻¹s

(10) L⁻¹ 1

2ln s²+β² (s−α)²

(20)

5.7. Partial Fractions Method

This method is powerful yet the idea lies in ability to seek the roots of certain poly- nomials and they could be independent and the same which uses different treatment.

L⁻¹

1

s²+ (α+β)s+αβ

=L⁻¹ A

s+α + B s+β

(36) Example 15. Find the inverse Laplace transform of the following functions of X(s)!

(1) L⁻¹

1 s²−5s+ 6

(2) L⁻¹

1 s²−7s+ 12

(3) L⁻¹

s−1 s²−6s+ 25

(4) L⁻¹

s+ 2 s²−4s+ 13

(5) L⁻¹

s+ 4 s(s−1) (s²+ 4)

(6) L⁻¹

3s+ 1 s(s−1) (s²+ 1)

(7) L⁻¹

s²

(s²+α²) (s²+β²)

(8) L⁻¹

s²+ 2s+ 6 s³

(9) L⁻¹

11s²−2s+ 5 2s³−3s²−3s+ 2

(10) L⁻¹

16 (s²+ 2s+ 5)²

5.8. Addendum of Inverse Theorems - Integral of F(s), Convolution, Heaviside Inverse Formula and Periodic Functions

This additional methods focus on these four methods, Integral, Convolution, Heavi- side inverse and Periodic functions with some supportive examples to strengthen the concepts. The first goes to theintegral formula, such that

L⁻¹ Z ∞

s

F(s)ds

= 1

tL⁻¹[F(s)] = xt

t L⁻¹[F(s)] =tL⁻¹

Z ∞ s

F(s)ds

(37) From Eq. (37), the inverse Laplace transform solution of the following functionX(s) is written as,

L⁻¹

2s (s²+ 1)²

=tL⁻¹ Z ∞

s

2s (s²+ 1)² ds

=tL⁻¹

− 1 s²+ 1

∞ s

=tL⁻¹

−0 + 1 s²+ 1

→tsint

The following is the convolution theorem as defined below. The individual inverse from the two xt and yt should be acknowledged before conducting the integral with certain functionτ other than the original (t), therefore

L⁻¹[X(s)·Y(s)] = Z t

0

x(τ)y(t−τ)dτ (38)

(21)

and the question to examine the method in Eq. (39) is

L⁻¹

s²

(s²+α²) (s²+β²)

where











X(s) = s

s²+α², xt= cosαt Y(s) = s

s²+β², yt= cosβt such that,

L⁻¹[X(s)·Y(s)] = Z t

0

cosατ cosβ(t−τ)dτ

= 1 2

Z t 0

[cos (ατ+βt−βτ) + cos (ατ−βt+βτ)]dτ

= 1 2

Z t 0

[cos (α−β)τ +βt]dτ+1 2

Z t 0

cos [(α+β)τ−βt]dτ

=

sin [(α−β)τ +βt]

2 (α−β) t

0

+

sin [(α+β)τ −βt]

2 (α+β) t

0

= sinαt−sinβt

2 (α−β) + sinαt+ sinβt

2 (α+β) → αsinαt−βsinβt α²−β²

The third isHeaviside formula of two functions ofX(s) andY(s). If the degree of the numeratorX(s) is less than that of denominator Y(s) and supposeϕ₁, ϕ₂, ϕ₃, . . . , ϕ_n become the roots of such denominatorY(s), then

L⁻¹ X(s)

Y(s)

=

n

X

i=1

X(ϕ_i)

Y⁰(ϕ_i)e^ϕⁱ^t (39) for instance,

L⁻¹

2s²+ 5s−4 s³+s²−2s

where











X(s) = 2s²+ 5s−4

Y(s) =s³+s²−2s−→Y⁰(s) = 3s²+ 2s−2 Roots of Y(s) = 0,1,−2 (n= 3)

such that,

L⁻¹ X(s)

Y(s)

= X(ϕ₁)

Y(ϕ1)e^ϕ¹^t+X(ϕ₂)

Y(ϕ2)e^ϕ²^t+X(ϕ₃) Y(ϕ3)e^ϕ³^t

= −4

−2e⁰+ 3

3e^t+−6

6 e^−2t →2 +e^t−e^−2t

The last comes to the Periodic function which is primarily based on the binomial series expansion of powern=−1,

(1−x)⁻¹ = 1 +x+x²+x³+· · ·+x^r+. . . (1 +x)⁻¹ = 1−x+x²−x³+· · ·+ (−1)^rx^r+. . .

(22)

With the question is presented below, its inverse Laplace transform would be L⁻¹

3 (1−e^−s) s(1−e^−3s)

=L⁻¹ 3

s 1−e^−s) 1 +e^−3s+e^−6s+e^−9s+. . .

= 3

s −3e^−s

s + 3e^−3s

s −3e^−4s

s +3e^−6s

s −3e^−7s s From that series, thext is

x_t= 3u(t)−3u(t−1) + 3u(t−3)−3u(t−4) + 3u(t−6)−3u(t−7) +. . . x_t=

(3, 0< t≤1

0, 1< t≤3 −→x_t+3=x_t

Figure 9.:Periodic function of the given question

Example 16. Find the inverse Laplace transform of the following functions ofX(s) using those four methods!

(1) L⁻¹

4s (s²−4)²

(I) (2) L⁻¹

1 s(s²+α²)

(C)

(3) L⁻¹

11s²−2s+ 5 2s³−3s²−3s+ 2

(H) (4) L⁻¹

1

2s² − 2e^−4s s(1−e^−4s)

(P)

6. Linear Systems of Differential Equations

This chapter emphasizes on the application of Laplace transform in solving various linear system in the form of differential equation. For example, solve ¨x+ 4 ˙x+ 13x = 2δ(t) where att= 0, x= 2 and ˙x₀= 0 and based on the given information, the Laplace transform is written as in the following,

L[¨x+ 4 ˙x+ 13x] =L[2δ(t)]

From this, the term we would like to operate is X(s) before switching back into the inverse Laplace transform. The term x0 should be considered and collected in the right-side of the equation and arranged as numerator and denominator. This is then inversed based on the insight from the preceding theorems

s²X(s)−sx₀−x˙₀

+ 4 [sX(s)−x₀] + 13X(s) = 2·1 s²+ 4s+ 13

X(s) = 2s+ 10

(23)

X(s) = 2s+ 10 s²+ 4s+ 13

= 2(s+ 2)

(s+ 2)²+ 9+ 6 (s+ 2)²+ 9 xt=L⁻¹

2(s+ 2)

(s+ 2)²+ 9+ 6 (s+ 2)²+ 9

= 2e^−2t(cos 3t+ sin 3t)

Another example is given in the following. Solve ¨x+ 5 ˙x+ 4x = 3δ(t−2) where at t = 0, x = 2 and ˙x₀ = −2. Compared to the previous, the ˙x₀ is not zero and the Laplace structure is defined below,

L[¨x+ 5 ˙x+ 4x] =L[3δ(t−2)]

After inputting the information, theX(s) is arranged as follows, s²X(s)−sx₀−x˙₀

+ 5 [sX(s)−x₀] + 4X(s) = 3e^−2s s²+ 5s+ 4

X(s) = 2s+ 8 + 3e^−2s such that the inverse Laplace transform is

X(s) = 2s+ 8 + 3e^−2s s²+ 5s+ 4

= 2(s+ 4)

(s+ 1) (s+ 4) +e^−2s 3 (s+ 1) (s+ 4) xt=L⁻¹

2

s+ 1+e^−2s 1

s+ 1− 1 s+ 4

= 2e^−t+e^−(t−2)u(t−2)−e^−4(t−2)u(t−2)

The last example with input impulse is presented by solving ¨x+ 6 ˙x+ 8x = 4δ(t−5) where at t= 0, x = 0 and ˙x₀ = 3. The Laplace transform and the collected X(s) is denoted as

L[¨x+ 6 ˙x+ 8x] =L[4δ(t−5)]

s²X(s)−sx₀−x˙₀

+ 6 [sX(s)−x₀] + 8X(s) = 4e^−5s s²+ 6s+ 8

X(s) = 3 + 4e^−5s and the inverse would be written in the following,

X(s) = 3 + 4e^−5s s²+ 6s+ 8

= 3

(s+ 2) (s+ 4) +e^−5s 4 (s+ 1) (s+ 4) xt=L⁻¹

3 2

1

s+ 2− 1 s+ 4

−2 e^−5s

s+ 2− e^−2s s+ 4

= 3

2 e^−2t−e^−4t + 2

e^−2(t−5)u(t−5)−e^−4(t−5)u(t−5)

(24)

These two instances involve more dynamic input other than impulse, such as (sin) and exponential (e). Solve the differential equation of ¨x+ 2 ˙x+ 2x= 5 sin twhere both x0

and ˙x₀ are zero, therefore the Laplace transform is written as L[¨x+ 2 ˙x+ 2x] =L[5 sint]

s²X(s)−sx0−x˙0

+ 2 [sX(s)−x0] + 2X(s) = 5 s²+ 1 s²+ 2s+ 2

X(s) = 5 s²+ 1

applying the partial fractions method, the inverse of suchX(s) is resulted in,

X(s) = 5

(s²+ 1) (s²+ 2s+ 2)

= −2s+ 1

s²+ 1 + 2s+ 3

s²+ 2s+ 2 (partial fraction, A and B) x_t=L⁻¹

−2s+ 1

s²+ 1 + 2s+ 3 s²+ 2s+ 2

=L⁻¹

−2s s²+ 1

+L⁻¹

1 s²+ 1

+L⁻¹

2(s+ 1) + 1 (s+ 1)²+ 1

=−2 cost+ sint+L⁻¹

2(s+ 1) (s+ 1)²+ 1

+L⁻¹

1 (s+ 1)²+ 1

= 2e^−tcost−e^−tsint−2 cost+ sin t

Another dynamic input is shown by proposing the combination of exponential (e) and (sin), such that ¨x+ 2 ˙x+ 5x=e^−tsin twherex₀= 0 and ˙x₀= 1, therefore the Laplace transform is formulated as

L[¨x+ 2 ˙x+ 5x] =L

e^−tsin t s²X(s)−sx0−x˙0

+ 2 [sX(s)−x0] + 5X(s) = 1 (s+ 1)²+ 1 s²+ 2s+ 5

X(s) = 1 + 1 (s+ 1)²+ 1

Likewise, implementing the partial fractions method, the inverse of theX(s) is yielded below

X(s) = s²+ 2s+ 3

(s²+ 2s+ 5) (s²+ 2s+ 2)

= 2 3

1

s²+ 2s+ 5+1 3

1 s²+ 2s+ 2 x_t=L⁻¹

2 3

1

s²+ 2s+ 5+ 1 3

1 s²+ 2s+ 2

= 1 3L⁻¹

2 (s+ 1)²+ (2)²

+1

3L⁻¹

1 (s+ 1)²+ 1

= 1

3e^−t(sin 2t+ sin t)

(25)

7. Harmonic Oscillators

Suppose a position at certain time (t) of a system is defined by the differential equation of xt where xt satisfies a¨xt+bxt = 0 with x0 = α and ˙x0 = β, then the Laplace transform of the system is denoted as

L[a¨xt+bxt] =L[0]

a

s²X(s)−sα−β

+bX(s) = 0 giving,

X(s) = sα+β as²+b X(s) =

sα a s²+ b

a +

β a s²+b

a

−→x_t= α acos

rb at+β

asin rb

at

The systems runs simple harmonic, the oscillatory motion of the following properties, such that

f = rb

a (rad/unit of time) T = 2π ra

b (period)

(Harmonic Oscillation) The following is the example to implement the formula above. Suppose the oscillatory harmonic motion being described as ¨x_t+ 16x_t= 0 with x0 = 1 and ˙x0= 0, then the Laplace and its inverse are shown below,

X(s) = s

s²+ 16 −→xt= cos 4t

where the frequency and the period are depicted in the following by considering the conversion of (Hz) to radian per unit time,

f = 4 (rad/unit of time) T = π 2

(Damped Motion) Consider the expression of the system 5¨xt+ 5 ˙xt+ 10xt= 0 with x₀ = 0 and ˙x₀= 4, then the Laplace transform and the inverse are

L[5¨xt+ 5 ˙xt+ 10xt] =L[0]

5

s²X(s)−4

+ 5 [sX(s)] + 10X(s) = 0

X(s) = 20

5s²+ 5s+ 10 = 4

s+1 2