Review of Laplace Transform
ECE generic        Han Q. Le (c)

Part 1

1 Introduction

1.1 Overview

This is only meant to be a review (a refresher) about Laplace Transform that you should have had in previous courses (example, Math3312 or ECE3337). It is quite important, hence we will invest some time to ensure we are on a good foundation.
Why are we interested in learning about Laplace transform (LT)?
- It is a very useful tool to solve certain class of linear differential equations (DE)
- Many physical problems involve the type of DE with an initial time condition for which, LT is the perfect tool to use: harmonic oscillator, damped oscillators, transient phenomena in circuits and problems in control engineering.
Our objectives in this chapter: 
- Learn LT as a tool, not necessarily as a mathematical topic: you don’t have to be a mathematician. You are learning it as an engineering tool. (Although any indepth mathematical understanding is of course, very valuable)
- How to use a tool properly: solve differential equations, especially with initial conditions.
- How to interprete the outcome of using the tool. Remember, this is about engineering, dealing with physical problems. Mathematical answers by themselves are NOT sufficient. One must interprete the mathematical result to describe the physical phenomenon.

1.2 Example 1 from circuit

Remember this? (LectSet 1)

We can write the mesh current Eq. for the right half after switch A is opened and B is closed: t>=0: 
(choose clockwise current loop)
                                     (Subscript[v, L])+(Subscript[v, C])=L ((di)/(dt))+(q/C)   =0  
Hence:                         L (((d^2)i)/(d(t^2))) +(1/C) ((dq)/(dt))=0               
Or:                                 L (((d^2)i)/(d(t^2)))+ (1/C) i=0   ⇔     (((d^2)i)/(d(t^2))) +(1/(L C)) i=0  
Define:              (1/(L C))=((Subscript[ω, 0])^2)
Then                                     (((d^2)i)/(d(t^2))) +((Subscript[ω, 0])^2) i=0  
The solution is             i=(Subscript[i, p]) (e^(i (Subscript[ω, 0]) t))+(Subscript[i, m]) (e^(-i (Subscript[ω, 0]) t)) 
Hence:                        (Subscript[v, out])=(Subscript[v, L])=L ((di)/(dt)) =i (Subscript[ω, 0]) L  ((Subscript[i, p]) (e^(i (Subscript[ω, 0]) t))-(Subscript[i, m]) (e^(-i (Subscript[ω, 0]) t)))  
To solve for a specific solution, we have to use the initial conditions:
1- For t<=0, from the left part of the circuit, we know that:
                                 (Subscript[-(Subscript[v, C])|, t=(SuperMinus[0])])=(Subscript[V, 0])  
Thus, after switch B is closed, the immediate output voltage:
                                 (Subscript[(Subscript[(Subscript[v, out])|, t=0])=(Subscript[v, L])|, t=0])=(Subscript[-(Subscript[v, C])|, t=(SuperMinus[0])])=(Subscript[V, 0])      
Hence:
                               (Subscript[(Subscript[v, L])|, t=(SuperPlus[0])])=i (Subscript[ω, 0]) L  ((Subscript[i, p]) -(Subscript[i, m]) ) =(Subscript[V, 0])     
then:                        ((Subscript[i, p]) -(Subscript[i, m]) ) =-i ((Subscript[V, 0])/((Subscript[ω, 0]) L))

2- We need another equation, and the initial condition is

Solve[(({{1, 1}, {1, -1}})) . (({{(Subscript[i, p])}, {(Subscript[i, m])}}))== (({{0}, {-I ((Subscript[V, 0])/((Subscript[ω, 0]) L))}})),{(Subscript[i, p]), (Subscript[i, m])}]

Hence:              (Subscript[v, out])=(Subscript[v, L])=L ((di)/(dt)) =i (Subscript[ω, 0]) L  ((Subscript[i, p]) (e^(i (Subscript[ω, 0]) t))-(Subscript[i, m]) (e^(-i (Subscript[ω, 0]) t)))

I (Subscript[ω, 0]) L  ((Subscript[i, p]) (E^(I (Subscript[ω, 0]) t))-(Subscript[i, m]) (E^(-I (Subscript[ω, 0]) t)))  /. {(Subscript[i, p])->-((I (Subscript[V, 0]))/(2 L (Subscript[ω, 0]))),(Subscript[i, m])->((I (Subscript[V, 0]))/(2 L (Subscript[ω, 0])))}

FullSimplify[%]

Which is to be expected:       (Subscript[v, out])=(Subscript[V, 0]) cos(t (Subscript[ω, 0]))

This problem is pretty straightforward and we can derive the solution without going through all the elementary steps above (just for the sake of thoroughness). 
But what if we have something slightly more complicated (not really, but relatively speaking) like the next example:

1.3 Example 2 from circuit

Now we see that we have 3 terms:
                  (Subscript[v, L])+(Subscript[v, R])+(Subscript[v, C])=L ((di)/(dt))+R i+(q/C)   =0  
                              L (((d^2)i)/(d(t^2)))+R ((di)/(dt))+ (1/C) i=0   
                   (Subscript[i, LT]) (L s+R )+(Subscript[i, LT]) (1/(C s))=(Subscript[V, 0])  
                 ((Subscript[V, 0])/(L s+R  +(1/(C s))))=

We will see how Laplace transform can be so useful for this type of problem. Although we will not get into details until in latter sections, we can have a brief view

(L s +R +(1/(C s)) )ℐ[s]  -((Subscript[V, 0])/s)=0  
           (L s +R +(1/(C s)) )ℐ[s]  =((Subscript[V, 0])/s)  
          ℐ[s]  =((Subscript[V, 0])/(s (L s +R +(1/(C s)) ))) =(C/(((s^2)/((Subscript[ω, 0])^2)) +τ s+1 ))  (Subscript[V, 0]) =((C  ((Subscript[ω, 0])^2))/((s^2) +τ  ((Subscript[ω, 0])^2) s+((Subscript[ω, 0])^2) ))   (Subscript[V, 0])

2 Basics of Laplace Transform Integral

2.1 Definition

Let x[t] be a real function (which usually represents some real physical quantity that is a function of time), the Laplace transform is defined as:
                  ℒ[x[t]]≡X[σ+i  ω]=(∫_0^∞) x[t]   (e^(-(σ+i  ω)  t))dt
                                          =(∫_0^∞) x[t]   (e^(-i  ω t -σ t))dt    ,    σ>0       (2.1a)
Notice that the integral starts from 0. This is known as single-sided LT. For many physical problems, x[t]=0 for t<0, i. e., the physical quantity is zero or constant before certain time. (If a constant, we just change the variable by subtracting this baseline constant).
Mathematically, LT can be extended for integration from -∞ to ∞ as well as long as the integral converges. This is known as double-sided LT.
More generally, we write:
                               ℒ[x[t]]≡X[s]=(∫_0^∞) x[t]   (e^(-s t))dt,  Re[s]>0.             (2.1b)
If Re[s]=0, we notice that the LT becomes Fourier transform (FT) as we can extend the integration to -∞ for the physical function that is 0 for t<0.

Illustration of Laplace transforms of some simple commonly used functions

2.1.1 Discussion

What are the applications of Laplace transform? Just like we have seen how useful FT is for analyzing steady state harmonic signals in numrous physical problems, LT is useful to deal with causal transient signals: signal with a known, definitive initial condition at a start time, which we can refer to as zero. We'll see that with LT, we can solve problems more easily:
1- Differential Eq. becomes algebraic manipulations
2- The total solution that involves both forced response (from an external circuit driver) and transient response can be obtained: both are included and one does not have to obtain separately the oscillatory part and exponential decay part.
3- As mentioned above, the initial conditions are included in the solutions.
Bottom line: it is a tool and technique that allows us solving these problems with ease!

2.1.2 Using Mathematica

Suppose we want to find the LT of the unit step function:

Plot[UnitStep[t],{t,-5,5}, PlotStyle-> {Thickness[0.02]}]

We can do this: 
                 ℒ[Θ[t]]≡(∫_0^∞)Θ[t](e^(-s t))dt=(∫_0^∞)(e^(-s t))dt
                                                 =-(1/s)(e^(-s t))(Subscript[|, 0]^∞)=(1/s)(1-(e^(-s ∞)))
                                                =(1/s) if (e^(-s t)) -> 0for t->∞,                   (2.2)
Or... we can just ask Mathematica to perform LT for us:

LaplaceTransform[UnitStep[t],t,s]

Notice also:

LaplaceTransform[1,t,s]

What does the complex function (1/s) look like

GraphicsGrid[{{Plot3D[Re[(1/(σ+I ω))],{σ,-5,5},{ω,-5,5}]
,Plot3D[Im[(1/(σ+I ω))],{σ,-5,5},{ω,-5,5}]}}]

GraphicsGrid[{{Plot3D[Abs[(1/(σ+I ω))],{σ,-5,5},{ω,-5,5}]
,Plot3D[Arg[(1/(σ+I ω))],{σ,-5,5},{ω,-5,5}]}}]

Another example:

Manipulate[Plot[{Cos[(Subscript[ω, 0]) t],Sin[(Subscript[ω, 0]) t]},{t,0,20}, PlotStyle-> {{Thick,Red},{Thick,Blue}}]
,{{(Subscript[ω, 0]),1},0,10}]

LaplaceTransform[Cos[(Subscript[ω, 0]) t],t,s]
LaplaceTransform[Sin[(Subscript[ω, 0]) t],t,s]

Let’s choose (Subscript[ω, 0])=1

GraphicsGrid[{{Plot3D[Abs[((σ+I ω)/((((σ+I ω))^2)+1))],{σ,-5,5},{ω,-5,5}]
,Plot3D[Arg[((σ+I ω)/((((σ+I ω))^2)+1))],{σ,-5,5},{ω,-5,5}]}}]

LaplaceTransform[((Cos[ω t])^2),t,s]
LaplaceTransform[((Sin[ω t])^2),t,s]

Let’s look at other functions:

LaplaceTransform[Log[t],t,s]

ℽ is known as Euler’s γ constant

N[EulerGamma,10]

LaplaceTransform[Sin[ω t] Log[t],t,s]

Finally, let’s look at this:

Manipulate[f[t_,α_,ω_]:=(E^(-α t)) Sin[ω t];
Plot[f[t,α,ω],{t,0,20}, PlotStyle-> {{Thick,Red},{Thick,Blue}}, PlotRange-> All]
,{{α,1},0,4},{{ω,1},0,10}]

This type of function is known as damped oscillator or damped harmonic.

LaplaceTransform[(E^(-α t)) Sin[ω t],t,s]

2.2 Inverse LT

Similar to FT with inverse FT, we can also obtain the inverse of LT:
                                                         (ℒ^(-1))[X[s]]=x[t]                          (2.3)
as follow. From Eq. (1.3a): 
                X[s]=(∫_0^∞) x[t]   (e^(-i  ω t -σ t))dt =(∫_(-∞)^∞) (x[t] u[t]  (e^(-σ t)) ) (e^(-i  ω t ))dt
Notice that this is the same as: 
                                          X[s]=ℱ[ (x[t] u[t]  (e^(-σ t)) ) ]    
Hence:                               x[t] u[t]  (e^(-σ t))=(ℱ^(-1))[ X[σ+i ω]  ]
Or:                                     x[t] u[t]  =(e^(σ t))(ℱ^(-1))[ X[σ+i ω]  ]                       (2.4a)
                                                         =(e^(σ t))(1/(2 π))(∫_(-∞)^∞) X[σ+i ω] (e^(i  ω t ))dω        (2.4b)
Since x[t]=0 for t<0, we can drop the u[t] term:
                                      x[t] =(1/(2 π))(∫_(-∞)^∞) X[σ+i ω] (e^((σ+i  ω)  t ))dω  with t>0
Substitute σ+i ω≡ s, with  dω=(1/i)ds
                                       x[t] =(1/(2 π i))(∫_(σ-i ∞)^(σ+i ∞)) X[s] (e^( s t ))ds                                   (2.5)
which is the inverse LT. Notice that the integration limit σ-i ∞ to σ+i ∞ is essentially a contour integration in the complex plane, which is a straight line parallel with the imaginary axis. In fact, contour integration is very useful to find the inverse LT. But we'll also see that we can just use rules and look up to obtain the inverse LT from simple functions.

2.2.1 Using Mathematica

Examples

InverseLaplaceTransform[(1/(s+a)),s,t]

InverseLaplaceTransform[(ω/((s^2)+(ω^2))),s,t]

InverseLaplaceTransform[(s/((s^2)+(ω^2))),s,t]

2.3 Convergence

Obviously, LT cannot be applied to solve problem where the integral (∫_0^∞) x[t]   (e^(-s t))dt does not exist, i. e. not converge. Thus, determine the conditions for the LT of a signal to exist is crucial. We think that (e^(-s t)) with Re[s]>0 is a very strong condition for integral convergence. But this is NOT guaranteed. x[t] can diverge, it can go to infinity, but its rate of divergence must be slower than that of (e^(-s t)) such that the integral converges.
So precisely, given a function x[t], how do we know when we have a valid LT? To put this question differently, let X[s] be the LT of x[t], do we know the range of value of s, i. e. the domain of s in the complex plane for which X[s] is valid? (i. e. the integral  (∫_0^∞) x[t] (e^(-s t))dt to converges?) We need to know this to obtain valid solutions to our problems. To examine this issue, let's consider a few examples.

2.3.1 Example 1

From 2.1.2 above:
                                                         ℒ[Θ[t]]=(1/s)  
The function (1/s) is analytic everywhere in the complex plane except at the pole at 0. But is it valid as a LT everywhere?
The answer is no. Obviously, Re[s] must be >0 for (e^(-s t)) to vamish as t->∞. Thus, the LT result  X[s]=(1/s) is valid ONLY on the real positive half of the plane, excluding the s=0 line and the pole at 0 (obviously).

2.3.2 Example 2

What is ℒ[(e^(-α t))]?

LaplaceTransform[(E^(-α t)),t,s]

We can also obtain it with straight forward integration:
                  ℒ[(e^(-α t))] =(∫_0^∞)(e^(-α t))(e^(-s t))dt=-(1/(s+α))(e^(-(s +α) t))(Subscript[|, 0]^∞)=(1/(s+α))(1-(e^(-(s +α) ∞)))   (2.6)
But here, the condition is Re[s +α]>0 or Re[s ]>-α. The domain of s now looks like this:

Bottom line: the LT of a function must include a specified domain where the LT is valid. 
Do we have to worry about this in solving physical problems? In many cases, no. But we must be aware to use physical common sense to obtain and interprete solutions that are physically meaningful, for example, a quantity can not grow exponentially to infinity, which means that the value of s of a solution cannot be in the invalid domain of LT.

2.4 Condition for convergence and absolute convergence

We can ask more generally what are the conditions for valid LT. Here, we have an important theorem about this. First, let's introduce the concept of absolute convergence.
A function x[t] is said to be absolutely convergent on the interval [0, ∞) iff: 
                                                       (∫_0^∞) |x[t]| dt  <∞
in other words, it is bounded.
It can be proven easily that if we have any function x[t] with the following behavior:
    i) there exist number L and K such that: (∫_0^L)|x[t]| dt  ≤ K<∞; and                              (2.2.7a)
    ii) |x[t]| ≤ A (e^(c t))  for t ≥ L                                                                                                (2.2.7b)
then, its Laplace transform integral will absolutely converge for Re[s]> c. c is often referred to as the abscissa of absolute convergence.
What this means is that as long as we can find a function bounded with (2.2.7a) and (2.2.7b) conditions, we can be assured that its LT exists at least for the right half of the plane beyond c. This applies even if x[t] can go to infinity any where. As discussed above, we don’t have to worry about it as long as our quantity x[t] is physically valid.

3 Laplace Transform of Important Functions

There are a number of functions we use in system analysis, and we often have to obtain their LT. See App above

3.1 Impulse (delta) function

What is the Laplace Transform of δ[t-(Subscript[t, 0])]?
If (Subscript[t, 0])<0, it is obvious that 
                                 ℒ[δ[t-(Subscript[t, 0])]] =(∫_0^∞)δ[t-(Subscript[t, 0])](e^(-s t))dt=0                                        (3.1a)
If (Subscript[t, 0])>0, then: 
                                 ℒ[δ[t-(Subscript[t, 0])]] =(∫_0^∞)δ[t-(Subscript[t, 0])](e^(-s t))dt=(e^(-s (Subscript[t, 0])))                                   (3.1b)
But what about (Subscript[t, 0])=0? There is ambiguity here because we cannot determine what happens at 0. So, we will choose a convention: the convention is that when we perform integration (∫_0^∞)x[t](e^(-s t))dt, the limit of 0 is on the left hand side, i. e. (SuperMinus[0]). With this, the result is simply:
                                 ℒ[δ[t]] =lim(∫_(SuperMinus[0])^∞)δ[t](e^(-s t))dt=1                                                (3.1c)

LaplaceTransform[DiracDelta[t-τ],t,s]

3.2 Step function (Heaviside function)

We have seen this example above:

Plot[UnitStep[t],{t,-5,5}, PlotStyle-> {Thickness[0.02]}]

ℒ[Θ[t]]≡(∫_0^∞)Θ[t](e^(-s t))dt=(1/s) if (e^(-s t)) -> 0for t->∞,                   (3.2)

LaplaceTransform[UnitStep[t],t,s]

3.3 Exponential function

Consider:                            f[t]=Θ[t] A (e^(-α t))                  (3.3)

Manipulate[Plot[(HeavisideThetaSeq[t]) (E^(-α t)) ,{t,-2,5}, PlotStyle-> {Thick,Red}, PlotRange-> {0,1}],{{α,1},0,5} ]

LaplaceTransform[(HeavisideThetaSeq[t]) (E^(-α t)),t,s]

ℒ[(HeavisideThetaSeq[t]) (e^(-α t))] =(1/(α+s))                                        (3.4)

The function (1/(α+s)) is just a rational of polynominals with a pole at -α. We will see that this form of LT functions are quite common in solving differential equations, and they can help finding solutions quite readily.

3.4 Ramp function

Consider:                            f[t]=Θ[t] a t                  (3.5)

Manipulate[Plot[(HeavisideThetaSeq[t]) a t ,{t,-1,1}, PlotStyle-> {Thick,Red}, PlotRange-> {0,1}],{{a,1},0,3} ]

LaplaceTransform[(HeavisideThetaSeq[t]) a t ,t,s]

ℒ[(HeavisideThetaSeq[t]) (e^(-α t))] =(a/(s^2))                                        (3.6)

Again, note that the LT is a polynomial rational.

3.5 Harmonic function

Consider:                            f[t]=Sin[ω t]                  (3.7a)
                                           g[t]=Cos[ω t]                  (3.7b)
                                           h[t]=(e^(i ω t))                         (3.7c)

Manipulate[Plot[{Cos[(Subscript[ω, 0]) t],Sin[(Subscript[ω, 0]) t]},{t,0,20}, PlotStyle-> {{Thick,Red},{Thick,Blue}}]
,{{(Subscript[ω, 0]),1},0,10}]

LaplaceTransform[Cos[(Subscript[ω, 0]) t],t,s]
LaplaceTransform[Sin[(Subscript[ω, 0]) t],t,s]
LaplaceTransform[(E^(I (Subscript[ω, 0]) t )),t,s]

Hence:
                                                         ℒ[Cos[(Subscript[ω, 0]) t]] =(s/((s^2)+(Subscript[ω, 0]^2)))            (3.8a)
                                                          ℒ[Sin[(Subscript[ω, 0]) t]] =((Subscript[ω, 0])/((s^2)+(Subscript[ω, 0]^2)))            (3.8b)
                                                          ℒ[(e^(i (Subscript[ω, 0]) t ))]        =(1/(s-i (Subscript[ω, 0])))           (3.8c)

What is:                                           (1/(s-i (Subscript[ω, 0])))±(1/(s+i (Subscript[ω, 0]))) =?

3.6 General transformed function

Suppose we know the LT of a given function f[t].
What if we perform some simple transformation of this function, such as;
        - translation:       Θ[t]  f[t] ->Θ[t-τ]  f[t-τ]  (time-shifted)                     (3.8)
        - scaling:              f[t] ->  f[(t/T)]  (changing the scale)            (3.9)
        - or a simple multiplication with an exponential:       f[t] ->  (e^(-α t ))f[t]            (3.10)
Can we find the LT of these transformed functions?
The answer is yes. It is not too complicated

3.6.1 Translated function

ℒ[Θ[t-τ] f[t-τ]]=(∫_0^∞) Θ[t-τ] f[t-τ] (e^(-s t))dt
                                                           =(e^(-s τ))(∫_0^∞) Θ[t-τ] f[t-τ] (e^(-s (t-τ)))dt
                                                           =(e^(-s τ))(∫_(-τ)^∞) Θ[t'] f[t'] (e^(-s t'))dt'   
                                                           =(e^(-s τ)) ((∫_(-τ)^0) Θ[t'] f[t'] (e^(-s t'))dt'+(∫_0^∞) Θ[t'] f[t'] (e^(-s t'))dt') 
If τ ≥ 0:              ℒ[Θ[t-τ] f[t-τ]]=(e^(-s τ)) (0+(∫_0^∞) Θ[t'] f[t'] (e^(-s t'))dt')
                                       =(e^(-s τ)) ℒ[Θ[t] f[t]]                 (3.11)                                                However, if τ<0:
                           ℒ[Θ[t-τ] f[t-τ]]=(e^(-s τ)) ((∫_(-τ)^0) Θ[t'] f[t'] (e^(-s t'))dt'+(∫_0^∞) Θ[t'] f[t'] (e^(-s t'))dt')
                                       =(e^(-s τ)) (-(∫_0^(-τ)) Θ[t'] f[t'] (e^(-s t'))dt'+(∫_0^∞) Θ[t'] f[t'] (e^(-s t'))dt')
                                       =(e^(-s τ)) (-(∫_0^(-τ)) f[t'] (e^(-s t'))dt'+(∫_0^∞) Θ[t'] f[t'] (e^(-s t'))dt') 
Or:
                           ℒ[Θ[t-τ] f[t-τ]]= (e^(-s τ)) (ℒ[Θ[t] f[t]] -(∫_0^(-τ)) f[t'] (e^(-s t'))dt')      (3.12)

f[t_]:=0.01*Cos[ 0.5 t]((((t^5)+3(t^4)-12(t^3)+2 (t^2)-15 t+3 ))/(Cosh[0.8 t])) ;
Manipulate[Plot[{f[t-τ],UnitStep[t-τ],f[t-τ]*UnitStep[t-τ]}, {t,-20,20}, PlotStyle->{ {Thick,Red},{Thick,Green},{Thick,Dashed,Blue}}, PlotRange-> All]
,{{τ,0},-10,10}]

3.6.2 Scaled function

ℒ[ f[(t/T)]]=(∫_0^∞)f[(t/T)] (e^(-s t))dt                      (T≠0)
                                                           =(∫_0^∞)  f[(t/T)] (e^(-s T ((t/T))))T d(t/T)
                                                           =T(∫_0^∞)  f[t'] (e^(-(s T) t'))dt'
                                                           =T ℒ[f[t]][s T] =T F[s T]                (3.13)

where F[s] is the LT of f[t]. This means that after we obtain the LT, we substitute s by sT.

3.6.3 Multiplied by an exponential (s-shift or complex frequency shift)

ℒ[ (e^(-α t))f[t]]=(∫_0^∞)(e^(-α t))f[t] (e^(-s t))dt                          
                                                           =(∫_0^∞)  f(e^(-(s+α) t))f[t] (e^(-s t))dt
                                                           =ℒ[f[t]][s +α] =F[s +α]                (3.14)

where F[s] is the LT of f[t]. This means that after we obtain the LT, we substitute s by s+α.

F[s_]:=LaplaceTransform[ ((t^3)-3 t+1) Cos[ω t], t,s]

F[s]

LaplaceTransform[ (E^(-α t)) ((t^3)-3 t+1) Cos[ω t], t,s]

F[s+α]

Another example, we recall the behavior of damped oscillator:
                                       x[t] =(e^(-α t ))Sin[ω t]
The LT is:
                             ℒ[ (e^(-α t ))Sin[ω t]] =(ω/((((s+α))^2)+(ω^2)))

4 Laplace Transform Theorems

There are a number of LT theorems that are useful to find the LT of various functions without excessive work. Below is a summary of these theorems.

4.1 Linear superposition:

ℒ[(Subscript[a, 1]) (Subscript[f, 1])+(Subscript[a, 2]) (Subscript[f, 2])] = (Subscript[a, 1])  ℒ[(Subscript[f, 1])]  +(Subscript[a, 2]) ℒ[(Subscript[f, 2])]
                                      = (Subscript[a, 1])  (Subscript[F, 1])[s]  +(Subscript[a, 2]) (Subscript[F, 2])[s]             (4.1)

LaplaceTransform[(E^(I ω  t)),t,s]

ℒ[(1/2)((e^(i ω  t))±(e^(-i ω  t)))]=(1/(s-i ω))±(1/(s+i ω))=((s or i ω)/((s^2)+ (ω^2)))
      or:             ℒ[Cos[ω  t]]=((s )/((s^2)+ (ω^2)))  ; ℒ[Sin[ω  t]]=((ω )/((s^2)+ (ω^2)))           (4.2)

4.2 Differentiation:

ℒ[ ((d^n)/(d(t^n)))f[t] ] = (s^n) F[s]-(∑_(k=1)^n)(s^(n-k))(f^((k-1)))[(SuperMinus[0])]                                   (4.3) 
Or                                 = (s^n) F[s]-(s^(n-1))f[(SuperMinus[0])]-(s^(n-2))(f^((1)))[(SuperMinus[0])]-...(f^((n-1)))[(SuperMinus[0])]
This theorem turns out to be the most useful theorem of the LT for differential equations. It convert all the derivatives into just polynomials, with the coefficients being the initial conditions of the function f[t].
Let’s recall Fourier transform, the equivalent theorem is just:
                                     ℱ[ ((d^n)/(d(t^n)))x[t] ] = (((i 2 π f))^n)χ[f]                     (4.4)
which does not depend on any point in time of x[t], eventhough for both, the theorem involves integration by part.

The reason is that for single-sided LT, the integration starts at 0:
           ℒ[ (d/(dt))x[t] ]=(∫_0^∞)((dx[t])/(dt)) (e^(-s t))dt=x[t] (e^(-s t))(Subscript[|, (SuperMinus[0])]^∞) +(∫_0^∞) x[t] s (e^(-s t))dt
                                                             =-x[(SuperMinus[0])]  +s(∫_0^∞) x[t] (e^(-s t))dt
                                                             =s χ[s]-x[(SuperMinus[0])]                             (4.5)
We can prove that, using integration by part again:
                                ℒ[ ((d^n)/(d(t^n)))x[t] ]=s ℒ[ ((d^(n-1))/(d(t^(n-1))))x[t] ]-(x^((n-1)))[(SuperMinus[0])]                (4.6)
Thus, by recursive proof using (4.5) and (4.6), the result in (4.3) is obtained.

Example1: an electrical circuit (transient response)

This example shows why LT is useful for problem with an initial condition, and find a transient response of the current i[t] or the output y[t] as the switch is activated.
The circuit KVL is:     x[t]=L((di)/(dt)) +R i[t]               (E4.1)
Take the LT of both side for t ≥ 0:
                                    0=L (s ℐ[s]- i[(SuperMinus[0])]) +R ℐ[s]   
                                    i[(SuperMinus[0])]=(s+(R/L) ) ℐ[s]            (E4.2)
Thus, we can solve for ℐ[s]:
                                       ℐ[s]=((i[(SuperMinus[0])])/((s+(R/L) )))                     (E4.3)
We can find the inverse Laplace transform:

InverseLaplaceTransform[(i0/((s+(R/L) ))), s,t]

We need only to determine i[(SuperMinus[0])] to obtain the complete solution, including the initial condition. For t<0, assume that we have a steady state i[t]=constant and  x[t]=V, hence (E3.1) is:
                                        V=R i[t]  , or    i[(SuperMinus[0])] =(V/R) 
Thus, the solution of the problem is:
                       i[t]=(V/(R ))(e^(-((R t)/L)))   ; or    y[t]=V (e^(-((R t)/L)))          (E4.4)

It appears that the ODE:   0=L((di)/(dt)) +R i[t]  is pretty simple, with an obvious solution:
                                             i[t]=(Subscript[C, 1]) (e^(-((R t)/L))) + (Subscript[C, 2])  
so, why do we bother with the LT? The answer is that the LT can be applied in much more complex ODE and at the same time, give solutions with unknown coefficients such as (Subscript[C, 1]) and (Subscript[C, 2]) embedded. In other words, we don't have to perform extra work to solve for the boundary (or initial) condition. This is the convenience of the LT method.

4.3 Integration:

Let                                       g[t]=(∫_(-∞)^t)f[t']dt'                      (4.7)
                                    ℒ[ g[t]] ≡ℒ[ (∫_(-∞)^t)f[t']dt'  ] = ((F[s] )/s)+((g[(SuperMinus[0])])/s)        (4.8)
This can be obtained in the same integration by part manner.

Example 2

Here, we consider a sudden switch on (from off) at t=0. We want to find the response of the circuit, which includes transient response.
The circuit KVL is:     x[t]=L((di)/(dt)) +((q[t])/C)+R i[t]    
                                    x[t]=L((di)/(dt)) +(1/C)(∫_(-∞)^t)i[t']dt'  +R i[t]            (E4.5)
Take the LT of both side:
                      χ[s] =L (s ℐ[s]-i[(SuperMinus[0])])+(1/C)(((ℐ[s] )/s)+((q[(SuperMinus[0])])/s)) +R ℐ[s]   
                      χ[s] =ℐ[s] (L s +(1/(C s))+R) -L i[(SuperMinus[0])] +((q[(SuperMinus[0])])/C) (1/s)          (E4.6)
We now insert initial conditions:
                           i[(SuperMinus[0])]=0   ;      ((q[(SuperMinus[0])])/C)=(Subscript[v, C])[(SuperMinus[0])]                                  (E4.7)
Then, we can solve for ℐ[s]:
                                       ℐ[s]=((s χ[s] -(Subscript[v, C])[(SuperMinus[0])])/((L (s^2) +R s +(1/C))))                              (E4.8)
To find the inverse Laplace transform, we need to know χ[s] and (Subscript[v, C])[(SuperMinus[0])].
Let's consider some simple examples:
Case a.  (Subscript[v, C])[(SuperMinus[0])]=0, and x[t] constant V. As we know,
                         ℒ[V]= (V/s)                                                       (E4.9)
We will further define some parameter:   L C=(1/((Subscript[ω, 0])^2)), and (R/L)=(1/τ)    (E4.10)
Then   (E3.8) becomes:      ℐ[s]=(V/(L ( (s^2) + s /τ+((Subscript[ω, 0])^2))))                      (E4.11)

InverseLaplaceTransform[(1/( ( (s^2) + s /τ+((Subscript[ω, 0])^2)))),s,t]

FullSimplify[%]

Or, to simplify it further, we define:
                        (Sqrt[1-4 (τ^2) (Subscript[ω, 0]^2)])=2 α τ                    (E4.12)

FullSimplify[((τ (E^(-((t ((Sqrt[1-4 (τ^2) (Subscript[ω, 0]^2)])+1))/(2 τ)))) ((E^((t (Sqrt[1-4 (τ^2) (Subscript[ω, 0]^2)]))/τ))-1))/(Sqrt[1-4 (τ^2) (Subscript[ω, 0]^2)])) /. (Sqrt[1-4 (τ^2) (Subscript[ω, 0]^2)]) -> 2 α τ]

i[t]=((τ ((e^(2 α t))-1) (e^(-t (α+(1/(2 τ))))))/(2 α τ))
        = (V/(L  α))(e^(-(t/(2 τ))))  sinh(α t)                        (E4.13)

Case b. Let (Subscript[v, C])[(SuperMinus[0])]=0, and x[t]= V Cos[ω t]
We know that (from above)    X[s]=V ℒ[Cos[ω t]]=V (s/((s^2)+(ω^2)))  
Thus, from (E3.8)                   ℐ[s]=((V (s^2))/((L (s^2) +R s +(1/C))((s^2)+(ω^2))))
                                               ℐ[s]=((V (s^2))/(L ( (s^2) + s /τ+((Subscript[ω, 0])^2))((s^2)+(ω^2))))

InverseLaplaceTransform[ ((s^2)/( ( (s^2) + s /τ+((Subscript[ω, 0])^2)) ((s^2)+(ω^2)))), s,t]

We notice here that eventhough Mathematica can give us inverse LT, the form is not necessarily convenient, and we have to simplify, which requires understanding. There are two basic steps of the technique:
- Partial fraction expansion that reduce a complicated rational function into a simple form, e. g. 
                          ((s^2)/( ( (s^2) + s /τ+((Subscript[ω, 0])^2)) ((s^2)+(ω^2))))=(∑_k)((Subscript[c, k;M])/(((s-(Subscript[α, k])))^(Subscript[M, k])))                    (E4.14)
where (Subscript[α, k]) are the poles; 
- Then, use contour integral in the complex plain to obtain the inverse LT for each simple fraction term. In practice, all we need to do is to look up in a table. The technique of finding inverse LT is quite simple for simple rational function and there is no need to repeat in each and every little problem.

Solve[ ((s^2)+(ω^2))==0,s]

Solve[ ( (s^2) + s /τ+((Subscript[ω, 0])^2))==0 ,s]

Apart[((s^2)/( (s-(Subscript[α, 1])) (s-(Subscript[α, 2]))  (s-I ω)(s+I ω))),s]

InverseLaplaceTransform[((Subscript[α, 1]^2)/(((Subscript[α, 1])-(Subscript[α, 2])) (ω-I (Subscript[α, 1])) (ω+I (Subscript[α, 1])) (s-(Subscript[α, 1]))))+((I ω)/(2 (ω-I (Subscript[α, 1])) (ω-I (Subscript[α, 2])) (s+I ω)))-((I ω)/(2 (ω+I (Subscript[α, 1])) (ω+I (Subscript[α, 2])) (s-I ω)))+((Subscript[α, 2]^2)/(((Subscript[α, 2])-(Subscript[α, 1])) (ω-I (Subscript[α, 2])) (ω+I (Subscript[α, 2])) (s-(Subscript[α, 2])))),s,t]

We see that it is in a neater form.

4.4 Convolution:

ℒ[ (Subscript[f, 1]) ⊗ (Subscript[f, 2])] =  (Subscript[F, 1])[s] (Subscript[F, 2])[s]                        (4.9)
The requirement that (Subscript[f, 1])[t] and (Subscript[f, 2])[t] must vanish for t<0 is essential. This is the most important theorem used in linear time-invariant system analysis.

4.5 Multiplication (Product):

This is the reverse of convolution. Although we can express the LT of a product of two functions similarly to a convolution, it involves complex integration and we can explicitly express it:
                  ℒ[ (Subscript[f, 1])[t]  (Subscript[f, 2])[t]] =  (1/(2 π i))(∫_(Re[λ]-i∞)^(Re[λ]+i∞))(Subscript[F, 1])[λ] (Subscript[F, 2])[s-λ]dλ          (4.10)                         
Notice here that the integration is in the complex plane where both  (Subscript[F, 1])[λ]  and (Subscript[F, 2])[s-λ] are defined and the integral exists.

4.6 Limit values:

There are two important theorems on limit values that are useful for problems with initial conditions:
   a. Initial value theorem:  (lim_(s-> ∞)) s F[s] =f[(SuperPlus[0])] if f[(SuperPlus[0])] exists;    (4.11)
and:
   b. Final value theorem:  (lim_(s-> 0)) s F[s] =f[∞] if  f[∞] exists.        (4.12)
   (or s F[s]  has no poles on the Im axis and analytic in the RHP)
Here, we denote: f[(SuperPlus[0])]=(Underscript[ lim, t-> (SuperPlus[0])]) f[t] and f[∞]=(Underscript[ lim, t-> ∞]) f[t]                   (4.13)

4.7 Lookup table of useful LT based on LT theorems

There are numerous tables of LT available on the web, but you can use Mathematica as a table.