1.1 Introduction
The Laplace transform, named after its inventor Pierre-Simon Laplace, is an integral transform. Laplace transform converts a function of a real variable (often time) to a function of a complex variable (complex frequency).
Laplace transforms are useful to solve linear ordinary differential equations and analyze frequency response and stability analysis. Process control feedback loops and their response properties and stability can thus be conveniently analyzed in the Laplace domain.
Also, depending on the boundary conditions of your problem, it can be judicious to use a Laplace transform to solve the diffusion equation, heat transfer equation, and Navier–Stokes. So, Laplace transforms will show up in many core engineering curricula, for example, mass transport, heat transport, fluid transport, and process controls. For the basic concepts of Laplace transform and its applications, one can refer to [2–6,8,9,13,14].
1.2 Definition of Laplace transforms
As we have explained in Introduction, the integral transform of functionft in the intervala≤x≤b is given by,
(1.1)ℓft=∫abKt,sftdt=∘ˆs,
where−∞≤a<b≤∞
Letft be a function defined fort≥00≤t≤∞ and kernelKt,s=e−st and then eq. (1.1) will be called Laplace transform under certain conditions to be explained later.
Thus, the Laplace transform of functionft which is defined fort≥0 (t is real variable), is formally defined as follows:
Lft=∫0∞e−stftdt=Fs, Res>0
wheres is the transform variable, which is a complex number.
Figure 1.1: The Laplace transforms as a mapping.
Therefore, the Laplace transform converts time domain functions and operations into frequency domainft→Fs t∈R, s∈CasshowninFigure1.1.
1.3 Laplace transform of some elementary functions using definition
Example 1.1
Find the Laplace transformft=1 fort≥0.
Solution
We know thatLft=∫0∞e−stftdt
Here,ft=1:
∴ Lft=1=∫0∞e−st1dt
⇒L1= e−st−st=0t=∞
⇒L1=1−slimt→∞e−st−1
⇒L1=1−s0−1=1s=Fs Thus,L1=1s, s>0.
Example 1.2
Find the Laplace transform off(t)=eat fort≥0, where “a” is a constant.
Solution
We know thatLft=∫0∞e−stftdt
Heref(t)=eat
∴Lft=eat=∫0∞e−steatdt
⇒ Leat= ∫0∞e−s−at dt
⇒ Leat= e−s−at−s−at=0t=∞
⇒Leat=1−s−alimt→∞e−s−at−1
⇒Leat=1−s−a0−1=1s−a=Fs
Thus,Leat=1s−a, s>a.
Note: Similarly,Le−at=1s+a, s>−a.
Example 1.3
Find the Laplace transform offt=sinat, where “a” is a real constant.
Solution
We know that
sinat=eiat−e−iat2i
Now,ft=sinat andLft= ∫0∞e−stft dt
∴Lft=sinat=∫0∞e−stsinatdt
⇒ Lsinat= ∫0∞e−st eiat−e−iat2i dt
=12i∫0∞e−steiat−e−ste−iatdt=12i∫0∞e−s−iat−e−s+iatdt=12ie−s−iat−s−ia+e−s+iats+iat=0t=∞=12i1s−ia−1s+ia=12is+ia−s+ias−ias+ia=as2−ia2∵i2=−1=as2+a2
Thus,Lsinat=as2+a2.
Example 1.4
Find the Laplace transform offt=cosat, where “a” is a real constant.
Solution
We know that
cosat=eiat+e−iat2
Now,ft=cosat and Lft=∫0∞e−stft dt ,
∴Lft=cosat=∫0∞e−stcosatdt
Lft=cosat=∫0∞e−stcosatdt⇒Lcosat=∫0∞e−steiat+e−iat2dt=12∫0∞e−steiat+e−ste−iatdt=12∫0∞e−s−iat+e−s+iatdt=12e−s−iat−s−ia−e−s+iats+iat=0t=∞=121s−ia+1s+ia=12s−ia+s+ias−ias+ia=ss2−ia2i2=−1=ss2+a2
Thus,Lcosat=ss2+a2.
1.3.1 Linearity of Laplace transforms
Functionsf1t andf2t have Laplace transformsF1s andF2s, respectively.
Also, ifc1 andc2 are any constants, then,
Lc1f1t+c2f2t=c1L[f1t+c2Lf2t]=c1F1s+c2F2s
Proof. We know that
{Lft= ∫0∞e−stft dt=Fs
∴Lf1t=∫0∞e−stf1tdt=F1s
Lf2t= ∫0∞e−stf2t) dt=F2s
Now,
Lc1f1t+c2f2t=∫0∞e−stc1f1t+c2f2tdt=∫0∞e−stc1f1t+e−stc2f2tdt=∫0∞e−stc1f1t+e−stc2f2tdt=c1∫0∞e−stf1tdt+c2∫0∞e−stf2tdt=c1L[f1t+c2Lf2t]=c1F1s+c2F2s
Example 1.5
Find the Laplace transform offt=sinhat, where “a” is a real constant.
Solution
We know that
sinhat=eat−e−at2
Now,
Lsinhat= Leat−e−at2=12Leat−Le−at=12 1s−a−1s+a=12 s+a−s+as−as+a= as2−a2
(by linearity property)
Thus,Lsinhat=as2−a2
Example 1.6
Find the Laplace transform offt=coshat, where “a” is a real constant.
Solution
We know that
coshat=eat+e−at2
Now,
Lcoshat= Leat+e−at2
=12Leat+Le−at =12 1s−a+1s+a=12 s+a+s−as−as+a=ss2−a2
(by linearity property)
Thus,Lcoshat=ss2−a2.
Note
The gamma function is defined by the improper integral∫0∞e−xxn−1 dx=|n‾ for n>0
Ifn is a positive integer, then|n+1‾=n!
|n+1‾=n|n‾
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