2.1: Laplace Transforms

Learning Objectives

After completing this section, you will be able to:

1. Provide definitions of the Laplace transform and inverse Laplace transform.
2. Compute Laplace transforms for given time-domain functions.
3. Compute inverse Laplace transforms for given frequency-domain functions.

The Laplace transform is an integral transform that converts a function of a real value variable in the time domain to a function of a complex variable in the frequency domain, also known as the s-domain. The transform has many applications in engineering, especially as a tool for solving linear ordinary differential equations (ODEs). The main advantage of using the Laplace transform to solve ODEs is that it converts ODEs into algebraic equations, making them easier to solve. For a given real variable function \(f(t)\) in the time domain, the Laplace transform is defined as follows.

\[F\left(s\right)\triangleq L\left[f\left(t\right)\right]=\int^{\infty }_0{f\left(t\right)e^{-st}dt} \label{2.1}\]

Here, \(F\left(s\right)\) is the Laplace transform of \(f\left(t\right)\) and \(s\) is a complex variable. The inverse Laplace transform converts a complex value function in the frequency domain into a real value function in the time domain defined as follows.

\[f\left(t\right)\triangleq L^{-1}\left[F\left(s\right)\right]=\dfrac{1}{2πi}{\mathop{\mathrm{l}\mathrm{im}}_{T\to \infty } \int^{γ+iT}_{γ-iT}{e^{st}F\left(s\right)ds}\ } \label{2.2}\]

Here \(γ\) is a real variable. The integration is performed along the vertical line of \(Re(s)\ =γ\) in the complex plane such that \(γ\) is greater than the real part of all singularities of \(F(s)\) and \(F(s)\ \)is bounded on the line. If all singularities are in the left half-plane, then \(γ\) can be set to zero; the above inverse integral formula then becomes identical to the inverse Fourier transform. In practice, this complex integral can be calculated by using the Cauchy residue theorem [13].

Example 2.1

Consider a unit step function \(u\left(t\right)\) defined below.

\[u\left(t\right)\triangleq \left\{ \begin{array}{cc} 0, & t<0 \\ 1, & t\ge 0 \end{array} \right. \label{2.3}\]

Its Laplace transform can be calculated by

\[U\left(s\right)\triangleq L\left[u\left(t\right)\right]=\int^{\infty }_0{e^{-st}dt}=\dfrac{-1}{s}{\left.e^{-st}\right|}^{\infty }_0=\dfrac{1}{s} \label{2.4}\]

Computing the inverse Laplace transform \(U(s)\) is a little complicated based on equation \ref{2.2}. By applying the Cauchy residue theorem [13], we have

\[u\left(t\right)\triangleq L^{-1}\left[U\left(s\right)\right]=\dfrac{1}{2πi}{\mathop{lim}_{T\to \infty } \int^{γ+iT}_{γ-iT}{e^{st}F\left(s\right)ds}\ }=Res\left(\dfrac{1}{s}\right)=1,\ \ t\ge 0 \label{2.5}\]

Because the Laplace transform is defined only over \(t\ge 0\), for \(t<0\), we can set \(u\left(t\right)=0\), resulting in the original unit step function.

Calculating the inverse Laplace transform is relatively complicated when using definition from equation \ref{2.2}. However, in practice, the inverse Laplace transform is often based on the Laplace transforms seen in Table 2.2 and the Laplace transform properties seen in Table 2.3, with the help of fractional expansion. The unit impulse function \(δ(t)\) is defined in equation \ref{2.6}.

\[δ\left(t\right)\triangleq \left\{ \begin{array}{cc} \infty , & 0\le t\le 0^+ \\ 0, & \mathrm{e}\mathrm{lsewhere} \end{array} \right.\]

and

\[ \int^{\infty }_0{δ\left(t\right)dt}=\int^{0^+}_0{δ\left(t\right)dt}=1 \label{2.6}\]

Example 2.2

Consider a Laplace transform \(F\left(s\right)\), defined below, that can be expanded using the partial fraction expansion technique.

\[F\left(s\right)=\dfrac{1}{(s+1)(s+2)}=\dfrac{1}{s+1}-\dfrac{1}{s+2} \label{2.7}\]

Table 2.2: Laplace transforms of typical time-domain functions.

Description	\(f\left(t\right),\ \ t\ge 0\)	\(F\left(s\right)\triangleq L\left[f\left(t\right)\right]\)
Unit impulse	\(δ\left(t\right)\)	\(1\)
Unit Step	\(u\left(t\right)\)	\(\dfrac{1}{s}\)
Unit ramp	\(t\)	\(\dfrac{1}{s^2}\)
\(n^{th}\) order ramp	\(t^n\)	\(\dfrac{n!}{s^{n+1}}\)
Exponential	\(e^{at}\)	\(\dfrac{1}{s-a}\)
Sinusoid	\({\mathrm{s}\mathrm{in} ωt\ }\)	\(\dfrac{ω}{s^2+ω^2}\)
Cosine	\({\mathrm{c}\mathrm{os} ωt\ }\)	\(\dfrac{s}{s^2+ω^2}\)

Its inverse Laplace transform can be calculated by using the exponential property in Table 2.3.

\[f\left(t\right)\triangleq L^{-1}\left[\dfrac{1}{s+1}-\dfrac{1}{s+2}\right]=L^{-1}\left[\dfrac{1}{s+1}\right]-L^{-1}\left[\dfrac{1}{s+2}\right]=e^{-t}-e^{-2t} \label{2.8}\]

Example 2.3

Consider the Laplace transform function \(F\left(s\right)\) defined below.

\[F\left(s\right)=\dfrac{e^{-2s}}{s+2} \label{2.9}\]

Its inverse Laplace transform can be calculated by using the time-shift (“t” shift) property in Table 2.3.

\[f\left(t\right)\triangleq L^{-1}\left[\dfrac{e^{-2s}}{s+2}\right]={\left.e^{-2τ}\right|}_{τ=t-2}\]

\[u(t-2)=u(t-2)e^{-2(t-2)} \label{2.10}\]

where \(u(t-2)\) is a unit step function with its time shifted by 2 seconds. That is,

\[ u\left(t-2\right)\triangleq \left\{ \begin{array}{cc} 0, & t<2 \\ 1, & t\ge 2 \end{array} \right.\label{2.11}\]

One of the Laplace transform applications is to solve ordinary differential equations (ODEs), as seen in Example 2.4.

Table 2.3: Laplace transform properties.

Linearity	\(L\left[af\left(t\right)+bg\left(t\right)\right]=aL\left[f\left(t\right)]+bL[g\left(t\right)\right]\)
“\(s\)” shift property	\(L\left[e^{at}f\left(t\right)\right]=F\left(s-a\right)={\left.L[f\left(t\right)]\right|}_{s=s-a}\)
“\(t\)” shift property	\(L\left[f\left(t-a\right)u(t-a)\right]=e^{-as}F\left(s\right)=e^{-as}L[f\left(t\right)]\)
Transform of derivatives	\(L\left[\dot{f}(t)\right]=sF\left(s\right)-f(0)\)\(L\left[\ddot{f}(t)\right]=s^2F\left(s\right)-sf\left(0\right)-\dot{f}(0)\)\(L\left[\mathop{f}\limits^{...}(t)\right]=s^3F\left(s\right)-s^2f\left(0\right)-s\dot{f}\left(0\right)-\ddot{f}(0)\)\(\vdots\)
Transform of integrals	\(L\left[\int^t_0{f\left(τ\right)dτ}\right]=\dfrac{1}{s}L[f\left(t\right)]=\dfrac{1}{s}F(s)\)
Convolution	\(L^{-1}\left[F\left(s\right)G\left(s\right)\right]=\int^t_0{f\left(τ\right)g\left(t-τ\right)dτ}=\int^t_0{g\left(τ\right)f\left(t-τ\right)dτ}\)

Example 2.4

Consider the following ODE with its initial condition \(\dot{x}\left(0\right)=x\left(0\right)=0\).

\[\ddot{x}\left(t\right)+3\dot{x}\left(t\right)+2x(t)=f(t) \label{2.12}\]

Applying the Laplace transform to both sides of equation \ref{2.12} and using the derivative property in Table 2.3 yields the following.

\[X\left(s\right)(s^2+3s+2)=F(s) \label{2.13}\]

or

\[X\left(s\right)=\dfrac{F(s)}{s^2+3s+2} \label{2.14}\]

Assuming \(F\left(s\right)=1\) to be a unit impulse (\(f\left(t\right)=δ\left(t\right)=L^{-1}[F(s)])\), the time response \(x\left(t\right)=L^{-1}[X(s)]\) can be solved by using the inverse Laplace transform below, as based on Example 2.2.

\[L^{-1}\left[X\left(s\right)\right]=L^{-1}\left(\dfrac{1}{s^2+3s+2}\right)=L^{-1}\left(\dfrac{1}{s+1}-\dfrac{1}{s+2}\right)=e^{-t}-e^{-2t} \label{2.15}\]

Laplace transforms significantly simplify the process of solving ODEs; they convert the problem to solving an algebraic equation in the frequency domain. Before concluding this section, let’s look at the concept of the transfer function that is often used to represent a linear system model. Let the input of a mass-damper-spring system be applied force \(f(t)\) and associated output be displacement \(x\left(t\right)\). Assuming that the mass equals \(1\), the damping coefficient equals \(3/(2\sqrt{2})\), and spring stiffness equals \(2\), the resulting equation of motion (EOM) is an ordinary differential equation, as defined in equation \ref{2.12}, and using the Laplace transform of equation \ref{2.13}. We can define the system transfer function \(G\left(s\right)\), representing the linear system model, as

\[G\left(s\right)=\dfrac{X(s)}{F(s)}=\dfrac{1}{s^2+3s+2} \label{2.16}\]

where \(F(s)\) is the Laplace transform of system input, \(X(s)\) is the Laplace transform of the system output, and \(G\left(s\right)\) is the system transfer function. Note that for any transfer function, the system initial condition is assumed to be zero. When \(G(s)\) is known, the system response \(x(t)\) can be solved by taking the inverse Laplace transform of \(G\left(s\right)F(s)\).

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