## What is meant by s-domain?

The concept of s-domain is one of the fundamental concepts in digital control systems in the area of electronics engineering and signal analysis. In many engineering applications such as real-time harmonic analysis or vibration analysis, raw real-time signals are acquired by the sensing systems. These are sent to the data acquisition control (DAC) modules or oscilloscopes to process the data and get a visual representation in the form of two-dimensional graphs, which can be seen on a computer screen. But the signals that are viewed on the screen are time domain-based, that is, they are signal amplitude vs time domain-based. Not much information can be extracted from this representation. For an in-depth analysis purpose, the time domain is converted into a frequency domain so that the representation is signal amplitude vs frequency-based. Two of the most popular mathematical approaches to achieve this are either using Laplace transforms or Fourier transforms. In this article, a brief emphasis is laid on the approach of Laplace transforms.

The s-domain is also known as s plane, or Laplace domain where the Laplace transforms are mapped. The Laplace transforms are mapped in a complex plane rather than in a real plane. The time-domain equations are mapped in the s-domain by the use of Laplace transforms. The application of the s-domain concept is ideal to analyze systems with transient responses of linear time-invariant systems (LTIS). In such systems, the output signals show linearity with the input signals. These systems are necessarily dynamic and are mostly governed by second-order differential equations. All the real systems encountered in control systems are approximately modeled in LTIS to simplify the problem. For ease of calculations, these equations thus evolved, are solved by a suitable math processing software such as MATLAB and SCILAB.

## Fundamentals of the Laplace transform

It is an integral transform that converts the function involving time domain or t-domain into a function of a complex variable (s domain) or complex frequency.

If $f\left(t\right)$ is any function in the t-domain, where $t\ge 0$. Then, the Laplace transform for $f\left(t\right)$ is given by $F\left(s\right)$.

$F\left(s\right)={\int}_{0}^{\infty}f\left(t\right){e}^{-st}dt......\left(1\right)$

The function instead of $F$ is written as $L\left\{f\right\}$, therefore equation $\left(1\right)$ can be re-written as,

$L\left\{f\right\}\left(s\right)={\int}_{0}^{\infty}f\left(t\right){e}^{-st}dt......\left(2\right)$

where $s$ represents the complex domain or complex number frequency parameter expressed as,

$s=\omega +i\psi $

**Laplace transform of sine and cosine functions**

The integral can be applied to derive the Laplace transform of simple trigonometric functions such as sine and cosine.

If $f\left(t\right)=\mathrm{sin}\left(at\right)$ and $g\left(t\right)=\mathrm{cos}\left(at\right)$ be two trigonometric functions in the t-domain. Then, their Laplace transform is given by,

$L\left[f\left(t\right)\right]=L\left[\mathrm{sin}\left(at\right)\right]$

where, the coefficient $L$ denotes Laplacian operator.

$\Rightarrow L\left[\mathrm{sin}\left(at\right)\right]=\frac{a}{{s}^{2}+{a}^{2}}$, and

$L\left[g\left(t\right)\right]=L\left[\mathrm{cos}\left(at\right)\right]$

$\Rightarrow L\left[\mathrm{cos}\left(at\right)\right]=\frac{s}{{s}^{2}+{a}^{2}}$

## Application of Laplace transform to the first-order differential equation

The Laplace transform forms a powerful tool for the solution of differential equations, an application of which is outlined below.

Considering the first-order differential equation below,

$\frac{dy}{dt}-2y={e}^{bx}$, with $y\left(0\right)=a$ as the boundary condition.

Introducing Laplacian operator on both sides gives,

$L\left[\frac{dy}{dt}-2y\right]=L\left[{e}^{bt}\right]$

$\Rightarrow L\left[\frac{dy}{dt}\right]-L\left[2y\right]=L\left[{e}^{bt}\right]$

Applying fundamental Laplace transform formulas or theorems, the above equation is written as,

$\Rightarrow sL\left[y\right]-y\left(0\right)-2L\left[y\right]=\frac{1}{s-b}$

The boundary condition is introduced, hence the equation reduces to,

$\Rightarrow sL\left[y\right]-a-2L\left[y\right]=\frac{1}{s-b}$

$\Rightarrow sL\left[y\right]-2L\left[y\right]=\frac{1}{s-b}+a$

$\Rightarrow \left(s-2\right)L\left[y\right]=\frac{1}{s-b}+a$

$\Rightarrow L\left[y\right]=\frac{\left(s-2\right)}{\left(s-b\right)}+\frac{a}{\left(s-2\right)}$

Hence, the solution of the given differential equation is given by performing an inverse Laplace transform, that is,

$y={L}^{-1}\left[\frac{\left(s-2\right)}{\left(s-b\right)}+\frac{a}{\left(s-2\right)}\right]$

## Mathematical modeling of control systems

Control systems are the systems that produce the desired amount of output by controlling the input parameters. A control system can be simple and complex. A complex control system needs to be reduced to simpler models for ease of analysis. There are three mathematical modeling procedures used, they are:

- Differential equation model
- Transfer function model
- State-space model

The differential equation model is specially discussed in this article.

Consider the below circuit (Figure. 1). The circuit consists of a capacitor $C$, a resistor $R$ of certain Ohms, an Inductor $L$, and a current *I* flowing in the circuit. ${V}_{i}$ and ${V}_{o}$ represent the input and output voltages respectively.

Using the mesh analysis, the mesh equation for the above circuit can be written as,

${V}_{i}=Ri+L\frac{di}{dt}+{V}_{o}......\left(3\right)$

The relation between voltage and current for a capacitor is given by,

$i=C\frac{d{V}_{o}}{dt}......\left(4\right)$

Therefore substituting equation $\left(4\right)$ in equation $\left(3\right)$ gives,

${V}_{i}=R\left(C\frac{d{V}_{o}}{dt}\right)+L\frac{d}{dt}\left(C\frac{d{V}_{o}}{dt}\right)+{V}_{o}......\left(5\right)$

Or, ${V}_{i}=RC\frac{d{V}_{o}}{dt}+LC\frac{{d}^{2}{V}_{o}}{d{t}^{2}}+{V}_{o}......\left(6\right)$

Dividing equation $\left(6\right)$ by $LC$ gives,

${V}_{i}\left(\frac{1}{LC}\right)=\left(\frac{R}{L}\right)\frac{d{V}_{o}}{dt}+\frac{{d}^{2}{V}_{o}}{d{t}^{2}}+\left(\frac{1}{LC}\right){V}_{o}......\left(7\right)$

Rearranging equation $\left(7\right)$ gives,

$\frac{{d}^{2}{V}_{o}}{d{t}^{2}}+\left(\frac{R}{L}\right)\frac{d{V}_{o}}{dt}+\left(\frac{1}{LC}\right){V}_{o}=\left(\frac{1}{LC}\right){V}_{i}......\left(8\right)$

The equation $\left(8\right)$ thus represents a simple control system circuit into a second order differential equation.

Now, from elementary mathematics, applying Laplace theorem in both the sides of equation $\left(8\right)$ gives,

${s}^{2}{V}_{o}\left(s\right)+\left(\frac{sR}{L}\right){V}_{o}\left(s\right)+\left(\frac{1}{LC}\right){V}_{o}\left(s\right)=\left(\frac{1}{LC}\right){V}_{i}\left(s\right)......\left(9\right)$

$\Rightarrow \left\{{s}^{2}+\left(\frac{R}{L}\right)s+\frac{1}{LC}\right\}{V}_{o}\left(s\right)=\left(\frac{1}{LC}\right){V}_{i}\left(s\right)......\left(10\right)$

$\Rightarrow \frac{{V}_{o}\left(s\right)}{{V}_{i}\left(s\right)}=\frac{{\displaystyle \frac{1}{LC}}}{{s}^{2}+\left({\displaystyle \frac{R}{L}}\right)s+{\displaystyle \frac{1}{LC}}}......\left(11\right)$

Equation $\left(11\right)$ is known as the transfer function of the control system, the solution of which can be done using the convolution theorem.

## Context and Applications

The topic is widely taught in various undergraduate and postgraduate degree courses of:

- Bachelors of Technology (Electronics Engineering)
- Bachelors of Technology (Mechanical Engineering)
- Master of Technology (Control System Engineering)
- Master of Science (Mathematics)

## Practice Problems

Q 1. In which of the following areas is Laplace transform important?

- Analysis of harmonic and vibration systems
- Analysis of LTIS
- Control system analysis
- All of these

Answer: Option d

Explanation: The Laplace transform is important in areas like control system analysis, analysis of harmonic systems and vibration analysis, and LTIS.

Q 2. Which of the following transforms in mathematics maps a t-domain function in the frequency domain?

- Laplace transforms
- Fourier transforms
- Hankel transforms
- Both a and b

Answer: Option d

Explanation: The Laplace transforms and Fourier transforms maps a given t-domain function into a frequency domain.

Q 3. Which of the following is the Laplace transform of $\mathrm{sin}\left(\varnothing t\right)$?

a. $\frac{s}{{s}^{2}+{\varnothing}^{2}}$

b. $\frac{\varnothing}{{s}^{2}+{\varnothing}^{2}}$

c. $\frac{\varnothing}{{s}^{2}-{\varnothing}^{2}}$

d. $\frac{s}{{s}^{2}-{\varnothing}^{2}}$

Answer: Option b

Explanation: The correct Laplace transform of $\mathrm{sin}\left(\varnothing t\right)$ is $\frac{\varnothing}{{s}^{2}+{\varnothing}^{2}}$

Q 4. What is $s$ in Laplace transforms?

a. $s=\psi $ a real-valued parameter

b. $s=a{y}^{2}$ a parabolic parameter

c. $s=\psi +\mathrm{i\omega}$ a complex parameter

d. $s=\mathrm{i\omega}$ an imaginary parameter

Answer: Option c

Explanation: The $s$ in the s-domain analysis or Laplace transforms denotes a complex parameter or plane given by $s=\psi +\mathrm{i\omega}$.

Q 5. What are the characteristics of a real-time signal acquired by sensors?

- They are time domain based
- They are frequency domain based
- Both a and b
- They are filtered signals without noises

Answer: Option a

Explanation: The characteristics of a real-time signal acquired by a sensor is that the signals are time domain-based, shown as signal vs time in a graphical representation.

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