Laplace Transform Comprehensive Notes

Definition of the Laplace Transform

The Laplace Transform is a powerful mathematical tool designed to convert differential or integral equations from the time domain $(t)$ into algebraic equations in the complex frequency domain ($s$ domain). For any real-valued function $f(t)$ defined for t>0, its Laplace Transform, denoted as $\mathcal{L}{f(t)}=F(s)$, is given by the integral definition: $F(s)=\displaystyle\int_{0}^{\infty} e^{-st}\,f(t)\,dt$. In this definition, $s$ represents a complex variable, expressed as $s = \sigma + j\omega$.

For the Laplace Transform of $f(t)$ to exist, certain conditions are considered sufficient. Firstly, $f(t)$ must be continuous or piecewise continuous across every finite sub-interval of $(0,\infty)$, allowing for a finite number of jump discontinuities within any finite interval. Secondly, $f(t)$ must be of exponential order; this means there exist constants M>0 and \alpha>0 such that $\mid f(t) \mid \leq Me^{\alpha t}\,\text{for all}\,t \geq T$ for some $T$. This condition is crucial for ensuring the convergence of the integral. A practical test derived from the exponential order condition states that if $\displaystyle\lim_{t\to\infty}f(t)e^{-\alpha t}$ is finite, then $\mathcal{L}{f(t)}$ exists.

Basic Laplace Transforms

These fundamental transforms serve as essential building blocks for obtaining the Laplace transforms of more complex functions. The basic transforms include: $\mathcal{L}{1}=\dfrac{1}{s}$ (for s>0), $\mathcal{L}{e^{at}}=\dfrac{1}{s-a}$ (for s>a), $\mathcal{L}{\sin at}=\dfrac{a}{s^{2}+a^{2}}$ (for s>0), and $\mathcal{L}{\cos at}=\dfrac{s}{s^{2}+a^{2}}$ (for s>0). Additionally, the transforms for hyperbolic functions are given by $\mathcal{L}{\sinh at}=\dfrac{a}{s^{2}-a^{2}}$ and $\mathcal{L}{\cosh at}=\dfrac{s}{s^{2}-a^{2}}$ (both for s>\mid a \mid), which are derived using their exponential definitions.

For power functions, the Laplace Transform is expressed using the Gamma function as $\mathcal{L}{t^{n}}=\dfrac{n!}{s^{n+1}}$ for $n\in\mathbb{N}\cup{0}$ and s>0. A proof sketch for $\mathcal{L}{t^{n}}=\dfrac{n!}{s^{n+1}}$ involves substituting $x=st$ into the integral definition, which means $dt=\dfrac{dx}{s}$. The integral then becomes $\displaystyle\int<em>{0}^{\infty} e^{-x}\left(\dfrac{x}{s}\right)^{n}\,\dfrac{dx}{s} = \dfrac{1}{s^{n+1}}\displaystyle\int</em>{0}^{\infty} e^{-x}x^{n}\,dx$. The integral term is the definition of the Gamma function, $\Gamma(n+1)$, which simplifies to $n!$ for integer values of $n$.

Properties of the Laplace Transform

Various properties of the Laplace Transform significantly simplify the process of solving differential equations by transforming them into the $s$-domain and then finding their inverse transforms.

Linearity

The Linearity property states that the Laplace transform of a linear combination of functions is equivalent to the same linear combination of their individual Laplace transforms: $\mathcal{L}{af<em>{1}(t)\pm bf</em>{2}(t)}=aF<em>{1}(s)\pm bF</em>{2}(s)$. This property is critical for decomposing and solving complex functions.

Change of Scale

The Change of Scale property states that if $\mathcal{L}{f(t)}=F(s)$, then $\mathcal{L}{f(at)}=\dfrac{1}{a}F\left(\dfrac{s}{a}\right)$. This property illustrates how scaling the time variable in the time domain corresponds to both scaling and multiplication in the $s$-domain. For example, if we know that $\mathcal{L}{\cos t} = \frac{s}{s^2+1}$, then applying this property, $\mathcal{L}{\cos(2t)} = \frac{1}{2} \frac{s/2}{(s/2)^2+1} = \frac{s}{s^2+4}$.

First Shifting Theorem (Multiplication by Exponential)

This theorem indicates that multiplying a function by an exponential term in the time domain results in a shift of the $s$ variable in the Laplace domain: $\mathcal{L}{e^{at}f(t)}=F(s-a)$. As an illustration, given that $\mathcal{L}{\sin 4t}=\dfrac{4}{s^{2}+16}$, then $\mathcal{L}{e^{2t}\sin 4t}=\dfrac{4}{(s-2)^{2}+16}$.

Second Shifting Theorem (Unit–Step)

The Second Shifting Theorem is particularly useful for handling time-delayed or piecewise functions. If $\mathcal{L}{f(t)}=F(s)$, then the Laplace transform of $f(t-a)U(t-a)$ is given by $\mathcal{L}{f(t-a)U(t-a)}=e^{-as}F(s)$. Here, $U(t-a)$ is the Heaviside unit step function, which effectively turns on $f(t-a)$ at $t=a$. This theorem highlights that a time delay of 'a' in the time domain corresponds to multiplication by $e^{-as}$ in the $s$-domain. For example, to find $\mathcal{L}{(t-2)^2U(t-2)}$, we identify $a=2$ and $f(t)=t^2$. Since $\mathcal{L}{t^2} = \frac{2!}{s^{2+1}} = \frac{2}{s^3}$, then $\mathcal{L}{(t-2)^2U(t-2)} = e^{-2s} \frac{2}{s^3}$.

Initial Value Theorem

The Initial Value Theorem provides a method to determine the initial value of a function (as $t\to0^{+}$) directly from its Laplace transform without requiring the inverse transform. The theorem states $\displaystyle\lim<em>{t\to0^{+}}f(t)=\lim</em>{s\to\infty}sF(s)$. This is valid provided that $sF(s)$ approaches a finite limit as $s\to\infty$.

Final Value Theorem

The Final Value Theorem allows for the determination of the steady-state value of a function (as $t\to\infty$) from its Laplace transform: $\displaystyle\lim<em>{t\to\infty}f(t)=\lim</em>{s\to0}sF(s)$. It is important to note that this theorem is only applicable if all poles of $sF(s)$ have negative real parts (meaning they lie in the left half of the complex plane), excluding the possibility of a simple pole at the origin.

Unit Step (Heaviside) Function $U(t-a)$

The unit step function, also known as the Heaviside function, is a fundamental discontinuous function. Its value is zero for arguments less than zero and one for arguments greater than or equal to zero. It is commonly denoted as $U(t-a)$ or $H(t-a)$ and is formally defined as:
$U(t-a) = \begin{cases} 0, &amp; t &lt; a \ 1, &amp; t \geq a \end{cases}$ This function is crucial for modeling signals that switch on or off at a specific time $t=a$. For instance, it can represent a voltage suddenly applied to a circuit at a given time or a force abruptly beginning to act on a system. The Laplace transform of a simple unit step function starting at $t=a$ is $\mathcal{L}{U(t-a)}=\dfrac{e^{-as}}{s}$, for s>0. This transform is derived directly from the integral definition: $\displaystyle\int<em>{0}^{\infty} e^{-st}\,U(t-a)\,dt = \displaystyle\int</em>{a}^{\infty} e^{-st}\,1\,dt = \left[-\dfrac{e^{-st}}{s}\right]_{a}^{\infty} = 0 - \left(-\dfrac{e^{-as}}{s}\right) = \dfrac{e^{-as}}{s}$.