Laplace Transforms

Laplace Transforms acquired a new importance when the English Electrical Engineer Oliver Heaviside(1850-1925 AD) made use of them in operational calculus(devised by him) and for the solution of ordinary differential equations with constant coefficients. The methods of Laplace transforms enabled him to solve some important practical problems that could not be dealt with by the classical methods. For instance we can solve differential equation representing the motion of a spring driven by a periodic or constant force, by usual methods, however these methods no more help if the force is impulsive. In other words, the usual classical methods can solve an ordinary differential equation $F\left(D\right)y=f\left(t\right)$ , when $f\left(t\right)$ is continuous but these methods become ineffective and of no use if $f\left(t\right)$ is discontinuous or piece-wise continuous. In such cases the methods of Laplace transforms come to our rescue and resolve the problems.

Initial value and boundary value problems can also be solved with the help of the Laplace transforms

input signal $u\left(t\right)$
system
output signal $x\left(t\right)$

The laws controlling a system’s behavior serve as the foundation for the correlation between its input and output. A linear differential equation with constant coefficients connects the output to the input if the system is linear and time-invariant. In actuality, it is an initial value problem that the Laplace transform approach may resolve.

Definition of the Laplace Transforms

One type of integral transform is the Laplace transform. It takes a function f(t), of one variable t (time) into a function $F\left(s\right)$ of another variable s (frequency). Let f(t) be a given function defined for all $t\ \ge 0$ . The Laplace transform of f(t) is denoted by $\mathcal{L}\left\{f\left(t\right)\right\}$ or $\mathcal{L}\left\{f\right\}, [latex]\mathcal{L}$ being Laplace transform operator. It is defined as:

$\mathcal{L}\left\{f\left(t\right)\right\}=\int _0^{\infty }^{ }e^{-st}f(t)\ dt=F\left(s\right)$ (1)

Example 1:

Determine the Laplace transformations of, $f\left(t\right)=1,\ f\left(t\right)=t,\ f\left(t\right)=e^{at}\ ,\ f\left(t\right)=\sin \ at,\ f\left(t\right)=\ \cos \ at.$

Solution

Using Eq. (1) we obtain,

For $f\left(t\right)=1,\ \ \mathcal{L}\left\{f\left(t\right)\right\}=\int _0^{\infty }e^{-st}.1\ dt=\left[\frac{e^{-st}}{-s}\right]_0^{\infty }=\frac{1}{s}$

For $\begin{array}{l}f\left(t\right)=t,\ \ \mathcal{L}\left\{f\left(t\right)\right\}=\ \int _0^{\infty }e^{-st\ }t\ dt\ =\ \left[\ \frac{te^{-st}}{s}-\ e\ \frac{^{-st}}{s^2}\right]_0^{\infty }\end{array}=\ \frac{1}{s^2}$

For $\begin{array}{l}f\left(t\right)=\ \sin \ at,\ \mathcal{L}\left\{f\left(t\right)\right\}=\ \int _0^{\infty }e^{-st}\ \sin \ at\ dt\ =\ \left[\frac{e^{-st\ }\left(-s\ \sin at-a\ \cos \ at\right)}{s^2+a^2}\right]_0^{\infty }=\frac{a}{s^2+a^2}\\ \end{array}$

$\begin{array}{l}f\left(t\right)=e^{st}\end{array}\ ,\mathcal{L}\left\{f\left(t\right)\right\}=\ \int _0^{\infty }e^{-st}e^{st\ \ }dt\ =\left[\frac{e^{-\left(s-a\right)t}}{-\left(s-a\right)}\right]_0^{\infty }=\frac{1}{s-a}$

$f\left(t\right)=\cos at,\ \mathcal{L}\left\{f\left(t\right)\right\}=\int _0^{\infty }e^{-st}\cos atdt=\left[\frac{e^{-st}\left(-s\ \cos at+a\sin at\right)}{s^2+a^2}\right]_0^{\infty }= \frac{s}{s^2+a^2}$

Enough Prerequisites for the Laplace Transform to Exist

The integral

$\int _0^{\infty }e^{-st}f\left(t\right)\ dt=F\left(s\right),\ \ \ t\ge 0,$

defining the Laplace transform of a function $f\left(t\right)$ may not always converge.For instance,

$\mathcal{L}\left\{\frac{1}{t}\right\}$ and $\ \ L\left\{e^{t^2}\right\}$ do not exist. Sufficient conditions that guarantee the existence of $\ \ L\left\{f\left(t\right)\right\}$ are as under:

a. f is piecewise continuous on $\left[0,\infty \right)$ , and

b. f is of exponential order for $t>0$ .

A function $f\left(t\right)$ is piecewise continuous on $\left[0,\infty \right)$ , if in any interval $\left[a,b\right]$ there are at most a finite number of discontinuous at points $t_{1,}t_{2,},t_{3,}...t_{k-1,}t_k$ and f is continuous in open intervals $\left(t_{1,}t_2\right),\left(t_{2,}t_3\right),...,\left(t_{k-1,}t_k\right)$ .An example of piecewise continuous function is as under:

$f(t)=\begin{cases}\ t^2,&0\le t\le 2\\ \ \ 1,&2<t\le 3\\ -1,&3<t\le 4\end{cases}$

A function $f(t)$ is said to be of exponential order if there exist numbers $c$ , M>0 and t>0, such that $\left|f\left(t\right)\right|\le Me^{ct\ },\ i.e.,\$ the graph of $f(t)$ does not grow faster than the graph of $Me^{ct\ }$ . For example, $f\left(t\right)=t\ ,\ f\left(t\right)=e^{t,\ }f\left(t\right)=2\cos t\$ are all of exponential order for t>0, since we have, respectively

$\left|t\right|\le e^{t,\ \ \ \ }\ \ \ \ \left|e^{-t}\right|\le e^t,\ \ \ \ \ \left|2\cos t\right|\le 2e^t.$

On the other hand $f\left(t\right)=e^{t^2}$ is not of exponential order, since it grows faster than $Me^{ct}$ for any choice of $M$ and .

Theorem (Existence of Laplace Transform)

Let $f\left(t\right)$ be piecewise continuous on the interval $\left[0,\infty \right)$ and of exponential order for $t>0$ , (i.e., there exist $c,\ M>0$ , such that $\left|f\left(t\right)\right|\le Me^{ct}$ for all $t\ge 0$ ).Then $L\left\{f\left(t\right)\right\}$ exists for $s>c$ .

Proof

Since, $f\left(t\right)$ is piecewise continuous on $\left[0,\infty \right)$ and of exponential order for $t>0,$ , i.e, $\left|f\left(t\right)\right|\le Me^{ct}\ \ \ \ for\ \ \ M>0,\ c>0,\ so$

$\mathcal{L}\left\{f\left(t\right)\right\}=\left|\int _0^{\infty }f\left(t\right)e^{-st\ }dt\right|\le \ \int _0^{\infty }\left|f\left(t\right)\right|e^{-st}dt\le \int _0^{\infty }M\ e^{ct}e^{-st}dt$

$M\int _0^{\infty }e^{-\left(s-c\right)t}dt=M\left[\frac{e^{-\left(s-c\right)t}}{-\left(s-c\right)}\right]_0^{\infty }=\frac{M}{s-c}\ \ for\ s>c$

This demonstrates the existence of the function’s piecewise continuous and exponentially ordered Laplace transforms. It should be mentioned, nevertheless, that these prerequisites are adequate—not essential—for a Laplace transform to exist. For example, the function $f\left(t\right)=\ t^{-\frac{1}{2}}$ is not piecewise continuous on the interval $\left[0,\infty \right)$ because of its behaviour at t=0, but its Laplace transform exists(see example below)

Example 2:

Determine the Laplace transform of $f\left(t\right)=t^{-\frac{1}{2}}$

Solution

$\mathcal{L}\left\{t^{-\frac{1}{2}}\right\}=\int _0^{\infty }e^{-st}\ t^{-\frac{1}{2}}\ dt,\ \ \ \ \ s>0$ (2)

Let $st=z^{2,\ \ \ \ \ \ \ \ }$ , i.e., $t=\frac{z^2}{s}$ . Differentiating the former equation w.r.t t we get,

$S\ dt=2z\ dz\$ i.e., $\ \ \ \ dt=\frac{2z}{s}dz$

Therefore, equation (2) becomes

$\mathcal{L}\left\{t^{-\frac{1}{2}}\right\}=\int _0^{\infty }e^{-z^2}\ \ \frac{\sqrt{s}2z}{zs}dz=\frac{2}{\sqrt{s}}\int _0^{\infty }e^{-z^3}dz$

But $\int _0^{\infty }e^{-z^2}dz=\ \frac{\sqrt{\pi }}{2}$ .

Therefore

$L\left\{t^{-\frac{1}{2}}\right\}=\sqrt{\frac{\pi }{s}},\ s>0$

Linearity of the Laplace Transform

The Laplace transform is a linear operator. This property of the former is stated as under:

If f(t) and g(t) are any two functions having Laplace transforms, and if a and b are any constants, then

$\mathcal{L}\left\{a\ f\left(t\right)+bg\left(t\right)\right\}=a\ \mathcal{L}\left\{f\left(t\right)+b\mathcal{L}g\left(t\right)\right\}$

Accordingly we say that Laplace transform operator $/mathcal{L}$ is a linear operator. Using the Laplace transform definition, the aforementioned may be easily demonstrated as follows:

$\begin{array}{l}\mathcal{L}\left\{af\left(t\right)+bg\left(t\right)\right\}=\int _0^{\infty }\left[af\left(t\right)+bg\left(t\right)\right]e^{-st}dt\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int _0^{\infty }af\left(t\right)e^{-st}+bg\left(t\right)e^{-st}dt\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =a\int _0^{\infty }f\left(t\right)e^{-st}dt+b\int _0^{\infty }g\left(t\right)e^{-st}dt\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =a\ \mathcal{L}\left\{f\left(t\right)\right\}+b\mathcal{L}\left\{g\left(t\right)\right\}\end{array}$

The First Shift Property of Laplace Transform

The First shift property of Laplace transform sometimes referred to as Exponential Modulation Theorem.

The First Shift Theorem

If f(t) is a function having Laplace transform $F\left(s\right),\ s>0$ , then

$\mathcal{L}\left\{e^{at}f\left(t\right)\right\}=F\left(s-a\right),\ s>a$

Proof

Using the definition of Laplace transform we get,

$\begin{array}{l}\mathcal{L}\left\{e^{at}f\left(t\right)\right\}=\int _0^{\infty }e^{at}f\left(t\right)e^{-st}dt\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int _0^{\infty }f\left(t\right)e^{-\left(s-a\right)}dt,\ s>a\ \ \ \end{array}$ (3)

Now, by definition

$\mathcal{L}\left\{f\left(t\right)\right\}=F\left(s\right)=\int _0^{\infty }f\left(t\right)e^{-st}dt.$ (4)

We see that integral (3) is exactly the same as integral (4), the Laplace transform of f(t) , with (s-a) in place of s. Thus integral (3) gives,

$\mathcal{L}\left\{e^{at}f\left(t\right)\right\}=F\left(s-a\right),\ s>a$ .

$\mathcal{L}\left\{e^{at}f\left(t\right)\right\}=\mathcal{L}\left\{f\left(t\right)\right\}_{s\longrightarrow s-a}$ .

Therefore, if we already know $\mathcal{L}\left\{f\left(t\right)\right\}=F\left(s\right)$ , we can easily compute $\mathcal{L}\left\{e^{at}f\left(t\right)\right\}$ by shifting or translating $F\left(s\right)\ to\ F\left(s-a\right)$ .If we consider the graphs of $F\left(s\right)$ and $F\left(s-a\right)$ , we see that the graph of $F\left(s-a\right)$ is the graph of $F\left(s\right)$ shifted on the s-axis by an amount of $\left|a\right|$ units to the right.