Main Tex

\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{mathrsfs}
\usepackage{graphicx}
\newtheorem{Theorem}{Theorem}[section]
\newtheorem{Example}{Example}[Theorem]
\begin{document}
\begin{center}
\huge \textbf{INTERNAL ASSESSMENT } \vspace{0.5 cm} \\
\large\textbf{M. Sc in Mathematics Semester - II Examination 2024} \vspace{0.1cm}\\
\textbf{Paper - MATH205}\\(Application of Type setting Software Latex) \
vspace{1cm}\\
\includegraphics[width=3cm]{download.png} \\
\vspace{1cm}
Submitted by \vspace{1cm}\\
Name :\textbf{ SUBHANKAR BARMAN}\vspace{0.1cm} \\
Roll -\textbf{002416 }\hspace{0.5cm} No. :\textbf{ 0035}\vspace{0.1cm} \\
Registration Number :\textbf{101-1112-0397-20} \vspace{0.1cm}\\
Session :\textbf{2020-2021} \vspace{0.1cm} \\
\vspace{1cm}
Department of Mathematics \vspace{0.1cm}\\
University of Gour Banga \\
Malda - 732103, \\West Bengal
\end{center}
\newpage
\setcounter{section}{1}
\setcounter{Theorem}{0}
\begin{Theorem}
If $\mathcal{L} \{ f(t) \} = \bar{f}(s)$, then the Second Shifting property holds:
\begin{equation}
\mathcal{L} \{ f(t-a)H(t-a) \} = e^{-as} \bar{f}(s) = e^{-as} \mathcal{L} \
{ f(t) \}, \quad a > 0.
\label{equation(1)}
\end{equation}
Or, equivalently,
\begin{equation}
\mathcal{L} \{ f(t)H(t-a) \} = e^{-as} \mathcal{L} \{ f(t+a) \}.
\label{eq2}
\end{equation}
where $H(t-a)$ is the Heaviside unit step function defined by (***).
It follows from the definition that

\begin{equation*}
\begin{aligned}
\mathcal{L} \{ f(t-a)H(t-a) \} &= \int_{0}^{\infty} e^{-st} f(t-a) H(t-a) \, dt \\
&= \int_{a}^{\infty} e^{-st} f(t-a) \, dt,
\end{aligned}
\end{equation*}
which is, by putting $t-a = \tau$,
\begin{equation*}
\begin{aligned}
&= e^{-as} \int_{0}^{\infty} e^{-s\tau} f(\tau) \, d\tau = e^{-as} \bar{f}(s).
\end{aligned}
\end{equation*}
We leave it to the reader to prove (\ref{eq2}). In particular, if $f(t) = 1$, then

\begin{equation}
\mathcal{L} \{ H(t-a) \} = \frac{1}{s} e^{-sa}.
\end{equation}
\end{Theorem}
\begin{Example}
Use the shifting property (\ref{equation(1)}) or (\ref{eq2}) to find the Laplace

transform of
(a) $f(t) =
\begin{cases}
1, & 0<t<1 \\
-1, & 1<t<2 \\
0, & t>2
\end{cases}
$,
\qquad(b) $g(t) = \sin t H(t-\pi)$.
To find $\mathcal{L}\{f(t)\}$, we write $f(t)$ as \\

$f(t)=1-2H(t-1) + H(t-2)$.
Hence,\\
$\bar{f}(s) = \mathcal{L}\{f(t)\} = \mathcal{L}\{1\} - 2\mathcal{L}\{H(t-1)\} + \
mathcal{L}\{H(t-2)\}$ \\
$=\frac{1}{s} - \frac{2e^{-s}}{s} + \frac{e^{-2s}}{s}$.
\newpage
To obtain $\mathcal{L}{g(t)}$, we use (\ref{eq2}) so that \begin{align*} \bar{g}(s)
&= \mathcal{L}{\sin t H(t - \pi)}\ &=-e^{-\pi s} \mathcal{L}{\cos t}\ &=-\
frac{se^{-\pi s}}{s^2+1}. \end{align*}
\textbf{Scaling Property:}
\begin{equation}
\mathcal{L}{f(at)} = \frac{1}{|a|} \bar{f} \left( \frac{s}{a} \right), \ a \neq 0.
\end{equation}
\end{Example}
\begin{Example}
Show that the Laplace transform of the square wave function $f(t)$ defined by \
begin{equation} f(t) = H(t)-2H(t-a) + 2H(t-2a)-2H(t-3a) +... \quad \end{equation}
is \begin{equation} f(s) = \frac{1}{s}\tanh\left(\frac{as}{2}\right). \quad \
end{equation} The graph of $f(t)$ is shown in Figure 3.1.
\begin{align*} f(t) &= H(t)-2H(t-a)=1-2\cdot0=1, \qquad &0<t<a\ \\ f(t) &= H(t)-

2H(t-a) + 2H(t - 2a) \ \\&=1-2\cdot1+2\cdot0=-1, \qquad &0<a<t<2a\end{align*}
Thus,\begin{align*}f(s) &= \frac{1}{s} - 2\cdot\frac{e^{-as}}{s} + 2\cdot\frac{e^{-

2as}}{s} - 2\cdot\frac{e^{-3as}}{s} +... \ \\&= \frac{1}{s}[1 - 2r(1-r + r^2
- ...)], \qquad \text{where } r = e^{-as} \ \\&= \frac{1}{s}\left[1-\frac{2r}{1+r}\
right] = \frac{1}{s}\left[1-\frac{2e^{-as}}{1+e^{-as}}\right] \ \\&= \frac{1}{s}\
left(\frac{1-e^{-as}}{1+e^{-as}}\right) =
\frac{1}{s}\left(\frac{e^{sa/2}-e^{-sa/2}}{e^{sa/2}+e^{-sa/2}}\right) = \frac{1}
{s}\tanh\left(\frac{as}{2}\right). \end{align*}
\end{Example}
\begin{Example}
(The Laplace Transform of a periodic function)\\
If $f(t)$ is a periodic function of period a,and if $\mathcal{L}\{f(t)\} $exists,
Show that
\begin{equation}
\mathcal{L}\{f(t)\} = [1 - \exp(-as)]^{-1} \int_0^a e^{-st} f(t)dt.
\label{eq7}
\end{equation}
\newpage
Clearly, the property (\ref{eq7}) gives \begin{align*} \mathcal{L} {f(t)} &= \
frac{a}{s^2 + a^2} \cdot \frac{1 + \exp\left(-\frac{s\pi}{a}\right)}{1 - \exp\
left(-\frac{s\pi}{a}\right)} \ \\&= \frac{a}{s^2 + a^2} \left[ \frac{\exp\left(\
frac{s\pi}{2a}\right) + \exp\left(-\frac{s\pi}{2a}\right)}{\exp\left(\frac{s\pi}
{2a}\right) - \exp\left(-\frac{s\pi}{2a}\right)} \right] \ \\&= \frac{a}{s^2 + a^2}
\coth \left(\frac{\pi s}{2a}\right). \end{align*}
\end{Example}
\begin{Theorem}
(Laplace Transforms of Derivatives)
If $\mathcal{L} \{f(t)\} = f(s),$then

\begin{equation}
\mathcal{L} \{f'(t)\} = s\mathcal{L} \{f(t)\} - f(0) = sf(s) - f(0),
\label{eq8}
\end{equation}
\begin{equation}
\mathcal{L} \{f''(t)\} = s^2\mathcal{L} \{f(t)\} - sf (0) - f'(0) = s^2 f(s) -s
f(0) - f'(0).
\end{equation}
More generally,
\begin{equation}
\mathcal{L}\{f^{(n)}(t)\} = s^n f(s) - s^{n-1} f(0)-s^{n-2} f'(0) - \dots - sf^{(n-
2)}(0) - f^{(n-1)}(0),
\end{equation}
where $f^{(r)}(0)$is the value of $f^{(r)}(t)$ at $t=0, r=0,1,..., (n - 1).$\\
\end{Theorem}
\begin{proof} We have, by definition,
\begin{align*}
\mathcal{L}\{f'(t)\} &= \int_0^{\infty} e^{-st} f'(t)dt,
\end{align*}
which is, integrating by parts,
\begin{align*}
&= [e^{-st} f(t)]_0^{\infty} + s\int_0^{\infty} e^{-st} f(t)dt \\
&= sf(s) - f(0),
\end{align*}
in which we assumed $f(t) e^{-st}\to 0$ as $t\to\infty$.
Similarly,
\begin{align*}
\mathcal{L} \{f''(t)\} &= s\mathcal{L} \{f'(t)\} - f'(0),\qquad by(\ref{eq8})\\
&= s[s f(s) - f(0)] - f'(0) \\
&=s^2f(s) - sf (0) - f'(0),
\end{align*}
\end{proof}
\end{document}

Main Tex

Uploaded by

Copyright:

Available Formats

Main Tex

Uploaded by

Document Information

Original Title

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

Main Tex

Uploaded by

Copyright:

Available Formats

\documentclass{article}

It follows from the definition that

We leave it to the reader to prove (\ref{eq2}). In particular, if $f(t) = 1$, then

Use the shifting property (\ref{equation(1)}) or (\ref{eq2}) to find the Laplace

To find $\mathcal{L}\{f(t)\}$, we write $f(t)$ as \\

\begin{align*} f(t) &= H(t)-2H(t-a)=1-2\cdot0=1, \qquad &0<t<a\ \\ f(t) &= H(t)-

Thus,\begin{align*}f(s) &= \frac{1}{s} - 2\cdot\frac{e^{-as}}{s} + 2\cdot\frac{e^{-

If $\mathcal{L} \{f(t)\} = f(s),$then

You might also like