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\section{Mathematical transformation methods}
\subsection{Laplace Transform}
\subsubsection{Historical review}


\includegraphics[scale=0.25,width=2in, height=2.25in]{Pierre-Simon_Laplace.jpg} \\

Figure 14 : Pieere Smon Laplace 

\cite{pic1}


\paragraph{}The laplace transform was builded by mathematician and astronomer Pierre Simon Laplace. Laplace transform method was used on probability theory in Pierre's works. And from 1744 Leonhard Eular applies integral with below formula.

\begin{gather}
z =  \int_a^b X(x)e^{-ax} *a^x\,dx. and z =  \int_a^b X(x)x^A\,dx.
\end{gather}

as solutions of differential equations \cite{history1}.

\subsubsection{Why Laplace Transform is used}
\paragraph{} Laplace transform is used to solve linear differential equation. Thanks to Laplace transform we can convert some common functions such as sinusoidal function, damped sinusoidal function, exponential function \footnote{i think you learn something more (besides defition) from Wolfram,
wikipedia or something else} into algebraic function of complex variable s \cite{Ref1}.



\subsubsection{How is it used}
\paragraph{} Differentiation and integration can be replaced by algebraic operation in the complex plane. \cite{Ref1}


\subsubsection{Complex variable and Complex function}  \cite{Ref1}
    \paragraph{Complex variable}
\paragraph{} Complex variable that is a variable that has a real part or imaginary part wihch both are constant \cite{Ref1}. 
\begin{gather} 
alpha=\frac{\beta+i\gamma}{} 
\end{gather}
or
\begin{gather}
alpha=\frac{\beta+j\gamma}{} 
\end{gather}
\nocite{*}
\paragraph{}Some formulas  
\paragraph{}Magitude of F(s)  
\begin{gather}
\sqrt{Fx^2+jFy^2}  
\end{gather}
\paragraph{}Angle of teta F)s)
\begin{gather}
 \tan^-1(F(y)/F(x) 
\end{gather}
\paragraph{Complex Function}
\paragraph{} Complex function F(s) is a function of s that has real and imaginary part \cite{Ref1}.


\paragraph{}The angel is measured counterclockwise from positive real axis.
Complex conjugate of f(s)
\begin{gather}
F(x)-jF(y)
\end{gather}
\paragraph{} Complex function commonly encountered in linear control system analysis are single-valued function of s and are uniquely determined for agiven value of s.



\paragraph{} A complex function is G(s) is analytic iff g(s) and its all derivatives of G(s)
is in that region.The derivative of G(s) is found in that way.
\begin{gather}
 g'(z)=\lim_{h\to0}\frac{g(z+h)-g(z)}h 
\end{gather}

\paragraph{} $\Delta s = \Delta \sigma + j \Delta w$ can approach a zero along
an infinite different path

\paragraph{} For a particular path $\Delta s = \Delta \sigma$  (path is in real axis)
\begin{gather}
d/d(s) G(s) = \mathop {\lim }\limits_{j \Delta_w \to \infty} (\Delta G_x/j\Delta \theta + j\Delta G_y/j\Delta \theta) = -j\phi G_x/\phi \theta + \Delta G_y/\phi \theta 
\end{gather}
\paragraph{} For another particular path $\Delta s =j  \Delta \sigma$ (path is in
imaginary axis)
\begin{gather}
d/d(s) G(s) = \mathop {\lim }\limits_{j \Delta_w \to \infty} (\Delta G_x/j\Delta \theta + j\Delta G_y/j\Delta \theta) = -j\phi G_x/\phi w + \Delta G_y/\phi w 
\end{gather}
\paragraph{} Both following condition can be preferred to show dG(s) uniquely deternined. In adittion these two conditions is known as Cauchy-Riemann condition and if these two conditions are satisfied, it is possible to tell system is analitic \cite{Ref1}.

\begin{gather}
\phi G_x/\phi \theta + \phi G_y/\phi \theta = \phi G_x/\phi w - j \phi G_y/\phi w
\end{gather}
\paragraph{} or
\begin{gather}
\phi G_x /\phi \theta = \phi G_y/\phi w  , \phi G_y/\phi \theta = -\phi G_x/\phi w
\end{gather}
\subsubsection{Euler's theorem}
\paragraph{} The power series of $\cosh$ and $\sinh$ \cite{Ref1}. 
\begin{gather}
\cosh = 1- h^2/2! + h^4/4!...
\end{gather}
\begin{gather}
\sinh = h - h^3/3! + h^5/5! - h^7/7!...
\end{gather}
\paragraph{} When add operation is carried. The result is
\begin{gather}
\cosh + j\sinh = 1 + (j\theta) + (j\theta)'2/2! + (j\theta)^3/3!...
\end{gather}
\paragraph{} there is a known condition. Which is;
\begin{gather}
e^x = 1 + x + x^2/2!...
\end{gather}
Namely,
\begin{gather}
\cosh + \sinh = e^{j\theta}
\end{gather}

\paragraph{} By using euler formula, we can generate conjugate function \footnote{sometimes in root Locus plot,  complex conjugate root has to be found. May be help of that method will be necessary} \cite{Ref1} .
\begin{gather}
\cosh + \sinh = e^{j\theta}
\end{gather}
\begin{gather}
\cosh - j\sinh = e'{-j\theta}
\end{gather}
\subsection{Laplace transform}
\paragraph{} In the beginning of this part, Something is written for Laplace Transform. Now it is time to go further over parameters of Laplace Transform\cite{Ref1}.
\paragraph{} Parameter of Laplace transform
\begin{itemize}
\item f(t) is function of t such that f(t) = 0 for t<0.
\item s is complex variable.
\item $L$ is operational syymbol that stands for laplace transfor.
\item f(s) is Laplace Transform of f(t) \footnote{t is for time domain, s is for frequency domain.} \cite{Ref1}.
\end{itemize}

\paragraph{} The formula is;
%\begin{gather}
%L[f(t) = F(s) = \int_0^\inf \! f(t)e^[-st] \, dt.
%\end{gather}
\paragraph{} The formula of inverse laplace transform is;
\begin{gather}
L^-1[f(s) = f(t) = 1/2\pi j \int_c-\inf^c+\inf \! f(s)e^{st} \, ds. t>0
\end{gather}

\subsubsection{Existance of Laplace Transform}
\paragraph{} The laplace transform exist, if the laplace integral converges.

\paragraph{} In the laplace transform, some function can be considered as part of laplace transform.
\subsubsection{Exponential function}
\begin{gather}
f(t)=0 for t<0 \\
f(t) = A e^{-at}
\end{gather}
\paragraph{} A and a is given constant.

\subsubsection{Sinusoidal fuction}
\begin{gather}
f(t) = 0 for t<0 \\ 
f(t) = A sinwt for t>=0
\end{gather}
%\begin{gather}
%sinwt = 1/2 * j * (e^[jwt] - e ^[-jwt]) \\ 
%L[Asinwt] = A/2j[\int_0^\inf \! (e^[jwt] - e'[-jwt]) * e^{-st} \, dt]. = As/s^2 + w^2
%\end{gather}
\subsubsection{Ramp function}
\begin{gather}
f(t) = 0 for t<0 \\ 
f(t) = At for t>= 0
\end{gather}
\subsubsection{Pulse function}
\begin{gather}
f(t) = A/t_0 for 0<t<t_0 \\ 
f(t) = 0 for t<0
\end{gather}
\paragraph{} So laplace transform can be obtained by;
\begin{gather}
L[f(t)] = L[1/t_0 1(t)] - L[A/t_0(t-t_0)] = A/t_0 s(1 - e^{-st_0})
\end{gather}
\subsubsection{Impulse function}
\paragraph{} Special limitting version of pulse fnction.
 \begin{align} \lim_{x\to\infty} \frac{1}{t_0} & \quad \text{for $0<t<t_0$} \\ g(t) = 0 & \quad \text{for $t<0,t_0<t$} \end{align}

\paragraph{} Laplace transform can be obtained stated as;
%\begin{gather}
%L[g(t)] = \lim_{x\to\infty}\frac{A}{t_0 s}(A_t_0 s (1 - e + (-e^{-st_0}))
%\end{gather}

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