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-\documentclass[a4paper,10pt]{report}
-\pagestyle{plain}
-\usepackage{graphicx}
-\usepackage{caption}
-\usepackage{algorithmic}
-% Title Page
-\title{Half-Wave Rectifier}
-\author{Generated by SMCSim}
-
-\begin{document}
-\maketitle
-\hrule\vspace{5mm}
-\begin{center} {\bf Simulation of ckt/HWRectifierFilter.ckt} \end{center}
-\hrule\vspace{5mm}
-
-{\bf Circuit Diagram:} \\
-\vspace{2mm}
-\hrule\vspace{5mm}
-
-{\bf NetList:} \\
-{\it * Half-Wave Rectifier} \\
-V1 1 0 sine (5 50) \\
-D1 1 2 mymodel (1e-8 0.026) \\
-R1 2 0 10000 \\
-C1 2 0 10e-3 \\
-.tran 0 100 0.5 \\
-.plot v(1) v(2) \\
-.end 
-\vspace{2mm}
-\hrule\vspace{5mm}
-
-{\bf System of Equations representing the electrical circuit:}
-\vspace{2mm}
-\begin{equation}
-     i_{V_1} + D_{1f}(v_1,v_2) = 0 
-\end{equation}
-\begin{equation}
-     (R_1)v_2 + (C_1)\frac{dv_2}{dt} + -D_{1f}(v_1,v_2) = 0 
-\end{equation}
-\begin{equation}
-     v_1 = V_1
-\end{equation}
-\vspace{2mm}
-$$ D_{nf}(v_a,v_b)=Is_n(1-e^{(v_a-v_b)/vt_n})$$
- where $Is_n$=reverse saturation current and $vt_n$=threshold voltage of diode $n$\\
-\hrule\vspace{5mm}
-
-{\bf Matrix form:}\\
-The system of equations $\mathbf{A}\mathbf{x}+\mathbf{D}_f(\mathbf{\widehat{x}})+\mathbf{C}(d\mathbf{x}/dt)=b$ (Symbolically)\\
-Where $\mathbf{A}$, $\mathbf{D}_f$ and $\mathbf{C}$ represent matrices corresponding to linear,
-  nonlinear and time dependent electrical elements respectively.
-  $\mathbf{b}$ represents the vector corresponding to sources.
-
-\begin{equation}
-\mathbf{A}=  
-\left[ 
-\begin{array}{ccc} 
-0 &0  &1  \\ 
-0 &\widehat{R}_1 &0  \\ 
-1 &0  &0  
-\end{array}
-\right]
-\end{equation} 
-\begin{equation}
-\mathbf{b}=  
-\left[ 
-\begin{array}{c} 
-0  \\ 
-0   \\ 
-V_1  
-\end{array}
-\right]
-\end{equation} 
-\begin{equation}
-\mathbf{D}_f=  
-\left[ 
-\begin{array}{c} 
-D_{1f}  \\ 
--D_{1f}   \\ 
-0  
-\end{array}
-\right]
-\end{equation} 
-\begin{equation}
-\mathbf{C}=  
-\left[ 
-\begin{array}{ccc} 
-0 &0  &0  \\ 
-0 &C_1 &0  \\ 
-0 &0  &0  
-\end{array}
-\right]
-\end{equation} 
-\begin{equation}
-\mathbf{x}=  
-\left[ 
-\begin{array}{c} 
-v_1   \\ 
-v_2    \\ 
-i_{V_1}  
-\end{array}
-\right]
-\end{equation} 
-\begin{equation}
-\mathbf{\widehat{x}}=  
-\left[ 
-\begin{array}{c} 
-(v_1,v_2)     
-\end{array}
-\right]
-\end{equation} 
-Note that the matrix contains $\widehat{R}$ entries (corresponding to resistors) whose values are equal to 1/$R$\\
-\hrule\vspace{2mm}
-The number of equations are $3$ \\
-Unknowns: \\
-  Node potentials: $2$ Current Variables: $1$ \\
-\hrule\vspace{5mm}
-
-{\bf Operating Point (DC) Analysis: } \\
-{\it All capacitors are open circuited and inductors are short circuited.} 
-\vspace{2mm}
-
-{\bf System of Equations representing the electrical circuit:}
-\begin{equation}
-     i_{V_1} + D_{1f}(v_1,v_2) = 0 
-\end{equation}
-\begin{equation}
-     (R_1)v_2 + -D_{1f}(v_1,v_2) = 0 
-\end{equation}
-\begin{equation}
-     v_1 = V_1
-\end{equation}
-\vspace{2mm}
-$$ D_{nf}(v_a,v_b)=Is_n(1-e^{(v_a-v_b)/vt_n})$$
- where $Is_n$=reverse saturation current and $vt_n$=threshold voltage of diode $n$\\
-\hrule\vspace{5mm}
-
-{\bf Application of Newton-Raphson method: }\\
-\vspace{2mm}
-{\it Nonliner models: }\\   
-See linearized model for diode $D_1$ in diode\_D1.eps
-\begin{figure}[h]
-\centering
-\includegraphics{diode_D1.eps}
-\caption{linearization of diode $D_1$}
-\end{figure}
-\vspace{2mm}
-
-{\bf System of Equations representing the electrical circuit:}\\
-\begin{equation}
-     (R_{D_1})v_1 + (-R_{D_1})v_2 + i_{V_1} = -i_{D_1} 
-\end{equation}
-\begin{equation}
-    (R_{D_1})v_1 + (R_{D_1}+R_1)v_2 = i_{D_1} 
-\end{equation}
-\begin{equation}
-     v_1 = V_1
-\end{equation}
-\hrule\vspace{5mm}
-
-{\bf Transient Analysis:} \\
-\hrule\vspace{5mm}
-
-{\bf Results:} \\
-\begin{figure}[h]
-\centering
-\includegraphics[scale=0.5]{output.eps}
-\caption{plot}
-\end{figure}
-
-
-\end{document}
-