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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}
+