MikTex 簡易安裝教學
如果是簡單純英文文件,個人建議使用 pdfLatex 來編輯比較方便(也比較容易處理圖檔嵌入),不管是用哪個編輯器幾乎套用內建的編譯設定就能很快做出你的第一份 Latex 文件。
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原始碼:
\documentclass[12pt]{article}
\title{Exp6: Polarization and Crystal Display\\
Final Report}
\author{Group1}
\linespread{1.1}
\setlength{\parskip}{5mm}
\begin{document}
\maketitle
\section*{Theoretical analysis}
If we put a $ \lambda /4$ waveplate between parallel linear polarizers which directions are $ x $ polarized and $ y $ polarized, and the incident laser beam passing the first polarizer denotes $ \mathbf{J}=[0,1]^T $ by Jones calculus. Assume $ \theta $ is the angle between fast axis(or slow axis$^\dag$ ) and laboratory coordinate $ x $-axis, and $ \beta $ is the rotating angle of the analyzer (the second polarizer).
After passing $ \lambda /4$ waveplate, the Jones vector becomes
\begin{eqnarry*}
$\mathbf{J}'$ &=& $\mathbf{M}$_{\lambda /4}\mathbf{J}=\mathbf{R}(\theta)\left[\begin{array}{cc}e^{-i\Gamma /2}&0\\0&e^{i\Gamma /2}\end{array}\right]\mathbf{R}(-\theta)\mathbf{J}\\
&=&\left[\begin{array}{cc}e^{-i\Gamma /2}\cos^2 \theta+e^{i\Gamma /2}\sin^2 \theta&-i\sin (\Gamma /2)\sin 2\theta&-i\sin (\Gamma/2)\sin 2\theta&e^{-i\Gamma /2}\sin^2 \theta+e^{i\Gamma /2}\cos^2 \theta\end{array}\right]\mathbf{J}
\end{eqnarray*}
With $ \Gamma = \pi /4 $ and $ \mathbf{J}=[0,1]^T $, we have
\begin{align}
$\mathbf{J}'$=\left[\begin{array}{c}-i\sin (\Gamma /2)\sin 2\theta&e^{-i\Gamma /2}\sin^2 \theta+e^{i\Gamma /2}\cos^2 \theta\end{array}\right]=\frac{1}{\sqrt{2}}\left[\begin{array}{c}-i\sin 2\theta &1+i\cos 2\theta\end{array}\right]
\end{align}
Passing the analyzer, the final Jones vector is
\begin{align}
$\mathbf{J}_o$=$\mathbf{R}$(-\beta)\left[\begin{array}{cc}1&0&0&0\end{array}\right]\mathbf{R}(\beta)\mathbf{J}=\frac{1}{\sqrt{2}}\left[ \begin{array}{c}
\sin\beta\cos\beta+i(\sin 2\beta\cos 2 \theta /2-\cos^2\beta\sin 2\theta)&\sin^2\beta+i(\sin^2\beta\cos 2\theta -\sin 2\beta\sin 2 \theta /2)\end{array}\right]
\end{align}
The optical intensity is proportional to $ \mathbf{J}^\dag\mathbf{J} $. Obviously the intensity is a function of $ \theta $ and $ \beta $.
\begin{align}
2I= 2$ \mathbf{J}_o^\dag\mathbf{J}_o$ = \sin^2\beta (1+\cos^2 2\theta)+\cos^2\beta \sin^2 2 \theta +\sin 2\theta\cos 2\theta\sin 2 \beta
\end{align}
With trivial caculations,
\begin{eqnarray*}
2I&=&\sin^2 \beta +(\cos\beta\sin 2 \theta +\sin \beta\cos 2\theta)^2=\sin^2\beta+\sin^2(\beta +2\theta)\\
&=&1-\frac{1}{2}(\cos 2\beta +\cos (2\beta +4\theta))=1-\frac{1}{2}\left\lbrace \cos 2\beta [1+\cos 4\beta]+\sin 2\beta\sin 4\beta\right\rbrace \\
&=&1-\frac{1}{2}A\cos (2\beta +\alpha)
\end{eqnarry*}
where
$$
A=\sqrt{(1+\cos 4\theta)^2+\sin^2 4\theta}=\sqrt{2+ 2\cos 4\theta}=2\cos 2\theta
$$
$$
\alpha =-\arctan \left(\frac{\sin 4\theta}{1+\cos 4\theta}\right)
$$
The output intensity is a sinusoidal function
$$
2I=1-\cos 2\theta\cos (2\beta +\alpha)
$$
The maximum corresponds with the long axis of polarization ellipse,
which means we could measure the elliptic polarized light produced by $ \lambda $/4 waveplate.
$ \alpha $ is the azumith of polarization ellipse which could be calculated from $ \theta $ or varing $ \beta $.
\section*{Consider special case(keep $ \lambda $/4 waveplate unmoved)}
\subsection*{$ \theta $ = (3n+1)$ \pi $/4}
$$
2I=\sin^2\beta +\cos^2\beta =1
$$
Since the output beam of $ \theta $ = $ \pi $/4 waveplate are cicular polarized,
the intensity drop to const with ratio 1/2 regardless of the angle $ \beta $.
\subsection*{$ \theta $ = n$ \pi $/2}
$$
2I=2\sin^2 \beta
$$
Since the input beam's polarizaion direction is parallel to slow axis or fast axis,
the output polariization suffers zerp phase different between two eigenbasis.
The intensity varies just as Mulus law.
\end{document}
加入圖檔與程式碼區塊的範本
使用 pdfLatex 來編譯很方便的一點在於可以加入 pdf 圖片,不管是向量格式或者點矩陣格式的圖檔,你只要把它轉成 pdf 統統可以餵給 pdfLaTeX 來編譯(通常向量繪圖軟體都可以匯出 pdf ,點矩陣圖也可以透過虛擬印表機印成 pdf 檔)
單欄位的圖片可以用下面類似的程式碼來引入:
單欄位的圖片可以用下面類似的程式碼來引入:
\begin{figure}
\centering
%圖片置中
\includegraphics[width=0.9\textwidth]{guassian-amp-equal-phase.pdf}
%將圖片寬度設為文字頁寬的 0.9 倍
%圖片連結此用相對路徑,換句話說 guassian-amp-equal-phase.pdf
%放在跟主文件 ex. alt-hw2.tex 同一個資料夾中
\caption{Gauassian distributed magnitude with equal phases}
%圖片標題
\end{figure}
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原始碼:
\documentclass[pra,onecolumn,reprint]{revtex4}
%使用 revTeX 的 onecolume 版本
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage{natbib}
%\usepackage{epstopdf}
%\pagestyle{empty} %去除頁碼
\begin{document}
\title{Advanced Laser Technologies Homework \#2}
\author{Chih-Han Lin~~~student ID: r99245002}
\affiliation{Graduate Institute of Applied Physics, National Taiwan University}
\maketitle
\section{Demonstration of mode-locking pulses}
Assume that complex amplitude has following expression,
\begin{align*}
A(t)=\sum_a A_q \exp\left(\frac{iq2\pi t}{T_f}\right)
\end{align*}
\noindent and pulse intensity $I(t)=|A(t)|^2$. We choose modes $M=11$ ($q$ sums from 1 to 11) with different complex coefficient
$A_q$ as follows,
\begin{itemize}
\item [(a)] Equal magnitude and equal phases, i.e. $A_q=1$.
\item [(b)] Magnitude that obey the Gaussian spectral profile and equal phase, i.e. $A_q=\exp\left[-\frac{1}{2}\left(\frac{q}{5}\right)^2\right]$.
\item [(c)] Equal magnitude and random phases (uniform distributing between 0 and $2\pi$), i.e. $A_q=\exp(rand()*2\pi)$.
\item [(d)] Equal magnitude and phase that obey the Gaussian distribution, i.e. $A_q=\exp \left[-i\cdot 2\pi\frac{1}{2}\left(\frac{q}{5}\right)^2\right]$
\end{itemize}
\section{Simulation result}
\subsection{Equal magnitude with equal phases}
FIG. 1. shows additive intensity function with equal relative phase in time domain and frequency domain respectively. Square Mask in frequency domain would generate ripples between pulses in time domain (oscillation with decay property of sinc function).
\begin{figure}
\centering
\includegraphics[width=0.9\textwidth]{equal-amp-equal-phase.pdf}
\caption{Equal magnitude and equal phases}
\end{figure}
\subsection{Gauassian distributed magnitude with equal phases}
First, we demonstrate Gaussian distributed magnitude with equal phase, i.e. $A_q=\exp\left[-\frac{1}{2}\left(\frac{q-6}{5}\right)^2\right]$, and the simulation result is shown in FIG. 2. Since the square mask effect is almost neglected comparing to standard deviation of Gaussian distributed spectral profile, the additive ultrashort pulse in time domain would have less ripple than previous case (modes with equal amplitude).
\begin{figure}
\centering
\includegraphics[width=0.9\textwidth]{guassian-amp-equal-phase.pdf}
\caption{Gauassian distributed magnitude with equal phases}
\end{figure}
We than increase the mode number ($M$=101) with symmetrical spectrum and observe the dependence between pulse duration and Gaussian spectral width by varying the complex mode's amplitude with:
\begin{align*}
A_q=\sum_a A_q \exp\left(-4\frac{q-50}{50k}\right),~~~\mbox{k}=1,2,\cdots ,5.
\end{align*}
The result shown in FIG. 3. says that pulse duration becomes shorter as $k$ increases. If the standard deviation $\sigma$ of the Gaussian spectrum is large compared with total mode spacing, the shortest pulse duration in time domain is limited by total mode spacing in frequency domain. We may slowly change the $\sigma$ via computer simulation (seeing FIG. 4.) to specify different operation regime of spectrum control.
\begin{figure}
\centering
\includegraphics[width=0.9\textwidth]{guassian-amp-equal-phase-multi.pdf}
\caption{pulse duration decrease with $\sigma$}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=0.9\textwidth]{pulse-duration-sigma.pdf}
\caption{pulse duration v.s. $\sigma$}
\end{figure}
\subsection{Equal magnitude with random phases}
If mode relative phase is a random number between $0$ and $2\pi$, the additive pulse would have lower AC component and generate randomly periodic signal which not specified as any usable pulses. The simulation result is shown in FIG. 5.
\begin{figure}
\centering
\includegraphics[width=0.9\textwidth]{equal-amp-rand-phase.pdf}
\caption{equal magnitude with random phases}
\end{figure}
\subsection{Equal magnitude with Gaussian distributed phases}
We may simulate Gaussian shape relative phase with parameters taking values $A_q=\exp \left[-i\cdot \theta_c\frac{1}{2}\left(\frac{q}{5}\right)^2\right]$ with $\theta_c=0.5\pi, \pi$ and $2\pi$. We observe the pulse duration broaden as neighbor mode relative phase getting larger (seeing FIG. 6.).
\begin{figure}
\centering
\includegraphics[width=0.7\textwidth]{equal-amp-guassian-phase.pdf}
\caption{equal magnitude with Gaussian distributed phases}
\end{figure}
\appendix
\section{Matlab code}
We use a for loop to directly sum the additive pulse in time domain and adjust parameter {\verb amp } and {\verb phase } to simulate different cases (Gaussian distribution or constant).
\begin{verbatim}
T_f=1;
samp_time=1E-4;
t=0:samp_time:3;
E_total=zeros(size(t));
for mode=1:11,
amp=1;
%amp=exp(-0.5*(mode/5)^2);
phase=-0.5*pi*(mode/5)^2;
%phase=rand()*2*pi;
%phase=0;
A_q=amp*exp(j*phase);
E_total=E_total+A_q*exp(j*mode*2*pi*t/T_f);
end
\end{verbatim}
To reconstruct the spectrum information, we use embedded {\verb fft } function in Matlab.
\begin{verbatim}
samp=size(t,2);
Spectrum=fft(E_total,samp);
f=1/samp_time*(1:samp)/samp;
I_f=Spectrum.*conj(Spectrum)/samp;
I_fphase=atan(imag(Spectrum)./real(Spectrum));
\end{verbatim}
Finally we plot intensity function with phase both in time and frequency domain via following codes:
\begin{verbatim}
subplot(221)
plot(t,abs(E_total).^2);
xlabel('time (arbitary unit)');
ylabel('Intensity (arbitary unit)')
subplot(223)
plot(t,atan(imag(E_total)./real(E_total)),'r')
xlabel('time (arbitary unit)');
ylabel('phase (radian)')
subplot(222)
plot(f(1:cutoff_freq),I_f(1:cutoff_freq));
xlabel('Frequency (arbitary unit)');
ylabel('Intensity (arbitary unit)')
subplot(224)
plot(f(1:cutoff_freq),I_fphase(1:cutoff_freq),'r');
xlabel('Frequency (arbitary unit)');
ylabel('phase (radian)')
\end{verbatim}
To generate data like FIG. 3., we have to add some analysis to automatically recognize the FWHM of the short pulse, the modified code shows as follows,
\begin{verbatim}
\end{verbatim}
\begin{verbatim}
T_f=1;
samp_time=1E-4;
t=0:samp_time:3;
num=5;
E_total=zeros(num,size(t,2));
for k=1:num,
for mode=1:101,
%amp=1;
%amp=exp(-4*((mode-50)/(50*k))^2);
phase=-2*pi*(mode/5)^2;
%phase=rand()*2*pi;
phase=0;
A_q=amp*exp(j*phase);
E_total(k,:)=E_total(k,:)+A_q*exp(j*mode*2*pi*t/T_f);
end
a0=0;
It=abs(E_total(k,:)).^2;
for i=1:size(E_total(k,:),2),
if It(1,i) > a0,
a0=abs(E_total(k,i)).^2;
end
end
a1=0.5*a0;
p=1;
for i=1:size(E_total(k,:),2)-1,
if (It(1,i)-a1)*(It(1,i+1)-a1)<0,
a3(1,p)=t(1,i);
a5(1,p)=i;
p=p+1;
end
end
a4(k)=a3(1,3)-a3(1,2);
clear a3;
samp=size(t,2);
Spectrum=fft(E_total(k,:),samp);
f=1/samp_time*(1:samp)/samp;
I_f=Spectrum.*conj(Spectrum)/samp;
I_fphase=atan(imag(Spectrum)./real(Spectrum));
\end{verbatim}
A little modification is needed when you want to generate data like FIG. 5. We only change the parameter {\verb amp } and neglect plotting procedure.
\end{document}