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\documentclass{article}
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\usepackage{graphicx}
\usepackage{amsmath}

\setcounter{MaxMatrixCols}{10}
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%TCIDATA{Version=5.00.0.2606}
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
%TCIDATA{BibliographyScheme=Manual}
%TCIDATA{Created=Mon Feb 28 15:45:07 2005}
%TCIDATA{LastRevised=Saturday, March 19, 2005 00:03:19}
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%TCIDATA{<META NAME="DocumentShell" CONTENT="General\Blank Document">}
%TCIDATA{CSTFile=LaTeX article (bright).cst}

\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}
\newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\input{tcilatex}

\begin{document}


\bigskip

\bigskip

$W_{ekviv}=\frac{\frac{0.24242\ast 55}{s(0.015s+1)}}{1+\frac{0.24242\ast 55}{%
s(0.015s+1)}\ast \frac{K_{4}T_{4}s}{T_{4}s+1}}=\frac{13.3331(T_{4}s+1)}{%
s(T_{4}s+1)(0.015s+1)+13.3331K_{4}T_{4}s}$

\bigskip

$D(s)=s(T_{4}s+1)(0.015s+1)+13.3331K_{4}T_{4}s=\allowbreak 0.0\,\allowbreak
15T_{4}s^{3}+s^{2}(T_{4}+0.0\,\allowbreak 15)+s(1+13.\,\allowbreak
333K_{4}T_{4})$

$\bigskip $

\bigskip $\Delta _{3}= 
\begin{array}{ccc}
a_{1} & a_{3} & 0 \\ 
a_{0} & a_{2} & 0 \\ 
0 & a_{1} & a_{3}%
\end{array}
$

$\Delta _{1}=a_{1}=T_{4}+0.015>0$ \ \ \ \ \ \ \ \ $=>T_{4}>-0.015$

$\Delta _{2}=a_{1}a_{2}-a_{0}a_{3}(T_{4}+0.0\,\allowbreak
15)(1+13.\,\allowbreak 3331K_{4}T_{4})=\allowbreak T_{4}+13.\,\allowbreak
333K_{4}T_{4}^{2}+0.0\,\allowbreak 15+0.\,\allowbreak 2\allowbreak
K_{4}T_{4}>0$

\bigskip

$K_{4}>-\frac{0.0\,\allowbreak 15+T_{4}}{13.\,\allowbreak
333T_{4}^{2}+0.\,\allowbreak 2\allowbreak T_{4}}$ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ $T_{4}>0$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $=>K_{4}=1$

\bigskip

$13.3331T_{4}^{2}+1.2T_{4}+0.015=0$

$\left\{ T_{4}=-0.07\,\allowbreak 500\,2\right\} ,\left\{
T_{4}=-0.0\,\allowbreak 15\right\} \allowbreak $ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ tak kak$\ \ T_{4}>0\ \ =>\ T_{4}=0.01$

\bigskip

$\Delta _{3}= 
\begin{array}{ccc}
0.025 & 0 & 0 \\ 
0.00015 & 1.133331 & 0 \\ 
0 & 0.025 & 0%
\end{array}
$, determinant: $0$

\bigskip

$W_{ekviv}=\frac{13.3331(0.01s+1)}{s(0.01s+1)(0.015s+1)+0.133331s}=\frac{%
13.3331(0.01s+1)}{s(0.00015s^{2}+0.025s^{2}+1.133331)}=\frac{11.765(0.01s+1)%
}{s(0.000132353s^{2}+0.022058868s+1)}$

\bigskip

===============================================================

\bigskip

$W_{n/sk}=\frac{11.765(0.01s+1)}{s(0.000132353s^{2}+0.022058868s+1)}\ast 
\frac{1.5}{0.04s^{2}+0.16s+1}\ast \frac{1}{s}$

\bigskip

$\omega _{1}=5$ c$^{-1}$

$\omega _{2}=87$ c$^{-1}$

$\omega _{3}=100$ c$^{-1}$

\bigskip

$20\lg K=20\lg 17.6475=25$ \ db

================================================================

\bigskip

$\varepsilon _{ust.}=1$

\bigskip

$T_{reg}=2\sqrt[2]{\frac{\varepsilon _{ust.}}{g^{\prime \prime }(t)_{\max }}}%
=2\sqrt[2]{\frac{1}{20}}=0.45$ \ c

\bigskip

$\omega _{cp}=\frac{b\pi }{T_{reg}}=\frac{1.7\pi }{0.45}=\allowbreak
11.\,\allowbreak 87$ \ c$^{-1}$ \ ( $\delta =15\%)$

\bigskip ===============================================================

\bigskip

$\Delta \theta =\pi -\frac{\pi }{2}\nu +\sum \arctan \frac{\omega _{cp}}{%
\omega _{i}^{zhel}}-\sum \arctan \frac{\omega _{cp}}{\omega _{j}^{zhel}}$

$\Delta \theta =\arctan \frac{12}{2.6}-4\arctan \frac{12}{87}+\arctan \frac{%
12}{100}=\allowbreak 53.2$

\bigskip

===============================================================

\bigskip $20lgK_{KU}=-32.5$

$K_{KU}=0.0237$

\bigskip

$T_{1}=\frac{1}{2.6}=0\allowbreak .\,\allowbreak 385$

$\bigskip $

$T_{2}=\frac{1}{87}=\allowbreak 0.01$

$W_{KU}=\frac{0.0237s(0.01s+1)^{4}}{0.385s+1}$

\bigskip

\end{document}
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