Пример выполнения курсовой работы / swp / WskorrParal

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%TCIDATA{Created=Saturday, March 19, 2005 00:03:38}
%TCIDATA{LastRevised=Saturday, March 19, 2005 20:58:56}
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\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]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\input{tcilatex}

\begin{document}


$W_{\U{441} \U{43a} \U{43e} \U{440} \U{440} \U{435} \U{43a} \U{442} .}^{%
\U{43f} \U{430} \U{440} \U{430} \U{43b} \U{43b} }=W_{\U{43d} \U{435} \U{441} 
\U{43a} .}\frac{1}{1+W_{\U{43e} \U{43a} \U{432} }W_{\U{43e} \U{441} }}=\frac{%
\frac{11.765(0.01s+1)}{s(0.00013s^{2}+0.02s+1)}\frac{1.5}{0.04s^{2}+0.165s+1}%
\frac{1}{s}}{1+\frac{11.765(0.01s+1)}{s(0.00013s^{2}+0.02s+1)}\frac{1.5}{%
0.04s^{2}+0.165s+1}\frac{0.0237s(0.01s+1)^{3}}{0.385s+1}}=$

$\bigskip $

$=\frac{17.6475(0.01s+1)(0.385s+1)}{s^{2}(2\ast 10^{-6}s^{5}+3.21\ast
10^{-4}s^{4}+1.75\ast 10^{-2}s^{3}+0.113s^{2}+0.582s+1.418)}$

\bigskip

\bigskip $\omega _{1}=3.7$ \ (-20)

$\omega _{2}=5$ \ \ \ \ (-40)

$\omega _{3}=2.6$ \ \ \ \ \ (+20)

$\omega _{4}=87.7$\ \ \ \ (-40)

$\omega _{5}=100$ \ \ \ \ \ (+20)

\bigskip

$20\lg k=20\lg 17.6475=25$ \ deb

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

\bigskip \U{416}\U{435}\U{43b}\U{430}\U{435}\U{43c}\U{430}\U{44f} \U{41b}%
\U{410}\U{425} \U{431}\U{443}\U{434}\U{435}\U{442} \U{43d}\U{435}\U{441}%
\U{43a}\U{43e}\U{43b}\U{44c}\U{43a}\U{43e} \U{438}\U{43d}\U{43e}\U{439}.

\bigskip $\omega _{\U{441} \U{43e} \U{43f} \U{440} _{1}}=0.9$ c$^{-1}$ \ \ \
(+20)

$\omega _{\U{441} \U{43e} \U{43f} \U{440} _{2}}=87.7$ c$^{-1}$ \ (-40)

$\omega _{\U{441} \U{43e} \U{43f} \U{440} _{3}}=100$ c$^{-1}$ \ \ (+20)

$\bigskip $

$\omega _{\U{441} \U{440} }=20$ c$^{-1}$

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

$\Delta \theta =\arctan \frac{20}{0.9}-2\arctan \frac{20}{87}+\arctan \frac{%
20}{100}=72.84^{\circ }$ \ \ \ \ \ \ - \U{447}\U{442}\U{43e} \U{441}\U{43e}%
\U{43e}\U{442}\U{432}\U{435}\U{442}\U{441}\U{442}\U{432}\U{443}\U{435}\U{442}
\U{442}\U{440}\U{435}\U{431}\U{43e}\U{432}\U{430}\U{43d}\U{438}\U{44f}\U{43c}
\U{443}\U{441}\U{442}\U{43e}\U{439}\U{447}\U{438}\U{432}\U{43e}\U{441}\U{442}%
\U{438}

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

$W_{\U{43a} \U{443} }^{\U{43f} \U{43e} \U{441} \U{43b} \U{435} \U{434} 
\U{43e} \U{432} }=\frac{1.778(0.27s+1)(0.2s+1)^{2}(1.11s+1)}{(0.385s+1)}$

\bigskip

$W=W_{\U{441} \U{43a} \U{43e} \U{440} \U{440} \U{435} \U{43a} \U{442} }^{%
\U{43f} \U{430} \U{440} \U{430} \U{43b} \U{43b} }W_{\U{43a} \U{443} }^{%
\U{43f} \U{43e} \U{441} \U{43b} \U{435} \U{434} \U{43e} \U{432} }=\frac{%
17.6475(0.01s+1)(0.385s+1)}{s^{2}(2\ast 10^{-6}s^{5}+3.21\ast
10^{-4}s^{4}+1.75\ast 10^{-2}s^{3}+0.113s^{2}+0.582s+1.418)}\frac{%
1.778(0.27s+1)(0.2s+1)^{2}(1.11s+1)}{(0.385s+1)}=$

\bigskip

$=\frac{1.448\ast
10^{-3}s^{6}+0.17s^{5}+2.65s^{4}+16.76s^{3}+50.16s^{2}+68.25s+31.38}{%
0.000\,000\,77\allowbreak s^{6}+0.000\,125\,585\,\allowbreak
s^{5}+0.007\,058\,5s^{4}+0.061\,005\,\allowbreak
s^{3}+0.337\,07s^{2}+1.\,\allowbreak 127\,93s+1.\,\allowbreak 418}$

\bigskip 

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

\bigskip \U{421}\U{442}\U{440}\U{43e}\U{438}\U{43c} \U{41b}\U{410}\U{425} 
\U{441}\U{43a}\U{43e}\U{440}\U{440}\U{435}\U{43a}\U{442}\U{438}\U{440}\U{43e}%
\U{432}\U{430}\U{43d}\U{43d}\U{43e}\U{439} \U{441}\U{438}\U{441}\U{442}%
\U{435}\U{43c}\U{44b} \U{438} \U{432}\U{438}\U{434}\U{438}\U{43c}, \U{447}%
\U{442}\U{43e} \U{437}\U{430}\U{43f}\U{430}\U{441} \U{443}\U{441}\U{442}%
\U{43e}\U{439}\U{447}\U{438}\U{432}\U{43e}\U{441}\U{442}\U{438} \U{43f}%
\U{43e} \U{444}\U{430}\U{437}\U{435} \U{441}\U{43e}\U{441}\U{442}\U{430}%
\U{432}\U{43b}\U{44f}\U{435}\U{442} 58 \U{433}\U{440}\U{430}\U{434}\U{443}%
\U{441}\U{43e}\U{432}.

$\sigma =\frac{h_{\max }-h_{\U{443} \U{441} \U{442} }}{h_{\U{443} \U{441} 
\U{442} }}\ast 100\%=\frac{1.174-1}{1}\ast 100\%=17.4\%$

\bigskip 

t$_{\U{440} \U{435} \U{433} }=$

\end{document}