Difference between revisions of "Sandbox"

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alpha: <math>1+1</math>
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{{LaTeX Shortcuts}}
<!--
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$$\newcommand{\blah}{\alpha\beta}$$
<center><math>
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$$\newcommand{\blaha}{\alpha\beta}$$
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$$\newcommand{\blahb}{\alpha\beta}$$
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$$\newcommand{\blahc}{\alpha\beta}$$
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$$\newcommand{\blahd}{\alpha\beta}$$
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There is $$\frac{g \beta \Delta T H^3}{\alpha \nu}$$ and then there is
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\[ \frac{g \beta \Delta T H^3}{\alpha \nu} \]
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$$
 
\begin{align}
 
\begin{align}
H&=\frac{j\omega+8}{(j\omega)^2+4*j\omega+13}\\
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\frac{g \beta \Delta T H^3}{\alpha \nu}
H&=\frac{s+8}{(s+2)^2+(3)^2}=\frac{A(s+2)+B(3)}{(s+2)^2+(3)^2}
 
 
\end{align}
 
\end{align}
</math></center>
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$$
From the <math>s</math> in the numerator, you can see that <math>A</math> is 1.  That means the numerator so far is <math>1(s+2)</math>  Therefore, there is already a constant 2 up top.  To get the total of 8, then <math>B</math> needs to be 2.  That is:
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\[
<center><math>
 
 
\begin{align}
 
\begin{align}
H&=\frac{s+8}{(s+2)^2+(3)^2}=\frac{A(s+2)+B(3)}{(s+2)^2+(3)^2}=\frac{1(s+2)+2(3)}{(s+2)^2+(3)^2}
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\frac{g \beta \Delta T H^3}{\alpha \nu}
 
\end{align}
 
\end{align}
</math></center>
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\]
meaning
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<center><math>
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$$\blah$$
h(t):=e^{-2t}\left(1*\cos(3t)+2*\sin(3t)\right)~u(t)
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</math></center>
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-->
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Latest revision as of 16:12, 11 February 2024

$$\newcommand{E}[2]{#1_{\mathrm{#2}}}$$ $$\newcommand{\blah}{\alpha\beta}$$ $$\newcommand{\blaha}{\alpha\beta}$$ $$\newcommand{\blahb}{\alpha\beta}$$ $$\newcommand{\blahc}{\alpha\beta}$$ $$\newcommand{\blahd}{\alpha\beta}$$ There is $$\frac{g \beta \Delta T H^3}{\alpha \nu}$$ and then there is \[ \frac{g \beta \Delta T H^3}{\alpha \nu} \] $$ \begin{align} \frac{g \beta \Delta T H^3}{\alpha \nu} \end{align} $$ \[ \begin{align} \frac{g \beta \Delta T H^3}{\alpha \nu} \end{align} \]

$$\blah$$


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