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Nathanael A. Fortune added EDITOR_S_NOTE_The_above__.tex
over 8 years ago
Commit id: 55325a2d449f393458257146be8bebefdc2fe722
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EDITOR'S NOTE:
The above equation has several errors.
\begin{enumerate}
\item a voltage cannot equal a voltage squared, so $V \neq <(V)^2>$
\item There are two independent (random) noise sources --- the resistor being measured (Johns) and the amplifier circuits (instrumentation) ---so you don't add their voltages $V_{Johnson}$ and $V_{instrumentation}$, you need to add $(V_{\textrm{Johnson Noise}})^2 $ and $(V_{\textrm{instrument noise}})^2$. See textbook and manual.
\item if $V_{Mult}$ means the \textit{multiplication} of two voltage signals $V_s \times V_s$ , it should be written $V^2_{mult}$
\item if $V_{Mult}$ means the voltage signal measured by a \textit{multimeter}, then it should read $V^2_{multimeter}$ = $(V_s \times V_s) / (10 \ V)$ since the output of the multiplier is divided by a 10 V reference signal before being sent to the multimeter
\end{enumerate}
A better set of equations would be
\begin{equation}
(V_{\textrm{total noise}})^2 = (V_{\textrm{Johnson Noise}})^2 + (V_{\textrm{instrument noise}})^2
\end{equation}
and
\begin{equation}
V_{\textrm{multimeter}} = \frac{(V_{\textrm{total noise}})^2 }{\textrm{10 V}} = \frac{ (V_{\textrm{Johnson Noise}})^2 + (V_{\textrm{instrument noise}})^2}{\textrm{10 V}}
\end{equation}
