\section{Discussion}
\label{sec:calibcant:discussion}
-\subsection{Fitting with a Lorentzian}
-\label{sec:calibcant:lorentzian}
-
-It is popular to refer to the thermal power spectral density as a
-``Lorentzian''\citep{howard88,hutter93,roters96,levy02,florin95} even
-though \cref{eq:model-psd} differs from the classic
-Lorentzian\citep{mathworld-lorentzian}.
-\begin{equation}
- L(x) = \frac{1}{\pi}\frac{\frac{1}{2}\Gamma}
- {(x-x_0)^2 + \p({\frac{1}{2}\Gamma})^2} \;,
- \label{eq:lorentzian}
-\end{equation}
-where $x_0$ sets the center and $\Gamma$ sets the width of the curve.
-It is unclear whether the references are due to uncertainty about the
-definition of the Lorentzian or to the fact that \cref{eq:model-psd}
-is also peaked and therefore \cref{eq:lorentzian} a potential
-substitute for \cref{eq:model-psd}. \citet{florin95}
-likely \emph{are} using \cref{eq:lorentzian}, as the slope of the
-fitted \PSD\ in their figure 2, has a slope at $f=0$.
-Using \cref{eq:model-psd}, the derivative would have been zero, as we
-can see by using the chain rule repeatedly,
-
-\begin{align}
- \deriv{f}{\PSD_f}
- &= \deriv{f}{}\p({\frac{G_{1f}}{(f_0^2-f^2)^2 + \beta_f^2 f^2}})
- = \frac{-G_{1f}}{\p({(f_0^2-f^2)^2 + \beta_f^2 f^2})^2}
- \deriv{f}{}\p({(f_0^2-f^2)^2 + \beta_f^2 f^2}) \\
- &= \frac{-G_{1f}}{\p({(f_0^2-f^2)^2 + \beta_f^2 f^2})^2}
- \p({2(f_0^2-f^2)\deriv{f}{}(f_0^2 - f^2) + 2\beta_f^2 f}) \\
- &= \frac{-G_{1f}}{\p({(f_0^2-f^2)^2 + \beta_f^2 f^2})^2}
- \p({-4f(f_0^2-f^2) + 2\beta_f^2 f}) \\
- &= \frac{2G_{1f}f}{\p({(f_0^2-f^2)^2 + \beta_f^2 f^2})^2}
- \p({2(f_0^2-f^2) - \beta_f^2}) \\
- \left.\deriv{f}{\PSD_f}\right|_{f=0} &= 0 \;.
- \label{eq:model-psd-df}
-\end{align}
-
-However, \citet{benedetti12} has a solid derivation of
-\cref{eq:DHO-psd}, which he then refers to as the ``Lorentzian''. In
-order to avoid any uncertainty, we leave \cref{eq:model-psd} unnamed.
-I encourage future researchers to explicitly list the model they use,
-ideally by citing their associated open source calibration package.
-
\subsection{Peak frequency}
\label{sec:calibcant:peak-frequency}
value for $V_p$ is given by
\begin{equation}
\avg{V_p(t)^2} = \frac{\pi G_{1f}}{2\beta_f f_0^2} \;.
- \label{eq:Vp-from-freq-fit}
+ \label{eq:avg-Vp-Gone-f}
\end{equation}
%
\nomenclature[PSDf]{$\PSD_f$}{Power spectral density in
\nomenclature{$f$}{Frequency (hertz)}
\nomenclature{$f_0$}{Resonant frequency (hertz)}
-Combining \cref{eq:equipart_k,eq:x-from-Vp,eq:Vp-from-freq-fit}, we
+Combining \cref{eq:equipart_k,eq:x-from-Vp,eq:avg-Vp-Gone-f}, we
have
\begin{align}
\kappa &= \frac{\sigma_p^2 k_BT}{\avg{V_p(t)^2}}
%as do
% see Gittes 1998 for more thermal noise details
% see Berg-Sorensen05 for excellent overdamped treament.
+
+\section{Fitting with a Lorentzian}
+\label{sec:calibcant:lorentzian}
+
+It is popular to refer to the thermal power spectral density as a
+``Lorentzian''\citep{howard88,hutter93,roters96,levy02,florin95}, but
+there is dissagreement on what this means. The classic Lorentzian
+function is\citep{mathworld-lorentzian}
+\begin{equation}
+ L(x) = \frac{1}{\pi}\frac{\frac{1}{2}\Gamma}
+ {(x-x_0)^2 + \p({\frac{1}{2}\Gamma})^2} \;,
+ \label{eq:lorentzian}
+\end{equation}
+where $x_0$ sets the center and $\Gamma$ sets the width of the curve.
+However, the correct \PSD\ for a damped harmonic oscillator in bath of
+white noise is given by \cref{eq:psd-Vp}\citep{burnham03,benedetti12}.
+
+These formula are fundamentally different.
+
+For example, the slope of \cref{eq:psd-Vp} is zero at $f=0$, as we can
+see by using the chain rule repeatedly,
+\begin{align}
+ \deriv{f}{\PSD_f}
+ &= \deriv{f}{}\p({\frac{G_{1f}}{(f_0^2-f^2)^2 + \beta_f^2 f^2}})
+ = \frac{-G_{1f}}{\p({(f_0^2-f^2)^2 + \beta_f^2 f^2})^2}
+ \deriv{f}{}\p({(f_0^2-f^2)^2 + \beta_f^2 f^2}) \\
+ &= \frac{-G_{1f}}{\p({(f_0^2-f^2)^2 + \beta_f^2 f^2})^2}
+ \p({2(f_0^2-f^2)\deriv{f}{}(f_0^2 - f^2) + 2\beta_f^2 f}) \\
+ &= \frac{-G_{1f}}{\p({(f_0^2-f^2)^2 + \beta_f^2 f^2})^2}
+ \p({-4f(f_0^2-f^2) + 2\beta_f^2 f}) \\
+ &= \frac{2G_{1f}f}{\p({(f_0^2-f^2)^2 + \beta_f^2 f^2})^2}
+ \p({2(f_0^2-f^2) - \beta_f^2})
+ \label{eq:model-psd-df} \\
+ \left.\deriv{f}{\PSD_f}\right|_{f=0} &= 0 \;.
+ \label{eq:model-psd-df-zero}
+\end{align}
+On the other hand, the slope of \cref{eq:lorentzian} is only zero at
+the peak (where $x=x_0$).
+\begin{align}
+ \deriv{x}{L(x)}
+ &= \frac{1}{\pi}\frac{\frac{-1}{2}\Gamma}
+ {\p({(x-x_0)^2 + \p({\frac{1}{2}\Gamma})^2})^2}
+ \cdot \deriv{x}{}\p({(x-x_0)^2 + \p({\frac{1}{2}\Gamma})^2}) \\
+ &= \frac{1}{\pi}\frac{\frac{-1}{2}\Gamma}
+ {\p({(x-x_0)^2 + \p({\frac{1}{2}\Gamma})^2})^2}
+ \cdot 2 (x-x_0) \\
+ &= \frac{1}{\pi}\frac{-\Gamma (x-x_0)}
+ {\p({(x-x_0)^2 + \p({\frac{1}{2}\Gamma})^2})^2}
+ \label{eq:lorentzian-dx}
+\end{align}
+
+It is unclear whether the ``Lorentzian'' references are due to
+uncertainty about the definition of the Lorentzian or to the fact that
+the two equations have similar behaviour near the
+peak. \citet{florin95} likely \emph{are} using \cref{eq:lorentzian},
+as the slope of the fitted \PSD\ in their \fref{figure}{2}, has a
+slope at $f=0$. If they were using \cref{eq:psd-Vp}, the derivative
+would have been zero (\cref{eq:model-psd-df-zero}).
+
+We have at least two models in use, one likely the
+``Lorentzian'' (\cref{eq:lorentzian}) and one that's not. Perhaps
+researchers claiming to use the ``Lorentzian'' are consistently
+using \cref{eq:lorentzian}? There is at least one
+counterexample: \citet{benedetti12} has a solid derivation of
+\cref{eq:DHO-psd}, which he then refers to as the ``Lorentzian''.
+Which formula are the remaining ``Lorentzian'' fitters using? What
+about groups that only reference their method as ``thermal
+calibration'' without specifying a \PSD\ model? In order to avoid any
+uncertainty, we leave \cref{eq:psd-Vp} unnamed. I encourage future
+researchers to explicitly list the model they use, ideally by citing
+their associated open source calibration package.
Plots of \cref{eq:psd-Vp-offset} fits look better than
\cref{eq:psd-Vp} fits (\cref{fig:calibcant:vibration}), but the
significance on the variance calculated with
-\cref{eq:Vp-from-freq-fit} depends on the amount of background noise
+\cref{eq:avg-Vp-Gone-f} depends on the amount of background noise
in the vibration data. With over an order of magnitude difference
between the power of the damped harmonic oscillator peak and the
background noise, the effect of $P_{0f}$ will be small. With noisier
-\section{Theory}
+\section{Power spectra of damped harmonic oscillators}
\label{sec:calibcant:theory}
-Our cantilever can be approximated as a damped harmonic
-oscillator\index{damped harmonic oscillator}
+As discussed in \cref{sec:calibcant:lorentzian}, the power spectral
+density for a Hookean cantilever is surprisingly ambiguous. In this
+section, I'll derive the frequency-space power spectra of the
+deflection voltage (\cref{eq:avg-Vp-Gone-f,eq:model-psd}), modeling
+the cantilever as a damped harmonic oscillator\index{damped harmonic
+ oscillator}.
\begin{equation}
m\ddt{x} + \gamma \dt{x} + \kappa x = F(t) \;,
\label{eq:DHO}
Plugging \cref{eq:GOdef} into \cref{eq:DHO-psd-F} we have
\begin{equation}
- \PSD(x, \omega) = \frac{G_0/m^2}{(\omega_0^2-\omega^2)^2 +\beta^2\omega^2}\;,
- \label{eq:model-psd} \;.
+ \PSD(x, \omega) = \frac{G_0/m^2}{(\omega_0^2-\omega^2)^2 +\beta^2\omega^2}\;.
+ \label{eq:model-psd}
\end{equation}
Integrating over positive $\omega$ to find the total power per unit
= \iOInf{f}{\frac{1}{2\pi}\PSD_f(x,f)2\pi\cdot}
= \iOInf{f}{\PSD_f(x,f)} \;.
\end{align}
-We can now extract \cref{eq:psd-Vp,eq:Vp-from-freq-fit} from
+We can now extract \cref{eq:psd-Vp,eq:avg-Vp-Gone-f} from
\cref{eq:psd-Vp-Gone,eq:avg-Vp-Gone}.
\begin{align}
\begin{split}
= \frac{2\pi G_1}{16\pi^4(f_0^2-f^2)^2 + \beta^2 4\pi^2f^2} \\
&= \frac{G_1/8\pi^3}{(f_0^2-f^2)^2 + \frac{\beta^2 f^2}{4\pi^2}}
= \frac{G_{1f}}{(f_0^2-f^2)^2 + \beta_f^2 f^2}
+ %\label{eq:psd-Vp}
\end{split} \\
\avg{V_p(t)^2}
&= \frac{\pi \frac{G_1}{(2\pi)^3}}
{2 \frac{\beta}{2\pi} \p({\frac{\omega_0}{2\pi}})^2}
= \frac{\pi G_{1f}}{2 \beta_f f_0^2} \;.
+ %\label{eq:avg-Vp-Gone-f}
\end{align}
where $f_0\equiv\omega_0/2\pi$, $\beta_f\equiv\beta/2\pi$, and
$G_{1f}\equiv G_1/8\pi^3$. Finally, we can generate
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