# Difference: Ctf3Sensitivity (15 vs. 16)

#### Revision 1611 Feb 2014 - FrankieCullinan

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# Cavity Sensitivity Estimates

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## Finite Decay Time

Because the signal decays with a finite decay time, the assumption that the root mean square voltage over one period is the peak voltage divided by is no longer valid. The corresponding value can be found by

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where f is the signal frequency, τ is the decay time and T is the time of one period in time t and is equal to 1/f. Evaluating the definite integral gives
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\begin{equation*} \frac{V_{rms}(\tau)}{\hat{V}}=\int_0^T e^{-\frac{2t}{\tau}}\sin^2 (\omega t) dt = \frac{1}{2}\int_0^T e^{-\frac{2t}{\tau}}(1-\cos (2 \omega t) dt \end{equation*} %ENDLATEX% where ω is the angular frequency, τ is the decay time and T is the time of one period in time t. Evaluating the integral gives
This is one case where the signal is given by a sine function. In the opposing case, where a cosine function is used instead, the expression evaluates to %BEGINLATEX{}%
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\begin{displaymath}
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\frac{V_{rms}(\tau)}{\hat{V}}=\frac{\tau}{4}\left(1+\frac{1}{1+\omega^2\tau^2}\right)\left(1-e^{-\frac{T}{\tau}}\right)\, .
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\left(\frac{V_{rms}(\tau)}{\hat{V}}\right)^2=\frac{\tau}{4T}\left(1-\frac{1}{1+(2 \pi f \tau)^2}\right)\left(1-e^{-\frac{2 T}{\tau}}\right)
\end{displaymath} %ENDLATEX%
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This is one case where the signal is given by a sine function. In the opposing case, where a cosine function is used instead, the expression evaluates to
The median of these two cases is
Another effect is the signal decaying before the first peak in the oscillation is seen. The peak time in the case of a

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