Difference: Ctf3Sensitivity (14 vs. 15)

Revision 1511 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time, the assumption that the root mean square voltage over one period is the peak voltage divided by $\sqrt{2}$ is no longer valid. The corresponding value can be found by

 \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*}
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
 \begin{equation*} \frac{V_{rms}(\tau)}{\hat{V}}=\frac{\tau}{4T}\left(1-\frac{1}{1+\omega^2\tau^2}\right)\left(1-e^{-\frac{T}{\tau}}\right) \end{equation*}
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
 \begin{displaymath} \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)\, . \end{displaymath}
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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