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Cavity Sensitivity Estimates

Position Cavity

The dipole mode sensitivity was estimated from the following formula, the measured external quality factor of 617 and R/Q from simulation in several codes, of which CST and GdfidL use a hexahedral mesh while Ace3p uses a tetrahedral mesh:

Code R/Q /Ω Sensitivity/V nC-1 mm-1
GdfidL 0.85 8.7
Ace3p 0.77 8.3
CST 0.77 8.3

Reference Cavity

The measured external quality factor of the reference cavity is 203.

Code R/Q /Ω Sensitivity/V nC-1
GdfidL 28.2 87
Ace3p 9.8 52
CST 10 52

possense.png refsense.png

Electronics Chain Propagation

Electronics gain/dB 5
Cable loss/dB -2.6
Position cavity attenuation/dB 6
Reference cavity attenuation/dB 20

Position signal (offset=1mm, charge=1nC) Reference signal (Charge=1nC)
Location Volts dBm Volts dBm
Cavity out 8.3 31.4 52.0 47.3
Electronics in 4.2 25.4 5.2 27.3
Electronics out 7.4 30.4 9.2 32.3
Digitiser in 5.5 27.8 6.9 29.7

  • Position scale factor: 1.25 mm

Correction, June 2013

The mode R/Qs were now calculated by integrating over the full geometry instead of just the cavity length:. The results for the position cavity are:

Code R/Q /Ω Sensitivity/V nC-1 mm-1
GdfidL 3.26 17.08
Ace3p 3.27 17.11
CST 4.55 20.2

The results for the reference cavity are:

Code R/Q /Ω Sensitivity/V nC-1
GdfidL 50.3 117
Ace3p 50.7 118

Position signal (offset=1mm, charge=1nC) Reference signal (Charge=1nC)
Location Volts dBm Volts dBm
Cavity out 14.7 36.4 118 54.4
Electronics in 3.37 23.6 5.39 27.6
Electronics out 5.99 9.60 9.2 32.6
Digitiser in 4.44 26.0 7.11 30.0

  • Position scale factor: 0.62 mm

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time, the assumption that the root mean square voltage over one period is the peak voltage divided by $\sqrt{2}$ is no longer valid.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

\[ \left(\frac{V_{rms}(\tau)}{\hat{V}}\right)^2=\frac{1}{T} \int_0^T e^{-\frac{2t}{\tau}}\sin^2 (2 \pi f t) dt = \frac{1}{2T}\int_0^T e^{-\frac{2t}{\tau}}(1-\cos (4 \pi f t)) dt \]
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
 \begin{displaymath} \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}
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
\[ \left(\frac{V_{rms}(\tau)}{\hat{V}}\right)^2=\frac{\tau}{4T}\left(1+\frac{1}{1+(2 \pi f \tau)^2}\right)\left(1-e^{-\frac{2T}{\tau}}\right)\, . \]
The median of these two cases is
\[ \frac{V_{rms}(\tau)}{\hat{V}}=\sqrt{\frac{\tau}{4T}\left(1-e^{-\frac{2T}{\tau}}\right)}\, . \]
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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Topic revision: r16 - 11 Feb 2014 - FrankieCullinan

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