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Beam position monitor handbook

Cavity Beam position monitors

As a beam passes through a cylindrical cavity with position $x$, beam trajectory angle $x^{\prime}$ and tilt $\theta$, the cavity output voltage for the different contributions is given by
  \begin{eqnarray*} V_x(t;\omega,\Gamma;x,q,C_x) & = & C_x q x \Theta(t-t_0) e^{-\Gamma t}\sin(\omega t) \\ V_{x^{\prime}}(t;\omega,\Gamma;x^{\prime},q,C_{x^{\prime}}) & = & C_{x^{\prime}} q \theta \Theta(t-t_0) e^{-\Gamma t}\cos(\omega t)) \\ V_{\theta}(t;\omega,\Gamma;\theta,q,C_{\theta}) & = & -C_{\theta} q \alpha \Theta(t-t_0) e^{-\Gamma t}\cos(\omega t)) \end{eqnarray*} (1)

Generally we can write the output as (using complex oscillating functions)

  \begin{displaymath}  V(x,x^{\prime},\theta,t;t_0,\omega,\Gamma) = q (x C_x + i C_{x^{\prime}} x^{\prime} - i C_{\theta}\theta) \Theta(t-t_0) \exp(-\Gamma t) \sin(\omega t) \end{displaymath}  (2)

So generally we can write the output of the cavity as

  \begin{displaymath}  V(x,x^{\prime},\theta,t;t_0,\omega,\Gamma) = A(x,x^{\prime},\theta) \Theta(t-t_0) \exp(-\Gamma t) \sin(\omega t) \end{displaymath}  (3)

where $A$ is a complex amplitude which contains the information on the beam position, direction and angle of attack. Alternatively we can consider the complex amplitude ($A$) consisting of in-phase $I$ and quadrature-phase $Q$ components, $A = I + i Q $, giving

  \begin{eqnarray*}  I & = & q x C_x \\ Q & = & q (x^{\prime} C_{x^{\prime}} - \theta C_{\theta})  \end{eqnarray*}  (4)

An example of a cavity signal is below

Simple signal processing

  1. Calculate RMS signal in background region
  2. Calculate pedestal, background subtract
  3. Set to zero waveform upto end of background region

Signal processing (mixing/filtering)

The analogue IF is digitized at some frequency $\f_{d}$ so that the IF can be written in sample time $k$, where $t = 1/f_{d}k$.
  \begin{equation*} V_x(k) = V_x(k/f_d) \end{equation*} (5)

The digital signal is mixed with a digital complex local oscillator given by

  \begin{equation*} V_{LO}(k) = e^{j\omega k}  \end{equation*} (6)

 \begin{equation*} V_x(k)V_{LO}(k) =  \end{equation*}

Time domain gaussian filter with

  \begin{eqnarray*} a_0 & = & 1 \\ b_i & = & \frac{\sqrt{2\pi}\Delta f}{f_s}\exp\left(\frac{-t_i^2(2\pi\Delta f)^2}{2f_s^2}\right) \end{eqnarray*}  (7)

Reference frequency phase jitter problem

Consider two cavity signals which are mixed with a digital LO
  \begin{eqnarray*} V_{IF,d}(k) & = & A_d\exp(-i (\omega_{d} k+\phi_d)) \exp(i\omega_d k) \\ V_{IF,r}(k) & = & A_r\exp(-i (\omega_{r} k+\phi_r)) \exp(i\omega_r k) \end{eqnarray*} (8)

If there is any timing jitter in the sample clock so then $k \rightarrow k+\delta k$ we find

  \begin{eqnarray*} V_{IF,d}(k) & = & A_d\exp(-i (\omega_{d} (k+\delta k)+\phi_d)) \exp(i\omega_d (k)) \\ V_{IF,r}(k) & = & A_r\exp(-i (\omega_{r} (k+\delta k)+\phi_r)) \exp(i\omega_r (k)) \end{eqnarray*} (9)

Looking at just the phase of the oscillators

  \begin{eqnarray*} \phi_{IF,d}(k) & = & - (\omega_{d} (k+\delta k)+\phi_d) + \omega_d (k) \\ \phi_{IF,r}(k) & = & - (\omega_{r} (k+\delta k)+\phi_r) + \omega_r (k) \end{eqnarray*} (10)

Simplifying we find

  \begin{eqnarray*} \phi_{IF,d}(k) & = & - \omega_{d}\delta k+\phi_d  \\ \phi_{IF,r}(k) & = & - \omega_{r}\delta k+\phi_r \end{eqnarray*} (11)

Subtracting the two phases $\phi_d-\phi_r$ gives

  \begin{equation*} \Delta \phi_{IF} = \phi_{IF,d}-\phi_{IF,r} = (\omega_{r}-\omega_{d})\delta k + (\phi_d-\phi_d) \end{equation*}  (12)

This is usally not a problem as the reference frequency is the same as the dipole frequency. To use a reference at a different frequency then the phase must be corrected by $(\omega_{r}-\omega_{d})\delta k$. Even if the digitizers both both reference and dipole change together. It is worse for the S-band system as the signal and mixed either side of the LO, so the $(\omega_{r}-\omega_{d})\delta k$ term becomes something like $(\omega_{r}+\omega{d})\delta k$


We can write I and Q as

  \begin{eqnarray*}  I & = & \frac{A_d}{A_r}\cos\left(\phi_d-\phi_r\right) \\ Q & = & \frac{A_d}{A_r}\sin\left(\phi_d-\phi_r\right) \end{eqnarray*}  (13)

Rotating I and Q so that position sensitive part

  \begin{eqnarray*}  I^{\prime} & = &  I\cos \theta_{IQ} + Q \sin \theta_{IQ} \\ Q^{\prime} & = &  -I\sin \theta_{IQ} + Q \cos \theta_{IQ}  \end{eqnarray*}  (14)

Fit line in I-Q plane for gradient $g$ so

  \theta_{IQ} = \tan^{-1}g (15)

The error on the rotation angle is

  \sigma_{\theta} = \cos^2\theta_{IQ} \sigma_{g} (16)

where $g$ is the gradient in the I-Q plane and $\sigma_{g}$ its error.

The error on the rotated $I$ and $Q$ values, $\sigma_{I^{\prime}}$ and $\sigma_{Q^{\prime}}$ is then given by

  \begin{eqnarray*}  \sigma_{I^{\prime}}^2 & = &  \cos^2\theta_{IQ} \sigma_{I}^2 + \sin^2\theta_{IQ} \sigma_{Q}^2 +(-I\sin\theta_{IQ}+Q\cos\theta_{IQ})^2\sigma_{\theta_{IQ}}^2 \\ \sigma_{Q^{\prime}}^2 & = &  \sin^2\theta_{IQ} \sigma_{I}^2 + \cos^2\theta_{IQ} \sigma_{Q}^2 +(-I\cos\theta_{IQ}-Q\sin\theta_{IQ})^2\sigma_{\theta_{IQ}}^2  \end{eqnarray*}  (17)

The BPM can be calibrated via

  \begin{eqnarray*} x      & = & S_x I^{\prime} \\ \theta & = & S_{\theta} Q^{\prime} \end{eqnarray*}  (18)

Calibration Tone

Noise sources

Thermal noise

Thermal noise from cavity at temperature.

  \begin{displaymath} P_{T} = k_{B} T \Delta f \end{displaymath}  (19)

Thermal power ($P_{T}$), temperature ($T$) and bandwidth ($\Delta f$)

Amplitude and phase noise

Extrapolation (error)

The signal at sample time $t_{s}$, $y(y_e)$ can be extrapolated to $t_{e}$ with the following equation :
  \begin{displaymath} y(t_e) = \exp\left[-\Gamma(t_e-t_s)\right] y(t_s) = \exp\left[-\Gamma \Delta_t\right] y(t_s) \end{displaymath}  (20)

The statistical error on the sample at the extrapolated time can be simply calculated via

  \begin{displaymath} \left(\frac{\sigma_y_e}{y_e}\right)^2 = \Delta t^2 \sigma_{\Gamma}^2 + \left(\frac{\sigma_y_s}{y_s}\right)^2 + \Gamma^2 \sigma_{\Delta t}^2 \end{displaymath}  (21)

The third term can be neglected as the error in the sample time is small


Electronics design

  \begin{eqnarray*} V_{signal} & = & k x \\  V_{thermal} & \propto & \sqrt\left(k_B T \Delta f\right)   \end{eqnarray*} (22)

So approximately

  \begin{displaymath} \sigma_x = \frac{\sqrt(k_B T \Delta f)}{k} \end{displaymath} (23)

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Topic revision: r27 - 17 Aug 2012 - StewartBoogert

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