autophase.tex 11.2 KB
 snuverink_j committed Sep 06, 2017 1 2 3 4 5 6 7 8 \input{header} \chapter{Auto-phasing Algorithm} \label{chp:autophasing} %\index{Autophase Algorithm} \section{Standing Wave Cavity} \index{Autophase Algorithm!Standing Wave Cavity}  snuverink_j committed Sep 07, 2017 9 In \textit{OPAL-t} the elements are implemented as external fields that are read in from a file. The fields are described by a 1D, 2D or 3D sampling (equidistant or non-equidistant). To get the actual field at any position a linear interpolation multiplied by $\cos(\omega t + \varphi)$, where $\omega$ is the frequency and $\varphi$ is the lag. The energy gain of a particle then is  snuverink_j committed Sep 06, 2017 10 11 12 13 14  \Delta E(\varphi,r) = q\,V_{0}\,\int_{z_\text{begin}}^{z_\text{end}} \cos(\omega t(z,\varphi) + \varphi) E_z(z, r) dz. To maximize the energy gain we have to take the derivative with respect to the lag, $\varphi$ and set the result to zero: \begin{multline}  snuverink_j committed Sep 11, 2017 15 16 17 \differential{\Delta E(\varphi,r)}{\varphi} = -\int_{z_\text{begin}}^{z_\text{end}} (1 + \omega \diffp{t(z,\varphi)}{\varphi}) \sin(\omega t(z,\varphi) + \varphi) E_z(z,r)\\ = -\cos(\varphi) \int_{z_\text{begin}}^{z_\text{end}} (1 + \omega \diffp{t(z,\varphi)}{\varphi}) \sin(\omega t(z,\varphi)) E_z(z,r) dz \\ -\sin(\varphi) \int_{z_\text{begin}}^{z_\text{end}} (1 + \omega \diffp{t(z,\varphi)}{\varphi}) \cos(\omega t(z,\varphi)) E_z(z,r) dz \stackrel{!}{=} 0.  snuverink_j committed Sep 06, 2017 18 19 20 21 22 23 24 25 \end{multline} Thus to get the maximum energy the lag has to fulfill \label{eq:rulelag} \tan(\varphi) = -\frac{\Gamma_1}{\Gamma_2}, where \label{eq:Gamma1}  snuverink_j committed Sep 11, 2017 26  \Gamma_1 = \sum_{i=1}^{N-1} (1 + \omega \diffp{t}{\varphi}) \int_{z_{i-1}}^{z_{i}} \sin\left(\omega (t_{i-1} + \Delta t_{i}\frac{z-z_{i-1}}{\Delta z_{i}})\right)\left(E_{z,i-1} + \Delta E_{z,i} \frac{z-z_{i-1}}{\Delta z_{i}}\right) dz  snuverink_j committed Sep 06, 2017 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42  and \label{eq:Gamma2} \Gamma_2 = \sum_{i=1}^{N-1} (1 + \omega \diffp{t}{\varphi}) \int_{z_{i-1}}^{z_{i}} \cos\left(\omega (t_{i-1} + \Delta t_{i}\frac{z-z_{i-1}}{\Delta z_{i}})\right)\left(E_{z,i-1} + \Delta E_{z,i} \frac{z-z_{i-1}}{\Delta z_{i}}\right) dz. Between two sampling points we assume a linear correlation between the electric field and position respectively between time and position. The products in the integrals between two sampling points can be expanded and solved analytically. We then find \begin{equation*} \Gamma_1 = \sum_{i=1}^{N-1} (1 + \omega \diff{t}{\varphi}) \Delta z_{i}(E_{z,i-1} (\Gamma_{11,i} - \Gamma_{12,i}) + E_{z,i}\, \Gamma_{12,i}) \end{equation*} and \begin{equation*} \Gamma_1 = \sum_{i=1}^{N-1} (1 + \omega \diffp{t}{\varphi}) \Delta z_{i}(E_{z,i-1} (\Gamma_{21,i} - \Gamma_{22,i}) + E_{z,i}\, \Gamma_{22,i}) \end{equation*} where \begin{align*}  snuverink_j committed Sep 11, 2017 43 44 45 46  \Gamma_{11,i} &= \int_0^1 \sin(\omega(t_{i-1} + \tau \Delta t_{i})) d\tau = - \frac{\cos(\omega t_{i}) - \cos(\omega t_{i-1})}{\omega \Delta t_{i}}\\ \Gamma_{12,i} &= \int_0^1 \sin(\omega(t_{i-1} + \tau \Delta t_{i})) \tau d\tau = \frac{-\omega \Delta t_{i} \cos(\omega t_{i}) + \sin(\omega t_{i}) - \sin(\omega t_{i-1})}{\omega^2 (\Delta t_{i})^2}\\ \Gamma_{21,i} &= \int_0^1 \cos(\omega(t_{i-1} + \tau \Delta t_{i})) d\tau = \frac{\sin(\omega t_{i}) - \sin(\omega t_{i-1})}{\omega \Delta t_{i}}\\ \Gamma_{22,i} &= \int_0^1 \cos(\omega(t_{i-1} + \tau \Delta t_{i})) \tau d\tau = \frac{\omega \Delta t_{i} \sin(\omega t_{i}) + \cos(\omega t_{i}) - \cos(\omega t_{i-1})}{\omega^2 (\Delta t_{i})^2}  snuverink_j committed Sep 06, 2017 47 48 \end{align*}  snuverink_j committed Sep 07, 2017 49 It remains to find the progress of time with respect to the position. In \textit{OPAL} this is done iteratively starting with  snuverink_j committed Sep 06, 2017 50 51 52 53 54 \begin{verbatim} K[i] = K[i-1] + (z[i] - z[0]) * q * V; b[i] = sqrt(1. - 1. / ((K[i] - K[i-1]) / (2.*m*c^2) + 1)^2); t[i] = t[0] + (z[i] - z[0]) / (c * b[i]) \end{verbatim}  snuverink_j committed Sep 06, 2017 55 By doing so we assume that the kinetic energy, K, increases linearly and proportional to the maximal voltage. With this model for the progress of time we can calculate $\varphi$ according to Equation~\ref{rulelag}. Next a better model for the kinetic Energy can be calculated using  snuverink_j committed Sep 06, 2017 56 57 58 59 60 61 62 63 64 65 66 67 68 \begin{alltt} K[i] = K[i-1] + q $$\Delta$$z[i](cos($$\varphi$$)(Ez[i-1]($$\Gamma\sb{21}$$[i] - $$\Gamma\sb{22}$$[i]) + Ez[i]$$\Gamma\sb{22}$$[i]) $$\,\,$$- sin($$\varphi$$)(Ez[i-1]($$\Gamma\sb{11}$$[i] - $$\Gamma\sb{12}$$[i]) + Ez[i]$$\Gamma\sb{12}$$[i])). \end{alltt} With the updated kinetic energy the time model and finally a new $\varphi$, that comes closer to the actual maximal kinetic energy, can be obtained. One can iterate a few times through this cycle until the value of $\varphi$ has converged. \section{Traveling Wave Structure} \index{Autophase Algorithm!Traveling Wave Structure} \begin{figure} \includegraphics[angle=0,width=0.9\linewidth]{figures/field_crop} \caption{Field map 'FINLB02-RAC.T7' of type 1DDynamic} \label{fig:tws} \end{figure}  snuverink_j committed Sep 06, 2017 69 Auto phasing in a traveling wave structure is just slightly more complicated. The field of this element is composed of a standing wave entry and exit fringe field and two standing waves in between, see Figure~\ref{tws}.  snuverink_j committed Sep 06, 2017 70 71 72 73 74 75 \begin{multline} \Delta E(\varphi,r) = q\, V_{0}\,\int_{z_\text{begin}}^{z_\text{beginCore}} \cos(\omega t(z,\varphi) + \varphi) E_z(z, r) dz \\ + q\, V_\text{core}\,\int_{z_\text{beginCore}}^{z_\text{endCore}} \cos(\omega t(z,\varphi) + \varphi_\text{c1} + \varphi) E_z(z, r) dz \\ + q\, V_\text{core}\,\int_{z_\text{beginCore}}^{z_\text{endCore}} \cos(\omega t(z,\varphi) + \varphi_\text{c2} + \varphi) E_z(z + s, r) dz \\ + q\, V_{0}\,\int_{z_\text{endCore}}^{z_\text{end}} \cos(\omega t(z,\varphi) + \varphi_\text{ef} + \varphi) E_z(z, r) dz, \end{multline}  snuverink_j committed Sep 06, 2017 76 where $s$ is the cell length. Instead of one sum as in Equation~\ref{Gamma1,Gamma2} there are four sums with different numbers of summands.  snuverink_j committed Sep 06, 2017 77 78 79 80 81 82 83 84 85 86 87 \subsection*{Example} \begin{fmpage} \begin{verbatim} FINLB02_RAC: TravelingWave, L=2.80, VOLT=14.750*30/31, NUMCELLS=40, FMAPFN="FINLB02-RAC.T7", ELEMEDGE=2.67066, MODE=1/3, FREQ=1498.956, LAG=FINLB02_RAC_lag; \end{verbatim} \end{fmpage} For this example we find \begin{align*}  snuverink_j committed Sep 11, 2017 88  V_\text{core} &= \frac{V_{0}}{\sin(2.0/3.0 \pi)} = \frac{2 V_{0}}{\sqrt{3.0}}\\  snuverink_j committed Sep 06, 2017 89 90  \varphi_\text{c1} &= \frac{\pi}{6}\\ \varphi_\text{c2} &= \frac{\pi}{2}\\  snuverink_j committed Sep 06, 2017 91  \varphi_\text{ef} &= - 2\pi \cdot(\text{\texttt{NUMCELLS}} - 1) \cdot \text{\texttt{MODE}} = 26\pi  snuverink_j committed Sep 06, 2017 92 93 94 95 96 97 98 99 100 101 102 \end{align*} \subsection{Alternative Approach for Traveling Wave Structures} If $\beta$ doesn't change much along the traveling wave structure (ultra relativistic case) then $t(z,\varphi)$ can be approximated by $t(z,\varphi)=\frac{\omega}{\beta c}z + t_{0}$. For the example from above the energy gain is approximately \begin{multline*} \Delta E(\varphi,r) = q\;V_0 \int_{0}^{1.5\cdot s} \cos\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \varphi\right) E_z(z,r)\, dz\\ + \frac{2 q\;V_{0}}{\sqrt{3}} \int_{1.5\cdot s}^{40.5\cdot s}\cos\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \frac{\pi}{6} + \varphi \right) E_z(z\;\;\quad,r) dz\\ + \frac{2 q\;V_{0}}{\sqrt{3}} \int_{1.5\cdot s}^{40.5\cdot s}\cos\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \frac{\pi}{2} + \varphi \right) E_z(z+s,r) dz \\ + q\;V_{0} \int_{40.5\cdot s}^{42\cdot s} \cos\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \varphi\right) E_z(z,r)\, dz. \end{multline*} Here $\beta c = 2.9886774\cdot10^8\;\text{m s}^{-2}$, $\omega = 2\pi\cdot 1.4989534\cdot10^9$~Hz and, the cell length, $s = 0.06\bar{6}$~m. To maximize this energy we have to take the derivative with respect to $\varphi$ and set the result to $0$. We split the field up into the core field, $E_z^{(1)}$ and the fringe fields (entry fringe field plus first half cell concatenated with the exit fringe field plus last half cell), $E_z^{(2)}$. The core fringe field is periodic with a period of $3\,s$. We thus find \begin{multline*}  snuverink_j committed Sep 11, 2017 103 104 105 106 0 \stackrel{!}{=} \int_{0}^{1.5\cdot s} \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \varphi\right) E_z^{(2)}(z,r)\, dz \\ + \frac{2}{\sqrt{3}} \int_{0}^{39\cdot s}\sin\left(\omega \left(\frac{z + 1.5\,s}{\beta c} + t_{0}\right) + \frac{\pi}{6} + \varphi \right) E_z^{(1)}(z \text{ mod}(3\,s)\;\;\qquad,r)\,dz \\ + \frac{2}{\sqrt{3}} \int_{0}^{39\cdot s}\sin\left(\omega \left(\frac{z + 1.5\,s}{\beta c} + t_{0}\right) + \frac{\pi}{2} + \varphi \right) E_z^{(1)}((z + s) \text{ mod} (3\,s),r)\, dz \\ + \int_{1.5\cdot s}^{3\cdot s} \sin\left(\omega\left(\frac{z + 39\,s}{\beta c} + t_{0}\right) + \varphi\right) E_z^{(2)}(z,r)\, dz  snuverink_j committed Sep 06, 2017 107 108 109 \end{multline*} This equation is much simplified if we take into account that $\omega / \beta c \approx 10\pi$. We then get \begin{multline*}  snuverink_j committed Sep 11, 2017 110 111 112 113 0 \stackrel{!}{=} \int_{0}^{3\cdot s} \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \varphi\right) E_z^{(2)}(z)\, dz \\ + \frac{26}{\sqrt{3}} \int_{0}^{3\cdot s}\left(\sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \frac{7\pi}{6} + \varphi \right) + \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \frac{5\pi}{6} + \varphi \right)\right) E_z^{(1)}(z)\, dz \\ = \int_{0}^{3\cdot s}\sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \varphi\right) \left(E_z^{(2)} - 26\cdot E_z^{(1)}\right)(z)\,dz  snuverink_j committed Sep 06, 2017 114 115 116 \end{multline*} where we used \begin{multline*}  snuverink_j committed Sep 11, 2017 117 118 119 \int_{0}^{3\cdot s} \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \frac{3\pi}{2} + \varphi\right) E_z^{(1)}((z + s) \text{ mod}(3\,s),r) dz \\ \stackrel{z' = z + s}{\longrightarrow} \int_{s}^{4\cdot s} \sin\left(\omega \left(\frac{z'-s}{\beta c} + t_{0} \right) + \frac{3\pi}{2} + \varphi\right)E_z^{(1)}(z' \text{ mod}(3\,s),r)dz' \\ = \int_{0}^{3\cdot s} \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \frac{5\pi}{6} + \varphi\right) E_z^{(1)}(z,r)\,dz.  snuverink_j committed Sep 06, 2017 120 \end{multline*}  snuverink_j committed Sep 11, 2017 121 In the last equal sign we used the fact that both functions, $\sin(\frac{\omega}{\beta c}z)$ and $E_z^{(1)}$ have a periodicity of $3\cdot s$ to shift the boundaries of the integral.  snuverink_j committed Sep 06, 2017 122 123 124 125 126 127 128 129 130 131  Using the convolution theorem we find \begin{equation*} 0 \stackrel{!}{=} \int_{0}^{3\cdot s} g(\xi - z) (G - 26 \cdot H)(z) \, dz = \mathcal{F}^{-1}\left(\mathcal{F}(g)\cdot(\mathcal{F}(G) - 26 \cdot \mathcal{F}(H))\right) \end{equation*} where \begin{align*} g(z) & = \begin{cases}  snuverink_j committed Sep 11, 2017 132  -\sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right)\right)\qquad & 0 \le z \le 3\cdot s\\  snuverink_j committed Sep 06, 2017 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149  0 & \text{otherwise} \end{cases}\\ G(z) & = \begin{cases} E_z^{(2)}(z) \qquad & 0 \le z \le 3\cdot s\\ 0 & \text{otherwise} \end{cases}\\ H(z) & = \begin{cases} E_z^{(1)}(z) \qquad & 0 \le z \le 3\cdot s\\ 0 & \text{otherwise} \end{cases} \intertext{and} -\frac{\omega}{\beta c} \xi &= \varphi. \end{align*} Here we also used some trigonometric identities: \begin{multline*}  snuverink_j committed Sep 11, 2017 150 151 152 153 154 155  \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \pi + \frac{\pi}{6} + \varphi \right) + \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \pi - \frac{\pi}{6} + \varphi \right) \\ = -\left(\sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \frac{\pi}{6} + \varphi\right) + \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) - \frac{\pi}{6} + \varphi\right)\right) \\ = -2\cdot \cos\left(\frac{\pi}{6}\right) \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \varphi\right) \\ = -\sqrt{3} \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \varphi\right)  snuverink_j committed Sep 06, 2017 156 157 158 \end{multline*} \input{footer}