In [2]:
import matplotlib.pyplot as plt
%matplotlib notebook
import numpy as np
import sympy
sympy.init_printing(use_unicode=True)
Doublet¶
Let's consider a doublet system which is form as $$Q_f-O-Q_d$$
and assume ${f_F}$ and $-{f_D}$ as the focal length of the two quad and a drift of length $L$ between them. Both $f_F$ and $f_D$ are positive numbers
Then the total transfer matrix gives:
\begin{equation} M_{dbl}= \left(\begin{array}{cc} 1 & 0\\ \frac{1}{f_D} & 1 \end{array}\right) \left(\begin{array}{cc} 1 & L\\ 0 & 1 \end{array}\right) \left(\begin{array}{cc} 1 & 0\\ -\frac{1}{f_F} & 1 \end{array}\right) \end{equation}In [3]:
l,fd,ff, f=sympy.symbols("L,f_D,f_F,f")
def mat_q(f):
return sympy.Matrix([[1,0],[1/f,1]])
def mat_d(d):
return sympy.Matrix([[1,d],[0,1]])
mat_dbl=sympy.simplify(mat_q(fd)*mat_d(l)*mat_q(-ff))
mat_dbl
Out[3]:
If we let the two focal length equal, then we got focusing effect that do not depend on horizontal and or vertical direction. And the trace of the matrix is 2!
In [4]:
mat_dbl.subs(fd,f).subs(ff,f)
Out[4]:
FODO cells¶
FODO cell is a type of widely used cell of magnets. It consist of only two quadrupoles and two dipoles or simply drifts between quads. To make a symmetric cell, we use the following sequence: $$\frac{1}{2}\text{QF}-\text{O}-\text{QD}-\text{O}-\frac{1}{2}\text{QF}$$
Betatron transfer matrix¶
We use the following approximation:
- Small bending angle approximation: the transfer matrix for a dipole reduces to a drift space, the space between quads are same, which is $L_1$
- Short quadrupole approximation, so that the quad can be modeled by its focusing length $f$
- Focusing length and defocusing length are same.
Then the matrix is
\begin{equation} M_{FODO}= \left(\begin{array}{cc} 1 & 0\\ -\frac{1}{2f} & 1 \end{array}\right) \left(\begin{array}{cc} 1 & L_1\\ 0 & 1 \end{array}\right) \left(\begin{array}{cc} 1 & 0\\ \frac{1}{f} & 1 \end{array}\right) \left(\begin{array}{cc} 1 & L_1\\ 0 & 1 \end{array}\right) \left(\begin{array}{cc} 1 & 0\\ -\frac{1}{2f} & 1 \end{array}\right) \end{equation}In [4]:
l,f,ff,fd,a=sympy.symbols(r"L_1,f,f_F,f_D,\alpha")
def mat_q(f):
return sympy.Matrix([[1,0],[1/f,1]])
def mat_d(d):
return sympy.Matrix([[1,d],[0,1]])
mat_fodo=sympy.simplify(mat_q(-2*ff)*mat_d(l)*mat_q(fd)*mat_d(l)*mat_q(-2*ff))
mat_fodo.subs(fd, ff)
Out[4]:
The transfer matrix to/from the mid-point of the focusing quad is
\begin{equation} M_{FODO}= \left(\begin{matrix}- \frac{L_{1}^{2}}{2 f^{2}} + 1 & \frac{L_{1}}{f} \left(L_{1} + 2 f\right)\\\frac{L_{1}}{4 f^{3}} \left(L_{1} - 2 f\right) & - \frac{L_{1}^{2}}{2 f^{2}} + 1\end{matrix}\right) \end{equation}The determinant of the matrix is always 1. The trace of the FODO cell is $2- \frac{L_{1}^{2}}{ f^{2}} $.
The starting point can be choose arbitrarily. If we use the midpoint of the defocusing quad:
In [5]:
mat_fodo_2=sympy.simplify(mat_q(2*f)*mat_d(l)*mat_q(-f)*mat_d(l)*mat_q(2*f))
mat_fodo_2
Out[5]:
The transfer matrix becomes:
\begin{equation} M_{FODO,midQD}= \left(\begin{matrix}- \frac{L_{1}^{2}}{2 f^{2}} + 1 & \frac{L_{1}}{f} \left(- L_{1} + 2 f\right)\\- \frac{L_{1}}{4 f^{3}} \left(L_{1} + 2 f\right) & - \frac{L_{1}^{2}}{2 f^{2}} + 1\end{matrix}\right) \end{equation}The trace remain unchanged.
In [6]:
mat_fodo_3=sympy.simplify(mat_d(0)*mat_q(f)*mat_d(l)*mat_q(-f)*mat_d(l))
mat_fodo_3
Out[6]:
If we set the starting point right after the focusing quadrupole, the matrix becomes
\begin{equation} M_{FODO,afterQF}= \left(\begin{matrix}\frac{1}{f} \left(- L_{1} + f\right) & \frac{L_{1}}{f} \left(- L_{1} + 2 f\right)\\- \frac{L_{1}}{f^{2}} & - \frac{L_{1}^{2}}{f^{2}} + \frac{L_{1}}{f} + 1\end{matrix}\right) \end{equation}Again the trace remain unchanged. That is a natural result since trace of the product of matrices is an invariant under cyclic permutation.
Use FODO cell in repeating structure¶
To use the FODO cell many time in a repeating structure, the trace has to be larger than $-2$. Therefore: $$- \frac{L_{1}^{2}}{2 f^{2}} + 1 > -1 $$ which reduces to: $$L_1<2f$$
Then we can get the optics parameters of the one-turn matrix: $$ M=\left( \begin{array}{cc} \cos\Phi+\alpha\sin\Phi & \beta\sin\Phi\\ -\gamma\sin\Phi & \cos\Phi-\alpha\sin\Phi \end{array}\right) $$
Then the betatron phase advance can be calculated from:
\begin{equation} \cos{\Phi}=\frac{1}{2}\operatorname{tr}(M)=1-\frac{L_1^2}{2f^2} \end{equation}or
\begin{equation} \sin{\frac{\Phi}{2}}=\frac{L_1}{2f} \end{equation}The trace is an invariant under changing the starting piont, so does the phase advance.
The optical function can be calculated at different locations by:
\begin{align} \beta &= \frac{M_{12}}{\sin\Phi}\\ \alpha &= \frac{M_{11}-M_{22}}{2\sin\Phi} \end{align}At the middle point of the QD or QF we have:
\begin{align} \beta_{F/D} &= \frac{2L_1\left(1\pm\sin(\Phi/2)\right)}{\sin\Phi}\\ \alpha &= 0 \end{align}Right after the focusing quad, we have:
\begin{align} \beta &= \frac{2L_1\left(1-\sin(\Phi/2)\right)}{\sin\Phi}\\ \alpha &= \frac{L_1}{f}\frac{\sin(\Phi/2)-1}{\sin\Phi} \end{align}Dispersion in FODO cells¶
Let's consider the FODO cells with small bending angle approximation, starting from the mid point of the focusing quad. The extended $3\times 3$ matrix $\mathcal{M}$ gives
\begin{equation} \mathcal{M}_{FODO}= \left(\begin{array}{cc} 1 & 0 & 0\\ -\frac{1}{2f} & 1 & 0\\ 0 & 0 & 1 \end{array}\right) \left(\begin{array}{cc} 1 & L_1 & L_1 \theta/2\\ 0 & 1 & \theta \\ 0 & 0 & 1 \end{array}\right) \left(\begin{array}{cc} 1 & 0 & 0\\ \frac{1}{f} & 1 & 0\\ 0 & 0 & 1 \end{array}\right) \left(\begin{array}{cc} 1 & L_1 & L_1 \theta/2\\ 0 & 1 & \theta \\ 0 & 0 & 1 \end{array}\right) \left(\begin{array}{cc} 1 & 0 & 0\\ -\frac{1}{2f} & 1 & 0\\ 0 & 0 & 1 \end{array}\right) \end{equation}In [7]:
t=sympy.symbols("\\theta")
def matd_q(f):
return sympy.Matrix([[1,0,0],[1/f,1,0],[0,0,1]])
def matd_d(d, ang):
return sympy.Matrix([[1,d,d*ang/2],[0,1,ang],[0,0,1]])
mat_fodo=sympy.simplify(matd_q(-2*f)*matd_d(l,t)*matd_q(f)*matd_d(l,t)*matd_q(-2*f))
mat_fodo
Out[7]:
In [8]:
s=sympy.symbols('\sin(\phi/2)')
sympy.simplify(sympy.Inverse(sympy.eye(2)-mat_fodo[:2,:2])*mat_fodo[:2,2]).subs(f, l/2/s)
Out[8]:
The matrix $\mathcal{M}$ of FODO cell gives
\begin{equation} \mathcal{M}_{FODO}= \left(\begin{matrix}- \frac{L_{1}^{2}}{2 f^{2}} + 1 & \frac{L_{1}}{f} \left(L_{1} + 2 f\right) & \frac{L_{1} \theta}{2 f} \left(L_{1} + 4 f\right)\\\frac{L_{1}}{4 f^{3}} \left(L_{1} - 2 f\right) & - \frac{L_{1}^{2}}{2 f^{2}} + 1 & \frac{\theta}{4 f^{2}} \left(- L_{1}^{2} - 2 L_{1} f + 8 f^{2}\right)\\0 & 0 & 1\end{matrix}\right) \end{equation}The dispersion function in periodical boundary condition gives:
\begin{equation} \left(\begin{array}{c} D\\ D' \end{array}\right) _{midQF}= \left(\begin{matrix}\frac{L_1\theta}{2 \sin(\Phi/2)} \left(1 + \frac{2}{\sin(\Phi/2)}\right)\\0\end{matrix}\right) \end{equation}One can easily get the dispersion function at the middle point of the defocusing quad as:
\begin{equation} \left(\begin{array}{c} D\\ D' \end{array}\right)_{midQD} = \left(\begin{matrix}\frac{L_1\theta}{2 \sin(\Phi/2)} \left(-1 + \frac{2}{\sin(\Phi/2)}\right)\\0\end{matrix}\right) \end{equation}In [9]:
epsilon=0.2
phi=np.linspace(epsilon,np.pi-epsilon,100)
beta_max=2*(1+np.sin(phi/2))/np.sin(phi)
eta_max=(np.sin(phi/2)+2)/2/np.sin(phi/2)/np.sin(phi/2)
beta_min=2*(1-np.sin(phi/2))/np.sin(phi)
eta_min=(-np.sin(phi/2)+2)/2/np.sin(phi/2)/np.sin(phi/2)
fig,ax=plt.subplots()
ax_m=ax.twinx()
ax.set_xlabel("Phase advance [rad]")
ax.set_ylabel(r"$\beta$ [in unit of $L$]")
ax_m.set_ylabel(r"$D$ [in unit of $L\theta$]")
ax.set_ylim(0,10)
ax.set_xlim(0,np.pi)
ax_m.set_ylim(-0,10)
l1=ax.plot(phi,beta_max,c='b',label=r'$\beta_{max}$')
l2=ax.plot(phi,beta_min,c='c',label=r'$\beta_{min}$')
l3=ax_m.plot(phi,eta_max, c='y', label=r'$D_{max}$')
l4=ax_m.plot(phi,eta_min, c='r', label=r'$D_{min}$')
lns = l1+l2+l3+l4
labs = [l.get_label() for l in lns]
ax.legend(lns,labs,loc='right')
ax.set_xticks(np.linspace(0,1,5,)*np.pi)
ax.set_xticklabels(["$0$", r"$\frac{1}{4}\pi$", r"$\frac{1}{2}\pi$",
r"$\frac{3}{4}\pi$", r"$\pi$",
])
fig.tight_layout()
In [2]:
import PyLatte.latticeIO.elegantIO as elegantIO
In [3]:
fodo_lattice=elegantIO.elegantLatticeFile()
with open('fodo.lte','r') as f:
fodo_lattice.parseFrom(f)
fodo_lattice.modifyElement('bend', l=1.8)
fodo_lattice.modifyAllElements('kquad', name_contain='qf', k1=0.5)
fodo_lattice.modifyAllElements('kquad', name_contain='qd', k1=-0.5)
fodo_lattice.setUseLine()
twiss_list, twiss_parameter=elegantIO.elegant_findtwiss(fodo_lattice,matched=1, closed_orbit=True, divide_element=5)
end_twiss=twiss_list.iloc[-1].to_numpy()[[1,2,4,5,6,7,9,10]]
end_phi=twiss_list.iloc[-1].to_numpy()[3]
end_twiss, end_phi
Out[3]:
In [9]:
fig,(ax1,ax2)=plt.subplots(2,gridspec_kw = {'height_ratios':[8, 1]}, sharex=True)
fodo_lattice.plotBeamline(ax2)
ax1.plot(twiss_list['s'],twiss_list['betax'], label='Horizontal beta')
ax1.plot(twiss_list['s'], twiss_list['betay'], label='Vertical beta')
ax1.plot(twiss_list['s'], twiss_list['etax']*10, label='10x Horizontal dispersion')
ax2.set_xlabel('s [m]')
ax1.set_ylabel(r'$\beta$ function/dispersion function [m]')
fig.tight_layout()
ax1.legend(loc='best')
plt.subplots_adjust(hspace=0.0)
In [10]:
fodo_copies=6
lines=[]
dispersion_suppress=elegantIO.elegantLatticeFile()
for i in range(fodo_copies):
line_name=dispersion_suppress.loadALine(fodo_lattice, 'fodo', suffix='_{}'.format(i+1))
lines.append(line_name)
dispersion_suppress.appendToBeamline('total_line', *lines)
angle=fodo_lattice.getElementProperties('bend','angle')
for i in range(fodo_copies-2, fodo_copies):
dispersion_suppress.modifyElement('bend_{}'.format(i+1), angle=angle/2.0)
dispersion_suppress.setUseLine('total_line')
twiss_list, twiss_parameter=elegantIO.elegant_findtwiss(dispersion_suppress, matched=0, initial_optics=end_twiss, divide_element=5)
In [6]:
fig,(ax1,ax2)=plt.subplots(2,gridspec_kw = {'height_ratios':[8, 1]}, sharex=True)
dispersion_suppress.plotBeamline(ax2)
ax1.plot(twiss_list['s'],twiss_list['betax'], label='Horizontal beta')
ax1.plot(twiss_list['s'], twiss_list['betay'], label='Vertical beta')
ax1.plot(twiss_list['s'], twiss_list['etax']*10, label='10x Horizontal dispersion')
ax2.set_xlabel('s [m]')
ax1.set_ylabel(r'$\beta$ function/dispersion function [m]')
fig.tight_layout()
ax1.legend(loc='best')
plt.subplots_adjust(hspace=0.0)
In [45]:
print(twiss_list[[1,2],-1]-end_twiss[[0,1]]) # beta, alpha
print(twiss_list[[4,5],-1]) # dispersion,dispersion'
In [7]:
fodo_copies=4
angle=fodo_lattice.getElementProperties('bend','angle')
phi=end_phi
param=1.0/2.0/(1-np.cos(phi))
angle1=angle*param
angle2=angle-angle1
print(param)
lines=[]
dispersion_suppress2=elegantIO.elegantLatticeFile()
for i in range(fodo_copies):
line_name=dispersion_suppress2.loadALine(fodo_lattice, 'fodo', suffix='_{}'.format(i+1))
lines.append(line_name)
dispersion_suppress2.appendToBeamline('total_line', *lines)
i=fodo_copies-1
dispersion_suppress2.modifyElement('bend_{}'.format(i+1), angle=angle1)
i=fodo_copies-2
dispersion_suppress2.modifyElement('bend_{}'.format(i+1), angle=angle2)
dispersion_suppress2.setUseLine('total_line')
twiss_list, twiss_parameter=elegantIO.elegant_findtwiss(dispersion_suppress2, matched=0, initial_optics=end_twiss, divide_element=5)
fig,(ax1,ax2)=plt.subplots(2,gridspec_kw = {'height_ratios':[8, 1]}, sharex=True)
dispersion_suppress2.plotBeamline(ax2)
ax1.plot(twiss_list['s'],twiss_list['betax'], label='Horizontal beta')
ax1.plot(twiss_list['s'], twiss_list['betay'], label='Vertical beta')
ax1.plot(twiss_list['s'], twiss_list['etax']*10, label='10x Horizontal dispersion')
ax2.set_xlabel('s [m]')
ax1.set_ylabel(r'$\beta$ function/dispersion function [m]')
fig.tight_layout()
ax1.legend(loc='best')
plt.subplots_adjust(hspace=0.0)
$\mathcal{H}$ function in FODO cell¶
In [13]:
fodo_lattice=elegantIO.elegantLatticeFile()
with open('fodo.lte','r') as f:
fodo_lattice.parseFrom(f)
fodo_lattice.modifyElement('bend', l=1.6)
fodo_lattice.modifyAllElements('kquad', name_contain='qf', k1=0.5)
fodo_lattice.modifyAllElements('kquad', name_contain='qd', k1=-0.4)
fodo_lattice.setUseLine()
twiss_list, twiss_parameter=elegantIO.elegant_findtwiss(fodo_lattice,matched=1,divide_element=5)
#end_twiss=twiss_list[[1,2,3,4,5],-1]
fig,(ax1,ax2)=plt.subplots(2,gridspec_kw = {'height_ratios':[8, 1]}, sharex=True)
fodo_lattice.plotBeamline(ax2)
Hx=twiss_list['betax']*twiss_list['etaxp']*twiss_list['etaxp']+2*twiss_list['alphax']*twiss_list['etax']*twiss_list['etaxp']+(
1+twiss_list['alphax']*twiss_list['alphax'])*twiss_list['etax']*twiss_list['etax']/twiss_list['betax']
ax1.plot(twiss_list['s'], twiss_list['betax'], label='Horizontal beta')
ax1.plot(twiss_list['s'], twiss_list['etax']*10, label='10x Horizontal dispersion')
ax1.plot(twiss_list['s'], Hx*100, label=r'100x $\mathcal{H}$ function')
ax2.set_xlabel('s [m]')
ax1.set_ylabel(r'$\beta$ function/dispersion function [m]')
fig.tight_layout()
ax1.legend(loc='best')
plt.subplots_adjust(hspace=0.0)
end_twiss
Out[13]:
We have optics at the middle of QF and QD
\begin{align} \beta_{F/D} &= \frac{2L_1\left(1\pm\sin(\Phi/2)\right)}{\sin\Phi}\\ D_{F/D} &= \frac{L_1\theta}{2 \sin(\Phi/2)} \left( \pm 1 + \frac{2}{\sin(\Phi/2)}\right) \end{align}Therefore the $\mathcal{H}$ at these two location gives:
\begin{align} \mathcal{H}_{F/D} &= \frac{L\theta^2 \cos(\Phi/2)}{\sin^3(\Phi/2)\left(1\pm\sin(\Phi/2) \right)} \left(1\pm\frac{1}{2}\sin\frac{\Phi}{2}\right)^2 \end{align}Let's replace the integral with the average of the value at both end of the dipole.
\begin{align} \rho\frac{I_5}{I_2} &= \rho\frac{\oint\frac{\mathcal{H}}{\rho^3}ds}{\oint\frac{1}{\rho^2}ds} \\ &=\frac{1}{2}\left({\mathcal{H}_{F}+\mathcal{H}_{D}}\right) \\ &=\frac{\rho\theta^3 \cos(\Phi/2)}{2\sin^3(\Phi/2)} \left(\frac{\left(1+\frac{1}{2}\sin\frac{\Phi}{2}\right)^2}{\left(1+\sin(\Phi/2) \right)}+ \frac{\left(1-\frac{1}{2}\sin\frac{\Phi}{2}\right)^2}{\left(1-\sin(\Phi/2) \right)}\right) \\ &=\rho\theta^3\frac{ \cos(\Phi/2)}{\sin^3(\Phi/2)\cos(\Phi/2)}\left(1-\frac{3}{4}\sin^2\frac{\Phi}{2}\right) \end{align}Also Let's look at the natural chromaticity of this lattice:
\begin{align} C_x&=-\frac{1}{4\pi}\int\beta_x K_x ds \\ &=-\frac{1}{4\pi f}\left(\beta_{max} - \beta_{min}\right) \\ &=-\frac{\tan(\Phi/2)}{\pi} \\ &=-\frac{\tan(\Phi/2)}{\Phi/2} \nu_x \end{align}In [19]:
epsilon=0.2
phi=np.linspace(epsilon,np.pi-epsilon,100)
beta_max=2*(1+np.sin(phi/2))/np.sin(phi)
eta_max=(np.sin(phi/2)+2)/2/np.sin(phi/2)/np.sin(phi/2)
beta_min=2*(1-np.sin(phi/2))/np.sin(phi)
eta_min=(np.sin(phi/2)-2)/2/np.sin(phi/2)/np.sin(phi/2)
H_min=eta_max*eta_max/beta_max
H_max=eta_min*eta_min/beta_min
loc=np.argmin(H_max+H_min)
print(phi[loc]/np.pi*180, (H_max+H_min)[loc]/2)
chrom=-np.tan(phi/2.0)/(phi/2.0)
fig,ax=plt.subplots()
ax_m=ax.twinx()
ax.set_xlabel("Phase advance [rad]")
ax.set_ylabel(r"$\mathcal{H}$ in unit of $\rho \theta^3$")
ax_m.set_ylabel(r"Specific chromaticity $\xi/\nu$")
ax.set_ylim(0,10)
ax.set_xlim(0,np.pi)
#ax_m.set_ylim(-10,10)
l1=ax.plot(phi,(H_max+H_min)/2.0, c='b', label=r'Average $\mathcal{H}$')
l2=ax.plot(phi,H_max,c='y',label=r'$\mathcal{H}_{max}$', alpha=0.4)
l3=ax.plot(phi,H_min,c='c',label=r'$\mathcal{H}_{min}$', alpha=0.4)
l4=ax_m.plot(phi,chrom, c='r', label=r'Chromaticity $\xi$')
lns = l1+l2+l3+l4
labs = [l.get_label() for l in lns]
ax.legend(lns,labs,loc=3)
ax.set_xticks(np.linspace(0,1,5,)*np.pi)
ax.set_xticklabels(["$0$", r"$\frac{1}{4}\pi$", r"$\frac{1}{2}\pi$",
r"$\frac{3}{4}\pi$", r"$\pi$",
])
ax.axvline(x=phi[loc], ls='--', lw=.5)
fodo_lattice=elegant.elegantLatticeFile()
with open('fodo.lte','r') as f:
fodo_lattice.parseFrom(f)
fodo_lattice.modifyAllElements('kquad', name_contain='qf', k1=0.5)
fodo_lattice.modifyAllElements('kquad', name_contain='qd', k1=-0.4)
I5I2list=[]
phi_list=[]
l_list=np.linspace(1.2, 2.9, 20)
for dipole_length in l_list:
fodo_lattice.modifyElement('bend', l=dipole_length)
fodo_lattice.setUseLine()
twiss_list, twiss_parameter=elegant.elegant_findtwiss(fodo_lattice,matched=1)
I5I2list.append(twiss_parameter['I5']/twiss_parameter['I2'])
phi_list.append(twiss_parameter['nux']*2*np.pi)
angle=fodo_lattice.getElementProperties('bend','angle')
l5=ax.plot(phi_list, np.array(I5I2list)/angle/angle/angle, label=r'$\mathcal{H}$ of FODO example')
lns=lns+l5
labs = [l.get_label() for l in lns]
ax.legend(lns,labs,loc=3)
fig.tight_layout()
print(np.min(np.array(I5I2list)/angle/angle/angle))
In [2]:
import PyLatte.latticeIO.elegantIO as elegantIO
In [3]:
fodo_lattice=elegantIO.elegantLatticeFile()
with open('fodo.lte','r') as f:
fodo_lattice.parseFrom(f)
fodo_copies=30
fodo_lattice.modifyElement('bend', l=1.8, angle=np.pi/fodo_copies)
fodo_lattice.modifyAllElements('kquad', name_contain='qf', k1=0.5)
fodo_lattice.modifyAllElements('kquad', name_contain='qd', k1=-0.5)
lines=[]
fodo_ring=elegantIO.elegantLatticeFile()
for i in range(fodo_copies):
line_name=fodo_ring.loadALine(fodo_lattice, 'fodo', suffix='_{}'.format(i+1))
lines.append(line_name)
fodo_ring.modifyElement('QDH_5', DX = 1e-4)
fodo_ring.modifyElement('QFH_9', DX = 0e-4)
fodo_ring.appendToBeamline('total_line', *lines)
fodo_ring.useline="total_line"
twiss_list, twiss_parameter=elegantIO.elegant_findtwiss(fodo_ring, matched=1, closed_orbit=True, divide_element=5)
fig,(ax1,ax2)=plt.subplots(2,gridspec_kw = {'height_ratios':[8, 1]}, sharex=True)
ax1m=ax1.twinx()
fodo_ring.plotBeamline(ax2)
centroid, _ = elegantIO.read_elegant_SDDS_content("temp.cen", ["Cx"])
#ax1.plot(twiss_list['s'],twiss_list['betax'], label='Horizontal beta')
#ax1.plot(twiss_list['s'], twiss_list['betay'], label='Vertical beta')
ax1.plot(twiss_list['s'], centroid['Cx']*1e3, label='Centroid', color='C1')
ax1m.plot(twiss_list['s'], twiss_list['x']*1e3, label='Closed orbit', color='C2')
ax1.set_ylim(-0.3,0.3)
ax1m.set_ylim(-0.3,0.3)
ax2.set_xlabel('s [m]')
ax1.set_ylabel(r'One pass trajectory [mm]')
ax1m.set_ylabel(r'closed orbit [mm]')
fig.tight_layout()
ax1.legend(loc='best')
ax1m.legend(loc='best')
plt.subplots_adjust(hspace=0.0)
In [5]:
elegantIO.elegantLatticeFile.elegantParameterNames['CSBEND']
Out[5]:
In [ ]: