Coverage for pySDC/implementations/problem_classes/BuckConverter.py: 100%
47 statements
« prev ^ index » next coverage.py v7.6.7, created at 2024-11-16 14:51 +0000
1import numpy as np
3from pySDC.core.problem import Problem
4from pySDC.implementations.datatype_classes.mesh import mesh, imex_mesh
7class buck_converter(Problem):
8 r"""
9 Example implementing the model of a buck converter, which is also called a step-down converter. The converter has two different
10 states and each of this state can be expressed as a nonhomogeneous linear system of ordinary differential equations (ODEs)
12 .. math::
13 \frac{d u(t)}{dt} = A_k u(t) + f_k (t)
15 for :math:`k=1,2`. The two states are the following. Define :math:`T_{sw}:=\frac{1}{f_{sw}}` as the switching period with
16 switching frequency :math:`f_{sw}`. The duty cycle :math:`d` defines the period of how long the switches are in one state
17 until they switch to the other state. Roughly saying, the duty cycle can be seen as a percentage [1]_. A duty cycle of one means
18 that the switches are always in only one state. If :math:`0 \leq \frac{t}{T_{sw}} \bmod 1 \leq d` [2]_:
20 .. math::
21 \frac{d v_{C_1} (t)}{dt} = -\frac{1}{R_s C_1}v_{C_1} (t) - \frac{1}{C_1} i_{L_1} (t) + \frac{V_s}{R_s C_1},
23 .. math::
24 \frac{d v_{C_2} (t)}{dt} = -\frac{1}{R_\ell C_2}v_{C_2} (t) + \frac{1}{C_2} i_{L_1} (t),
26 .. math::
27 \frac{d i_{L_1} (t)}{dt} = \frac{1}{L_1} v_{C_1} (t) - \frac{1}{L_1} v_{C_2} (t) - \frac{R_\pi}{L_1} i_{L_1} (t),
29 Otherwise, the equations are
31 .. math::
32 \frac{d v_{C_1} (t)}{dt} = -\frac{1}{R_s C_1}v_{C_1} (t) + \frac{V_s}{R_s C_1},
34 .. math::
35 \frac{d v_{C_2} (t)}{dt} = -\frac{1}{R_\ell C_2}v_{C_2} (t) + \frac{1}{C_2} i_{L_1} (t),
37 .. math::
38 \frac{d i_{L_1} (t)}{dt} = \frac{R_\pi}{R_s L_1} v_{C_1} (t) - \frac{1}{L_1} v_{C_2} (t) - \frac{R_\pi V_s}{L_1 R_s}.
40 using an initial condition.
42 Parameters
43 ----------
44 duty : float, optional
45 Duty cycle :math:`d` between zero and one indicates the time period how long the converter stays on one switching
46 state until it switches to the other state.
47 fsw : int, optional
48 Switching frequency, it is used to determine the number of time steps after the switching state is changed.
49 Vs : float, optional
50 Voltage at the voltage source :math:`V_s`.
51 Rs : float, optional
52 Resistance of the resistor :math:`R_s` at the voltage source.
53 C1 : float, optional
54 Capacitance of the capacitor :math:`C_1`.
55 Rp : float, optional
56 Resistance of the resistor in front of the inductor :math:`R_\pi`.
57 L1 : float, optional
58 Inductance of the inductor :math:`L_1`.
59 C2 : float, optional
60 Capacitance of the capacitor :math:`C_2`.
61 Rl : float, optional
62 Resistance of the resistor :math:`R_\pi`
64 Attributes
65 ----------
66 A : np.2darray
67 Coefficient matrix of the ODE system.
69 Note
70 ----
71 The duty cycle needs to be a value in :math:`[0,1]`.
73 References
74 ----------
75 .. [1] J. Sun. Pulse-Width Modulation. 25-61. Springer, (2012).
76 .. [2] J. Gyselinck, C. Martis, R. V. Sabariego. Using dedicated time-domain basis functions for the simulation of
77 pulse-width-modulation controlled devices - application to the steady-state regime of a buck converter. Electromotion (2013).
78 """
80 dtype_u = mesh
81 dtype_f = imex_mesh
83 def __init__(self, duty=0.5, fsw=1e3, Vs=10.0, Rs=0.5, C1=1e-3, Rp=0.01, L1=1e-3, C2=1e-3, Rl=10):
84 """Initialization routine"""
86 # invoke super init, passing number of dofs
87 nvars = 3
88 super().__init__(init=(nvars, None, np.dtype('float64')))
89 self._makeAttributeAndRegister(
90 'nvars', 'duty', 'fsw', 'Vs', 'Rs', 'C1', 'Rp', 'L1', 'C2', 'Rl', localVars=locals(), readOnly=True
91 )
93 self.A = np.zeros((nvars, nvars))
95 def eval_f(self, u, t):
96 """
97 Routine to evaluate the right-hand side of the problem.
99 Parameters
100 ----------
101 u : dtype_u
102 Current values of the numerical solution.
103 t : float
104 Current time of the numerical solution is computed.
106 Returns
107 -------
108 f : dtype_f
109 The right-hand side of the problem.
110 """
111 Tsw = 1 / self.fsw
113 f = self.dtype_f(self.init, val=0.0)
114 f.impl[:] = self.A.dot(u)
116 if 0 <= ((t / Tsw) % 1) <= self.duty:
117 f.expl[0] = self.Vs / (self.Rs * self.C1)
118 f.expl[2] = 0
120 else:
121 f.expl[0] = self.Vs / (self.Rs * self.C1)
122 f.expl[2] = -(self.Rp * self.Vs) / (self.L1 * self.Rs)
124 return f
126 def solve_system(self, rhs, factor, u0, t):
127 r"""
128 Simple linear solver for :math:`(I-factor\cdot A)\vec{u}=\vec{rhs}`.
130 Parameters
131 ----------
132 rhs : dtype_f
133 Right-hand side for the linear system.
134 factor : float
135 Abbrev. for the local stepsize (or any other factor required).
136 u0 : dtype_u
137 Initial guess for the iterative solver.
138 t : float
139 Current time (e.g. for time-dependent BCs).
141 Returns
142 -------
143 me : dtype_u
144 The solution as mesh.
145 """
146 Tsw = 1 / self.fsw
147 self.A = np.zeros((3, 3))
149 if 0 <= ((t / Tsw) % 1) <= self.duty:
150 self.A[0, 0] = -1 / (self.C1 * self.Rs)
151 self.A[0, 2] = -1 / self.C1
153 self.A[1, 1] = -1 / (self.C2 * self.Rl)
154 self.A[1, 2] = 1 / self.C2
156 self.A[2, 0] = 1 / self.L1
157 self.A[2, 1] = -1 / self.L1
158 self.A[2, 2] = -self.Rp / self.L1
160 else:
161 self.A[0, 0] = -1 / (self.C1 * self.Rs)
163 self.A[1, 1] = -1 / (self.C2 * self.Rl)
164 self.A[1, 2] = 1 / self.C2
166 self.A[2, 0] = self.Rp / (self.L1 * self.Rs)
167 self.A[2, 1] = -1 / self.L1
169 me = self.dtype_u(self.init)
170 me[:] = np.linalg.solve(np.eye(self.nvars) - factor * self.A, rhs)
171 return me
173 def u_exact(self, t):
174 r"""
175 Routine to compute the exact solution at time :math:`t`.
177 Parameters
178 ----------
179 t : float
180 Time of the exact solution.
182 Returns
183 -------
184 me : dtype_u
185 The exact solution.
186 """
187 assert t == 0, 'ERROR: u_exact only valid for t=0'
189 me = self.dtype_u(self.init)
191 me[0] = 0.0 # v1
192 me[1] = 0.0 # v2
193 me[2] = 0.0 # p3
195 return me