Question

我正在尝试使用matplotlib和scipy在Python中对Chau's Circuit进行建模，这涉及求解常微分方程组。

这已经完成了in matlab，我只是想在python中尝试这个问题。链接的matlab代码有点令人困惑;左边的代码似乎与解决描述Chua's Circuit的system of ode's（第3页，等式（2）（3）和（4））没什么关系，而右边的代码超出了那就是按组件对电路元件进行建模。

我不熟悉scipy的odeint函数，因此我使用scipy cookbook中的一些示例作为指导。

任何人都可以帮我解决我的系统问题;为什么我会得到一个如下图：

与一个看起来像这样的人相反？

我的代码附在下面：

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint

def fV_1(V_1, G_a, G_b, V_b):
    if V_1 < -V_b:
        fV_1 = G_b*V_1+(G_b-G_a)*V_b
    elif -V_b <= V_1 and V_1 <=V_b:
        fV_1 = G_a*V_1
    elif V_1 > V_b:
        fV_1 = G_b*V_1+(G_a-G_b)*V_b
    else:
        print "Error!"
    return fV_1

def ChuaDerivatives(state,t):
    #unpack the state vector
    V_1 = state[0]
    V_2 = state[1]
    I_3 = state[2]

    #definition of constant parameters
    L = 0.018 #H, or 18 mH
    C_1 = 0.00000001 #F, or 10 nF
    C_2 = 0.0000001 #F, or 100 nF
    G_a = -0.000757576 #S, or -757.576 uS
    G_b = -0.000409091 #S, or -409.091 uS
    V_b = 1 #V (E)
    G = 0.000550 #S, or 550 uS VARIABLE

    #compute state derivatives
    dV_1dt = (G/C_1)*(V_2-V_1)-(1/C_1)*fV_1(V_1, G_a, G_b, V_b)
    dV_2dt = -(G/C_2)*(V_2-V_1)+(1/C_2)*I_3
    dI_3dt = -(1/L)*V_2

    #return state derivatives
    return dV_1dt, dV_2dt, dI_3dt

#set up time series
state0 = [0.1, 0.1, 0.0001]
t = np.arange(0.0, 53.0, 0.1)

#populate state information
state = odeint(ChuaDerivatives, state0, t)

# do some fancy 3D plotting
from mpl_toolkits.mplot3d import Axes3D
fig = plt.figure()
ax = fig.gca(projection='3d')
ax.plot(state[:,0],state[:,1],state[:,2])
ax.set_xlabel('V_1')
ax.set_ylabel('V_2')
ax.set_zlabel('I_3')
plt.show()

Answer 1

所以我设法在一些摆弄后为自己解决;我正在解释odeint函数错误;更仔细地阅读docstring并从头开始阻止我按照困难的方法解决它。代码如下：

import numpy as np
import scipy.integrate as integrate
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

#define universal variables
c0 = 15.6
c1 = 1.0
c2 = 28.0
m0 = -1.143
m1 = -0.714

#just a little extra, quite unimportant
def f(x):
    f = m1*x+(m0-m1)/2.0*(abs(x+1.0)-abs(x-1.0))
    return f

#the actual function calculating
def dH_dt(H, t=0):
    return np.array([c0*(H[1]-H[0]-f(H[0])),
                  c1*(H[0]-H[1]+H[2]),
                  -c2*H[1]])



#computational time steps
t = np.linspace(0, 30, 1000)
#x, y, and z initial conditions
H0 = [0.7, 0.0, 0.0]

H, infodict = integrate.odeint(dH_dt, H0, t, full_output=True)

print infodict['message']

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot(H[:,0], H[:,1], H[:,2])
plt.show()

这给了我这个：

使用python和graphing解决ODE系统（Chua的电路）

1 个答案: