Question

为经典的霍奇金·赫x黎ODE增加高斯白噪声以提高膜电位

嗨，

我已经找到了通过霍奇金·赫x黎方程评估膜电位的代码。

import matplotlib.pyplot as plt
import numpy as np

from scipy.integrate import odeint

#runtime in milliseconds

#print("Type in the runtime-interval (in milliseconds)")

#t_0 = input()
#t_1 = input()

t_0 = 0.0
t_1 = 200.0
print("The interval of runtime is:")
print(t_0, t_1)

#Potassium (alpha_n, beta_n) and Sodium (alpha_m, beta_m, alpha_h, beta_h) ion-channel rate functions


def alpha_n(V):
    return (0.01*(10.0-V))/(np.exp((10.0-V)/10.0)-1.0)

def alpha_m(V):
    return (0.1*(25.0-V))/(np.exp((25.0-V)/10.0)-1)

def beta_n(V):
    return 0.125*np.exp(-V/80.0)

def beta_m(V):
    return 4.0*np.exp(-V/18.0)

def alpha_h(V):
    return 0.07*np.exp(-V/20.0)

def beta_h(V):
    return (1.0)/((np.exp((30.0-V)/10.0))+1)

#mostly used parameters

#Sodium potential (mV)
V_NA = 50.0

#Potassium Potential
V_K = -77.0

#Leak Potential
V_L = -54.4

#Potassium channel conductance
g_k = 36.0

#Sodium channel conductance
g_NA = 120.0

#Leak channel conductance
g_L = 0.3

#Membrane capacitance
C = 1.0

#equally ('distributed') time values

T = np.linspace(t_0,t_1,10000)

#please input the stimulus (first of all with a fixed stimulus)
#one spike below
def stim(t):
    if 0.0 < t < 1.0:
        return 150.0
    elif 35.0 < t < 36.0 :
        return 5000.0
    return 0.0



#steady - state values

def n_infty(V = 0.0):
    return alpha_n(V)/(alpha_n(V)+beta_n(V))

def m_infty(V = 0.0):
    return alpha_m(V)/(alpha_m(V)+beta_m(V))

def h_infty(V = 0.0):
    return alpha_h(V)/(alpha_h(V)+beta_h(V))

def tau_m(V = 0.0):
    return 1.0/(alpha_m(V)+beta_m(V))

def tau_n(V = 0.0):
    return 1.0/(alpha_n(V)+beta_n(V))

def tau_h(V = 0.0):
    return 1.0/(alpha_h(V)+beta_h(V))

#derivatives


def derivatives(x, z):

    dx = np.zeros((4,))

    V = x[0]
    m = x[1]
    n = x[2]
    h = x[3]

    dx[0] = (stim(z)-g_NA*np.power(m,3.0)*h*(V-V_NA)-g_k*np.power(n,4.0)*(V-V_K)-g_L*(V-V_L))/1.0

    dx[1] = (m_infty(V)-m)/tau_m(V)

    dx[2] = (n_infty(V)-n)/tau_n(V)

    dx[3] = (h_infty(V)-h)/tau_h(V)

    return dx

X = np.array([0.0,m_infty(),n_infty(),h_infty()])

V_x = odeint(derivatives,X,T)

print(V_x)





plt.plot(T,V_x)
plt.xlabel('Time in ms')
plt.ylabel('Membrane Potential in mV')

我想添加一个高斯白噪声处理-但我不确定如何做，因为我不是Python的好用户-

也许

noise = numpy.random.normal(mean=0.0, std=1.0, size=T)

和

 dx[0] =
 (stim(z)-g_NA*np.power(m,3.0)*h*(V-V_NA)-g_k*np.power(n,4.0)*(V-V_K)-g_L*(V-V_L))/1.0
 + noise

我可以简单地将其添加到微分方程中吗？

亲切的问候，并在此先感谢

Answer 1

首先，您需要将noise参数添加到函数中，

def derivatives(x, z,noisyInput):

您已经以正态分布创建了噪声。

noisyInput = np.ones(length)*7 + np.random.normal(0,3,length)

您只需在容量确定的dvdt项中添加噪声项。

dx[0] = (noisyInput + stim(z)-g_NA*np.power(m,3.0)*h*(V-V_NA)-g_k*np.power(n,4.0)*(V-V_K)-g_L*(V-V_L))/1.0

您应该将噪声传递给函数。

V_x = odeint(derivatives,X,T,args=(noisyInput,))

您还可以检查我编写的代码： https://github.com/nosratullah/HodgkinHuxely

高斯白噪声到Hodgkin Huxley方程

1 个答案: