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我正在运行 LTI 状态空间模型的模拟，我需要在不同的时间以不同的输入运行它。换句话说，从 t0 = 0 模拟到 t1=1 秒，根据 t1 的结果对系统的输入进行更改，然后从 t1= 1 秒继续到 t2= 2 秒。

我尝试在 t[0] 处使用初始条件 X0=array([0,0]) 运行，然后将 xout1 的最后一个元素作为下一次运行的初始条件，并给它一个从 t 开始的新时间序列[0] = 1

import numpy as np
import scipy.signal as sig
import matplotlib.pyplot as plt

r_wh = 60e-3 / 2.0              # m
m_v = 20                        # kg
R_m = 21.2                      # Ohms
L_m = 1.37e-3                   # Henries
K_t = 21.2e-3                   # Nm.A^-1
K_v = 60 / (450 * 2 * np.pi)    # V.rad^-1.s^-1
I_m = 4.2e-7                    # kg.m^2
I_gb = 0.4e-7                   # kg.m^2
I_wh = 65275e-9                 # kg.m^2
gr = 1.0/19.0

I_eq = ((I_m + I_gb) / gr) + (gr  * (I_wh + 0.25 * m_v * r_wh ** 2))

A = np.array([[-R_m/L_m, (-K_v/L_m) / gr],[K_t/I_eq, 0]])
B = np.array([[1/L_m,0],[0,-1/I_eq]])
C = np.array([[0,0],[0,1]])
D = np.zeros((2,2))

SS = sig.StateSpace(A,B,C,D)


T1 = np.arange(0,1,0.01)
T2 = np.arange(1,2,0.01)

U1 = np.array([12*np.ones_like(T1),np.zeros_like(T1)]).transpose()
U2 = np.array([12*np.ones_like(T2),np.zeros_like(T2)]).transpose()

tout1, yout1, xout1 = sig.lsim(SS,U1,T1)
tout2, yout2, xout2 = sig.lsim(SS,U2,T2,X0=xout1[-1])

plt.plot(T1,xout1[:,1],T2,xout2[:,1])

您希望状态向量输出数组“xout2”中的第一个元素与 X0 条件匹配，但事实并非如此。此函数“lsim”是否要求第一个时间点为 0 ？

绘图输出

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lsim assumes that X0 is the state at time 0, not at time T[0]. You can get the behavior that you expected (almost) by using

tout2, yout2, xout2 = sig.lsim(SS, U2, T2 - T2[0], X0=xout1[-1])

I say "almost", because there will still be a small gap in the plot at the transition from xout1 to xout2. That is because the values in T1 are [0., 0.01, 0.02, ..., 0.98, 0.99]. Note that 1.0 is not in T1. So the last value in xout1 is the state at t=0.99, not t=1.0. One way to fix that is to include the final t values in T1 and T2, for example by using np.linspace instead of np.arange:

T1 = np.linspace(0, 1, 101)
T2 = np.linspace(1, 2, 101)
于 2019-01-29T02:29:58.067 回答