def stepSig_demo():
from scipy.signal import lti, step2, impulse2
import matplotlib.pyplot as plt
s1 = lti([3], [1, 2, 10]) # 以分子分母的最高次幂降序的系数构建传递函数，s1=3/(s^2+2s+10）
s2 = lti([1], [1, 0.4, 1]) # s2=1/(s^2+0.4s+1)
s3 = lti([5], [1, 2, 5]) # s3=5/(s^2+2s+5)
t1, y1 = step2(s1) # 计算阶跃输出，y1是Step response of system.
t2, y2 = step2(s2)
t3, y3 = step2(s3)
t11, y11 = impulse2(s1)
t22, y22 = impulse2(s2)
t33, y33 = impulse2(s3)
f, ((ax1, ax2, ax3), (ax4, ax5,
ax6)) = plt.subplots(2,
3,
sharex='col',
sharey='row') # 开启subplots模式
ax1.plot(t1, y1, 'r', label='s1 Step Response', linewidth=0.5)
ax1.set_title('s1 Step Response', fontsize=9)
ax2.plot(t2, y2, 'g', label='s2 Step Response', linewidth=0.5)
ax2.set_title('s2 Step Response', fontsize=9)
ax3.plot(t3, y3, 'b', label='s3 Step Response', linewidth=0.5)
ax3.set_title('s3 Step Response', fontsize=9)
ax4.plot(t11, y11, 'm', label='s1 Impulse Response', linewidth=0.5)
ax4.set_title('s1 Impulse Response', fontsize=9)
ax5.plot(t22, y22, 'y', label='s2 Impulse Response', linewidth=0.5)
ax5.set_title('s2 Impulse Response', fontsize=9)
ax6.plot(t33, y33, 'k', label='s3 Impulse Response', linewidth=0.5)
ax6.set_title('s3 Impulse Response', fontsize=9)
##plt.xlabel('Times')
##plt.ylabel('Amplitude')
# plt.legend()
plt.show()
def plot_time(analog_system, discrete_system, response='step', num_samples=100, sample_rate=48e3, plot_error=False):
n = np.arange(num_samples)
t = n/sample_rate
fig, ax1 = plt.subplots()
if response == 'impulse':
ta, ya = signal.impulse2(analog_system, T=t)
_, yd = signal.dimpulse((*discrete_system,1/sample_rate), t=t)
ax1.set_title('Resposta ao impulso')
elif response == 'step':
ta, ya = signal.step2(analog_system, T=t)
_, yd = signal.dstep((*discrete_system,1/sample_rate), t=t)
ax1.set_title('Resposta ao degrau')
elif response == 'ramp':
ramp_factor = 1
ta = np.arange(int(num_samples*ramp_factor))/(ramp_factor*sample_rate)
print(ta.shape)
ta, ya, xa = signal.lsim2(analog_system, U=ta, T=ta, printmessg=True)
_, yd = signal.dlsim((*discrete_system,1/sample_rate), u=t, t=t, x0=None)
ax1.set_title('Resposta a rampa')
ax1.plot(ta*sample_rate,ya,'b')
ax1.plot(n,np.ravel(yd),'k.')
ax1.set_ylabel('Amplitude', color='b')
ax1.set_xlabel('Amostras')
ax1.grid(True)
ax1.legend(['Analógico', 'Discreto'])
if plot_error:
ax2 = ax1.twinx()
ax2.plot(n, np.abs(ya-np.ravel(yd)), 'r')
ax2.set_ylabel('Erro', color='r')
ax2.axis('tight')
from scipy import signal system = ([1.0], [1.0, 2.0, 1.0]) t, y = signal.impulse2(system) print t print y
def test_impulse_invariant(self, sys, sample_time, samples_number):
time = np.arange(samples_number) * sample_time
_, yout_cont = impulse2(sys, T=time, **self.tolerances)
_, yout_disc = dimpulse(c2d(sys, sample_time, method='impulse'),
n=len(time))
assert_allclose(sample_time * yout_cont.ravel(), yout_disc[0].ravel())
C = 1e-6 # define the capacitor value
# calculate the other parts of the circuit
L = 1 / ((2 * np.pi * f0)**2 * C)
R = 2 * alpha * L
print("R=%.2f Ohm" % R)
print("L=%.2e Henry" % L)
print("C=%.2e Farad" % C)
system = signal.TransferFunction([R * C, 0], [L * C, R * C, 1])
# the sampling frequency to simulate the time-continuous system in discrete time
Fs = 1e6
T = np.arange(0, 1e-3, 1 / Fs)
t, h = signal.impulse2(system, T=T)
plt.figure(figsize=(8, 4))
plt.plot(t * 1000, h, 'navy')
plt.grid(True)
plt.xlabel('$t$ [ms]')
plt.ylabel('$h(t)$')
plt.title('Impulse response of RLC circuit')
# plt.savefig('1_Impulse response of RLC circuit.png', dpi=600)
plt.show()
# calculte frequency response
w, H = signal.freqresp(system, 2 * np.pi * np.logspace(0, 6, 10000))
# find indices for both test frequencies and evaluate the amplitude
indf0 = np.argmin(abs(w - 2 * np.pi * f0))
indf1 = np.argmin(abs(w - 2 * np.pi * f1))
# The initial conditions provided are $x(0)=1$ and $\dot{x}(0)= \dot{y}(0)=y(0)=0$. Working out the math using unilateral laplace transforms,
# \begin{align}
# X(s)&= \frac{s^2+2}{s^3+3s}\\
# Y(s)&= \frac{2}{s^3+3s}\\
# \end{align}
# These are passed to the $sp.impulse$ functions to study the behaviour of the system.
#
# In[198]:
num= [1,0,2]
den=[1,0,3,0]
X= sp.lti(num,den)
T= np.linspace(0,50,301)
t,y=sp.impulse2(X,None,T)
plotfigure(size,"t","$x$","$x(t)$",t,y,"b-")
# In[199]:
num= [2]
den=[1,0,3,0]
X= sp.lti(num,den)
T= np.linspace(0,50,301)
t,y=sp.impulse2(X,None,T)
plotfigure(size,"t","$y$","$y(t)$",t,y,"b-")
# ## Problem 4
plt.plot(tp2, mp2, 'mo')
plt.plot(tp3, mp3, 'mo')
plt.plot([ts1, ts1], [0.0, val_ts1], 'ko')
plt.plot([ts2, ts2], [0.0, val_ts2], 'ko')
plt.plot([ts3, ts3], [0.0, val_ts3], 'ko')
plt.legend(['Escalón Unitario', '$Butter$', '$Cheby$', '$Bessel$'])
plt.xlabel('time [seg]', fontsize=12)
plt.ylabel('Step Response $c(t)$ [V]', fontsize=12)
plt.grid()
plt.xlim(t0, tf * 9 / 10)
# ----------------------------------------------------------------------------
# Respuesta al Impulso: h(t):
t, resp1_h = sig.impulse2(my_tf_bw, None, t)
t, resp2_h = sig.impulse2(my_tf_ch, None, t)
t, resp3_h = sig.impulse2(my_tf_be, None, t)
fig12 = plt.figure()
plt.suptitle('Comparación: Respuesta al Impulso h(t)')
plt.plot(t, resp1_h, 'b')
plt.plot(t, resp2_h, '#ff7f0e')
plt.plot(t, resp3_h, 'g')
plt.legend(['$Butter$', '$Cheby$', '$Bessel$'])
plt.xlabel('time [seg]', fontsize=12)
plt.ylabel('Impulse Response $h(t)$', fontsize=12)
plt.grid()
plt.xlim(-1, 45)
def test_impulse_invariant(self, sys, sample_time, samples_number):
time = np.arange(samples_number) * sample_time
_, yout_cont = impulse2(sys, T=time, **self.tolerances)
_, yout_disc = dimpulse(c2d(sys, sample_time, method='impulse'),
n=len(time))
assert_allclose(sample_time * yout_cont.ravel(), yout_disc[0].ravel())
[-2,-3]
])
B = np.array([
[0],
[1]
])
C = np.array([
[1,0]
])
D = 0
sys = signal.StateSpace(A,B,C,D) #State space to time domain conversion
t1,y1= signal.impulse2(sys) # time and output axis for natural(impulse) response
a,b = signal.ss2tf(A,B,C,D) #Importing Num. and Den. of T.F. from State Space Rep.
#rounding off
num = np.around(a[0],decimals = 0)
den = np.around(b,decimals = 0)
s = sp.symbols('s')
H_s = sp.Poly(num,s)/sp.Poly(den,s) # Getting polynomial expressions
print("THE TRANSFER FUNCTION OF THE SYSTEM ")
print("H(s) =",H_s)
plt.plot(t1,y1,label='impulse response')
plt.legend()
plt.grid()
plt.show()
# Second order system with a repeated root: x''(t) + 2*x(t) + x(t) = u(t) from scipy import signal system = ([1.0], [1.0, 2.0, 1.0]) t, y = signal.impulse2(system) import matplotlib.pyplot as plt plt.plot(t, y)
from scipy.signal import lti, impulse2
import matplotlib.pyplot as plt
from pylab import mpl
import numpy as np
a = [0, 1, 0]
b = [1, 5, 6]
sys = lti(a, b) #以分子分母的最高次幂降序的系数构建传递函数
t, y = impulse2(sys)
plt.plot(t, y, 'r', label="s1 Step", linewidth=0.5)
mpl.rcParams['font.sans-serif'] = ['FangSong'] # 指定默认字体
mpl.rcParams['axes.unicode_minus'] = False # 解决保存图像是负号'-'显示为方块的问题
plt.xlabel('时间(t)')
plt.ylabel('h(t)')
plt.title('例2.6-1的单位冲击响应')
plt.xlim(-0.2, 4) #设置x轴的区间
plt.ylim(-0.2, 1) #设置y轴的区间
plt.vlines(0, -0.2, 1) #x=0
plt.hlines(0, -0.2, 4) #y=0
plt.show()
def get_impulse(self):
return signal.impulse2(self.denorm_sys)
