I/P频率为8000 Hz,那么为什么以下代码的输出在2000 Hz处显示峰值?
I/P频率为8000 Hz。为什么以下代码的O/P显示了频谱图在2000 Hz处的峰值?我是频谱图的新手,有人可以解释吗?
from scipy import signal
import matplotlib.pyplot as plt
import numpy as np
time = np.linspace(start = 0, stop = 10, num = 100000)
t1 = time[0,]
t2 = time[1,]
del_t = t2-t1
fs = 1e4 #np.ceil(1/del_t)
ip_f = 8000
x = np.sin(2*np.pi*ip_f*time)
# plt.plot(time,x)
op_f, op_t, op_Sxx = signal.spectrogram(x, fs)
plt.pcolormesh(op_t, op_f, op_Sxx, shading='gouraud')
plt.ylabel('Frequency [Hz]')
plt.xlabel('Time [sec]')
plt.show()
The i/p frequency is 8000 Hz. Why does the o/p of the following code shows, a peak in spectrogram at 2000 Hz? I am new to the spectrogram, can anyone please explain?
from scipy import signal
import matplotlib.pyplot as plt
import numpy as np
time = np.linspace(start = 0, stop = 10, num = 100000)
t1 = time[0,]
t2 = time[1,]
del_t = t2-t1
fs = 1e4 #np.ceil(1/del_t)
ip_f = 8000
x = np.sin(2*np.pi*ip_f*time)
# plt.plot(time,x)
op_f, op_t, op_Sxx = signal.spectrogram(x, fs)
plt.pcolormesh(op_t, op_f, op_Sxx, shading='gouraud')
plt.ylabel('Frequency [Hz]')
plt.xlabel('Time [sec]')
plt.show()
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如果您的样本率仅为10 kHz,则8 kHz在 nyquist频率(5 kHz)(5 kHz) 。它将别名 如您所见。
If your sample rate is only 10 kHz then 8 kHz is above the Nyquist Frequency (5 kHz). It will alias to 2 kHz, as you have seen.