引用@hotpaw,在this相似回答:
「實數和虛數組合在一起時可以表示一個複雜的數組。在頻域中的複雜陣列的每一個複雜的元件可被認爲是頻率係數,並具有大小SQRT(R R +我 I))」。
所以,係數是在複雜的元件這個函數返回的是fft函數,此外,重要的是使用FFT函數的bin的大小(數量),測試一堆值並選擇一個更有意義的值通常情況下,它的樣本數量是相同的,這與大多數給出的答案一樣,並且產生了很好的和合理的結果,如果需要探究的話,這裏是我的代碼版本:
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
import scipy.fftpack
fig = plt.figure(figsize=[14,4])
N = 600 # Number of samplepoints
Fs = 800.0
T = 1.0/Fs # N_samps*T (#samples x sample period) is the sample spacing.
N_fft = 80 # Number of bins (chooses granularity)
x = np.linspace(0, N*T, N) # the interval
y = np.sin(50.0 * 2.0*np.pi*x) + 0.5*np.sin(80.0 * 2.0*np.pi*x) # the signal
# removing the mean of the signal
mean_removed = np.ones_like(y)*np.mean(y)
y = y - mean_removed
# Compute the fft.
yf = scipy.fftpack.fft(y,n=N_fft)
xf = np.arange(0,Fs,Fs/N_fft)
##### Plot the fft #####
ax = plt.subplot(121)
pt, = ax.plot(xf,np.abs(yf), lw=2.0, c='b')
p = plt.Rectangle((Fs/2, 0), Fs/2, ax.get_ylim()[1], facecolor="grey", fill=True, alpha=0.75, hatch="/", zorder=3)
ax.add_patch(p)
ax.set_xlim((ax.get_xlim()[0],Fs))
ax.set_title('FFT', fontsize= 16, fontweight="bold")
ax.set_ylabel('FFT magnitude (power)')
ax.set_xlabel('Frequency (Hz)')
plt.legend((p,), ('mirrowed',))
ax.grid()
##### Close up on the graph of fft#######
# This is the same histogram above, but truncated at the max frequence + an offset.
offset = 1 # just to help the visualization. Nothing important.
ax2 = fig.add_subplot(122)
ax2.plot(xf,np.abs(yf), lw=2.0, c='b')
ax2.set_xticks(xf)
ax2.set_xlim(-1,int(Fs/6)+offset)
ax2.set_title('FFT close-up', fontsize= 16, fontweight="bold")
ax2.set_ylabel('FFT magnitude (power) - log')
ax2.set_xlabel('Frequency (Hz)')
ax2.hold(True)
ax2.grid()
plt.yscale('log')
輸出: