2
我有一個函數，我應該繪製10MHz和11.4MHz之間的幅度頻譜。這是一個函數：cos（6.72 *（10 ** 7 * t）+ 3.2 * sin（5 *（10 ** 5 * t）））我知道情節應該是什麼樣子，但我的完全錯誤。我很確定這是一個編程錯誤，而不是數學/理論錯誤，因爲我正在做一個MatLab示例正在展示的內容，但是我覺得Python正在做一些我不理解的事情。我應該在相距10ns的情況下對該函數進行8192次採樣，乘以漢明窗口，並在10MHz和11.4MHz之間繪製以dB爲單位的幅度譜。載波（10.7MHz）應該在55dB左右。第二對邊帶是60 dB處的最大幅度。然後，第五對邊帶是20 dB點，因此大約爲40 dB。任何人都可以發現這個代碼錯了，解釋了爲什麼我的不顯示這個？我的fft（）在Python中有些問題
import numpy
import scipy
import scipy.fftpack
import matplotlib
import matplotlib.pyplot as plt
from scipy import pi
import pylab
from pylab import *
import cmath
import sys
sys.setrecursionlimit(10000)
samples = 8192
#Defining the test function
t = scipy.linspace(0.01, 32/390625, samples, False) #Sample samples times at 10ns apart
#The samples is non-inclusive, it goes from 0 to samples-1.
x_t = cos(6.72*(10**7*t) + 3.2*sin(5*(10**5*t))) #Define x(t)
acc = lambda t: (x_t) #Define x[t] in terms of t as a variable in Python's eyes
signal = acc(t) #Make signal a function of t
plt.subplot(211) #Set up a plot
plt.xlabel('Ohmega (Hz)') #Label x axis
plt.ylabel('Amplitude of sampled x(t) (dB)') #Label y axis
window = numpy.hamming(samples) #Make a hamming window
w = scipy.linspace(0, 100*10**6, samples, False) #Create Ohmega between 1/(10ns)
signal = signal * window #Multiply by the hamming window
signal = scipy.fft(signal) #Take the FFT
signal = abs(20*log10(signal)) #Get the magnitude in dB scale
plt.xlabel('Ohmega') #Label x axis
plt.ylabel('|F[w]|') #Label y axis
#marker, stemlines, baseline = stem(w,signal, 'b-..') #Plot with stemlines
plot(w,signal, 'b-')
plt.xlim(10*10**6, 11.4*10**6) #Set x-limits
plt.show() #Show the plot
感謝您的幫助！
編輯：
現在我的代碼：
import numpy
import scipy
import scipy.fftpack
import matplotlib
import matplotlib.pyplot as plt
from scipy import pi
import pylab
from pylab import *
import cmath
import sys
sys.setrecursionlimit(10000)
samples = 8192
#Defining the test function
t = scipy.linspace(0.01, 32/390625, samples, False) #Sample samples times at 10ns apart
#The samples is non-inclusive, it goes from 0 to samples-1.
x_t = cos(6.72*(10**7*t) + 3.2*sin(5*(10**5*t))) #Define x(t)
acc = lambda t: cos(6.72*(10**7*t) + 3.2*sin(5*(10**5*t))) #Define x[t] in terms of t as a variable in Python's eyes
#signal = acc(t) #Make signal a function of t
plt.subplot(211) #Set up a plot
plt.xlabel('Ohmega (Hz)') #Label x axis
plt.ylabel('Amplitude of sampled x(t) (dB)') #Label y axis
window = numpy.hamming(samples) #Make a hamming window
w = scipy.linspace(0.01, 100*10**6, samples, False) #Create Ohmega between 1/(10ns)
signal = lambda t: abs(20*log10(scipy.fft(acc*window)))
#acc = acc * window #Multiply by the hamming window
#acc = scipy.fft(acc) #Take the FFT
#acc = abs(20*log10(acc)) #Get the magnitude in dB scale
plt.xlabel('Ohmega') #Label x axis
plt.ylabel('|F[w]|') #Label y axis
#marker, stemlines, baseline = stem(w,signal, 'b-..') #Plot with stemlines
plot(w,signal, 'b-')
plt.xlim(10*10**6, 11.4*10**6) #Set x-limits
plt.show() #Show the plot
和錯誤...：
Traceback (most recent call last):
File "/home/hollis/Documents/ELEN 322/ELEN_322_#2.py", line 39, in <module>
plot(w,signal, 'b-')
File "/usr/lib/pymodules/python2.7/matplotlib/pyplot.py", line 2467, in plot
ret = ax.plot(*args, **kwargs)
File "/usr/lib/pymodules/python2.7/matplotlib/axes.py", line 3893, in plot
for line in self._get_lines(*args, **kwargs):
File "/usr/lib/pymodules/python2.7/matplotlib/axes.py", line 322, in _grab_next_args
for seg in self._plot_args(remaining, kwargs):
File "/usr/lib/pymodules/python2.7/matplotlib/axes.py", line 300, in _plot_args
x, y = self._xy_from_xy(x, y)
File "/usr/lib/pymodules/python2.7/matplotlib/axes.py", line 240, in _xy_from_xy
raise ValueError("x and y must have same first dimension")
ValueError: x and y must have same first dimension
雖然我已經看到了之前修正了這個錯誤，我不知道現在如何解決它。顯然，一個採樣8192次的信號和確定採樣位置的變量是不同的尺寸。我這樣就失去了在這一個...
32/390625 == 0，它是整數除法。寫32.0/390625 – 2013-05-04 20:44:55