首先，需要注意的是，这个问题与1D和2D fft之间的差异无关，而是与总功率和平均功率如何随阵列中元素的数量而变化有关。在

当您说9的因子来自a中的元素比b中的元素多9倍时，您说得非常正确。令人困惑的是，你注意到你已经通过除以np.fft.fft2(a)/3000./3000.和{}进行了标准化，事实上，这些规范化对于得到空间和频率域中的总(非平均)功率是相等的是必要的。为了得到平均值，你必须再除以数组的大小。在

你的问题实际上是关于Parseval定理的，它指出两个领域(空间/时间和频率)的总功率是相等的。它的语句，对于DFT是this。注意，尽管右边是1/N，但这不是平均功率，而是总功率。1/N的原因是DFT的规范化约定。在

放在Python中，这意味着对于一个时间/空间信号sig，Parseval等价可以表述为：

np.sum(np.abs(sig)**2) == np.sum(np.abs(np.fft.fft(sig))**2)/sig.size

下面是一个完整的例子，从一些玩具箱开始(一维和二维数组填充一个1)，最后是你自己的箱子。注意，我使用了的.size属性努比·恩达雷，返回数组中元素的总数。它相当于你的/1000./1000.等。希望这有帮助！在import numpy as np

print 'simple examples:'

# 1-d, 4 elements:

ones_1d = np.array([1.,1.,1.,1.])

ones_1d_f = np.fft.fft(ones_1d)

# compute total power in space and frequency domains:

space_power_1d = np.sum(np.abs(ones_1d)**2)

freq_power_1d = np.sum(np.abs(ones_1d_f)**2)/ones_1d.size

print 'space_power_1d = %f'%space_power_1d

print 'freq_power_1d = %f'%freq_power_1d

# 2-d, 4 elements:

ones_2d = np.array([[1.,1.],[1.,1.]])

ones_2d_f = np.fft.fft2(ones_2d)

# compute and print total power in space and frequency domains:

space_power_2d = np.sum(np.abs(ones_2d)**2)

freq_power_2d = np.sum(np.abs(ones_2d_f)**2)/ones_2d.size

print 'space_power_2d = %f'%space_power_2d

print 'freq_power_2d = %f'%freq_power_2d

# 2-d, 9 elements:

ones_2d_big = np.array([[1.,1.,1.],[1.,1.,1.],[1.,1.,1.]])

ones_2d_big_f = np.fft.fft2(ones_2d_big)

# compute and print total power in space and frequency domains:

space_power_2d_big = np.sum(np.abs(ones_2d_big)**2)

freq_power_2d_big = np.sum(np.abs(ones_2d_big_f)**2)/ones_2d_big.size

print 'space_power_2d_big = %f'%space_power_2d_big

print 'freq_power_2d_big = %f'%freq_power_2d_big

print

# asker's example array a and fft af:

print 'askers examples:'

a = np.random.randn(3000,3000)

af = np.fft.fft2(a)

# compute the space and frequency total powers:

space_power_a = np.sum(np.abs(a)**2)

freq_power_a = np.sum(np.abs(af)**2)/af.size

# mean power is the total power divided by the array size:

freq_power_a_mean = freq_power_a/af.size

print 'space_power_a = %f'%space_power_a

print 'freq_power_a = %f'%freq_power_a

print 'freq_power_a_mean = %f'%freq_power_a_mean

print

# the central 1000x1000 section of the 3000x3000 original array:

b = a[1000:2000,1000:2000]

bf = np.fft.fft2(b)

# we expect the total power in the space and frequency domains

# to be about 1/9 of the total power in the space frequency domains

# for matrix a:

space_power_b = np.sum(np.abs(b)**2)

freq_power_b = np.sum(np.abs(bf)**2)/bf.size

# we expect the mean power to be the same as the mean power from

# matrix a:

freq_power_b_mean = freq_power_b/bf.size

print 'space_power_b = %f'%space_power_b

print 'freq_power_b = %f'%freq_power_b

print 'freq_power_b_mean = %f'%freq_power_b_mean