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Generating new 2D data using power spectrum density function from spatial frequency domain via ifft?

This is my first post so apologies for any formatting related issues.
So I have a dataset which was obtained from an atomic microscope. The data looks like a 1024x1024 matrix which is composed of different measurements taken from the sample in units of meters, eg.
data = [[1e-07 ... 4e-08][ ... ... ... ][3e-09 ... 12e-06]]
np.size(data) == (1024,1024)
From this data, I was hoping to 1) derive some statistics about the real data; and 2) using the power spectrum density (PSD) distribution, hopefully create a new dataset which is different, but statistically similar to the characteristics of the original data. My plan to do this was 2a) take a 2d fft of data, calculate the power spectrum density 2b) some method?, 2c) take the 2d ifft of the modified signal to turn it back into a new sample with the same power spectrum density as the original.
Moreover, regarding part 2b) this was the closest link I could find regarding a time series based solution; however, I am not understanding exactly how to implement this so far, since I am not exactly sure what the phase, frequency, and amplitudes of the fft data represent in this 2d case, and also since we are now talking about a 2d ifft I'm not exactly sure how to construct this complex matrix while incorporating the random number generation, and amplitude/phase shifts in a way that will translate back to something meaningful.
So basically, I have been having some trouble with my intuition. For this problem, we are working with a 2d Fourier transform of spatial data with no temporal component, so I believe that methods which are applied to time series data could be applied here as well. Since the fft of the original data is the 'frequency in the spatial domain', the x-axis of the PSD should be either pixels or meters, but then what is the 'power' in the y-axis describing? I was hoping that someone could help me figure this problem out.
My code is below, hopefully someone could help me solve my problem. Bonus if someone could help me understand what this shifted frequency vs amplitude plot is saying:
here is the image with the fft, shifted fft, and freq. vs aplitude plots.
Fortunately the power spectrum density function is a bit easier to understand
Thank you all for your time.
data = np.genfromtxt('asample3.0_00001-filter.txt')
x = np.arange(0,int(np.size(data,0)),1)
y = np.arange(0,int(np.size(data,1)),1)
z = data
npix = data.shape[0]
#taking the fourier transform
fourier_image = np.fft.fft2(data)
#Get power spectral density
fourier_amplitudes = np.abs(fourier_image)**2
#calculate sampling frequency fs (physical distance between pixels)
fs = 92e-07/npix
freq_shifted = fs/2 * np.linspace(-1,1,npix)
freq = fs/2 * np.linspace(0,1,int(npix/2))
print("Plotting 2d Fourier Transform ...")
fig, axs = plt.subplots(2,2,figsize=(15, 15))
axs[0,0].imshow(10*np.log10(np.abs(fourier_image)))
axs[0,0].set_title('fft')
axs[0,1].imshow(10*np.log10(np.abs(np.fft.fftshift(fourier_image))))
axs[0,1].set_title('shifted fft')
axs[1,0].plot(freq,10*np.log10(np.abs(fourier_amplitudes[:npix//2])))
axs[1,0].set_title('freq vs amplitude')
for ii in list(range(npix//2)):
axs[1,1].plot(freq_shifted,10*np.log10(np.fft.fftshift(np.abs(fourier_amplitudes[ii]))))
axs[1,1].set_title('shifted freq vs amplitude')
#constructing a wave vector array
## Get frequencies corresponding to signal PSD
kfreq = np.fft.fftfreq(npix) * npix
kfreq2D = np.meshgrid(kfreq, kfreq)
knrm = np.sqrt(kfreq2D[0]**2 + kfreq2D[1]**2)
knrm = knrm.flatten()
fourier_amplitudes = fourier_amplitudes.flatten()
#creating the power spectrum
kbins = np.arange(0.5, npix//2+1, 1.)
kvals = 0.5 * (kbins[1:] + kbins[:-1])
Abins, _, _ = stats.binned_statistic(knrm, fourier_amplitudes,
statistic = "mean",
bins = kbins)
Abins *= np.pi * (kbins[1:]**2 - kbins[:-1]**2)
print("Plotting power spectrum of surface ...")
fig = plt.figure(figsize=(10, 10))
plt.loglog(fs/kvals, Abins)
plt.xlabel("Spatial Frequency $k$ [meters]")
plt.ylabel("Power per Spatial Frequency $P(k)$")
plt.tight_layout()

How to find period of signal (autocorrelation vs fast fourier transform vs power spectral density)?

Suppose one wanted to find the period of a given sinusoidal wave signal. From what I have read online, it appears that the two main approaches employ either fourier analysis or autocorrelation. I am trying to automate the process using python and my usage case is to apply this concept to similar signals that come from the time-series of positions (or speeds or accelerations) of simulated bodies orbiting a star.
For simple-examples-sake, consider x = sin(t) for 0 ≤ t ≤ 10 pi.
import numpy as np
from scipy import signal
import matplotlib.pyplot as plt
## sample data
t = np.linspace(0, 10 * np.pi, 100)
x = np.sin(t)
fig, ax = plt.subplots()
ax.plot(t, x, color='b', marker='o')
ax.grid(color='k', alpha=0.3, linestyle=':')
plt.show()
plt.close(fig)
Given a sine-wave of the form x = a sin(b(t+c)) + d, the period of the sine-wave is obtained as 2 * pi / b. Since b=1 (or by visual inspection), the period of our sine wave is 2 * pi. I can check the results obtained from other methods against this baseline.
Attempt 1: Autocorrelation
As I understand it (please correct me if I'm wrong), correlation can be used to see if one signal is a time-lagged copy of another signal (similar to how cosine and sine differ by a phase difference). So autocorrelation is testing a signal against itself to measure the times at which the time-lag repeats said signal. Using the example posted here:
result = np.correlate(x, x, mode='full')
Since x and t each consist of 100 elements and result consists of 199 elements, I am not sure why I should arbitrarily select the last 100 elements.
print("\n autocorrelation (shape={}):\n{}\n".format(result.shape, result))
autocorrelation (shape=(199,)):
[ 0.00000000e+00 -3.82130761e-16 -9.73648712e-02 -3.70014208e-01
-8.59889695e-01 -1.56185995e+00 -2.41986054e+00 -3.33109112e+00
-4.15799070e+00 -4.74662427e+00 -4.94918053e+00 -4.64762251e+00
-3.77524157e+00 -2.33298717e+00 -3.97976240e-01 1.87752669e+00
4.27722402e+00 6.54129270e+00 8.39434617e+00 9.57785701e+00
9.88331103e+00 9.18204933e+00 7.44791758e+00 4.76948221e+00
1.34963425e+00 -2.50822289e+00 -6.42666652e+00 -9.99116299e+00
-1.27937834e+01 -1.44791297e+01 -1.47873668e+01 -1.35893098e+01
-1.09091510e+01 -6.93157447e+00 -1.99159756e+00 3.45267493e+00
8.86228186e+00 1.36707567e+01 1.73433176e+01 1.94357232e+01
1.96463736e+01 1.78556800e+01 1.41478477e+01 8.81191526e+00
2.32100171e+00 -4.70897483e+00 -1.15775811e+01 -1.75696560e+01
-2.20296487e+01 -2.44327920e+01 -2.44454330e+01 -2.19677060e+01
-1.71533510e+01 -1.04037163e+01 -2.33560966e+00 6.27458308e+00
1.45655029e+01 2.16769872e+01 2.68391837e+01 2.94553896e+01
2.91697473e+01 2.59122266e+01 1.99154591e+01 1.17007613e+01
2.03381596e+00 -8.14633251e+00 -1.78184255e+01 -2.59814393e+01
-3.17580589e+01 -3.44884934e+01 -3.38046447e+01 -2.96763956e+01
-2.24244433e+01 -1.26974172e+01 -1.41464998e+00 1.03204331e+01
2.13281784e+01 3.04712823e+01 3.67721634e+01 3.95170295e+01
3.83356037e+01 3.32477037e+01 2.46710643e+01 1.33886439e+01
4.77778141e-01 -1.27924775e+01 -2.50860560e+01 -3.51343866e+01
-4.18671622e+01 -4.45258983e+01 -4.27482779e+01 -3.66140001e+01
-2.66465884e+01 -1.37700036e+01 7.76494745e-01 1.55574483e+01
2.90828312e+01 3.99582426e+01 4.70285203e+01 4.95000000e+01
4.70285203e+01 3.99582426e+01 2.90828312e+01 1.55574483e+01
7.76494745e-01 -1.37700036e+01 -2.66465884e+01 -3.66140001e+01
-4.27482779e+01 -4.45258983e+01 -4.18671622e+01 -3.51343866e+01
-2.50860560e+01 -1.27924775e+01 4.77778141e-01 1.33886439e+01
2.46710643e+01 3.32477037e+01 3.83356037e+01 3.95170295e+01
3.67721634e+01 3.04712823e+01 2.13281784e+01 1.03204331e+01
-1.41464998e+00 -1.26974172e+01 -2.24244433e+01 -2.96763956e+01
-3.38046447e+01 -3.44884934e+01 -3.17580589e+01 -2.59814393e+01
-1.78184255e+01 -8.14633251e+00 2.03381596e+00 1.17007613e+01
1.99154591e+01 2.59122266e+01 2.91697473e+01 2.94553896e+01
2.68391837e+01 2.16769872e+01 1.45655029e+01 6.27458308e+00
-2.33560966e+00 -1.04037163e+01 -1.71533510e+01 -2.19677060e+01
-2.44454330e+01 -2.44327920e+01 -2.20296487e+01 -1.75696560e+01
-1.15775811e+01 -4.70897483e+00 2.32100171e+00 8.81191526e+00
1.41478477e+01 1.78556800e+01 1.96463736e+01 1.94357232e+01
1.73433176e+01 1.36707567e+01 8.86228186e+00 3.45267493e+00
-1.99159756e+00 -6.93157447e+00 -1.09091510e+01 -1.35893098e+01
-1.47873668e+01 -1.44791297e+01 -1.27937834e+01 -9.99116299e+00
-6.42666652e+00 -2.50822289e+00 1.34963425e+00 4.76948221e+00
7.44791758e+00 9.18204933e+00 9.88331103e+00 9.57785701e+00
8.39434617e+00 6.54129270e+00 4.27722402e+00 1.87752669e+00
-3.97976240e-01 -2.33298717e+00 -3.77524157e+00 -4.64762251e+00
-4.94918053e+00 -4.74662427e+00 -4.15799070e+00 -3.33109112e+00
-2.41986054e+00 -1.56185995e+00 -8.59889695e-01 -3.70014208e-01
-9.73648712e-02 -3.82130761e-16 0.00000000e+00]
Attempt 2: Fourier
Since I am not sure where to go from the last attempt, I sought a new attempt. To my understanding, Fourier analysis basically shifts a signal from/to the time-domain (x(t) vs t) to/from the frequency domain (x(t) vs f=1/t); the signal in frequency-space should appear as a sinusoidal wave that dampens over time. The period is obtained from the most observed frequency since this is the location of the peak of the distribution of frequencies.
Since my values are all real-valued, applying the Fourier transform should mean my output values are all complex-valued. I wouldn't think this is a problem, except for the fact that scipy has methods for real-values. I do not fully understand the differences between all of the different scipy methods. That makes following the algorithm proposed in this posted solution hard for me to follow (ie, how/why is the threshold value picked?).
omega = np.fft.fft(x)
freq = np.fft.fftfreq(x.size, 1)
threshold = 0
idx = np.where(abs(omega)>threshold)[0][-1]
max_f = abs(freq[idx])
print(max_f)
This outputs 0.01, meaning the period is 1/0.01 = 100. This doesn't make sense either.
Attempt 3: Power Spectral Density
According to the scipy docs, I should be able to estimate the power spectral density (psd) of the signal using a periodogram (which, according to wikipedia, is the fourier transform of the autocorrelation function). By selecting the dominant frequency fmax at which the signal peaks, the period of the signal can be obtained as 1 / fmax.
freq, pdensity = signal.periodogram(x)
fig, ax = plt.subplots()
ax.plot(freq, pdensity, color='r')
ax.grid(color='k', alpha=0.3, linestyle=':')
plt.show()
plt.close(fig)
The periodogram shown below peaks at 49.076... at a frequency of fmax = 0.05. So, period = 1/fmax = 20. This doesn't make sense to me. I have a feeling it has something to do with the sampling rate, but don't know enough to confirm or progress further.
I realize I am missing some fundamental gaps in understanding how these things work. There are a lot of resources online, but it's hard to find this needle in the haystack. Can someone help me learn more about this?

Let's first look at your signal (I've added endpoint=False to make the division even):
t = np.linspace(0, 10*np.pi, 100, endpoint=False)
x = np.sin(t)
Let's divide out the radians (essentially by taking t /= 2*np.pi) and create the same signal by relating to frequencies:
fs = 20 # Sampling rate of 100/5 = 20 (e.g. Hz)
f = 1 # Signal frequency of 1 (e.g. Hz)
t = np.linspace(0, 5, 5*fs, endpoint=False)
x = np.sin(2*np.pi*f*t)
This makes it more salient that f/fs == 1/20 == 0.05 (i.e. the periodicity of the signal is exactly 20 samples). Frequencies in a digital signal always relate to its sampling rate, as you have already guessed. Note that the actual signal is exactly the same no matter what the values of f and fs are, as long as their ratio is the same:
fs = 1 # Natural units
f = 0.05
t = np.linspace(0, 100, 100*fs, endpoint=False)
x = np.sin(2*np.pi*f*t)
In the following I'll use these natural units (fs = 1). The only difference will be in t and hence the generated frequency axes.
Autocorrelation
Your understanding of what the autocorrelation function does is correct. It detects the correlation of a signal with a time-lagged version of itself. It does this by sliding the signal over itself as seen in the right column here (from Wikipedia):
Note that as both inputs to the correlation function are the same, the resulting signal is necessarily symmetric. That is why the output of np.correlate is usually sliced from the middle:
acf = np.correlate(x, x, 'full')[-len(x):]
Now index 0 corresponds to 0 delay between the two copies of the signal.
Next you'll want to find the index or delay that presents the largest correlation. Due to the shrinking overlap this will by default also be index 0, so the following won't work:
acf.argmax() # Always returns 0
Instead I recommend to find the largest peak instead, where a peak is defined to be any index with a larger value than both its direct neighbours:
inflection = np.diff(np.sign(np.diff(acf))) # Find the second-order differences
peaks = (inflection < 0).nonzero()[0] + 1 # Find where they are negative
delay = peaks[acf[peaks].argmax()] # Of those, find the index with the maximum value
Now delay == 20, which tells you that the signal has a frequency of 1/20 of its sampling rate:
signal_freq = fs/delay # Gives 0.05
Fourier transform
You used the following to calculate the FFT:
omega = np.fft.fft(x)
freq = np.fft.fftfreq(x.size, 1)
Thhese functions re designed for complex-valued signals. They will work for real-valued signals, but you'll get a symmetric output as the negative frequency components will be identical to the positive frequency components. NumPy provides separate functions for real-valued signals:
ft = np.fft.rfft(x)
freqs = np.fft.rfftfreq(len(x), t[1]-t[0]) # Get frequency axis from the time axis
mags = abs(ft) # We don't care about the phase information here
Let's have a look:
plt.plot(freqs, mags)
plt.show()
Note two things: the peak is at frequency 0.05, and the maximum frequency on the axis is 0.5 (the Nyquist frequency, which is exactly half the sampling rate). If we had picked fs = 20, this would be 10.
Now let's find the maximum. The thresholding method you have tried can work, but the target frequency bin is selected blindly and so this method would suffer in the presence of other signals. We could just select the maximum value:
signal_freq = freqs[mags.argmax()] # Gives 0.05
However, this would fail if, e.g., we have a large DC offset (and hence a large component in index 0). In that case we could just select the highest peak again, to make it more robust:
inflection = np.diff(np.sign(np.diff(mags)))
peaks = (inflection < 0).nonzero()[0] + 1
peak = peaks[mags[peaks].argmax()]
signal_freq = freqs[peak] # Gives 0.05
If we had picked fs = 20, this would have given signal_freq == 1.0 due to the different time axis from which the frequency axis was generated.
Periodogram
The method here is essentially the same. The autocorrelation function of x has the same time axis and period as x, so we can use the FFT as above to find the signal frequency:
pdg = np.fft.rfft(acf)
freqs = np.fft.rfftfreq(len(x), t[1]-t[0])
plt.plot(freqs, abs(pdg))
plt.show()
This curve obviously has slightly different characteristics from the direct FFT on x, but the main takeaways are the same: the frequency axis ranges from 0 to 0.5*fs, and we find a peak at the same signal frequency as before: freqs[abs(pdg).argmax()] == 0.05.
Edit:
To measure the actual periodicity of np.sin, we can just use the "angle axis" that we passed to np.sin instead of the time axis when generating the frequency axis:
freqs = np.fft.rfftfreq(len(x), 2*np.pi*f*(t[1]-t[0]))
rad_period = 1/freqs[mags.argmax()] # 6.283185307179586
Though that seems pointless, right? We pass in 2*np.pi and we get 2*np.pi. However, we can do the same with any regular time axis, without presupposing pi at any point:
fs = 10
t = np.arange(1000)/fs
x = np.sin(t)
rad_period = 1/np.fft.rfftfreq(len(x), 1/fs)[abs(np.fft.rfft(x)).argmax()] # 6.25
Naturally, the true value now lies in between two bins. That's where interpolation comes in and the associated need to choose a suitable window function.

Inverse FFT returns negative values when it should not

I have several points (x,y,z coordinates) in a 3D box with associated masses. I want to draw an histogram of the mass-density that is found in spheres of a given radius R.
I have written a code that, providing I did not make any errors which I think I may have, works in the following way:
My "real" data is something huge thus I wrote a little code to generate non overlapping points randomly with arbitrary mass in a box.
I compute a 3D histogram (weighted by mass) with a binning about 10 times smaller than the radius of my spheres.
I take the FFT of my histogram, compute the wave-modes (kx, ky and kz) and use them to multiply my histogram in Fourier space by the analytic expression of the 3D top-hat window (sphere filtering) function in Fourier space.
I inverse FFT my newly computed grid.
Thus drawing a 1D-histogram of the values on each bin would give me what I want.
My issue is the following: given what I do there should not be any negative values in my inverted FFT grid (step 4), but I get some, and with values much higher that the numerical error.
If I run my code on a small box (300x300x300 cm3 and the points of separated by at least 1 cm) I do not get the issue. I do get it for 600x600x600 cm3 though.
If I set all the masses to 0, thus working on an empty grid, I do get back my 0 without any noted issues.
I here give my code in a full block so that it is easily copied.
import numpy as np
import matplotlib.pyplot as plt
import random
from numba import njit
# 1. Generate a bunch of points with masses from 1 to 3 separated by a radius of 1 cm
radius = 1
rangeX = (0, 100)
rangeY = (0, 100)
rangeZ = (0, 100)
rangem = (1,3)
qty = 20000 # or however many points you want
# Generate a set of all points within 1 of the origin, to be used as offsets later
deltas = set()
for x in range(-radius, radius+1):
for y in range(-radius, radius+1):
for z in range(-radius, radius+1):
if x*x + y*y + z*z<= radius*radius:
deltas.add((x,y,z))
X = []
Y = []
Z = []
M = []
excluded = set()
for i in range(qty):
x = random.randrange(*rangeX)
y = random.randrange(*rangeY)
z = random.randrange(*rangeZ)
m = random.uniform(*rangem)
if (x,y,z) in excluded: continue
X.append(x)
Y.append(y)
Z.append(z)
M.append(m)
excluded.update((x+dx, y+dy, z+dz) for (dx,dy,dz) in deltas)
print("There is ",len(X)," points in the box")
# Compute the 3D histogram
a = np.vstack((X, Y, Z)).T
b = 200
H, edges = np.histogramdd(a, weights=M, bins = b)
# Compute the FFT of the grid
Fh = np.fft.fftn(H, axes=(-3,-2, -1))
# Compute the different wave-modes
kx = 2*np.pi*np.fft.fftfreq(len(edges[0][:-1]))*len(edges[0][:-1])/(np.amax(X)-np.amin(X))
ky = 2*np.pi*np.fft.fftfreq(len(edges[1][:-1]))*len(edges[1][:-1])/(np.amax(Y)-np.amin(Y))
kz = 2*np.pi*np.fft.fftfreq(len(edges[2][:-1]))*len(edges[2][:-1])/(np.amax(Z)-np.amin(Z))
# I create a matrix containing the values of the filter in each point of the grid in Fourier space
R = 5
Kh = np.empty((len(kx),len(ky),len(kz)))
#njit(parallel=True)
def func_njit(kx, ky, kz, Kh):
for i in range(len(kx)):
for j in range(len(ky)):
for k in range(len(kz)):
if np.sqrt(kx[i]**2+ky[j]**2+kz[k]**2) != 0:
Kh[i][j][k] = (np.sin((np.sqrt(kx[i]**2+ky[j]**2+kz[k]**2))*R)-(np.sqrt(kx[i]**2+ky[j]**2+kz[k]**2))*R*np.cos((np.sqrt(kx[i]**2+ky[j]**2+kz[k]**2))*R))*3/((np.sqrt(kx[i]**2+ky[j]**2+kz[k]**2))*R)**3
else:
Kh[i][j][k] = 1
return Kh
Kh = func_njit(kx, ky, kz, Kh)
# I multiply each point of my grid by the associated value of the filter (multiplication in Fourier space = convolution in real space)
Gh = np.multiply(Fh, Kh)
# I take the inverse FFT of my filtered grid. I take the real part to get back floats but there should only be zeros for the imaginary part.
Density = np.real(np.fft.ifftn(Gh,axes=(-3,-2, -1)))
# Here it shows if there are negative values the magnitude of the error
print(np.min(Density))
D = Density.flatten()
N = np.mean(D)
# I then compute the histogram I want
hist, bins = np.histogram(D/N, bins='auto', density=True)
bin_centers = (bins[1:]+bins[:-1])*0.5
plt.plot(bin_centers, hist)
plt.xlabel('rho/rhom')
plt.ylabel('P(rho)')
plt.show()
Do you know why I'm getting these negative values? Do you think there is a simpler way to proceed?
Sorry if this is a very long post, I tried to make it very clear and will edit it with your comments, thanks a lot!
-EDIT-
A follow-up question on the issue can be found [here].1

The filter you create in the frequency domain is only an approximation to the filter you want to create. The problem is that we are dealing with the DFT here, not the continuous-domain FT (with its infinite frequencies). The Fourier transform of a ball is indeed the function you describe, however this function is infinitely large -- it is not band-limited!
By sampling this function only within a window, you are effectively multiplying it with an ideal low-pass filter (the rectangle of the domain). This low-pass filter, in the spatial domain, has negative values. Therefore, the filter you create also has negative values in the spatial domain.
This is a slice through the origin of the inverse transform of Kh (after I applied fftshift to move the origin to the middle of the image, for better display):
As you can tell here, there is some ringing that leads to negative values.
One way to overcome this ringing is to apply a windowing function in the frequency domain. Another option is to generate a ball in the spatial domain, and compute its Fourier transform. This second option would be the simplest to achieve. Do remember that the kernel in the spatial domain must also have the origin at the top-left pixel to obtain a correct FFT.
A windowing function is typically applied in the spatial domain to avoid issues with the image border when computing the FFT. Here, I propose to apply such a window in the frequency domain to avoid similar issues when computing the IFFT. Note, however, that this will always further reduce the bandwidth of the kernel (the windowing function would work as a low-pass filter after all), and therefore yield a smoother transition of foreground to background in the spatial domain (i.e. the spatial domain kernel will not have as sharp a transition as you might like). The best known windowing functions are Hamming and Hann windows, but there are many others worth trying out.
Unsolicited advice:
I simplified your code to compute Kh to the following:
kr = np.sqrt(kx[:,None,None]**2 + ky[None,:,None]**2 + kz[None,None,:]**2)
kr *= R
Kh = (np.sin(kr)-kr*np.cos(kr))*3/(kr)**3
Kh[0,0,0] = 1
I find this easier to read than the nested loops. It should also be significantly faster, and avoid the need for njit. Note that you were computing the same distance (what I call kr here) 5 times. Factoring out such computation is not only faster, but yields more readable code.

Just a guess:
Where do you get the idea that the imaginary part MUST be zero? Have you ever tried to take the absolute values (sqrt(re^2 + im^2)) and forget about the phase instead of just taking the real part? Just something that came to my mind.

power spectrum by numpy.fft.fft

The figure I plot via the code below is just a peak around ZERO, no matter how I change the data. My data is just one column which records every timing points of some kind of signal. Is the time_step a value I should define according to the interval of two neighbouring points in my data?
data=np.loadtxt("timesequence",delimiter=",",usecols=(0,),unpack=True)
ps = np.abs(np.fft.fft(data))**2
time_step = 1
freqs = np.fft.fftfreq(data.size, time_step)
idx = np.argsort(freqs)
pl.plot(freqs[idx], ps[idx])
pl.show()

As others have hinted at your signals must have a large nonzero component. A peak at 0 (DC) indicates the average value of your signal. This is derived from the Fourier transform itself. This cosine function cos(0)*ps(0) indicates a measure of the average value of the signal. Other Fourier transform components are cosine waves of varying amplitude which show frequency content at those values.
Note that stationary signals will not have a large DC component as they are already zero mean signals. If you do not want a large DC component then you should compute the mean of your signal and subtract values from that. Regardless of whether your data is 0,...,999 or 1,...,1000, or even 1000, ..., 2000 you will get a peak at 0Hz. The only difference will be the magnitude of the peak since it measures the average value.
data1 = arange(1000)
data2 = arange(1000)+1000
dataTransformed3 = data - mean(data)
data4 = numpy.zeros(1000)
data4[::10] = 1 #simulate a photon counter where a 1 indicates a photon came in at time indexed by array.
# we could assume that the sample rate was 10 Hz for example
ps1 = np.abs(np.fft.fft(data))**2
ps2 = np.abs(np.fft.fft(data))**2
ps3 = np.abs(np.fft.fft(dataTransformed))**2
figure()
plot(ps1) #shows the peak at 0 Hz
figure()
plot(ps2) #shows the peak at 0 Hz
figure()
plot(ps3) #shows the peak at 1 Hz this is because we removed the mean value but since
#the function is a step function the next largest component is the 1 Hz cosine wave.
#notice the order of magnitude difference in the two plots.

Here is a bare-bones example that shows input and output with a peak as you'd expect it:
import numpy as np
from scipy.fftpack import rfft, irfft, fftfreq
time = np.linspace(0,10,2000)
signal = np.cos(5*np.pi*time)
W = fftfreq(signal.size, d=time[1]-time[0])
f_signal = rfft(signal)
import pylab as plt
plt.subplot(121)
plt.plot(time,signal)
plt.subplot(122)
plt.plot(W,f_signal)
plt.xlim(0,10)
plt.show()
I use rfft since, more than likely, your input signal is from a physical data source and as such is real.

If you make your data all positive:
ps = np.abs(np.fft.fft(data))**2
time_step = 1
then most probably you will create a large 'DC', or 0 Hz component. So if your actual data has little amplitude, compared to that component, it will disappear from the plot, by the autoscaling feature.

Improving wave detection algorithm

I am new to python and programming and working on the wave peak detection algorithm on my spline - interpolated plot. I have used the code given on this link : https://gist.github.com/endolith/250860 . I have to make this algorithm work on any type of waveform i.e. low as well as high amplitude, baseline not aligned, etc. The goal is to calculate the no of waves in the plot. But my peak detection calculates "invalid" peaks and so gives the wrong answers. By "invalid" peaks I mean if there are two notches close to each other at the wave peak, the program detects 2 peaks i.e. 2 waves when actually its just 1 wave. I have tried changing the 'delta' parameter defined in the peak detection function given on the link but that doesn't solve the generalisation goal which I am working on..Please suggest any improvement on the algorithm or any other approach which I should be using. Any kind of help is welcomed. Thanks in advance.
P.S. I am unable to upload an image of the wrong peak-detected wave plot. I hope my explanation is sufficient enough...
Code is as follows
wave = f1(xnew)/(max(f1(xnew))) ##interpolated wave
maxtab1, mintab1 = peakdet(wave,.005)
maxi = max(maxtab1[:,1])
for i in range(len(maxtab1)):
if maxtab1[i,1] > (.55 * maxi) : ## Thresholding
maxtab.append((maxtab1[i,0],maxtab1[i,1]))
arr_maxtab = np.array(maxtab)
dist = 1500 ## Threshold defined for minimum distance between two notches to be considered as two separate waves
mtdiff = diff(arr_maxtabrr[:,0])
final_maxtab = []
new_arr = []
for i in range(len(mtdiff)):
if mtdiff[i] < dist :
new_arr.append((arr_maxtab[i+1,0],arr_maxtab[i+1,1]))
for i in range(len(arr_maxtab)):
if not arr_maxtab[i,0] in new_arr[:,0]:
final_maxtab.append((arr_maxtab[i,0], arr_maxtab[i,1]))
else:
final_maxtab = maxtab

The ability to distinguish notches from true peaks implies you have a fundamental spacing between peaks. Said differently, there is a minimum frequency resolution at which you would like to run your peak detection search. If you zoom into a signal at which you are well narrower than the noise floor, you'll observe zig zags that seem to 'peak' every few samples.
What it sounds like you want to do is the following:
Smooth the signal.
Find the 'real' peaks.
Or more precisely,
Run the signal through a low pass filter.
Find the peaks within your acceptable peak widths with sufficient signal to noise ratio.
Step 1: Low-Pass Filtering
To do step 1, I recommend you use the signal processing tools provided by scipy.
I'm adapted this cookbook entry, which shows how to use FIR filters to lowpass a signal using scipy.
from numpy import cos, sin, pi, absolute, arange, arange
from scipy.signal import kaiserord, lfilter, firwin, freqz, find_peaks_cwt
from pylab import figure, clf, plot, xlabel, ylabel, xlim, ylim, title, grid, axes, show, scatter
#------------------------------------------------
# Create a signal. Using wave for the signal name.
#------------------------------------------------
sample_rate = 100.0
nsamples = 400
t = arange(nsamples) / sample_rate
wave = cos(2*pi*0.5*t) + 0.2*sin(2*pi*2.5*t+0.1) + \
0.2*sin(2*pi*15.3*t) + 0.1*sin(2*pi*16.7*t + 0.1) + \
0.1*sin(2*pi*23.45*t+.8)
#------------------------------------------------
# Create a FIR filter and apply it to wave.
#------------------------------------------------
# The Nyquist rate of the signal.
nyq_rate = sample_rate / 2.0
# The desired width of the transition from pass to stop,
# relative to the Nyquist rate. We'll design the filter
# with a 5 Hz transition width.
width = 5.0/nyq_rate
# The desired attenuation in the stop band, in dB.
ripple_db = 60.0
# Compute the order and Kaiser parameter for the FIR filter.
N, beta = kaiserord(ripple_db, width)
# The cutoff frequency of the filter.
cutoff_hz = 10.0
# Use firwin with a Kaiser window to create a lowpass FIR filter.
taps = firwin(N, cutoff_hz/nyq_rate, window=('kaiser', beta))
# Use lfilter to filter x with the FIR filter.
filtered_x = lfilter(taps, 1.0, wave)
Step 2: Peak Finding
For step 2, I recommend you use the wavelet transformation peak finder provided by scipy. You must provide as an input your filtered signal and a vector running from the minimum to maximum possible peak widths. This vector will be used as the basis of the wavelet transformation.
#------------------------------------------------
# Step 2: Find the peaks
#------------------------------------------------
figure(4)
plot(t[N-1:]-delay, filtered_x[N-1:], 'g', linewidth=1)
peakind = find_peaks_cwt(filtered_x, arange(3,20))
scatter([t[i] - delay for i in peakind], [filtered_x[i] for i in peakind], color="red")
for i in peakind:
print t[i] + delay
xlabel('t')
grid(True)
show()