How can I do a proper FFT of an interpolated signal?

Question

I have a signal $3.5 \cdot 10^{-5}s$ long as shown below.

It was unevenly sampled so I did cubic spine interpolation.

My signal as shown in time domain about 0.9V and 1MHz frequency.

I want to make an accurate FFT picture with $F_s=10\texttt{Mhz}$ so $T_s=10^{-7}s$, hence
$N = 3.5*10^{-5}/10^{-7}=350$

I want an accurate FFT with $1\texttt{Hz}$ bins, so $350$ samples is too little (I'll get spectral leakage.)

What could be done to perform a good FFT in this case?

from scipy.fftpack import fft
#import plotly
#import chart_studio.plotly as py
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from matplotlib.widgets import Cursor
from gekko import GEKKO
#%matplotlib qt
new_x=np.arange(0,3.5e-05,1/(1e7))
dataset_fft=pd.read_table("sinus_1mhz.txt")
array_fft=dataset_fft.values
m=GEKKO()
m.x=m.Param(new_x)
m.y=m.Var()
m.cspline(m.x,m.y,array_fft[:,0],array_fft[:,1])
m.options.IMODE=2
m.solve(disp=False)
fig=plt.figure()
ax=fig.subplots()
ax.grid()
cursor=Cursor(ax, horizOn=True,vertOn=True,useblit=True,color='r',linewidth =1)
#plt.plot(array_fft[:,0],array_fft[:,1])
plt.plot(m.x,m.y)
plt.xlabel("time")
plt.ylabel("amp")
plt.show()

Answer 1

The OP appears to want to have exactly one bin for every 1 Hz spacing in the FFT result. This is arbitrary and there is no requirement of such for a "proper FFT". What is a driving consideration in this regard is the desired frequency resolution of the result. The bin spacing and the frequency resolution are not necessarily related (see what occurs with zero padding for example). The primary driver to frequency resolution is the total time duration of the actual samples captured, independent of what sampling rate is used. (The sampling rate sets the range of the result in the frequency domain, as the resulting samples will extend from $0$ to $N-1$ where $0$ corresponds to DC and $N$ corresponds to the sampling rate.). Consideration for frequency resolution would be our needs to distinguish two signals closely spaced in the frequency domain- this is demonstrated at the link on zero padding given above.

Having sufficient resolution bandwidth, and not having spectral leakage effects, are two independent considerations. The resolution bandwidth is set by the total time duration of the capture, and minimizing spectral leakage is done by using more advanced windows (multiply the time domain capture by a window with tapers the beginning and the end of the capture, prior to taking the FFT).

The frequency resolution as given by the equivalent noise BW (ENBW) or resolution BW (RBW) in Hz is $1/T$ where $T$ is the capture duration in seconds, regardless of sampling rate, for an FFT without any further windowing to reduce spectral leakage. Such windowing only increases RBW (typically on the order of 2x, see this post for an accurate computation and further intuition) so, for example, if “accurate FFT” means being able to resolve 1 Hz in frequency, a capture length of at least 2 seconds would be required.

Thus for “accurate FFT” I recommend using a capture length commensurate with the desired RBW allowing for use of additional windowing, and then window to reduce the effects of spectral leakage. Thus assumes the waveform is sufficiently stationary over the duration of the capture since the average frequency over that duration will be presented. In this case with 350 samples at the 10 MHz sampling rate, $T = 350/10e6 = 35 \mu s$. $1/T = 28.6$ KHz. So the resolution bandwidth without use of further windowing is 28.6 KHz. Given the rectangular window in this case, spectral leakage will be dominant- and since there is a large DC offset, the leakage from this offset will swamp out much of the lower frequency content. Using a more advanced window will make the resolution bandwidth wider; to reduce that requires a longer $T$.

For computing accurate power spectrums (power spectral density), I recommend the Welch method.