# 54Fourier-based derivatives

This tutorial is based on Pelliccia (2019).

When we learned about Fourier transforms, we saw the following equation:

f(t) = \int_{-\infty}^{\infty} F(k) e^{2\pi i k t}dk. .

What happens if we take the time derivative of the expression above?

\begin{split} \frac{d}{dt}f(t) &= \frac{d}{dt}\int_{-\infty}^{\infty} F(k) e^{2\pi i k t}dk \\ &= \int_{-\infty}^{\infty} F(k) \frac{d}{dt} e^{2\pi i k t}dk \\ &= \int_{-\infty}^{\infty} F(k) (2\pi i k) e^{2\pi i k t}dk \end{split}

We found something interesting! The derivative of f(t) can be calculated by taking the Inverse Fourier Transform of (2\pi i k)F(k).

Let’s see this in action!

First let’s get used to taking the Fourier transform and then reconstituting the signal by taking the inverse Fourier transform.

import stuff
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
import scipy
import seaborn as sns
sns.set(style="ticks", font_scale=1.5)  # white graphs, with large and legible letters
from matplotlib.dates import DateFormatter
import matplotlib.dates as mdates
# %matplotlib widget
filename = "dead_sea_1d.csv"
df['date'] = pd.to_datetime(df['date'])
df = df.set_index('date')
fft = scipy.fft.fft(df['level'].values)
reconstituted = scipy.fft.ifft(fft).real
plot
fig, ax = plt.subplots(1, 1, figsize=(8,6))
ax.plot(df['level'], ls="--", color="tab:blue", label="signal")
ax.plot(df['level'].index, reconstituted, color="tab:orange", label=r"$F^{-1}[F[f(t)]]$")
ax.legend(frameon=False)
ylabel="level (m)");

Now let’s apply some smoothing, since we don’t want a choppy derivative.

fft = scipy.fft.fft(df['level'].values)
N = len(df['level'])
xi = scipy.fft.fftfreq(N)
cutoff_T = 3000.0 # days
cutoff_xi = 2.0 * np.pi / cutoff_T
fft_filtered = fft.copy()
reconstituted = scipy.fft.ifft(fft_filtered).real
plot
fig, ax = plt.subplots(2, 1, figsize=(8,8))
ax[0].plot(df['level'], ls="--", color="tab:blue", label="signal")
ax[0].plot(df['level'].index, reconstituted, color="tab:orange", label="Fourier-smoothed")
ax[0].legend(frameon=False)
ax[0].set(ylabel="level (m)")
ax[1].plot(xi, np.abs(fft), '.')
ax[1].set(ylim=[0,100],
xlabel=r"frequency (day$^{-1}$)",
ylabel=r"$|F|$");

There’s a problem now! We eliminated high frequencies and now the smoothed signal is really bad at the edges.

The reason for this is that one assumption of the fft tool is that we have a periodic signal. Our signal is clearly not periodic, and when we treat it as it were periodic, we have a very broad power spectrum, that has really high values even for the highest frequencies. We need to solve this.

One trick to is to make our signal periodic by subtracting from it a straight line that joins the first and last points. See below.

define useful function
def line_chord(p1, p2, t):
"""
given two points p1 and p2, return equation for the line that connects between them
"""
p1x, p1y = p1
p2x, p2y = p2
slope = (p1y-p2y) / (p1x-p2x)
intercept = (p1x*p2y - p1y*p2x) / (p1x-p2x)
return slope*t + intercept
r = np.arange(len(df))
p1 = (r[0], df['level'][0])
p2 = (r[-1], df['level'][-1])
line = line_chord(p1, p2, r)
slope_mperday = (p2[1] - p1[1]) / (p2[0] - p1[0])
df['periodic_level'] = df['level'] - line
plot
fig, ax = plt.subplots(2, 1, figsize=(8,8), sharex=True)
ax[0].plot(df['level'], ls="--", color="tab:blue", label="signal")
ax[0].plot(df.index, line, color="gray", label="straight line")
ax[0].legend(frameon=False)
ax[0].set(ylabel="level (m)")
ax[0].text(df.index[0], -425, f"slope = {slope_mperday:.3e} m/day", color="gray")
ax[1].plot(df['periodic_level'], ls="--", color="tab:blue", label='"periodic" signal')
ax[1].legend(frameon=False)
ax[1].set(ylabel="level (m)");

We can now check that after removing the highest frequencies from the signal, the smoothed reconstitution doesn’t suffer from weird boundary artifacts.

fft = scipy.fft.fft(df['periodic_level'].values)
N = len(df)
xi = scipy.fft.fftfreq(N)
cutoff_T = 3000.0
cutoff_xi = 2.0 * np.pi / cutoff_T
fft_filtered = fft.copy()
reconstituted = scipy.fft.ifft(fft_filtered).real
plot
fig, ax = plt.subplots(1, 1, figsize=(8,6), sharex=True)
ax.plot(df['periodic_level'], ls="--", color="tab:blue", label='"periodic" signal')
ax.plot(df.index, reconstituted, color="tab:orange", label="Fourier-smoothed")
ax.legend(frameon=False)
ax.set(ylabel="level (m)");

That looks great! We can finally calculate the derivative of the smoothed signal by using the nice property we saw at the top of this page.

derivative = scipy.fft.ifft(2.0*np.pi*1j*xi*fft_filtered).real
derivative_correct = derivative + slope_mperday

We have to correct for the fact that we subtracted a straight line from our original signal:

\begin{split} \frac{d}{dt}\text{signal} &= \frac{d}{dt}\left[\text{periodic'' signal + line}\right] \\ &= \frac{d}{dt}\left[\text{periodic'' signal}\right] + \text{slope} \end{split}

plot
fig, ax = plt.subplots(1, 1, figsize=(8,6))
ax.plot(df['level'].index, derivative, color="tab:orange", label='derivative of the "periodic" signal')
ax.plot(df['level'].index, derivative_correct, color="tab:blue", label="derivative of original signal")
ax.legend(frameon=False)
ax.set(ylim=[-0.005, 0.010],
ylabel="derivative (m/day)");

How does our solution compare to the previous method, the gradient?

df['level_smooth_yr'] = df['level'].rolling('365D', center=True).mean()
ax.set(ylabel="derivative (m/day)");