Real and apparent seismic frequency

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We often use Ricker wavelets to model seismic, for example when making a synthetic seismogram with which to help tie a well. One simple way to guesstimate the peak or central frequency of the wavelet that will model a particlar seismic section is to count the peaks per unit time in the seismic. But this tends to overestimate the actual frequency because the maximum frequency of a Ricker wavelet is more than the peak frequency. The question is, how much more?

To investigate, let's make a Ricker wavelet and see what it looks like in the time and frequency domains.

>>> T, dt, f = 0.256, 0.001, 25

>>> import bruges
>>> w, t = bruges.filters.ricker(T, dt, f, return_t=True)

>>> import scipy.signal
>>> f_W, W = scipy.signal.welch(w, fs=1/dt, nperseg=256)

When we count the peaks in a section, the assumption is that this apparent frequency — that is, the reciprocal of apparent period or distance between the extrema — tells us the dominant or peak frequency.

To help see why this assumption is wrong, let's compare the Ricker with a signal whose apparent frequency does match its peak frequency: a pure cosine:

>>> c = np.cos(2 * 25 * np.pi * t)
>>> f_C, C = scipy.signal.welch(c, fs=1/dt, nperseg=256)

Notice that the signal is much narrower in bandwidth. If we allowed more oscillations, it would be even narrower. If it lasted forever, it would be a spike in the frequency domain.

Let's overlay the signals to get a picture of the difference in the relative periods:

The practical consequence of this is that if we estimate the peak frequency to be \(f\ \mathrm{Hz}\), then we need to reduce \(f\) by some factor if we want to design a wavelet to match the data. To get this factor, we need to know the apparent period of the Ricker function, as given by the time difference between the two minima.

Let's look at a couple of different ways to find those minima: numerically and analytically.

Find minima numerically

We'll use scipy.optimize.minimize to find a numerical solution. In order to use it, we'll need a slightly different expression for the Ricker function — casting it in terms of a time basis t. We'll also keep f as a variable, rather than hard-coding it in the expression, to give us the flexibility of computing the minima for different values of f.

Here's the equation we're implementing:

$$ w(t, f) = (1 - 2\pi^2 f^2 t^2)\ e^{-\pi^2 f^2 t^2} $$

In Python:

>>> def ricker(t, f):
>>>     return (1 - 2*(np.pi*f*t)**2) * np.exp(-(np.pi*f*t)**2)

Check that the wavelet looks like it did before, by comparing the output of this function when f is 25 with the wavelet w we were using before:

>>> f = 25
>>> np.allclose(w, ricker(t, f=25))
True

Now we call SciPy's minimize function on our ricker function. It itertively searches for a minimum solution, then gives us the x (which is really t in our case) at that minimum:

>>> import scipy.optimize
>>> f = 25
>>> scipy.optimize.minimize(ricker, x0=0, args=(f))

fun: -0.4462603202963996
 hess_inv: array([[1]])
      jac: array([-2.19792128e-07])
  message: 'Optimization terminated successfully.'
     nfev: 30
      nit: 1
     njev: 10
   status: 0
  success: True
        x: array([0.01559393])

So the minimum amplitude, given by fun, is -0.44626 and it occurs at an x (time) of \(\pm 0.01559\ \mathrm{s}\).

In comparison, the minima of the cosine function occur at a time of \(\pm 0.02\ \mathrm{s}\). In other words, the period appears to be \(0.02 - 0.01559 = 0.00441\ \mathrm{s}\) shorter than the pure waveform, which is...

>>> (0.02 - 0.01559) / 0.02
0.22050000000000003

...about 22% shorter. This means that if we naively estimate frequency by counting peaks or zero crossings, we'll tend to overestimate the peak frequency of the wavelet by about 22% — assuming it is approximately Ricker-like; if it isn't we can use the same method to estimate the error for other functions.

This is good to know, but it would be interesting to know if this parameter depends on frequency, and also to have a more precise way to describe it than a decimal. To get at these questions, we need an analytic solution.

Find minima analytically

Python's SymPy package is a bit like Maple — it understands math symbolically. We'll use sympy.solve to find an analytic solution. It turns out that it needs the Ricker function writing in yet another way, using SymPy symbols and expressions for \(\mathrm{e}\) and \(\pi\).

import sympy as sp
t, f = sp.Symbol('t'), sp.Symbol('f')
r = (1 - 2*(sp.pi*f*t)**2) * sp.exp(-(sp.pi*f*t)**2)

Now we can easily find the solutions to the Ricker equation, that is, the times at which the function is equal to zero:

>>> sp.solvers.solve(r, t)
[-sqrt(2)/(2*pi*f), sqrt(2)/(2*pi*f)]

But this is not quite what we want. We need the minima, not the zero-crossings.

Maybe there's a better way to do this, but here's one way. Note that the gradient (slope or derivative) of the Ricker function is zero at the minima, so let's just solve the first time derivative of the Ricker function. That will give us the three times at which the function has a gradient of zero.

>>> dwdt = sp.diff(r, t)
>>> sp.solvers.solve(dwdt, t)
[0, -sqrt(6)/(2*pi*f), sqrt(6)/(2*pi*f)]

In other words, the non-zero minima of the Ricker function are at:

$$ \pm \frac{\sqrt{6}}{2\pi f} $$

Let's just check that this evaluates to the same answer we got from scipy.optimize, which was 0.01559.

>>> np.sqrt(6) / (2 * np.pi * 25)
0.015593936024673521

The solutions agree.

While we're looking at this, we can also compute the analytic solution to the amplitude of the minima, which SciPy calculated as -0.446. We just plug one of the expressions for the minimum time into the expression for r:

>>> r.subs({t: sp.sqrt(6)/(2*sp.pi*f)})
-2*exp(-3/2)

Apparent frequency

So what's the result of all this? What's the correction we need to make?

The minima of the Ricker wavelet are \(\sqrt{6}\ /\ \pi f_\mathrm{actual}\ \mathrm{s}\) apart — this is the apparent period. If we're assuming a pure tone, this period corresponds to an apparent frequency of \(\pi f_\mathrm{actual}\ /\ \sqrt{6}\ \mathrm{Hz}\). For \(f = 25\ \mathrm{Hz}\), this apparent frequency is:

>>> (np.pi * 25) / np.sqrt(6)
32.06374575404661

If we were to try to model the data with a Ricker of 32 Hz, the frequency will be too high. We need to multiply the frequency by a factor of \(\sqrt{6} / \pi\), like so:

>>> 32.064 * np.sqrt(6) / (np.pi)
25.00019823475659

This gives the correct frequency of 25 Hz.

To sum up, rearranging the expression above:

$$ f_\mathrm{actual} = f_\mathrm{apparent} \frac{\sqrt{6}}{\pi} $$

Expressed as a decimal, the factor we were seeking is therefore \(\sqrt{6}\ /\ \pi\):

>>> np.sqrt(6) / np.pi
0.779696801233676

That is, the reduction factor is 22%.

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Curious coincidence: in the recent Pi Day post, I mentioned the Riemann zeta function of 2 as a way to compute \(\pi\). It evaluates to \((\pi / \sqrt{6})^2\). Is there a million-dollar connection between the humble Ricker wavelet and the Riemann hypothesis?

I doubt it.

Happy π day, Einstein

It's Pi Day today, and also Einstein's 139th birthday. MIT celebrates it at 6:28 pm — in honour of pi's arch enemy, tau — by sending out its admission notices.

And Stephen Hawking died today. He will leave a great, black hole in modern science. I saw him lecture in London not long after A Brief History of Time came out. It was one of the events that inspired me along my path to science. I recall he got more laughs than a lot of stand-ups I've seen.

But I can't really get behind 3/14. The weird American way of writing dates, mixed-endian style, really irks me. As a result, I have previously boycotted Pi Day, instead celebrating it on 31/4, aka 31 April, aka 1 May. Admittedly, this takes the edge off the whole experience a bit, so I've decided to go full big-endian and adopt ISO-8601 from now on, which means Pi Day is on 3141-5-9. Expect an epic blog post that day.

Transcendence

Anyway, I will transcend the bickering over dates (pausing only to reject 22/7 and 6/28 entirely so don't even start) to get back to pi. It so happens that Pi Day is of great interest in our house this year because my middle child, Evie (10), is a bit obsessed with pi at the moment. Obsessed enough to be writing a book about it (she writes a lot of books; some previous topics: zebras, Switzerland, octopuses, and Settlers of Catan fan fiction, if that's even a thing).

I helped her find some ways to generate pi numerically. My favourite one uses Riemann's zeta function, which we'd recently watched a Numberphile video about. It's the sum of the reciprocals of the natural numbers raised to increasing powers:

$$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$$

Leonhard Euler solved the Basel problem in 1734, proving that \(\zeta(2) = \pi^2 / 6\), so you can compute pi slowly with a naive implementation of the zeta function:

def zeta(s, terms=1000):
    z = 0
    for t in range(1, int(terms)):
        z += 1 / t**s
    return z

(6 * zeta(2, terms=1e7))**0.5

Which returns pi, correct to 6 places:

3.141592558095893

Or you can use one of the various optimized versions of the zeta function, for example this one from the floating point math library mpmath (which I got from this awesome list of 100 ways to compute pi):

>>> from mpmath import *
>>> mp.dps = 50
>>> mp.pretty = True
>>>
>>> sqrt(6*zeta(2))
3.1415926535897932384626433832795028841971693993751068

...which is correct to 50 decimal places.

Here's the bit of Evie's book where she explains a bit about transcendental numbers.

Evie's book shows the relationships between the sets of natural numbers (N), integers (Z), rationals (Q), algebraic numbers (A), and real numbers (R). Transcendental numbers are real, but not algebraic. (Some definitions also let them be complex.)

I was interested in this, because while I 'knew' that pi is transcendental, I couldn't really articulate what that really meant, and why (say) √2, which is also irrational, is not also transcendental. Succinctly, transcendental means 'non-algebraic' (equivalent to being non-constructible). Since √2 is obviously the solution to \(x^2 - 2 = 0\), it is algebraic and therefore not transcendental. 

Weirdly, although hardly any numbers are known to be transcendental, almost all real numbers are. Isn't maths awesome?

Have a transcendental pi day!

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The xkcd comic is by Randall Munroe and licensed CC-BY-NC.

Jounce, Crackle and Pop

I saw this T-shirt recently, and didn't get it. (The joke or the T-shirt.)

It turns out that the third derivative of displacement \(x\) with respect to time \(t\) — that is, the derivative of acceleration \(\mathbf{a}\) — is called 'jerk' (or sometimes, boringly, jolt, surge, or lurch) and is measured in units of m/s³. 

So far, so hilarious, but is it useful? It turns out that it is. Since the force \(\mathbf{F}\) on a mass \(m\) is given by \(\mathbf{F} = m\mathbf{a}\), you can think of jerk as being equivalent to a change in force. The lurch you feel at the onset of a car's acceleration — that's jerk. The designers of transport systems and rollercoasters manage it daily.

$$ \mathrm{jerk,}\ \mathbf{j} = \frac{\mathrm{d}^3 x}{\mathrm{d}t^3}$$

Here's a visualization of velocity (green line) of a Tesla Model S driving in a parking lot. The coloured stripes show the acceleration (upper plot) and the jerk (lower plot). Notice that the peaks in jerk correspond to changes in acceleration.

The snap you feel at the start of the lurch? That's jounce  — the fourth derivative of displacement and the derivative of jerk. Eager et al (2016) wrote up a nice analysis of these quantities for the examples of a trampolinist and roller coaster passenger. Jounce is sometimes called snap... and the next two derivatives are called crackle and pop. 

What about momentum?

If the momentum \(\mathrm{p}\) of a mass \(m\) moving at a velocity \(v\) is \(m\mathbf{v}\) and \(\mathbf{F} = m\mathbf{a}\), what is mass times jerk? According to the physicist Philip Gibbs, who investigated the matter in 1996, it's called yank:

“Momentum equals mass times velocity.
Force equals mass times acceleration.
Yank equals mass times jerk.
Tug equals mass times snap.
Snatch equals mass times crackle.
Shake equals mass times pop.”

There are jokes in there, help yourself.

What about integrating?

Clearly the integral of jerk is acceleration, and that of acceleration is velocity, the integral of which is displacement. But what is the integral of displacement with respect to time? It's called absement, and it's a pretty peculiar quantity to think about. In the same way that an object with linearly increasing displacement has constant velocity and zero acceleration, an object with linearly increasing absement has constant displacement and zero velocity. (Constant absement at zero displacement gives rise to the name 'absement': an absence of displacement.)

Integrating displacement over time might be useful: the area under the displacement curve for a throttle lever could conceivably be proportional to fuel consumption for example. So absement seems to be a potentially useful quantity, measured in metre-seconds.

Integrate absement and you get absity (a play on 'velocity'). Keep going and you get abseleration, abserk, and absounce. Are these useful quantities? I don't think so. A quick look at them all — for the same Tesla S dataset I used before — shows that the loss of detail from multiple cumulative summations makes for rather uninformative transformations: 

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You can reproduce the figures in this article with the Jupyter Notebook Jerk_jounce_etc.ipynb. Or you can launch a Binder right here in your browser and play with it there, without installing a thing!

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References

David Eager et al (2016). Beyond velocity and acceleration: jerk, snap and higher derivatives. Eur. J. Phys. 37 065008. DOI: 10.1088/0143-0807/37/6/065008

Amarashiki (2012). Derivatives of position. The Spectrum of Riemannium blog, retrieved on 4 Mar 2018.

The dataset is from Jerry Jongerius's blog post, The Tesla (Elon Musk) and
New York Times (John Broder) Feud. I have no interest in the 'feud', I just wanted a dataset.

The T-shirt is from Chummy Tees; the image is their copyright and used here under Fair Use terms.

The vintage Snap, Crackle and Pop logo is copyright of Kellogg's and used here under Fair Use terms.