The basic low-cut filter

Many geophysical measurements contain very low-frequency noise called ``drift.'' For example, it might take some months to survey the depth of a lake. Meanwhile, rainfall or evaporation could change the lake level so that new survey lines become inconsistent with old ones. Likewise, gravimeters are sensitive to atmospheric pressure, which changes with the weather. A magnetic survey of an archeological site would need to contend with the fact that the earth's main magnetic field is changing randomly through time while the survey is being done. Such noises are sometimes called ``secular noise.''

The simplest way to eliminate low frequency noise is to take a time derivative. A disadvantage is that the derivative changes the waveform from a pulse to a doublet (finite difference). Here we examine the most basic low-cut filter. It preserves the waveform at high frequencies; it has an adjustable parameter for choosing the bandwidth of the low cut; and it is causal (uses the past but not the future).

We make our causal lowcut filter (highpass filter) by two stages which can be done in either order.

1.

Apply a time derivative, actually a finite difference, convolving the data with (1,-1).

2.

Integrate, actually to do a leaky integration, to deconvolve with $(1,-\rho)$ where numerically, $\rho$ is slightly less than unity.

The convolution ensures that the zero frequency is removed. The leaky integration almost undoes the differentiation (but does not restore the zero frequency). Adjusting the numerical value of $\rho$ adjusts the cutoff frequency of the filter. To learn the impulse response of the combined processes, we need to convolve the finite difference (1,-1) with the leaky integration $(1, \rho, \rho^2, \rho^3, \rho^4, \cdots)$.The result is $(1, \rho, \rho^2, \rho^3, \rho^4, \cdots)$ minus $(0, 1, \rho, \rho^2, \rho^3, \cdots)$.We can think of this as $(1, 0, 0, 0, 0, \cdots)$ minus $(1-\rho) (1, \rho, \rho^2, \rho^3, \cdots)$.In other words the impulse response is an impulse followed by the negative of a weak $(1-\rho)$ decaying exponential $\rho^t$.Roughly speaking, the cutoff frequency of the filter corresponds to matching one wavelength to the exponential decay time.

Some exercise with Fourier transforms or Z-transforms, shows the Fourier transform of this highpass filter filter to be  

$$H(Z) \eq {1-Z \over 1-\rho Z} \eq 1 -(1-\rho) [ Z^1 +\rho Z^2 +\rho^2 Z^3 +\rho^3 Z^4 \cdots ]$$ (26)

where the unit-delay operator is $Z=e^{i\omega\Delta t}$and where $\omega$ is the frequency. A symmetical (noncausal) lowcut filter would filter once forward with H(Z) and once backwards (adjoint) with H(1/Z).

Seismological data is more complex. A single ``measurement'' consists of an explosion and echo signals recorded at many locations. As before, a complete survey is a track (or tracks) of explosion locations. Thus, in seismology, data space is higher dimensional. Its most troublesome noise is not simply low frequency; it is low velocity. We will do more with multidimensional data in later chapters.

EXERCISES:

1. Give an analytic expression for the waveform of equation (26).
2. Define a low-pass filter as 1-H(Z). What is the low-pass impulse response?
3. Put Galilee data on a coarse mesh. Consider north-south lines as one-dimensional signals. Find the value of $\rho$ for which H is the most pleasing filter.
4. Find the value of $\rho$ for which $\bar HH$ is the most pleasing filter.
5. Find the value of $\rho$ for which H applied to Galilee has minimum energy. (Experiment with a range of about ten values around your favorite value.)
6. Find the value of $\rho$ for which $\bar HH$ applied to Galilee has minimum energy.
7. Repeat above for east-west lines.

 

galocut90
Figure 6 The depth of the Sea of Galilee after roughening.