Polarity-preserving low-cut filter

The simplest way to eliminate low frequency noise is to take a time derivative. A disadvantage is that the signal changes at high frequencies from a pulse to a doublet. Here we look at a slightly more complicated low-cut filter which preserves the wave shape at high frequencies and also has an adjustable parameter for choosing the bandwidth of the low cut.

A low-cut filter is one that removes from X the zero frequency $\omega=0$ component, and suppresses low frequency components in the output Y.  

$$Y \eq {\omega^2 \over \omega^2 + \epsilon^2} \ X$$ (24)

To exercise various popular notations we rewrite this as

$$( \omega^2 + \epsilon^2) Y \eq \omega^2 X$$ (25)

Recalling the fact that the negative of a second derivative operator $-(-i\omega)^2$corresponds to convolution with the coefficients (-1,2,-1) we can express the same thought using a convolution notation

$$(-1, 2+\epsilon^2, -1) \ {\tt \ast} \ Y \eq (-1, 2, -1) \ {\tt \ast} \ X$$ (26)

which you can think of as a convolution matrix equation where both sides have tridiagonal matrices. We can express the same thought in Z-transform notation as

$$\left({-1\over Z} +(2+\epsilon^2) -Z\right) Y \eq \left({-1\over Z} +2 -Z\right) X$$ (27)

and we can factor the polynomials  

$${1\over\rho} \left(1-{\rho\over Z}\right) \left(1-{\rho Z}\right) Y(Z) \eq \left(1-{1\over Z}\right) \left(1-{ Z}\right) X(Z)$$ (28)

where $2+\epsilon^2 = (1+\rho^2)/\rho$.(Factoring the polynomial amounts to factoring tridiagonal matrices into a product of upper and lower bidiagonal matrices.) We can group the causal filter parts together thereby defining a causal operator polynomial ratio H(Z):  

$$Y(Z) \eq \rho \ {\left(1-{1\over Z}\right) \over \left(1-{\r... ...-{\rho Z}\right)} \ X(Z) \eq \rho \ \bar H(1/Z) \ H(Z) \ \ X(Z)$$ (29)

Here $H=\bar H$ because the filter coefficients are not complex numbers. H(Z) is called a causal filter because it uses past values of the filter inputs to create the output. Likewise H(1/Z) is called anticausal because the present output comes from future inputs. If you filter with both, $\bar H(1/Z) H(Z)$,the filter response is $\omega^2/(\omega^2+\epsilon^2$(to the extent that a time derivate is the same as a finite difference).

For any application we have a choice between a symmetrical filter $\bar H(1/Z) H(Z)$ whose amplitude response is $\bar H(1/Z) H(Z)$,or the causal filter H(Z) whose energy response is $\bar H(1/Z) H(Z)$.The causal filter  

$$H(Z) \eq {1-Z \over 1-\rho Z}$$ (30)

obviously kills zero frequency because of the time derivative (1-Z). The denominator $1/(1-\rho Z)$ looks like the response of leaky integration. The combination of numerator and denominator has an impulse response that is a positive impulse, followed by a damped exponential of negative polarity. The area of the pulse equals the area of the damped exponential, assuring that the zero-frequency response vanishes and the high-frequency response is an impulse, i.e. it does not distort high frequencies.

EXERCISES:

1. Give an analytic expression for the waveform of equation (30).
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.

 

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