# The inverse of the source wavelet

Computation of the inverse of the source wavelet is accomplished mathematically by using the z-transform. For example, let the basic wavelet be a two-point time series given by w(t) : (1, - 12). The z-transform of this wavelet is defined by the following polynomial:

 ${\displaystyle f(t)=\delta (t)\ast {\frac {1}{w(t)}}.}$ (8)
 ${\displaystyle {W(z)=1-{\frac {1}{2}}z}.}$ (9)

The power of variable z is the number of unit time delays associated with each sample in the series. The first term has zero delay, so z is raised to zero power. The second term has unit delay, so z is raised to first power. Hence, the z-transform of a time series is a polynomial in z, whose coefficients are the values of the time samples.

A relationship exists between the z-transform and the Fourier transform. The z-variable is defined as

 ${\displaystyle {z=\exp(-iw\Delta t)},}$ (10)

where ω is angular frequency and Δt is sampling interval.

The convolutional relation in the time domain given by equation (8) means that the z-transform of the inverse filter, F(z), is obtained by polynomial division of the z-transform of the input wavelet, W(z), given by equation (9):

 ${\displaystyle {F(z)={\frac {1}{1-{\frac {1}{2}}z}}=1+{\frac {1}{2}}z+{\frac {1}{4}}z^{2}+\ldots }.}$ (11)

The coefficients of ${\displaystyle F\left(z\right):\left(1,\ {\frac {1}{2}},\ {\frac {1}{4}},\ \ldots \right)}$ represent the time series associated with the filter operator f(t). Note that the series has an infinite number of coefficients, although they decay rapidly. As in any filtering process, in practice the operator is truncated.

First consider the first two terms in equation (11) which yield a two-point filter operator (1, 12). The design and application of this operator is summarized in Table 2-3. The actual output is (1, 0, - 14), whereas the ideal result is a zero-delay spike (1, 0, 0). Although not ideal, the actual result is spikier than the input wavelet, (1, - 12).

Can the result be improved by including one more coefficient in the inverse filter? As shown in Table 2-4, the actual output from the three-point filter is (1, 0, 0, - 18). This is a more accurate representation of the desired output (1, 0, 0, 0) than that achieved with the output from the two-point filter (Table 2-3). Note that there is less energy leaking into the nonzero lags of the output from the three-point filter. Therefore, it is spikier. As more terms are included in the inverse filter, the output is closer to being a spike at zero lag. Since the number of points allowed in the operator length is limited, in practice the result never is a perfect spike.

 Filter Design Input Wavelet ${\displaystyle w\left(t\right):\left(1,\ -{\frac {1}{2}}\right)}$ The z-Transform ${\displaystyle W\left(z\right)=1-\ {\frac {1}{2}}z}$ The Inverse ${\displaystyle F\left(z\right)=1\ +\ {\frac {1}{2}}z\ +\ {\frac {1}{4}}z^{2}\ +\ \cdots }$ The Inverse Filter ${\displaystyle f\left(t\right):\left(1,\ {\frac {1}{2}},\ {\frac {1}{4}},\ \cdots \right)}$ Filter Application Truncated Inverse Filter ${\displaystyle \left(1,\ {\frac {1}{2}}\right)}$ Input Wavelet ${\displaystyle \left(1,\ -{\frac {1}{2}}\right)}$ Actual Output ${\displaystyle \left(1,\ 0,\ -{\frac {1}{4}}\right)}$ Desired Output (1, 0, 0) Convolution Table: 1 - 12 Output 12 1 1 12 1 0 12 1 - 14
 Filter Design Input Wavelet ${\displaystyle w\left(t\right):\left(1,\ -{\frac {1}{2}}\right)}$ The z-Transform ${\displaystyle W\left(z\right)=1-\ {\frac {1}{2}}z}$ The Inverse ${\displaystyle F\left(z\right)=1\ +\ {\frac {1}{2}}z\ +\ {\frac {1}{4}}z^{2}\ +\ \cdots }$ The Inverse Filter ${\displaystyle f\left(t\right):\left(1,\ {\frac {1}{2}},\ {\frac {1}{4}},\ \cdots \right)}$ Filter Application Truncated Inverse Filter ${\displaystyle \left(1,\ {\frac {1}{2}}\ {\frac {1}{4}}\right)}$ Input Wavelet ${\displaystyle \left(1,\ -{\frac {1}{2}}\right)}$ Actual Output ${\displaystyle \left(1,\ 0,\ 0,\ -{\frac {1}{8}}\right)}$ Desired Output (1, 0, 0, 0) Convolution Table: 1 - 12 Output 14 12 1 1 14 12 1 0 14 12 1 0 14 12 1 - 18

The inverse of the input wavelet ${\displaystyle w\left(t\right):\left(1,\ -{\frac {1}{2}}\right)}$ has coefficients that rapidly decay to zero (equation 11). What about the inverse of the input wavelet ${\displaystyle w\left(t\right):\left(-{\frac {1}{2}},\ 1\right)?}$ Again, define the z–transform:

 ${\displaystyle {W(z)=-{\frac {1}{2}}+z}.}$ (12)

The z–transform of its inverse is given by the polynomial division:

 ${\displaystyle {F(z)={\frac {1}{-{\frac {1}{2}}+z}}=-2-4z-8z^{2}-\cdots }}$ (13)

As a result, the inverse filter coefficients are given by the divergent series f(t) : (−2, −4, −8, …). Truncate this series and convolve the two-point operator with the input wavelet (- 12, 1) as shown in Table 2-5. The actual output is (1, 0, −4), while the desired output is (1, 0, 0). Not only is the result far from the desired output, but also it is less spiky than the input wavelet (- 12, 1). The reason for this poor result is that the inverse filter coefficients increase in time rather than decay (equation 13). When truncated, the larger coefficients actually are excluded from the computation.

If we kept the third coefficient of the inverse filter in the above example (equation 13), then the actual output (Table 2-6) would be (1, 0, 0, −8), which also is a bad approximation to the desired output (1, 0, 0, 0).

 Filter Design Input Wavelet ${\displaystyle w\left(t\right):\left(-{\frac {1}{2}},\ 1\right)}$ The z-Transform ${\displaystyle W\left(z\right)=-{\frac {1}{2}}+z}$ The Inverse ${\displaystyle F\left(z\right)=-2\ -\ 4z\ -\ 8z^{2}\ -\ \cdots }$ The Inverse Filter f(t): (−2, −4, −8, …) Filter Application Truncated Inverse Filter (−2, −4) Input Wavelet ${\displaystyle \left(-{\frac {1}{2}},\ 1\right)}$ Actual Output (1, 0, −4) Desired Output (1, 0, 0) Convolution Table: - 12 1 Output −4 −2 1 −4 −2 0 −4 −2 −4
 Filter Design Input Wavelet ${\displaystyle w\left(t\right):\left(-{\frac {1}{2}},\ 1\right)}$ The z-Transform ${\displaystyle W\left(z\right)=-{\frac {1}{2}}+z}$ The Inverse ${\displaystyle F\left(z\right)=-2\ -\ 4z\ -\ 8z^{2}\ -\ \cdots }$ The Inverse Filter f(t): (−2, −4, −8, …) Filter Application Truncated Inverse Filter (−2, −4, −8) Input Wavelet ${\displaystyle \left(-{\frac {1}{2}},\ 1\right)}$ Actual Output (1, 0, 0, −8) Desired Output (1, 0, 0, 0) Convolution Table: - 12 1 Output −8 −4 −2 1 −8 −4 −2 0 −8 −4 −2 0 −8 −4 −2 −8