REFERENCES

Bolondi, C. D., Loinger, E., and Rocca, F., 1982, Offset continuation of seismic sections: Geophysical Prospecting, 30, 813-828.

Forel, D., and Gardner, G. H. F., 1988, A three dimensional perspective on two-dimensional dip moveout: Geophysics, 53, 604-610.

Notfors, C. D., and Godfrey, R. J., 1987, Dip moveout in the frequency-wavenumber domain: Geophysics, 52, 1718-1721.

The time stretch (1) amounts to time stretch proportional to time:  

$${\Delta t} = {t_{1}-t} = {t_{1} (1-{1\over{\alpha}})}.$$ (7)

The logarithmic resampling is defined by:  

$${s} = {\log({t\over{t_{\min}}})}.$$ (8)

After logarithmic variable transformation from t to s the time stretch of equation (1) and (A-1) is:  

$${s_{1}} = {s + \log(\alpha)},$$ (9)

 

$${\Delta s} = {\log(\alpha)}.$$ (10)

Having Fourier transformed the time series p'(s) this constant shift translates to a complex scaling:  

$${P_{1}(\sigma_{1})} = {P(\sigma_{1}) e^{i \sigma_{1} \log(\alpha)}}.$$ (11)

$\sigma_{1}$ denotes the $\alpha$ stretched variable of the ${\bf FL}$ domain.

After Fourier transformation from variable t to $\omega$ the original stretch relation (1) becomes  

$${\omega_{1}} = {{1\over{\alpha}} \omega}.$$ (12)

After logarithmic transformation (A-2) of the frequency trace we find relationships equivalent to equation (A-3) and (A-4):  

$${\nu_{1}} = {\nu + \log({1\over{\alpha}})},$$ (13)

 

$${\Delta \nu} = {\log({1\over{\alpha}})}.$$ (14)

$\nu$ describes the new variable in the logarithmic domain. An additional Fourier transformation allows us to express the intended time stretch as a simple scaling (4) of the transformed trace.

We want to map our input trace onto a logarithmic mesh. How many samples do we need in the logarithmic domain?

According to equation (A-2) the relation between interval $\Delta s$ and $\Delta t$ is:  

$${\Delta s}= {\Delta t {ds\over{dt}}} = {\Delta t\over{t}}.$$ (15)

$\Delta t$ is the sampling interval for which the time trace is properly sampled. $t_{\max}$ is the maximum time of p(t). At $t_{\max}$ $\Delta s$is at a minimum:  

$${\Delta s} = {{\Delta t}\over{t_{\max}}} = {1\over{2 t_{\max} f_{\max}}},$$ (16)

where $f_{\max}$ is the Nyquist frequency corresponding to $\Delta t$.The number of sample points in the log domain has to be  

$${n_{s}} = {{s_{\max}-s_{\min}\over{\Delta{s}}}} = {2 t_{\max} f_{\max} \log({t_{\max}\over{t_{\min}}})}.$$ (17)