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It follows directly from convolution theorem and Formula 15 that
Moreover, if and
, then
| |
(16) |

Proof: Let us express Fourier-transformations of *f*_{1}(*t*) and *f*_{2}(*t*)
according to equation (2),
taking *N*=*r*=*q* (for *f*_{1}(*t*)) and *N*=*r*=*p*
(for *f*_{2}(*t*)). Performing multiplication, we get:
That is exactly the same that equation (16) expresses in temporal domain.
The convolution
is the *q*^{th} integration of the *f*(*t*).

** Next:** Discontinuities and high-frequency band
** Up:** 2: THE STANDARD DISCONTINUITIES
** Previous:** Fourier-transformation
Stanford Exploration Project

1/13/1998