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### Whittaker-Shannon Sampling Theorem: The Infinite Interval Version

Remark: Note that even though the expansion coefficients can be determined from these integrals, it is not necessary to do so. Instead, one can obtain the expansion coefficients from directly. One need not evaluate the integral at all. The key to success lies in the sifting property, Eq.(2.70).

Suppose one knows that has a Fourier transform which is non-zero only in the interval . This is no severe restriction because is square integrable, and one can set , provided we make large enough. This implies that

(Why?) Consequently, we have

where the wave packets are given by Eq.(2.72). It is easy to determine the expansion coefficients. Using the sifting property, Eq.(2.70), one obtains

This means that the expansion coefficients are determined from the values of sampled at the equally spaced points . These sampled values of determined its representation

in terms of the set of orthonormal wave packets. This representation of in terms of its sampled values is accurate. It is called the Whittaker-Shannon sampling theorem. It is a generalization of the special case, Eq.(2.17) mentioned on page .

Exercise 24.2 (WAVE PACKET TRAINS)
Consider the wave packet

Express the summed wave packets
(a)
(b)
(c)

in terms of appropriate Dirac delta functions, if necessary.

Lecture 16

Next: Phase Space Representation Up: Orthonormal Wave Packets: Definition Previous: Four Properties   Contents   Index
Ulrich Gerlach 2010-12-09