20.12.2013 Views

booklet format - inaf iasf bologna

booklet format - inaf iasf bologna

booklet format - inaf iasf bologna

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

A.A. 2011/2012<br />

Temporal Data Analysis<br />

3.4 The Sampling Theorem<br />

When discussing the Nyquist frequency, we already mentioned that we need at least two samples<br />

per period to show cosine oscillations at the Nyquist frequency. Now we’ll turn the tables and<br />

claim that as a matter of principle we won’t be looking at anything but functions f (t) that are<br />

“bandwidth-limited”, meaning that outside the interval [−Ω Nyq ,Ω Nyq ] their Fourier transforms<br />

F (ω) are 0. In other words: we’ll refine our sampling to a degree where we just manage to<br />

capture all the spectral components of f (t). Now we’ll skillfully use formulas we’ve learned<br />

when dealing with the Fourier series expansion and the continuous Fourier trans<strong>format</strong>ion with<br />

each other, and then pull the sampling theorem out of the hat. For this purpose we will recall<br />

(2.16) and (2.17) which show that a periodic function f (t) can be expanded into an (infinite)<br />

Fourier series:<br />

f (t) =<br />

C k = 1 T<br />

∞∑<br />

C k e iω k t<br />

k=−∞<br />

∫ +T /2<br />

−T /2<br />

ω k = 2πk<br />

T<br />

f (t)e −iω k t dt for k = 0,±1,±2,...<br />

Since F (ω) is 0 outside [−Ω Nyq ,Ω Nyq ], we can continue this function periodically and expand it<br />

into an infinite Fourier series. So we replace: f (t) → F (ω), t → ω, T /2 → Ω Nyq , and get<br />

∞∑<br />

F (ω) = C k e iπkω/Ω Nyq<br />

k=−∞<br />

C k = 1 ∫ +ΩNyq<br />

F (ω)e −iπkω/Ω Nyq<br />

dω<br />

2Ω Nyq<br />

−Ω Nyq<br />

(3.29)<br />

A similar integral also occurs in the defining equation for the inverse continuous Fourier trans<strong>format</strong>ion<br />

(see (2.35)):<br />

f (t) = 1 ∫ +ΩNyq<br />

F (ω)e iωt dω (3.30)<br />

2π −Ω Nyq<br />

The integrations boundaries are ±Ω Nyq , as F (ω) is band-limited. By comparing (3.30) with (3.29)<br />

we have<br />

∫ +ΩNyq<br />

and the two integrals are the same if<br />

−Ω Nyq<br />

F (ω)e −iπkω/Ω Nyq<br />

dω<br />

∫ +ΩNyq<br />

= 2C k Ω Nyq<br />

−Ω Nyq<br />

F (ω)e iωt dω = 2π f (t)<br />

M.Orlandini 53

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!