Fourier Integrals and the Sampling Theorem

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Fourier Integrals and the Sampling Theorem Fourier integrals and the sampling theorem Anna-Karin Tornberg Mathematical Models, Analysis and Simulation Fall semester, 2013 Fourier Integrals Read: Strang, Section 4.3. Fourier series are convenient to describe periodic functions (or functions with support on a finite interval). What to do if the function is not periodic? - Consider f : R C.TheFourier transform fˆ = (f ) is given by ! F 1 ikx fˆ(k)= f (x)e− dx, k R. 2 Z1 Here, f should be in L1(R). - Inverse transformation: 1 1 f (x)= fˆ(k)eikx dk 2⇡ Z1 Note: Often, you will se a more symmetric version by using a di↵erent scaling. - Theorem of Plancherel: f L2(R) fˆ L2(R)and 2 , 2 1 1 1 f (x) 2dx = fˆ(k) 2dk. | | 2⇡ | | Z1 Z1 Fourier Integrals: The Key Rules x\ df\/dx = ikfˆ(k) f (⇠)d⇠ = fˆ(k)/(ik) Z1 ikd icx f (\x d)=e− fˆ(k) e f = fˆ(k c) − − Examples of Fourier transforms: d f (x)=sin(!x) fˆ(k)= i⇡(δ(k !) δ(k + !)) − − − f (x)=cos(!x) fˆ(k)=⇡(δ(k !)+δ(k + !)) − f (x)=δ(x) fˆ(k)=1forallk R. 2 1, if L x L, sin kL f (x)= − fˆ(k)=2 =2Lsinc(kL), 0, if x > L. k ( | | where sinc(t)=sin(t)/t is the Sinus cardinalis function. Sampling of functions - The Nyquist sampling rate How many samples do we need to identify sin !t or cos !t? Assume equidistant sampling with step size T . Definition: Nyquist sampling rate: T = ⇡/!. The Nyquist sampling rate, is exactly 2 equidistant samples over a full period of the function. (period is 2⇡/!). For a given sampling rate T , frequencies higher than the Nyquist frequency !N = ⇡/T cannot be detected. A higher frequency harmonic is mapped to a lower frequency one. This e↵ect is called aliasing. Figure from Strang. Band-limited functions – When the continuous signal is a combination of many frequencies !, the largest frequency !max sets the Nyquist sampling rate at T = ⇡/!max . – No aliasing at any faster rate. (Do not set sampling rate equal to the Nyquist sampling rate. Consider sin !T to see why.) – A function f (x)iscalledband-limitedifthereisafiniteW such that its Fourier integral transform fˆ(k)=0forall k W . | | ≥ The Shannon-Nyquist sampling theorem states that such a function f (x) can be recovered from the discrete samples with sampling frequency T = ⇡/W . (Note that relating to above, W = !max + ", " > 0. ) The Sampling Theorem Theorem: (Shannon-Nyquist) Assume that f is band-limited by W ,i.e., fˆ(k)=0forall k W .LetT = ⇡/W be the Nyquist rate. Then it holds | | ≥ 1 f (x)= f (nT )sinc(⇡(x/T n)) − n= X1 where sinc(t)=sin(t)/t is the Sinus cardinalis function. Note: The sinc function is band-limited: 1, if ⇡ k ⇡, sinc(k)= − (0, elsewhere. d Also note: This interpolation procedure is not used in practice. Slow decay, summing an infinite number of terms. Approximations to the sinc function are used, introducing interpolation errors. The Sampling Theorem: Proof Assume for simplicity W = ⇡. By the inverse Fourier transform, ⇡ 1 1 1 f (x)= fˆ(k)eikx dk = fˆ(k)eikx dk. 2⇡ 2⇡ ⇡ Z1 Z− Define fˆ(k), if ⇡ < k < ⇡, f˜(k)= − periodic continuation, if k ⇡. ( | | ≥ f˜ can be represented as a Fourier series in k-space, 1 ˆ ink f˜(k)= f˜ne , n= X1 where ⇡ ⇡ ˆ 1 ink 1 ink f˜n = f˜(k)e− dk = fˆ(k)e− dk = f ( n). 2⇡ ⇡ 2⇡ ⇡ − Z− Z− Hence, for ⇡ < k < ⇡, − 1 fˆ(k)=f˜(k)= f ( n)eink . − n= X1 The Sampling Theorem: Proof, contd. Therefore, 1 ⇡ f (x)= fˆ(k)eikx dk 2⇡ ⇡ Z− ⇡ 1 1 = f ( n)eink eikx dk 2⇡ − ⇡ n= ! Z− X1 ⇡ 1 1 = f ( n) eik(x+n)dk − 2⇡ n= ⇡ X1 Z− 1 sin ⇡(x + n) = f ( n) − ⇡(x + n) n= X1 1 sin ⇡(x n) = f (n) − ⇡(x n) n= X1 − If W di↵erent from ⇡,rescalex by ⇡/W and k by W /⇡ to complete the proof..
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