Fourier Integrals and the Sampling Theorem

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..

View Full Text

Details

  • File Type
    pdf
  • Upload Time
    -
  • Content Languages
    English
  • Upload User
    Anonymous/Not logged-in
  • File Pages
    4 Page
  • File Size
    -

Download

Channel Download Status
Express Download Enable

Copyright

We respect the copyrights and intellectual property rights of all users. All uploaded documents are either original works of the uploader or authorized works of the rightful owners.

  • Not to be reproduced or distributed without explicit permission.
  • Not used for commercial purposes outside of approved use cases.
  • Not used to infringe on the rights of the original creators.
  • If you believe any content infringes your copyright, please contact us immediately.

Support

For help with questions, suggestions, or problems, please contact us