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  Fourier integrals and thesampling theorem Anna-Karin TornbergMathematical Models, Analysis and SimulationFall semester, 2011 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 . The  Fourier transform  ˆ f    =  F  ( f    ) is given byˆ f    ( k  ) = ￿   ∞−∞ f    ( x  ) e  − ikx  dx  ,  k   ∈ R . Here,  f    should be in  L 1 ( R ). -  Inverse transformation: f    ( x  ) = 12 π ￿   ∞−∞ ˆ f    ( k  ) e  ikx  dk  Note: Often, you will se a more symmetric version by using adi ff  erent scaling. -  Theorem of Plancherel:  f    ∈ L 2 ( R ) ⇔ ˆ f    ∈ L 2 ( R ) and ￿   ∞−∞ | f    ( x  ) | 2 dx   = 12 π ￿   ∞−∞ | ˆ f    ( k  ) | 2 dk  .  Fourier Integrals: The Key Rules    df    / dx   =  ik  ˆ f    ( k  )    ￿   x −∞ f    ( ξ  ) d  ξ   = ˆ f    ( k  ) / ( ik  )    f    ( x   − d  ) =  e  − ikd  ˆ f    ( k  )  󿿿 e  icx  f    = ˆ f    ( k  − c  ) Examples of Fourier transforms: f    ( x  ) =  sin ( ω x  ) ˆ f    ( k  ) = − i  π ( δ  ( k  − ω ) − δ  ( k   +  ω )) f    ( x  ) =  cos  ( ω x  ) ˆ f    ( k  ) =  π ( δ  ( k  − ω ) +  δ  ( k   +  ω )) f    ( x  ) =  δ  ( x  ) ˆ f    ( k  ) = 1 for all  k   ∈ R . f    ( x  ) = 􏿿 1 ,  if   − L ≤ x   ≤ L ,0 ,  if   | x  |  >  L .ˆ f    ( k  ) = 2sin kLk   = 2 L sinc( kL ) , 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 ω t   or cos ω t  ?Assume equidistant sampling with  step size T  . Definition:  Nyquist sampling rate:  T   =  π / ω .The Nyquist sampling rate, is exactly 2 equidistant samples over a fullperiod of the function. (period is 2 π / ω ).For a given sampling rate  T  , frequencies higher than the  Nyquist frequency   ω N   =  π / T   cannot be detected. A higher frequency harmonic ismapped to a lower frequency one. This e ff  ect is called  aliasing  .Figure from Strang.  Band-limited functions –  When the continuous signal is a combination of many frequencies  ω ,the largest frequency  ω max   sets the Nyquist sampling rate at T   =  π / ω max  . –  No aliasing at any faster rate. (Do not set sampling rate equal tothe Nyquist sampling rate. Consider sin ω T   to see why.) –  A function  f    ( x  ) is called band-limited if there is a finite  W   such thatits Fourier integral transform ˆ f    ( k  ) = 0 for all  | 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   =  π / W  .(Note that relating to above,  W   =  ω max   +  ε ,  ε  >  0. ) The Sampling Theorem Theorem:  (Shannon-Nyquist) Assume that  f    is band-limited by  W  , i.e.,ˆ f    ( k  ) = 0 for all  | k  | ≥ W  . Let  T   =  π / W   be the Nyquist rate. Then itholds f    ( x  ) = ∞ 󟿿 n = −∞ f    ( nT  )sinc( π ( x  / T   − n ))where sinc( t  ) = sin( t  ) / t   is the  Sinus cardinalis   function.Note: The sinc function is band-limited: 󿿿 sinc( k  ) = 􏿿 1 ,  if   − π  ≤ k   ≤ π ,0 ,  elsewhere.Also note: This interpolation procedure is not used in practice. Slowdecay, summing an infinite number of terms. Approximations to the sincfunction are used, introducing interpolation errors.  The Sampling Theorem: Proof  Assume for simplicity  W   =  π .By the inverse Fourier transform, f    ( x  ) = 12 π ￿   ∞−∞ ˆ f    ( k  ) e  ikx  dk   = 12 π ￿   π − π ˆ f    ( k  ) e  ikx  dk  . Define˜ f    ( k  ) = 􏿿 ˆ f    ( k  ) ,  if   − π  <  k   <  π , periodic continuation, if   | k  | ≥ π .˜ f    can be represented as a Fourier series in  k  -space,˜ f    ( k  ) = ∞ 󟿿 n = −∞ ˆ˜ f   n e  ink  , whereˆ˜ f   n  = 12 π ￿   π − π ˜ f    ( k  ) e  − ink  dk   = 12 π ￿   π − π ˆ f    ( k  ) e  − ink  dk   =  f    ( − n ) . Hence, for  − π  <  k   <  π ,ˆ f    ( k  ) = ˜ f    ( k  ) = ∞ 󟿿 n = −∞ f    ( − n ) e  ink  . The Sampling Theorem: Proof, contd. Therefore, f    ( x  ) = 12 π ￿   π − π ˆ f    ( k  ) e  ikx  dk  = 12 π ￿   π − π   ∞ 󟿿 n = −∞ f    ( − n ) e  ink   e  ikx  dk  = ∞ 󟿿 n = −∞ f    ( − n ) 12 π ￿   π − π e  ik  ( x  + n ) dk  = ∞ 󟿿 n = −∞ f    ( − n )sin π ( x   +  n ) π ( x   +  n )= ∞ 󟿿 n = −∞ f    ( n )sin π ( x   − n ) π ( x   − n )If   W   di ff  erent from  π , rescale  x   by  π / W   and  k   by  W  / π  to complete theproof.

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Jul 23, 2017
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