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					                 Discrete Fourier Transform- Applications
      Application 1 : Detecting periodic signals

Assume you have a long series of data

                                                   u(t)               t= 0 ,1,…., N                   (1)
containing an oscillating function

                                                    u(t) = A e i2 pt +  (t) (2)

with a non rational frequency p, i.e.

                                       n      n 1                                          n 
                                          p      ,                             p                              (3)
                                       N       N                                            N N

If we take the DFT of u(t) we get
                                                                                 Ae                    
                                                        N 1
                                                                                                   t 
                                                   1
                                     u ( m) 
                                     ˆ
                                                   N
                                                        e
                                                        l 0
                                                                   i 2ml / N          i 2pt
                                                                                                                   (4)


we want to study the DFT of the oscillating part

                                           A N 1 i 2ml / N i 2 ( n  )l / N A N 1 i 2 ( n  m )l / N
                            u (m; p) 
                            ˆ                e
                                           N l 0
                                                             e                   e
                                                                                 N l 0
                                                                                                              
                                                             l
                                         i 2 ( n   m )     N
                   A 1 e                                    N
                                                                 1  e i 2
                                                 A
                   N       i 2 ( n   m )
                                             l
                                                                   i 2 ( n   m ) 
                                                                                     l
                      1 e                   N
                                                        N 1  e
                                                          
                                                                                     N 
                                                                                       
                                                                                      
When m  n we can put
                                                               
                                                                                     
                                                     l
                                   i 2 ( n   m )      i 2
                                e                    N
                                                        e N  1  i 2
                                                                                     N
and we get
                                                                                        
                                                                                 i 2
                                                              1 e                      N
                                               u n ; p   A
                                               ˆ                                                            (5)
                                                               i 2

while when         mn

                                                                             1
                                                               u n ; p   O  .
                                                               ˆ
                                                                             N

Ex: compute u n ; p  when  = ½ ,  = ¼ ,
            ˆ

Hint : use
                                                    1  e i  1  cos   i sin 
and
                                                            1
                                                  cos  1   2                            sin   
                                                            2
when  is small.

Conclusion: the presence of ei2pt in u(t) produces a rising of the amplitude of the DFT around
                                      n
m  n , i.e. around the frequency f = .
                                      N


   Application 2 : Filtering a bounded signal

Assume we have sample data
                               y0n = Sn + n , n = 0 , ....N-1       (6)
where Sn is a deterministic signal and n a noise ; to fix the ideas we assume {n} to be a white
noise with a normal distribution

                                               
                                       n  N 0, 2
                                      
                                      
                                             
                                                                       ;         (7)
                                           ˆ ˆ
                                       E  n m   2 nm
                                      

Then we have
                                 1 N 1
                          k   n e i 2kn / N
                           ˆ
                                N n 0
                         
                          E ˆk   0
                                                                                  ;         (8)
                         
                          E j k      e i 2 ( k  j ) n / N     kj
                                           2 N 1                        2
                             
                         
                                        N 2 n 0                     N

In addition we assume to know that the signal {Sn} has a spectrum contained into a certain band of
frequencies.
For instance we assume
                                        
                                        Sk  0 , k  b ,
i.e. we assume that Sn can be represented as

                                        b 1                                N
                               Sn      S e
                                      k   b 1
                                                   k
                                                       i 2kn / N
                                                                    , b 
                                                                             2
                                                                                       (9)

                                                    
although we don’t know exactly whether we have all Sk  0 when k  b.
This problem has already been analysed and solved by testing theory, after a solution via least
squares of the model (6), (9).

Here we go back to the same problem and give a simple solution to it, directly in the frequency
domain, i.e. by representing (6) in terms of DFT
                                     
                               yk  S k  k , k = - M,…0,…M (10)
                               ˆ          ˆ
                              
Since we know for sure that S k = 0 when |k| > b , we can use as an estimator of 2/N the
expression
                            2 
                             ˆ          1       M         2 ( M b )
                                                           2

                           N   2M  b    k  2M  b  (11)
                          
                                                       2
                                                    ˆ
                                 stim      k b  M
Recalling that
                                                      
                                                  E n  n
                                                     2
                                                                     ,

we can conclude that (11) is indeed an unbiased estimator of 2/N, due to (8) and the hypotheses
(7)
                 N
Whenever M         b , as we assume, we can even claim that the estimator (11) has a negligible
                 2
variance, because  2  n   2n and then  2  n / n   2 / n .
                         2                        2




Even if this is not the case, we can proceed as follows.
Since
                                                    2     2
                                                j 
                                                ˆ                (22) ,
                                                N
even for | j |  b we can always consider the hypothesis
                                   H0 : S j = 0 | j |  N                     ,

which can be tested by the ratio
                          1 2                          1 2
                            j
                             ˆ                             ( 2)
                          2                  2 2                      F2, 2 ( M b )   (12)
                        1        M
                                              2 ( M b ) / 2M  b 
                                 ˆk
                                        2

                     2M  b  k b1

This implies that if F is the critical value of the F2,2(M-b) variable, with significance , then all y j
                                                                                                       ˆ
such that
                                                 2    2 
                                             yˆ j  2   F
                                                      N 
                                                          
can be considered as pure noise, or at least we cannot refuse the simple hypothesis that they are
noise.

Accordingly we can determine a threshold L  b such that
                                         2     2 
                                      y j  2   F
                                      ˆ        N           j L
                                               
and consider all the other Fourier coefficients as pure noise, as shown in Fig.1.
                 2
           ˆ
           yk




       2 
        ˆ
  2   v  F
      N 
       


          v2 
           ˆ
         N 
              
              

                                   L                        b                                          M   K




Fig.1: signal and noise spectrum with critical value to determine the maximum frequency L below
which there is signal; notice that only positive frequencies are used because yk  yk
                                                                                 2     2
                                                                              ˆ     ˆ


Therefore a filter adapted to Fig.1 would be obtained by setting to zero all the coefficients above
| j | >L , i.e. the filtered signal is
                                                     L
                                            Sn    y e
                                                    ˆ
                                                   k  L
                                                            k
                                                                i 2kn / N
                                                                                     (13)


It is also interesting to observe that since (41) can be considered as the anti-DFT of the product of
                                                     ˆ
                                                      yk   k L
                                          y k 1,k  
                                          ˆ
                                                     0
                                                          k L

then we can claim, thanks to the convolution theorem, that the same estimator (13) can be
expressed as
                                                   N 1
                                            S m   Fm  n y n                       (14)
                                                   n 0



where

                                     L                                L
                               1                                1
                        Fl 
                               N
                                     L,k e i 2kl / N 
                                   k  L                       N
                                                                    e
                                                                    k  L
                                                                              i 2kl / N
                                                                                            

                                              l           
                                       sin 2                                                  (15)
                         1
                         cos2
                                 Ll
                                              N sin 2 Ll 
                                                 l         
                         N      N                      N
                                      1  cos2
                                                N         
    Application 3 : Computing a convolution
We have a function u(t) defined on R1 and a kernel K(t-t’) and we want to compute the so called
integral convolution
                                        
                              v(t )   K (t  t ' )u (t ' )dt '           (16)
                                        
but all what we know is a sample of u(t)
                             u(0), u(), u(2)…..u((N-1)) .              (17)

First put T = N and approximate (16) as
                                        
                             v(t )      K (t  n)u (n)
                                       n  
                                                                   ;       (18)

if K() is a function going to zero fast enough




                                  -b                                   b


                           Fig.2 an example of symmetric kernel K()


so that we can assume
                           K()  0 , | | > b = L  , L << N ,              (19)

then (18) can be written in t = m 
                                        m L
                            v ( m)      K (m  n)u (n)
                                       nm L
                                                                            (20)


It is obvious that if L<<N we can compute (20) with the values of u(t) that we have, for every m
such that
                                L m N–1–L ,              (21)

i.e. the values
                                          m L
                            v(m)          K (m  n)u (n)
                                         nm L
                                                                            (22)


do coincide with (19) when m satisfies (20).
            0                LΔ                      ≤mΔ≤                 (N-1-L) Δ      N Δ=T


                       Fig.3 : the support of K() is contained in [0,T] when
                       moving the center of K() between L and T - (L+1)


Now take a periodic repetition of u(t) as in Fig.4; the periodization of u(t), ũ(n), will coincide
with u(t) for (0  k  N-1, t=k) , but it will be different from u(t) when k < 0 and k  N.




                                                     u(t)

                                                            u(t) = u(0)




                                  0                           T
                              Fig.4 : periodization and sampling of u(t)


Let us now compute
                                        m L
                           ~
                           v (m)       K (m  n)u (n)
                                       nm L
                                                      ~             .     (23)


Then we have
                    ~
                   v (m)  v(m)              L  m  N 1  L
                   ~                                                             (24)
                   v (m)  v(m)              0  m  L ; N 1  L  m  N 1

                 ~
in other words v (m) coincides with v(m ) apart from a border of width L.
This is the so called border effect.

In order to better understand formula (23) let us agree to put

                                      H n  K (n)  K n
                                      ~ ~
                                      u n  u (n)         ;       (25)
                                            ~
                                      ~  v (m)
                                      vm
indeed when K() is a symmetric function then basically H is the same as K , otherwise (25)
helps in visualizing (23) which now writes

                                                m L
                                      ~
                                      vm           H
                                               nm L
                                                         nm
                                                               ~
                                                               un           .         (26)


Let us take a value m < L and split (26) into two pieces

                                      1                      m L
                            ~
                            vm       H nm u n    H nm u n 
                                    nm L
                                             ~               ~
                                                               n 0
                                                                                       ;          (27)


since ũ is by definition periodical, we have

                                           ũn = ũN+n              -L  n < 0 .

Likewise we define a periodized H ,

                                    ~
                                   H n  H n N
                                                                     Ln0
                                  ~                                                        (28)
                                  H n  H n
                                                                     0n L

and we notice, also inspecting Fig.5,




                                          Hn                                                       ~
                                                                                                   Hn




               -L                                         L                        N-L                    N

                                             Fig.5 : the periodized H n


that, if L < N – L , 2L < N , then there is no overlapping of Hn with its replica and in the
interval between L and N – L , Hn is equal to zero because of the hypothesis (19).

With such definitions, (27) can be written as

                                  m L                         N 1
                           ~
                           vm    H
                                   n 0
                                           nm n
                                                ~
                                                u            H
                                                          n N  m L
                                                                        nm n
                                                                                 ~
                                                                                 u                (29)


and, because of the above remark, we arrive at (cfr.Fig.6)

                                        N 1
                                             ~ ~
                                   vm   H n  m u n 
                                   ~                                     .                 (30)
                                             n 0


                                                 ~
If, in analogy with (28), we define a periodized K n as
                                       ~    ~
                                       Kn  H n                  (31)

we get the final formula

                                          N 1
                                               ~ ~
                                     vm   K m  n u n 
                                     ~                              (32)
                                             n 0


which can be interpreted as wrapping up equation (18) on the circle, because of the periodization of
Kn and un.

                                              ~
As we said, when (19) holds true, (32) yields v m values which practically coincide with vm in the
central interval L < m < N – L.




              ~
              un                     H nm                  un                    ~
                                                                                  H nm




                   m-L n-m 0    m                           N+m-L   N N+m                       n
                                                               N+n-m


                                    Fig.6 : the wrapped convolution


Equation (32) constitutes the so called discrete convolution for periodic functions; for such an
operation a fundamental theorem in DFT theory holds.

Theorem 1 (convolution theorem):

Let fn , gn, (n=0,1,…., N-1) be two regular samples of periodic functions, on the interval [0,1], and
let

                                              1 N 1
                                      hm        f nm g n
                                              N n 0
                                                              (33)

                                              m = 0, 1, … N-1

                                                                                      ˆ ˆ ˆ
be its discrete convolution product ; then the corresponding coefficients of the DFT, f k , g k , hk
satisfy the simple relation

                                         ˆ    ˆ ˆ
                                         hk  f k gk    .        (34)

Proof: we prove (34) for the complex DFT coefficients, assuming, for the sake of simplicity, that
N = 2 M + 1.
In fact we have

                                       N 1                         N 1
                               1                             1
                          ˆ
                          hk 
                               N
                                        ei 2km/ N
                                       m 0                  N
                                                                    f
                                                                    n 0
                                                                               mn   gn 
                                                                                                                        ;          (35)
                            1   N 1
                                     1         N 1
                                                                                     
                          
                            N
                                 N
                                n 0 
                                                e
                                                m 0
                                                        i 2k ( m  n ) / N
                                                                               f mn e i 2kn / N g n
                                                                                     

Now observing that ei2k(m-n)/N and fm-n are both periodic of period N in the index m , we have

                                         N 1                                              n  N 1
                                  1                                                    1
                                  N
                                          ei 2k ( mn) / N f mn 
                                         m 0                                          N
                                                                                            e
                                                                                            mn
                                                                                                           i 2km / N
                                                                                                                        fm 
                                              N 1
                                        1
                                  
                                        N
                                              e
                                              m 0
                                                       i 2km / N        ˆ
                                                                    fm  fm                        ,


so that (35) implies (34).


Corollary 1 : we can prove Parceval’s identity as a corollary of the convolution theorem

Proof: in fact take in (33) (34)

                                                             fn = g-n                     (36)

so that
                                                           N 1
                                               1
                                          ˆ
                                          fk 
                                               N
                                                           g
                                                           n 0
                                                                     n
                                                                                         ˆ
                                                                          e i 2kn / N  g k                   . (37)


Then we have

                                                            1 N 1
                                                  hm          g nm g n
                                                            N n 0
                                                                                                           (38)


and on the same time (recalling (34) and (37) )

                                              M                                    M
                                hm            e i 2km/ N hk 
                                                             ˆ
                                                                                  g               e i 2km / N
                                                                                               2
                                                                                   ˆ       k                                (39)
                                          k  M                                 k  M



By taking m = 0 in (38) (39) we conclude that

                                                            N 1                     M
                                                       1
                                          h0                      gn              g
                                                                           2                           2
                                                                                      ˆ        k                (40)
                                                       N    n 0                  k  M


i.e. Perceval identity.

Corollary 2 : when f, g are real functions, as we have always assumed, we might be willing to
represent the convolution theorem in terms of real DFT and real coefficients.

If we put
                         ˆ
                        f k  ak  ibk                          ˆ
                                                        ak  Re f k                                    
                                                                                                        ˆ
                                                                                             bk  Im  f k
                        g k  ck  id k
                        ˆ                               ck  Re g k
                                                                ˆ                            d k  Im g k 
                                                                                                         ˆ
                        ˆ
                        hk  pk  iqk                            ˆ
                                                         pk  Re hk                                    
                                                                                                       ˆ
                                                                                             qk  Im  hk

so that

                                                            M
                                                                                      kn 
                    N  2M  1           ,   f n  a0  2  ak cos 2
                                                                         kn
                                                                             bk sin 2 
                                                           k 1         N             N

and similarly for the other functions, from (34) we have directly

                          pk  ak ck  bk d k                 qk  ak d k  bk ck                     (41).

To conclude we can go back to the original problem of computing (16) on the basis of the sample
(17).
                                                                ~                     ~
The answer is that we approximate v(t) with a periodic function v (t ) , with samples v m
computed as (cfr.(32))
                                              T N 1 ~ ~
                                        vm   K m  n u n
                                        ~
                                              N n 0

so that
                                       M                              M
                                                                                         ˆ ˆ
                          v t  
                                                                                         ~ ~
                          ~
                                       e i 2kt / T vk  T
                                     k  M
                                                      ˆ
                                                      ~
                                                                      e
                                                                 k  M
                                                                           i 2kt / T
                                                                                         K k uk .     (42)


and in particular
                                                 M
                                                                      ˆ ˆ
                                                                      ~ ~
                                     ~
                                     vm  T     e
                                               k  M
                                                        i 2km / T
                                                                      K k uk             .     (43)


                     ˆ
                     ~
We can observe that u k are exactly the DFT coefficients that we are able to compute from the
                 ˆ
                 ~
sample (17) and K k can be computed as well from the analytic expression of K() after we put

                         K (t )  K (t )  K (T  t ) , t  0, T 
                         ~
                                                                                                  .    (44)


It has to be stressed that the idea of computing the convolution (16) through a DFT becomes
extremely fruitful when we take a sample of N = 2Q points because in this case we can use the so
called Fast Fourier Transform, which is an algorithm performing the DFT and its inverse in an
extremely short time, i.e. a time of the order of O(N logN) .

				
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