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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 2ml / N i 2pt (4) we want to study the DFT of the oscillating part A N 1 i 2ml / 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 mn 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 ei2pt 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 2kn / 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 2kn / 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 2M b k 2M 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 ) / 2M b ˆk 2 2M b k b1 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 yk 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 2kn / 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 2kl / N k L N e k L i 2kl / 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) nm 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) nm 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) nm 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 nm L nm ~ un . (26) Let us take a value m < L and split (26) into two pieces 1 m L ~ vm H nm u n H nm u n nm 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 Ln0 ~ (28) H n H n 0n 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 nm n ~ u H n N m L nm 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 nm un ~ H nm 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 nm 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 2km/ N m 0 N f n 0 mn gn ; (35) 1 N 1 1 N 1 N N n 0 e m 0 i 2k ( m n ) / N f mn e i 2kn / N g n Now observing that ei2k(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 2k ( mn) / N f mn m 0 N e mn i 2km / N fm N 1 1 N e m 0 i 2km / 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 2kn / N g k . (37) Then we have 1 N 1 hm g nm g n N n 0 (38) and on the same time (recalling (34) and (37) ) M M hm e i 2km/ N hk ˆ g e i 2km / 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 2kt / T vk T k M ˆ ~ e k M i 2kt / T K k uk . (42) and in particular M ˆ ˆ ~ ~ ~ vm T e k M i 2km / 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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Discrete-Fourier-Transform

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