DISCRETE FOURIER TRANSFORM

DISCRETE FOURIER TRANSFORM 1. Introduction The sampled discrete-time fourier transform (DTFT) of a finite length, discrete-time signal is known as the discrete Fourier transform (DFT). The DFT contains a finite number of samples equal to the number of samples N in the given signal. Computationally efficient algorithms for implementing the DFT go by the generic name of fast Fourier transforms (FFTs). This chapter describes the DFT and its properties, and its relationship to DTFT. 2. Definition of DFT and its Inverse Lest us consider a discrete time signal x (n) having a finite duration, say in the range 0 n. The DTFT of the signal is Σ X ( ω) = n-0 x (n)e-jwn (1) Let us sample X using a total of N equally spaced samples in the range : ω (0,2π), so the sampling interval is 2π That is, we sample X(ω) using the frequencies. N ω= ωk = 2πk, 0 k. -jwn (2) Thus X (k) = Σ x (n)e n-0 The result is, by definition the DFT. That is, X (k) = Σ n-0 x (n)ej2πkn (2) Equation (0.2) is known as N-point DFT analysis equation. Fig 0.1 shows the Fourier transform of a discrete time signal and its DFT samples. N

x(w) 0 π 2π w Fig.1 Sampling of X(w) to get x(k) While working with DFT, it is customary to introduce a complex quantity W N = e-j2π /N Also, it is very common to represent the DFT operation N=1 X(k) = DFT ( x(n)) = Σ x(n) W N kn, 0 n The complex quantity Wn is periodic with a period equal to N. That is, W N a+n = e -j+2π/n(a+n) = e -j2π /N n = W N a where a is any integer. Figs. 0.2(a) and (b) shows the sequence for 0 n in the z-plane for N being even and odd respectively. 6 5 5 7 4 6 4 0 0 3 1 3 1 2 2 (a) (b) Fig.2 The Sequence for even N (b) The sequence for odd N. The sequence W k N N for 0 n lies on a circle of unit radius in the complex plane and the phases are equally spaced, beginning at zero. The formula given in the lemma to follow is a useful tool in deriving and analyzing various DFT oriented results.

2.1. Lemma Σ Wkn = N δ (k) = { N, k = 0 (3) n-0 N 0, k 0 Proof : Σ an = 1 - a N : a 1 n-0 1 - a We know that Applying the above result to the left side of equation (3.3), we get Σ (W k N ) n = 1- W k N N = 1- e-j2π k N N n= 0 1- W k N N 1- e-j2π k N N : k 0 = 1-1 1- e- j2π k N N = 0, k 0 when k = 0, the left side of equation (3.3) becomes Σ W 0xn N = Σ 1 = N n= 0 n =0 Hence, we may write 2.2 Inverse DFT N, k = 0 Σ W 0xn N = 0, k 0 = N δ (k), 0 k The DFT values (X(k), 0 k ), uniquely define the sequence x(n) through the inverse DFT formula (IDFT) : x (n) = IDFT (X(k) = 1 Σ X(k) W N -kn, 0 k The above equation is known as the synthesis equation.

Proof : 1 Σ X(k) W N -kn = 1 Σ [Σ x(m) W N km ] = WN -kn m=0 = 1 Σ x(m) [Σ W N -(n-m)k ] m=0 It can be shown that Σ W (n-m)k N = N, n = m 0, n m Hence, 1 Σ x(m) Nδ (n-m) N = 1 x Nx (m) ( sifting property) N m=n = x(n) 2.3 Periodicity of X (k) and x (n) The N-point DFT and N-point IDFT are implicit period N. Even though x (n) and X (k) are sequences of length N each, they can be shown to be periodic with a period N because the exponentials W N ±kn in the defining equations of DFT and IDFT are periodic with a period N. For this reason, x (n) and X (k) are called implicit periodic sequences. We reiterate the fact that for finite length sequences in DFT and IDFT analysis periodicity means implicit periodicity. This can be proved as follows : X (k) = Σ x(n) W N kn N-p=0 X (k+n) = Σ x(n) W N (k+n)n Since, W Nn N = e -j2πnnn = 1, we get X (k+n) = Σ x(n) W -kn N = X (k)

Similarly, x (n) Σ X (k) W N -kn k=0 x (n+n) = 1 Σ X(k) W N -k(n+n) = 1 Σ X(k) W N -kn W N -kn Since, W N -kn = e -j2π/n kn = e -j2πk = 1, we get x(n+n) = 1 Σ X(k) W -kn N = x (n) Since, DFT and its inverse are both periodic with period N, it is sufficient to compute the results for one period (0 to ). We want to emphasize that both x (n) and X (k) have a starting index of zero. A very important implication of x (n), being periodic is, if we wish to find DFT of a periodic signal, we extract one period of the periodic signal and then compute its DFT. Example 1 Compute the 8 point DFt of the sequence x (n) given below : x (n) = (1,1,1,1,0,0,0,0) Solution The complex basis functions, W for 0 n 7 lie on a circle of unit radius as shown in Fig. Ex.3 W8 6 W8 5 W8 7 W8 4 1.0 Re(z) W83 W81 W8 2 Fig. 3 Sequence W 8 0 for 0 n 8.

Since N = 8 we get W = e j2π/8 Thus, W 8 0 = 1 W 1 8 = e -jπ/4 = 1 j 1 W 2 8 = e -jπ/2 = j W 3 8 = e -j3π/4 = 1 j 1 W 4 8 = -W 0 8 = -1 By definition, the DFT of x (n) is X ( k) = DFTI (x (n)) = W kn 8 = 1+1 x W8 k + W 8-2k + W 8 3k. = 1 + W8 k + W 8 2k + W 8 3k k = 0, 1 7 X(0) = 1+1+1+1 = 4 X(1) = 1+ W8 1 + W 2 + 8 W 3 = 1 j2.414 8 X(2) = 1+ W8 2 + W 8 4+ W 8 6 = 0 X(3) = 1+ W8 3 + W 8 6+ W 8 1 = 1 j0.414 X(4) = 1 + W8 4 + W 8 0+ W8 4 = 0 X(5) = 1+ W8 5 + W 8 2 + W 8 7 = 1+j0.414 X(6) = 1+ W8 6 + W 8 4 + W 8 2 = 0. X(7) = 1+ W8 7 + W8 6 + W 8 5 = 1+j2.414 Please note the periodic property : W N a = WN a+n where a is any integer. Example 2 : Compute the DFT of the sequence defined by x (n) = (-1) n for a. = N= 3 b. N = 4, c. N odd, d. N even. Solution X (k) = DFT (x-n) = Σ (-1) n W N nk n-0

Σ (-1) n [W k] N n = 1 (1) N for W k -1 N 1+ W k N a. N = 3 X(k) = 2 = 2 0 k 2 1+ W k N 1+ cos (2kπ/3) j sin ((2πk/3) b. N = 4 X (k) = 0 for W 4 k -1 or k 2 With k = 2 we get X(2) Σ (-1) n W 4 2n = 1 - W 4 2 + W 4 4 - W 4 6 = 1 (-1) + (-1) 2 (-1) 2 = 4 Hence, c. We know that X (k) = 48(k-2) W 4 2n e -j2π / N k If N = 2k, we get W N k = -1. Since N is odd no k exists. This means to say that W k N -1 for all k from 0 to. Therefore, X(k) = 2 0 k 1+ W k N d. N even W N k = - 1, if k = N/2. X ( k) = 0 for k N/2 With k = 2, we get And x (N/2) = Σ [- W N k] n =Σ [1] = N Hence X (k) = N δ (k-n/2)

Example 3. Compute the inverse DFT of the sequence, X(k) = (2,1+j,0,1 j) Solution x (n) = IDFT (X(k)) 1 Σ x(n) W N -kn, 0 n N -1 N Please note that : W N -kn = [ WN kn]* Since, N = 4, we get x (n) = 1 Σ X(k) W 4 -kn, 0 n 3 4 = 1 [X(0) W 4-0xn + X(1) W 4 -n + X (2) W 4-2n + X (3) W 4-3n ] 4 = 1 [2 + (1+j) W 4 -n +0 + (1-j) + X (3) W 4-3n] 4 Hence, x (0) = 1 [2 + (1+j) +(1-j)] = 1 4 x (1) = 1 [2 + (1+j) W -1 4 +(1-j) W 4-3] = 1 4 x (2) = 1 [2 + (1+j) W -2 4 +(1-j) W -6 4 ] = 1 4 Because of periodicity, W -6 4 = W -2 4 Hence, x (2) = 1 [2 + (1+j) (-1) + (1-j) (-1)] = 0 4 x (3) = 1 [2 + (1+j) W 4-3 +(1-j) W 4-9] = 1 4 Because of periodicity, W 4-9 = W 4-5 = W 4-1

Hence, x (3) = 1 [2 + (1+j) W -3 4 +(1-j) W 4-1] 4 = 1 [2 + (1+j) (-j) (-j) + (1-j) ] = 1 4 Hence, x(n) = (1,0,0,1) 3. Matrix Relation for Computing DFT The defining relation for DFT of a finite length sequence x (n) is X (k) = Σ x(n) W n kn, 0.1, Let us evaluate X (k) for different values of k in the range (0, ) as given below : X (0) = W N 0 x (0) + W N 0 x (1) + + W N 0 x () X (1) = W N 0 x (0) + W N 1 x (1) + + W N () x () X (2) = W N 0 x (0) + W N 2 x (1) + + W N 2() x () X () = W N 0 x (0) + W N () x (1) +.. =+ W N () () x () Putting the N DFT equations in N unknowns in the matrix form, we get X = W N x Here X and x are (N x 1) matrices, and Wn is an (N x N) square matrix called the DFT matrix. The full matrix form is described by X(0) W N 0 W N 0 W N 0 W N 0 x(0) X(1) W N 0 W N 1 W N 2 W N () x(1) X(2) = W 0 N W 2 N W 4 N W kn x(2) N : : : : : : X() W 0 N W N W 2() N W ()() N x() The elements W N kn of W N are called complex basis functions or twiddle factors. Example : Compute the 4-point DFT of the sequence, x(n) = (1,2,1,0). Solution With N = 4, W 4 = e We know that -j2 π /4 = -j.

X = W N x X(0) W 0 4 W 0 4 W 0 4 W 0 4 x(0) X(1) W 0 4 W 1 4 W 2 4 W 3 4 x(1) X(2) = W 0 4 W 2 4 W 4 4 W 6 4 x(2) X(3) W 0 4 W 3 4 W 6 4 W 0 4 x(3) Exploiting the periodic property W N 0 = W N n+n where a is any integer the above matrix relation becomes. X(0) W 4 0 W 4 0 W 4 0 W 4 0 x(0) X(1) W 4 0 W 4 1 W4 2 W 4 3 x(1) X(2) = W 4 0 W 4 2 W 4 0 W 4 2 x(2) X(3) W 4 0 W 4 3 W 4 2 W 4 1 x(3) X(0) 1 1 1 1 1 X(1) 1 -j -1 j 2 X(2) = 1-1 1-1 1 X(3) 1 j -1 -j 0 Hence, X(k) = (4, -j2, 0, j2) 4. Matrix Relation for Computing IDFT We know that x = W 1 N X Premultiplying both the sides of the above equation by we get W N -1 -X = W N -1 W N x Or W N -1 X = x x = W N -1 X In the above equation (.8) x = W N -1 is called IDFT matrix. The defining equation for finding IDFT of a sequence X (k) is

X (n) = 1 Σ X(k) W, 0 n = 1 X Σ (k) [W N kn ]* The first set of N IDFT equation in N unknowns may be expressed in the matrix form as x = 1 W* N X N Where W* N denotes the complex conjugate of W N. Comparision of equation (3.8) and (3.9) leads us to conclude that W N -1 = 1/N W* N This very important result shows that W -1 N requires only conjugation of Wn multiplied by 1/N. an obvious computational advantage. The matrix relations (.7) and (.9) together define DFT as a linear transformation. 5. Using the DFT to Find the IDFT We know that x* (n) = 1 Σ X(k) W N -kn, n= 0,1. Taking complex conjugates on both the sides of the above equation, we get x*(n) = 1 Σ X(k) W -kn N *

x*(n) = 1 Σ X*(k) W N -kn * (10) N k =0 The right hand side of equation (3.10) is recognized as the DFT of X* (k), so we can rewrite equation (3.10) as follows : x*(n) = 1 DFT (X*(k)) N Taking complex conjugates on both the sides of equation (11), we get x*(n) = 1 [DFT (X*(k))] N The above results suggests the DFT algorithm itself can be used to find IDFT. In practice, this is indeed what is done. 6. Properties of DFT In the following section, we shall discuss some of the important properties of the DFt. They are strikingly similar to other frequency domain transforms, but must always be used in keeping with implied periodicity for both DFT and IDFT in time and frequency domains. 6.1 Linearity DFT (ax 1 (n) + bx 2 (n)j = ax 1 (k) + bx 2 (k), k = 0, N =1 If X 1 (k) and X 2 (k) are the DFTs of the sequence x 1 (n) and x 2 (n), respectively, both of lengths N. Proof : We know that DFT [x(n)] = Σ x(n) W N kn Letting x (n) = ax 1 (n) + bx 2 (n) we get DFT = ax 1 (n) + bx 2 (n) = Σ (ax 1 (n) +bx 2 (n)] W N kn = a Σ x 1 (n) W N kn + b Σ x 1 (n) W N kn

= ax 1 (k) + bx 2 (k), 0 k Sometimes we represent the linearity property as given below DFT ax 1 (n) + bx 2 (n) ax 1 (k) + bx 2 (k) Example : Find the 4- point DFT of the sequence X(n) = cos ( π n) + sin ( π n) 4 4 Use linearity property. Solution Given N = 4, W N = e j2π /n W = e We know that, W 4 0 = 1 Hence, W 4 0 = 1 Let W 4 1 = e = -j W 3 4 = e = -1 x 1 (n) = cos (π /4 n ) x 2 (n) = sin (π /4 n ) jπ /2 and Then, the values of x (n) and x2 (n) for 0 <n <3 tabulated below : N x 1 (n) = cos (π /4 n ) x 2 (n) = sin (π /4 n ) 0 1 0 1 1 1 2 0 1 3-1 1 X 1 (k) = DFT (x 1 (n)) 3 Σ x 1 (n) W 4 kn, k = 0,1,2,3 X 1 (k) = 1 + 1 W 4 k +0 1 W 4 3k

Hence, X 1 (0) = 1 + 1 1 = 1 X 1 (1) = 1 + 1 W 1 4 1 W 3 4 = 1 j1.414 X 1 (2) = 1 + 1 W 2 4 1 W 6 4 Similarly, = 1 + 1 W 2 4 1 W 4 2 = 1 X 1 (3) = 1 + 1 W 3 4 1 W 9 4 = 1 + 1 W 3 4 1 W 1 4 X 2 (k) = DFT (x 2 (n)) 3 Σ x 2 (n) W kn 4 X 2 (k) = 1 W k 4 + W 2k 4 + 1 W 3k 4 Hence, X 2 (0) = 1 + 1 + 1 = 2.414 X 2 (1) = 1 W 1 4 + W 2 4 + 1 W 3 4 = -1 X 2 (2) = 1 W 2 4 + W 0 4 + 1 W 9 4 = 1 W 3 4 + W 6 4 + 1 W 9 4 = -0.414 X 2 (3) = 1 W 4 3 + W 4 6 + 1 W 4 9 = 1 W 3 4 + W 2 4 + 1 W 1 4 = -1 Finally, applying the linearity property, we get X (k) = DFT ( x 1 (n) + x 2 (n)) = X 1 (k) + X 2 (k) = ( X 1 (0) + X 2 (0), X 1 (1) + X 2 (1). X 1 (2), X 2 (2), X 1 (3) + X 2 (3) ) = (3.414. j1.414.0.586. j1.414) k=0

It may be noted that the arrow, explicitly represents the position index of k = 0 or n = 0 of a given sequence. The absence of this arrow also implicitly means that the first element in a sequence always has the index k = 0 or n = 0. Example Compute DFT (x(n)) of the sequence given below using the linearity property. x (n) = cosh an, 0 n Solution Given x(n) = cosh an, 0 n Then the N point DFT of the sequence x (n) is X (k) = DFT [x(n)] = DFT ( cosh an) = DFT 1 e an + 1 e- an 2 2 Applying linearity property, we get X (k) = 1 DFT [e an ] + 1 DFT [e -an ], 0 n 2 2 We know from Example 3.5, that DFT (b N ) = b N -1, 0 k b W k N -1 Hence, X (k) = 1 e a(n) - 1 + e a(n) - 1 2 e a(n) W k N -1 e -a W k N - 1 = W N -kn ( e a() + e -a() - e - a + e a ] - e an - e an +2 2[1- W k N (e a e a ) + W k N ] = 1 cosh Na + WN k [cosh ( )a cosh a], 0 k 6.2 Circular time shift 1-2 W N k cosha + W N k If DFT [x(n)] = X (k). Then DFt [x(n-m))] = X(k), 0 n Proof : x (n) = N=1 1 Σ [ X (k) W N -kn

N=1 x (n-m) = 1 Σ [ X (k) W N k(n-m) Since, the time shift is circular, we can write the above equation as N=1 x (n-m) = 1 Σ [ X (k) W N km ] W N -kn x (n-m) N = IDFt [ X(k) W N km] or DFT [x(n-m) N ] = W N km X (k) In terms of the transform pair, we can write the above equation is DFT x (n-m) N W N km X (k) Example Find the 4- point DFT of the sequence, x(n) = (1, -1, 1, -1) Also, using time shift property, find the DFT of the sequence, y(n) = x (n-2) 4. Solution Given N = 4 We know that W 4 0 = 1, W 4 1 = -j W 2 4 = -1, W 3 4 = j Hence, X (k) = DFT (x (n) ) 3 = Σ x(n) W kn 4, 0 k 3 = 1 W 0k 4 x 1 x W k 4 +1 x W 2k 4-1 x W 3k 4 = 1- W 4 1+W 2k 4 - W 3k 4 X(0) = 1-1 +1-1 = 0 X(1) = 1 - W 4 1+ W 4 2- W 3 4 = 0 X(2) = 1 - W 2 4 + W 4 4 - W 6 4 = 1- W 4 2 + W 4 0 - W 4 2 = 4 X(3) = 1 - W 4 3 + W 4 6- W 4 9 Given y (n) = x(n-2) 4 = 1- W 4 3 + W 4 2 - W 4 1 = 0 Applying circular time shift property, we get Y(k) = W 4 2k X (k), k = 0,1,2,3 Y(0) = W 4 0 X(0) = 0

Y(1) = W 4 2 X (1) = 0 Y(2) = W 4 4 = W 4 0 x(2) = 4 Y(3) = W 4 6 x (3) = W 4 2 x (3) = 0 Hence, Y(k) = (0,0,4,0) k-0 Example Suppose x(n) is a sequence defined on 0-7 only as ( 0,1,2,3,4,5,6,7), a. Illustrate x(n-2) is b. If DFT (x(n)) = X (k), what is the DFT (x (n-2)s) Solution a. Given To generate x (n-2) move the last 2 samples of x (n) to the beginning. That is, x (n-2) = (6,7,0,1,2,3,4,5) s(n-2)g 7 6 4 2 3 1 0 1 2 3 4 5 6 7 n It should be noted that x (n-2) is implicity periodic with a period N = 8. b. Let y(n)=x(n-2) 8 Applying circular time shift property, we get Y(k) = W 3 2k X(k) Example : Let X (k) donate a 6-point DFT of a length 6 real sequence, x(n). The sequence is shown in Fig. Ex 3.17, without computing the IDFT, determine the length -6 sequence, y(n) whose 6-point DFT is given by, Y (k) = W 3 2k X(k) 5 1 2 3 x(n) 0 4 5-1 Fig. Sequence x(n)

Solution We may write W 3 2k = e- j 2π/3n x2k = e- j 2π/3n x2k Hence, W 3 2k = W6 4k It is given in the problem that Y(k) = W 3 j 2k X(k) Y(k) = W 6 4k X(k) We know that DFT (x(n=m) N = W N mk X(k) IDFT W N mk X(k) = x(n-m) N Hence, y(n) = x(n-4) 6 Since, x(n) = (1,-1,2,3,0,0) We get x(n-4) is by moving the last 4 samples of x(n) to the beginning y(n) = x(n-4) 6 = (2,3,0,0,1-1) Circular frequency shift ( Multiplication by exponential in time-domain) If DFT (x(n)) = X (k), then DFT = X (k-1) N. Proof : X (k) = DFT (x (n)) = Σ x(n) W N kn, 0 k X(k-1) = Σx(n) W N (k-1)n Since, the shift is frequency is circular, we may write the above equation as X(k-1) 8 = Σ { x(n) W N -ln } W N kn Hence, DFT {x(n) W N -ln } = X (k=1) N Example Compute the 4-point DFT of the sequence x (n) = (1,0,1,0), Also, find y (n) if Y (k) = X (k-2) 4.

Solution Given N = 4. Also W 4 0 = 1, W4 l =-j, W 4 2 =-1, W4 3 =j, The DFT of the sequence, x (n) is 3 X(k) = Σ x(n) W 4 kn, 0 k 3 = 1x W 4 0k + 0+1 x W4 2k =0 = 1 +W 4 2k X(0) = 1+1= 2 X(1) = 1+W2 4 = 0 X(2) = 1+W0 4 = 2 X(3) = 1+W2 4 = 0 X(0) = 1+1= 2 X(k) = X(k-2)) 4 Given Y(k) = X(k-2) 4 We know that, DFT (W N -ln x (n)) = X(k-l)N That is, Hence, y(n) = W N -ln x (n) DFT Y(k) = X(k-1)N y(n) = W 4-2n x (n) y(0) = W4-0 x (0) = 1 y(1) = W 4-2 x (1) = 0 y (2) = W4-4 x (2) = W 4-0 x (2) = 1x 1 =1 y(3) = W 4-6 x (3) = W4-2 x (3) = 0 That is, y(n) = (1,0,1,0)

Circular convolution Unlike DFT convolution in DFT in circular consider two sequence x(n) and y(n) the circular convolution of x(n) and y(n) in given by Let f (n) = x (n) x y(n) x(n+n) = 1 Σ x(n-m) N h (m) 0 n m=0 = Σ x(m) h (n-m) n Point to be noted here in that x(n) and y(n) should be of same length Example : Let x (n) = 1,1,1 y (n) = 1,-2,2 Retain x (n) as it is and circularly fold y (n) i.e. y(n) = 1,2,-2. N x(m) y(n-m) N f(n) 0 1,1,1 1,2,-2 1 x 1+1x2+1x-2 = 1 1 1,1,1-2,1,2 1 x -2 + 1x1 + 1x2 = 1 2 1,1,1 2,-2,1 1 x 2, +1x-2 + 1x1 = 1 h (n) = 1,1,1

Summary 1) x (k) = Σ x(n) e-jwn = Σ x(n) Wn kn 2) x (n) = 1 Σ x(n) Wn kn 0 n 3) Periodicity of W n kn W 4 8 W 5 8 -W 7 8 = W3 8 W 6 8 W 2 8 W 7 8 W 0 8 =W4 6 W 1 8 = -W5 8 = W 8 8 4) DFT { ax 1 (n) + bx 2 (n) } = ax 1 (k) + bx 2 (k) 5) DFT {x(n-m) N } = W N mk x (k) 6) DFT {W n -ln x(n) } = x(k-l)n 7) X (k) = X y (N-k) 8) DFT {x(n-n)} = x(n-k) 9) DFT { xe (n) } = 1 DFT {x(n) + 1 DFT { x(-n) N )} 2 = 1 x (k) + 1 x (-k) N 2 2 10) DFT { x1 (n) x2(n) } = 1 x 1(k) x2 (k) N W α N = W α+n N