7.16 Discrete Fourier Transform

Transcription

1 38 Signals, Systems, Transforms and Digital Signal Processing with MATLAB i.e. F ( e jω) = F [f[n]] is periodic with period 2π and its base period is given by Example 7.17 Let x[n] = 1. We have Π B (Ω) = u (Ω + B) u (Ω B), π Ω π. X(e jω ) = n= From the duality property we may write i.e. 2π n= n= e jωn = 2π k= δ(t 2nπ) F.S.C. 1 δ(t 2nπ) F.S.C. 1/ (2π) δ (Ω 2kπ). which are the expected Fourier series coefficients of the impulse train Discrete Fourier Transform Let x[n] be an N-point finite sequence that is generally non-nil for 0 n N 1 and nil otherwise. The z-transform of x[n] is given by Its Fourier transform is given by X(z) = x[n]z n. (7.112) ( X e jω) = x[n]e jωn. (7.113) We note that being the z-transform evaluated on the unit circle, X(e jω ) is periodic in Ω with period 2π. In fact, for k integer ( X e j(ω+2kπ)) = x[n]e j(ω+2kπ)n = ( x [n]e jωn = X e jω). (7.114) Similarly to the analysis of finite duration or periodic signals by Fourier series the analysis of finite duration or periodic sequences is the role of the Discrete Fourier Transform DFT. Moreover, in the same way that for continuous time signals the Fourier series is a sampling of the Fourier transform, for discrete time signals the DFT is a sampling of their Fourier transform. In particular, for an N-point finite duration sequence or a sequence that is periodic with a period N, the DFT is in fact a uniform sampling of the Fourier transform such that the unit circle is sampled into N points with an angular spacing of 2π/N, as shown in Fig for the case N = 16. The continuous angular frequency Ω is replaced by the discrete N values Ω k = 2πk/N, k = 0, 1,..., N 1. Denoting the DFT by the symbol X[k] we have its definition in the form ( X [k] = X e j2πk/n) = x[n]e j2πnk/n, k = 0, 1, 2,..., N 1 (7.115)

2 Discrete-Time Fourier Transform 39 FIGURE 7.28 Unit circle divided into 16 points. We note that the DFT is periodic in k with period N. This is the case since it s a sampling of the Fourier transform around the unit circle and The periodic sequence that is the periodic repetition of the DFT e j2π(k+mn)/n = e j2πk/n. (7.116) X[k], k = 0, 1, 2,... (7.117) is called the Discrete Fourier Series DFS and may be denoted by the symbol X[k]. The DFT is therefore only one period of the DFS as obtained by setting k = 0, 1,..., N 1. From the definition of the DFT: X[k] = x[n]e j2πnk/n, k = 0, 1,..., N 1 (7.118) the inverse transform can be evaluated by multiplying both sides of the equation by e j2πr/n. We obtain X[k]e j2πkr/n = Effecting the sum of both sides with respect to k x[n]e j2πk(n r)/n. (7.119) X[k]e j2πkr/n = x[n]e j2πk(n r)/n = x[n] e j2πk(n r)/n. (7.120) For integer m we have whence i.e. { e j2πkm/n N, for m = pn, p integer = 0, otherwise { e j2πk(n r)/n N, for n = r + pn, p integer = 0, otherwise Replacing r by n we have the inverse transform x[n] = 1 N (7.121) (7.122) X[k]e j2πkr/n = Nx[r]. (7.123) X[k]e j2πnk/n. (7.124)

3 40 Signals, Systems, Transforms and Digital Signal Processing with MATLAB Example 7.18 Evaluate the DTFT and the DFT of the sequence The z-transform is given by Let a = e jb X (z) = X (z) = The transform X(z) can be re-written The Fourier transform is written x[n] = cos Bn R N (n) cos nbz n = 1 2 ( e jbn + e jbn) z n. ( a n z n + a n z n) = 1 ( 1 a N z N + 1 ) a N z N 2 1 az 1 1 a z 1 X (z) = 1 cos B z 1 cos NB z N + cos[(n 1) B] z (N+1) 1 2cos B z 1 + z 2. ( X e jω) = 1 [ 1 a N e jnω 2 1 ae + 1 ] a N e jnω. jω 1 a e jω The student can verify that X ( e jω) can be written in the form X ( e jω) { = 0.5 e j(b Ω)()/2 Sd N [(B Ω) /2] } + e j(b+ω)()/2 Sd N[(B + Ω)/2] or, alternatively, ( X e jω) = N 2 {Φ (Ω B) + Φ(Ω + B)} where Φ(Ω) = sin (NΩ/2) N sin (Ω/2) e j()ω/2. The absolute value and phase angle of the function Φ (Ω) are shown in Fig for N = 8. We note that the Fourier transform X ( e jω) closely resembles the transform of a continuous time FIGURE 7.29 The Sd N function and transform.

4 Discrete-Time Fourier Transform 41 truncated sinusoid. The DFT is given by ( X[k] = X e j2πk/n) = N 2 { ( ) ( )} 2π 2π Φ N k B + Φ N k + B. For the special case where the interval N contains an integer number of cycles we have B = 2π m, m = 0, 1, 2,... N [ { } { }] 2π 2π X [k] = (N/2) Φ (k m) + Φ (k + m) { N N N/2, k = m and k = N m = 0, otherwise. The DFT is thus composed of two discrete impulses, one at k = m, the other at k = N m. Note that in the well behaved case B = 2πm/N we can evaluate the DFT directly by writing cos(bn) = 1 2 { e j 2π N mn + e j 2π mn} N = 1 X [k]e j 2π N nk, n = 0, 1,..., N 1.. N Equating the coefficients of the exponentials we have { N/2, k = m, k = N m X [k] = 0, otherwise. We recall from Chapter?? that the Fourier series of a truncated continuous time sinusoid contains in general two discrete sampling functions and that when the analysis interval is equal to the period of the sinusoid or to a multiple thereof the discrete Fourier series spectrum contains only two impulses. We see the close relation between the Fourier series of continuous time signals and the DFT of discrete time signals Discrete Fourier Series We shall use the notation x[n] to denote a periodic sequence of period N, i.e. x [n] = x [n + kn], k integer. (7.125) We shall write X [k] = DFS [ x [n]] meaning x[n] DF S X [k]. Let x [n] be an aperiodic sequence. A periodic sequence x[n] may be formed thereof in the form x[n] = x[n] k= δ [n + kn] = k= x[n + kn], k integer. (7.126) If x [n] is of finite duration 0 n N 1, i.e. a sequence of length N the added shifted versions thereof, forming x [n], do not overlap, and we have x[n] = x[n mod N] (7.127) where n mod N means n modulo N; meaning the remainder of the integer division n N. For example 70 mod 32 = 8. If the sequence x [n] is of length L < N, again no overlapping occurs and in the range 0 n N 1 the value of x[n] is the same as x[n] followed by (N L) zeros. If on the other hand the length of the sequence x[n] is L > N, overlap occurs leading to superposition ( aliasing ) and we can no more write x [n] = x[n mod N].

5 42 Signals, Systems, Transforms and Digital Signal Processing with MATLAB 7.18 DFT of a Sinusoidal Signal Given a finite-duration sinusoidal signal x c(t) = sin(βt + θ)r T(t) of frequency β and duration T, sampled with a sampling interval T s and sampling frequency f s = 1/T s Hz, i.e. ω s = 2π/T s r/s and the signal period is τ = 2π/β. For simplicity of presentation we let θ = 0, the more general case of θ 0 being similarly developed. We presently consider the particular case where the window duration T is a multiple m of the signal period τ i.e. T = mτ, as can be seen in Fig for the case m = 3. x( t), x [ n] 0 T s T N-1 t, n FIGURE 7.30 Sinusoid with 3 cycles during analysis window. The discrete-time signal is given by x[n] = x c(nt s) = sin(bn)r N[n], where B = βt s. We also note that the N-point DFT analysis corresponds to the signal window duration We may write Hence sin(bn) = 1 2j T = mτ = NT s. (7.128) B = βt s = 2π τ Ts = 2π m. (7.129) N { e j 2π N mn e j 2π mn} N = 1 X [k]e j2πnk/n, n = 0, 1,..., N 1. N { jn/2, k = m, k = N m X [k] = 0, otherwise. We note that the fundamental frequency of analysis in the continuous-time domain, which may be denoted by ω 0 is given by ω 0 = 2π/T. The sinusoidal signal frequency β is a multiple m of the fundamental frequency ω 0. In particular β = 2π τ = 2π T/m = m2π = mω0. (7.130) T B = βt s = m 2π T Ts = m2π N. (7.131) The unit circle is divided into N samples denoted k = 0, 1, 2,..., N 1 corresponding to the frequencies Ω = 0, 2π/N,, 4π/N,..., (N 1)π/N. The k = 1 point is the fundamental frequency

6 Discrete-Time Fourier Transform 43 Ω 0 = 2π/N. Since B = m2π/n its falls on the m th point of the circle as the m th harmonic. Its conjugate falls on the point k = N m. The following example illustrates these observations. Example 7.19 Given the signal x c(t) = sin βtr T(t), where β = 250π r/s and T = 24 ms. A C/D converter samples this signal at a frequency of 1000 Hz. At what values of k does the DFT X[k] display its spectral peaks? The signal period is τ = 2π/β = 8 ms. The Rectangular window of duration T contains m = T/τ = 24/8 = 3 cycles of the signal as can be seen in Fig The sampling period is T s = 1 ms. The sampled signal is the sequence x[n] = sin BnR N[n], where B = βt s = π/4 and N = T/T s = 24. The fundamental frequency of analysis is ω 0 = 2π/T, and the signal frequency is β = 2π/τ = (T/τ)ω 0 = mω 0. In the discrete-time domain B = βt s = 2π 2π Ts = τ T/m Ts = 2π N m = mω0. The spectral peak occurs at k = m = 3 and at k = N m = 24 3 = 21, which are the pole positions of the corresponding infinite duration signal, as can be seen in Fig N = 24 =3, 0 B, 12 0,, s 18 FIGURE 7.31 Unit circle divided into 24 points. Example 7.20 Let v (t) = cos(25πt) R T (t) Assuming a sampling frequency f s of 100 samples per second, evaluate the DFT if T = 1.28 sec. Let T s be the sampling interval. f s = 100 Hz, T s = 1 f s = 0.01 sec, N = T T s = 1.28 = v [n] = cos (25π nt s) R N (n) = cos (0.25πn) R N (n) = cos (Bn) R N (n). Writing B = 0.25π = (2π/N)m, we have m = 16. The DFT X[k] has a peak on the unit circle at k = 16 and k = = 112. as seen in Fig V [k] = { N/2 = 64, k = 16, k = 112 0, otherwise.

7 44 Signals, Systems, Transforms and Digital Signal Processing with MATLAB FIGURE 7.32 DFT of a sequence Deducing the Z-Transform from the DFT Consider a finite duration sequence x[n] which is in general non-nil for 0 n N 1 and nil otherwise, and its periodic extension x[n] with a period of repetition N x[n] = x[n + kn]. (7.132) k= Since x[n] is of length N its periodic repetition with period N produces no overlap; hence x[n] = x[n], 0 n N 1. The z-transform of the sequence x [n] is given by X (z) = and its discrete Fourier series DFS is given by x[n] z n (7.133) X [k] = x[n] e j2πkn/n = x[n] W kn N (7.134) where W N = e j2π/n is the N th root of unity. The inverse DFS is x[n] = x[n] = 1 N X [k] e j2πkn/n = 1 N X [k] W kn N, 0 n N 1 (7.135) and the z-transform may thus be deduced from the DFS and hence from the DFT. We have X (z) = = 1 N x [n] z n = = 1 z N N X [k] 1 N X [k] W kn N z n W kn N z n = 1 z N N X [k] 1 W k N z 1 X [k] 1 W k N z 1 (7.136) which is an interpolation formula reconstructing the z-transform from the N-point DFT on the z plane unit circle. We can similarly obtain an interpolation formula reconstructing the Fourier

8 Discrete-Time Fourier Transform 45 transform X ( e jω) from the DFT. To this end we replace z by e jω in the above obtaining X ( e jω) = 1 N = 1 N = 1 N = 1 N X [k] 1 W kn N e jωn 1 W k N e jω X [k] 1 ej(2π/n)kn e jωn 1 W k N e jω X [k]e j(ω 2πk/N)()/2 sin {(Ω 2πk/N)N/2} sin {(Ω 2πk/N) /2} X [k]e j(ω 2πk/N)()/2 Sd N [(Ω 2πk/N) /2] (7.137) The function Sd N (Ω/2) = sin (NΩ/2)/ sin (Ω/2) is depicted in Fig for the case N = 8. Note that over one period the function has zeros at values of Ω which are multiples of 2π/N = 2π/8. In fact { sin (rπ) N, r = 0 Sd N (rπ/n) = sin (rπ/n) = (7.138) 0, r = 1, 2,..., N 1. Hence X(e jω ) Ω=2πm/N = 1 N X [k]e j(2π/n)(m k)()/2 Sd N [π (m k) /N] = X [m] confirming that the Fourier transform X ( e jω) curve passes through the N points of the DFT. (7.139) N FIGURE 7.33 The function Sd N (Ω/2) DFT vs DFS The DFS is but a periodic repetition of the DFT. Consider a finite duration sequence x[n] of length N, i.e. a sequence that is nil except in the interval 0 n N 1, we may extend it periodically

9 46 Signals, Systems, Transforms and Digital Signal Processing with MATLAB with period N, obtaining the sequence x[n] = k= The DFS of x[n] is X [k] and the DFT is simply x[n + kn]. (7.140) X [k] = X [k], 0 k N 1. (7.141) In other words the DFT is but the base period of the DFS. We may write the DFT in the form The inverse DFT is X [k] = x [n] = 1 N x [n] e j 2π N nk = x [n] e j 2π N nk, 0 k N 1. (7.142) X [k] e j 2π N nk, n = 0, 1,..., N 1. (7.143) In summary, as we have seen in Chapter??, here again in evaluating the DFT of a sequence x[n] we automatically perform a periodic extension of x[n] obtaining the sequence x[n]. This in effect produces the sequence seen by the DFS. We then evaluate the DFS and deduce the DFT by extracting the DFS coefficients in the base interval 0 k N 1. It is common in the literature to emphasize the fact that X [k] = X [k] R N [k] (7.144) where R N [k] is the N-point rectangle R N [k] = u [k] u [k N] (7.145) that is, the DFT X [k] is an N-point rectangular window truncation of the periodic DFS X [k]. The result is an emphasis on the fact that X [k] is nil for values of k other than 0 k N 1. Such distinction, however, adds no new information than that provided by the DFS, and is therefore of little significance. In deducing and applying properties of the DFT a judicious approach is to perform a periodic extension, evaluate the DFS and finally deduce the DFT as its base period. Example 7.21 Let x[n] be the rectangle x[n] = R 4 [n]. Evaluate the Fourier transform, the 8-point DFS and 8-point DFT of the sequence and its periodic repetition. FIGURE 7.34 Rectangular sequence and periodic repetition. Referring to Fig we have ( X e jω) = 3 e jωn = 1 e j4ω e j2ω sin (4Ω/2) = = e j3ω/2 Sd 1 e jω e jω/2 4 (Ω/2) sin Ω/2

10 Discrete-Time Fourier Transform 47 The DFS, with N = 8, of the periodic sequence x[n] = ( X [k] = X e jω) Ω=(2π/N)k = k= x[n + 8k] is 3 e j 2π N kn = e j3(π/8)k Sd 4 (πk/8) The magnitude spectrum is 4, k = 0 X 2.613, k = 1, 7 [k] = 0, k = 2, 4, , k = 3, 5 which is plotted in Fig The DFT is the base period of the DFS, i.e. FIGURE 7.35 Periodic discrete amplitude spectrum. X [k] = X [k] R N [k] = e j3πk/8 Sd 4 (πk/8), k = 0, 1,..., Properties of DFS and DFT The following are basic properties of Discrete Fourier Series. Linearity The linearity property states that if x 1 [n] and x 2 [n] are periodic sequences of period N each then x 1 [n] + x 2 [n] DF S X 1 [k] + X 2 [k]. (7.146) Shift in Time The shift in time property states that x[n m] DF S W km N X [k]. (7.147) Shift in Frequency The dual of the shift in time property states that x [n] W nm N Duality From the definition of the DFS and its inverse we may write x[ n] = 1 N DF S X [n m]. (7.148) X [k] W nk N. (7.149)

11 48 Signals, Systems, Transforms and Digital Signal Processing with MATLAB Replacing n by k and vice versa we have x [ k] = 1 N X [n] W nk N = 1 [ ] N DFS X [n]. (7.150) In other words if x[n] DF S X [k] then X [n] DF S N x [ k]. This same property applies to the DFT where, as always, operations such as reflection are performed on the periodic extensions of the time and frequency sequences. The DFT is then simply the base period of the periodic sequence, extending from index 0 to index N 1. Example 7.22 We have evaluated the DFT X[k] and DFS X [k] of the rectangular sequence x[n] of example 7.21 and its periodic extension x [n] with a period N = 8, respectively, shown in Fig From the duality property we deduce that given a sequence i.e. y[n] = X [n] = X [n] R N [n] = e j3πn/8 Sd 4 (πn/8)r N [n] y[n] = e j3πn/8 Sd 4 (πn/8), n = 0, 1,..., 7. and its periodic repetition ỹ [n], the DFS of the latter is Ỹ [k] = N x[ k] and the DFT of y[n] is Y [k] = N x [ k] R N[k]. To visualize these sequences note that the complex periodic sequence ỹ [n] = X [n] = e j3πn/8 Sd 4 (πn/8), of which the absolute value is ỹ [n] = X [n], has the same absolute value as the spectrum shown in Fig with the index k replaced by n. The sequence y[n] has an absolute value which is the base N = 8-point period of this sequence and is shown in Fig y [ n] = X [ n] n FIGURE 7.36 Base-period y[n] of periodic absolute value sequence ỹ [n] The transform Ỹ [k] = N x [ k] is visualized by reflecting the sequence x [n] of Fig about the vertical axis and replacing the index n by k. The transform Y [k] is simply the N-point base period of Ỹ [k], as shown in Fig Periodic Convolution Given two periodic sequences x[n] and ṽ [n] of period N each, multiplication of their DFS X [k] and Ṽ [k] corresponds to periodic convolution of x [n] and ṽ [n]. Let w [n] denote the periodic convolution, written in the form w [n] = x[n] ṽ [n] = m=0 x[m] ṽ [n m]. (7.151)

12 Discrete-Time Fourier Transform 49 Y [k]= N X [-k] k Y [k] k FIGURE 7.37 Reflection of a periodic sequence and base-period extraction. The DFS of w [n] is given by W [k] = { m=0 x [m] ṽ [n m] } e j(2π/n)nk = x[m] ṽ [n m] e j(2π/n)nk. m=0 (7.152) Let n m = r W [k] = m=0 m+ x[m] r= m ṽ [r] e j(2π/n)(r+m)k (7.153) = x[m] e j(2π/n)mk ṽ [r] e j(2π/n)rk = X [k] Ṽ [k]. m=0 r=0 In other words x[n] ṽ [n] DF S X [k] Ṽ [k]. (7.154) The dual of this property states that x [n] ṽ [n] DF S (1/N) X [k] Ṽ [k]. (7.155) Example 7.23 Evaluate the periodic convolution z [n] = x[n] ṽ [n] for the two sequences x[n] and ṽ [n] shown in Fig.7.38 Proceeding graphically as shown in the figure we fold the sequence ṽ [n] about its axis and slide the resulting sequence ṽ [n m] to the point m = n along the m axis evaluating successively the sum of the product x[m] ṽ [n m] for each value of n. We obtain the value of z [n], of which the base period has the form shown in the following table. n z [n] The periodic sequence z [n] is depicted in Fig. 7.39

13 50 Signals, Systems, Transforms and Digital Signal Processing with MATLAB FIGURE 7.38 Example of periodic convolution. FIGURE 7.39 Circular convolution result.

14 Discrete-Time Fourier Transform Circular Convolution Circular convolution of two finite duration sequences, each N points long, is simply periodic convolution followed by retaining only the base period. Symbolically we may write x[n] v [n] = { x[n] ṽ [n]} R N [n]. (7.156) x[n] v [n] DF T X [k] V [k] (7.157) x[n] v [n] DF T (1/N)X [k] V [k]. (7.158) The practical approach is therefore to simply perform periodic convolution then extract the base period, obtaining the circular convolution. In other words, circular convolution is given by x[n] v [n] = [ m=0 x[m] ṽ [n m] ] R N[n] = [ m=0 ṽ [m] x[n m] ] R N[n] (7.159) For the case of the two sequences defined in the last example, circular convolution would be evaluated identically as periodic convolution, as seen in the example, followed by retaining only the base period of the evaluate periodic convolution z [n], i.e. x[n] v [n] = z [n] R N[n] x[n] v [n] = {16, 12, 7,3, 6,10, 13, 17}, for {n = 0, 1, 2, 3, 4, 5, 6, 7.} Circular convolution is not the same but can be related to the usual linear convolution. Let y [n] be the N-point linear convolution of two finite length sequences x[n] and v [n] circular convolution is given by x[n] v [n] = y [n] = x[n] v [n], (7.160) { k= y [n + kn] } R N [n]. (7.161) Circular convolution is therefore the aliasing (superposition of shifted versions) of the linear convolution sequence y [n]. Note that if the length of y [n] is less than or equal to N then the sum on the right hand side includes in the interval 0 n N 1 only the component y [n], while all other components such as y [n N], y [n + N] fall outside that interval so that y [n + kn] R N [n] = 0 for k 0 and x[n] v [n] = x [n] v [n] (7.162) i.e. in this case circular convolution is the same as linear convolution. If the sequences x[n] and v [n] are of lengths N 1 and N 2, the linear convolution sequence y [n] is of length N 1 + N 2 1. If an N-point circular convolution is effected the result would be the same as linear convolution if and only if N N 1 + N 2 1. Example 7.24 Evaluate the linear convolution y [n] = x[n] v [n] of the sequences x [n] and v [n] which are the base periods of x[n] and ṽ [n] of the last example. Proceeding similarly, as shown in Fig. 7.40, we obtain the linear convolution y [n] which may be listed in the form of the following table. n y [n]

15 52 Signals, Systems, Transforms and Digital Signal Processing with MATLAB FIGURE 7.40 Linear convolution of two sequences. The sequence y [n] is depicted in Fig To deduce the value of circular convolution z [n] from the linear convolution y [n] we construct the following table, where z[n] = y[n] + y[n + 8]. obtaining the circular convolution z [n] as found y [n] y [n + 8] z [n] above Circular Convolution using the DFT The following example illustrates circular convolution using the DFT. Example 7.25 Consider the circular convolution of the two sequences x[n] and ṽ [n] of the last

16 Discrete-Time Fourier Transform y [ n] n FIGURE 7.41 Linear convolution results. example. We evaluate X [k], Ṽ [k] and their product and verify that the circular convolution z [n] = x[n] ṽ [n] has the DFS X [k] = X [k] Ṽ [k]. By extracting the N-point base period we conclude that the DFT relation Z [k] = X [k] V [k] also holds. The sequences x[n] and ṽ [n] are periodic with period N = 8. For 0 n 7 we have Letting w = e j(2π/8)k x[n] = δ [n] + δ [n 1] + 2δ [n 2] + 2δ [n 3] + 3δ [n 4] + 3δ [n 5] ṽ [n] = δ [n 2] + 2 {δ [n 3] + 2δ [n 4] + 2δ [n 5]} X [k] = x[n] e j(2π/n)nk = 1 + e j(2π/8)k + 2e j(2π/8)2k + 2e j(2π/8)3k + 3e j(2π/8)4k + 3e j(2π/8)5k X [k] = X [k] R N [k] = X [k], k = 0, 1,..., N 1 Ṽ [k] = e j(2π/8)2k + 2 {e j(2π/8)3k + e j(2π/8)4k + e j(2π/8)5k} V [k] = Ṽ [k] RN [k] = Ṽ [k], k = 0, 1,..., N 1. we have X [k] = 1 + w + 2w 2 + 2w 3 + 3w 4 + 3w 5 Ṽ [k] = 2w 2 + 3w 3 + 3w 4 + 3w 5. Multiplying the two polynomials noticing that w k = w k mod 8, we have The inverse transform of Z [k] is Z [k] = X [k] Ṽ [k] = w + 7w2 + 3w 3 + 6w w w w 7, 0 k N 1 X [k] V [k] = Z [k] R N [k] = Z [k], 0 k 7. z [n] = 16δ [n] + 12δ [n 1] + 7δ [n 2] + 3δ [n 3] + 6δ [n 4] + 10δ [n 5] + 13δ [n 6] + 17δ [n 7], 0 n N 1 and z [n] = z [n], 0 n N 1. This is the same result obtained above by performing circular convolution directly in the time domain. as stated.

17 54 Signals, Systems, Transforms and Digital Signal Processing with MATLAB 7.24 Sampling the Spectrum Let x [n] be an aperiodic sequence with z-transform X (z) and Fourier transform X ( e jω). X (z) = ( X e jω) = n= n= x[n] z n (7.163) x[n] e jωn. (7.164) Sampling the z-transform on the unit circle uniformly into N points, that is at Ω = [2π/N] k, k = 0, 1,..., N 1 we obtain the periodic DFS X [k] = n= We recall, on the other hand, that the same DFS X [k] = x [n] e j(2π/n)nk. (7.165) x[n] e j(2π/n)nk (7.166) is the expansion of a periodic sequence x [n] of period N. To show that x[n] is but an aliasing of the aperiodic sequence x [n] we use the inverse relation Now wherefrom x[n] = 1 N = 1 N m= X [k]e j(2π/n)nk = 1 N x[m] m= e j(2π/n)(n m)k. 1 e j(2π/n)(n m)k = N x [n] = l= x[m]e j(2π/n)mk e j(2π/n)nk { 1, m n = ln 0, otherwise (7.167) (7.168) x[n + ln] (7.169) confirming that sampling the Fourier transform of an aperiodic sequence x [n], leading to the DFS, has for effect aliasing in time of the sequence x [n], which results in a periodic sequence x[n] that can be quite different from x[n]. If on the other hand x[n] is of length N or less the resulting sequence x[n] is a simple periodic extension of x[n]. Since the DFT is but the base period of the DFS these same remarks apply directly to the DFT Table of Properties of DFS and DFT Table 7.3 summarizes basic properties of the DFS expansion. Since the DFT of an N-point sequence x[n] is but the base period of the DFS expansion of x [n], the periodic extension of x [n], the same properties apply to the DFT. We simply replace the sequence x[n] by its periodic extension x[n], apply the DFS property and extract the base period of the resulting DFS. The following example illustrates the approach in applying the shift in time property, which states that if x [n] X [k] then x [n m] e j(2π/n)km X [k].

18 Discrete-Time Fourier Transform 55 FIGURE 7.41 Discrete Fourier series DFS properties. Time n x [n] x [n] x [ n] x [n m] Frequency k X [k] X [ k] X [k] e j(2π/n)km X [k] e j(2π/n)mn x[n] X [k m] x[n] ṽ [n] x[n] ṽ [n] X [k] Ṽ [k] (1/N) X [k] Ṽ [k] Proof of Shift in Time Property Let ṽ [n] = x [n m] Let n m = r i.e. Ṽ [k] = m+ r= m Ṽ [k] = x[n m] e j(2π/n)km. (7.170) x[r] e j(2π/n)k(r+m) (7.171) = e j(2π/n)km x[r] e j(2π/n)kr = e j(2π/n)km X [k] r=0 x [n m] e j(2π/n)km X [k]. (7.172) Note that if the amount of shift m is greater than N the resulting shift is by m mod N since the sequence x[n] is periodic of period N Shift in Time and Circular Shift Given a periodic sequence x [n] of period N the name circular shift refers to shifting the sequence by say m samples followed by extracting the base period, that is, the period 0 n N 1. If we consider the result of the shift on the base period before and after the shift we deduce that the result is a rotation, a circular shift, of the N samples. For example if the periodic sequence has a period N = 8 and has the values {..., 2, 9, 8, 7, 6, 5, 4, 3, 2, 9, 8, 7, 6, 5, 4, 3,...} with the base period x[n] = 2, 9, 8, 7, 6, 5, 4, 3 for n = 0, 1, 2,..., 7, as shown in Fig (a) then if it is shifted one point to the right the resulting base period is x[n 1] = 3, 2, 9, 8, 7, 6, 5, 4 as shown in Fig (b). If shifted instead by two points to the right the resulting sequence is x [n 2] = 4, 3, 2, 9, 8, 7, 6, 5, as shown in Fig (c). We note that the effect is a simple

19 56 Signals, Systems, Transforms and Digital Signal Processing with MATLAB rotation to the right by the number of shifts. If the shift of x[n] is to the left by one point the result is x[n + 1] = 9, 8, 7, 6, 5, 4, 3, 2, as shown in Fig (d). See Fig The base period of x[n] FIGURE 7.42 Circular shift operations. is given by x[n] R N [n], that of x[n m] is x[n m] R N [n] as shown in the figure. The arrow in the figure is the reference point. Shifting the sequence x[n] to the right by k points corresponds to the unit circle as a wheel turning anti-clockwise k steps and reading the values starting from the reference point and vice versa. Note: The properties listed are those of the DFS, but apply equally to the DFT with the proper interpretation that x [n] and X [k] are periodic extensions of the N-point sequences x[n] and X [k], that X [k] = X [k] R N [k] and x [n] = x[n] R N [n]. The shift in time producing x[n m] is equivalent to circular shift and the periodic convolution x [n] ṽ [n] is equivalent to cyclic convolution in the DFT domain Table of DFT Properties Properties of the DFT are listed in Table 7.4. In this table the notation x [ ((k)) N ] stands for x[k mod N]. As noted above, the properties are the same as those of the DFS except for a truncation of a periodic sequence to extract its base period.

20 Discrete-Time Fourier Transform 57 FIGURE 7.42 DFT properties. Time n x[n] Frequency k X [k] x [n] X [(( k)) N ] x [(( n)) N ] x[((n m)) N ] X [k] e j(2π/n)km X [k] e j(2π/n)mn x[n] X [((k m)) N ] x[n] v [n] x[n] v [n] X [k] V [k] (1/N)X [k] V [k] 7.28 Zero Padding Consider a sequence of x [n] of length N defined over the interval 0 n N 1 and zero elsewhere, and its periodic repetition x[n]. We study the effect on the DFT of annexing N zeros, called padding with zeros leading to a sequence x 2 [n] of length 2N. More generally, we consider padding the sequence x [n] with zeros leading to a sequence x 4 [n] say of length 4N, x 8 [n] of length 8N and higher. The addition of N zeros to the sequence implies that the new periodic sequence x 2 [n] is equivalent to a convolution of the original N-point sequence x [n] with an impulse train of period 2N. The result is a sequence of a period double the original period N of the sequence x[n], which is but a convolution of the sequence x [n] with an impulse train of period N. The effect of doubling the period is that in the frequency domain the DFS X 2 [k] and the DFT X 2 [k] are but finer sampling of the unit circle; into 2N points rather than N points. Similarly zero padding leading to a sequence x 4 [n] of length 4N produces a DFT X 4 [k] that is a finer still sampling of the unit circle into 4N points, and so on. We conclude that zero padding leads to finer sampling of the Fourier transform X ( e jω), that is, to an interpolation between the samples of X [k]. The duality between time and frequency domains implies moreover that given a DFT X [k] and DFS X [k] of a sequence x [n], zero padding of the X [k] and, equivalently, X [k], leading to a DFT sequence X 2 [k] corresponds to convolution in the frequency domain of X [k] with an impulse train of period 2N. This implies multiplication in the time domain of the sequence x [n] by an impulse train of double the frequency such that the resulting sequence x 2 [n] is a finer sampling, by a factor of two, of the original sequence x [n]. Similarly, zero padding X [k] leading to a sequence X 4 [k] of length 4N has for effect a finer sampling, by a factor of 4, i.e. interpolation, of the original sequence x[n]. Example 7.26 Let x [n] = R N [n] X ( e jω) = e jωn = 1 e jωn e jωn/2 = 1 e jω e jω/2 = e jω()/2 sin (NΩ/2) sin (Ω/2) 2j sin (ΩN/2) 2j sin (Ω/2) = e jω()/2 Sd N (Ω/2). (7.173)

21 58 Signals, Systems, Transforms and Digital Signal Processing with MATLAB We consider the case N = 4 sot that x [n] = R 4 [n] and then the case of padding x[n] with zeros obtaining the 16 point sequence y [n] = { x [n], n = 0, 1, 2, 3 0, n = 4, 5,..., 15. (7.174) We have 3 X [k] = e j(2π/4)nk = X ( e jω) Ω=(2π/4)k = e j(3/2)(π/2)k Sd 4 (kπ/4) { N = 4, k = 0 = 0, k = 1, 2, 3 Y [k] = 3 e j(2π/16)nk = 1 e j(2π/16)4k 1 e j(2π/16)k = e j(2π/16)2k e j(2π/16)k/2 sin [(2π/16) 2k] sin [(2π/16) k/2] = e j(2π/16)(3k/2) Sd 4 (kπ/16) which is a four times finer sampling of X ( e jω) than in the case of X [k]. (7.175) (7.176) Example 7.27 Consider a sinusoid x c(t) = sin(ω 1t), where ω 1 = 2πf 1), sampled at a frequency f s = Hz. The sinusoid is sampled for a duration of τ = 2.5 msec into N 1 samples. The frequency f 1 of x c(t) is such that in the time interval (0, τ) there are 8.5 cycles of the sinusoid. a) Evaluate the 64-point FFT of the sequence x[n] = x c(nt s, where T s is the sampling interval T s = 1/f s. b) Apply zero padding by annexing 192 zeros to the samples of the sequence x[n]. Evaluate the 256-point FFT of the padded signal vector. Observe the interpolation and the higher spectral peaks that appear thanks to zero padding. The following Matlab program evaluates the FFT of the signal x[n] and subsequently that of the zero-padded vector x z[n]. % Zero padding example. Corinthios 2008 fs=25600 % sampling frequency Ts=1/fs % sampling period Ts = x10 ( 5) tau= % duration of sinusoid N1=0.0025/Ts % N1=64 t=(0:n1-1)*ts % time in seconds % tau contains 8.5 cycles of sinusoid and 64 samples. tau1=tau/8.5 % tau1 is the period of the sinusoid. % f1 is the frequency of the sinusoid in Hz. f1=1/tau1 w1=2*pi*f1 x=sin(w1*t); figure(1) stem(t,x) title( x[n] ) X=fft(x); % N1=64 samples on unit circle cover the range 0 to fs Hz freq=(0:63)*fs/64; Xabs=abs(X); figure(2) stem(freq,xabs) title( Xabs[k] ) % Add = 192 zeros. N = 2 8

22 Discrete-Time Fourier Transform 59 T = N Ts % Duration of zero-padded vector. xz=[x zeros(1,192)]; % xz is x with zero padding t = (0 : N 1) Ts % t=(0:255)*ts figure(3) stem(t,xz) title( Zero-padded vector xz[n] ) Xz=fft(xz); Xzabs=abs(Xz); freqf = (0 : 255) fs/256; % frequency finer-sampling vector figure(4) stem(freqf,xzabs) title( Xzabs[k] ) See Matlab program zero padding example sinusoid.m The signal x[n] is depicted in Fig (a). The modulus X[k] of its DFT can be seen in Fig (b). We note that the signal frequency falls in the middle between two samples on the unit circle. Hence the peak of the spectrum X[k] which should equal N 1/2 = 32 falls between two samples and cannot be seen. The zero-padded signal x z[n] is shown in Fig (c). The modulus X z[k] of the DFT of the zero padded signal can be seen in Fig (d). 1 x[n] 25 Xabs[k] (a) x (b) x Zero-padded vector xz[n] 35 Xzabs[k] (c) (d) x 10 4 FIGURE 7.43 Zero-padding: (a) A sinusoidal sequence, (b) 64-point DFT, (c) zero padded sequence to 256 points, (d) 256-point DFT of padded sequence. We note that interpolation has been effected, revealing the spectral peak of N 1/2 = 32, which now falls on one of the N = 256 samples around the unit circle. By increasing the sequence length

23 60 Signals, Systems, Transforms and Digital Signal Processing with MATLAB through zero padding to N = 4N 1 an interpolation of the DFT spectrum by a factor of 4 has been achieved Discrete Z-Transform A Discrete z Transform DZT may be defined as the result of sampling a circle in the z plane centered about the origin. Note that the DFT is a special case of the DZT obtained if the radius of the circle is unity. An approach to system identification pole-zero modeling employing DZT evaluation and a weighting of z transform spectra has been proposed as an alternative to the well know Prony s Approach [5]. A system is given as a black box and the objective is to evaluate its poles and zeros by applying a finite duration input sequence or an impulse and observing its finite duration output. The approach is based on the fact that knowing only a finite duration of the impulse response the evaluation of the DZT on a circle identifies fairly accurately the frequency of the least damping poles. However, identification of the components damping coefficients, i.e. the radius of the pole or pole-pair cannot be deduced through radial z transforms since spectra along a radial contour passing through the pole-zero rises exponentially toword the origin of the z plane due to a multiple pole at the origin of the transform of such a finite duration sequence. The proposed weighting of spectra unmasks the poles identifying their location in the z plane both in angle and radius as shown in Fig [?]. Once the pole locations and their residues are found the zeros are deduced. The least damped poles are thus deleted deflating the system, i.e. reducing its order. The process is repeated identifying the new least damped poles and so on until all the poles and zeros have been identified. In [?] an example is given showing the identification of a system of the 14 th order. Example 7.28 Given the sequence x[n] = a n {u [n] u [n N]} with a = 0.7 and N = 16. a) Evaluate the z-transform X (z) of x[n], stating its ROC. b) Evaluate and sketch the poles and zeros of X (z) in the z plane. c) Evaluate the z-transform on a circle of radius a in the z plane. d) Evaluate X a [k], the Discrete z Transform DZT along the circle of radius a, by sampling the z-transform along the circle at frequencies Ω = 0, 2π/N, 3π/N,..., (N 1) π/n, similarly to the sampling the DFT effects along the unit circle. We have x [n] = a n R N [n] b) Zeros a) X (z) = a n z n = 1 an z N 1 az, z 0 1 = zn a N z (z a). a N z N = 1 = e j2πk z N = a N e j2πk z = ae j2πk/n = 0.7e j2πk/16 implying a coincidence pole-zero at z = a, pole of order N 1 at z = 0. See Fig c) X ( ae jω) = 1 e jωn 1 e jω = e jω()/2 sin (NΩ/2) sin (Ω/2) = e jω()/2 Sd N (Ω/2).

24 Discrete-Time Fourier Transform 61 FIGURE D plot of weighted z spectrum unmasking a pole pair. d) X a [k] = 1 { e j2πk N, k = 0 1 e = j2πk/n 0, k = 1, 2,..., N 1. Example 7.29 Evaluate the Fourier transform of the sequence { x[n] = 1 n N, N n N 0, otherwise. where N is odd. Using duality deduce the corresponding Fourier series expansion and Fourier transform. Evaluate the Fourier transform of the sequence x 1[n] = x[n N. We may write x[n] = v[n] v[n] where v[n] = Π ()/2 [n] ( X e jω) [ ( = V e jω)] { } 2 2 = Sd 2 sin (NΩ/2) N (Ω/2) =. sin (Ω/2)

25 62 Signals, Systems, Transforms and Digital Signal Processing with MATLAB FIGURE 7.45 Sampling a circle of general radius. Using duality we have Sd 2 N (t/2) F SC V n = { 1 n N, N n N 0, otherwise and Sd 2 N (t/2) F V (jω) = 2π N n= N (1 n /N) δ (ω n). X 1 (e jω) = e jωn Sd 2 N (Ω/2) Fast Fourier Transform The Fast Fourier Transform FFT is an efficient algorithm that reduces the computations required for the evaluation of the DFT. In what follows the derivation of the FFT is developed starting with a simple example of the DFT of N = 8 points. The DFT can be written in matrix form. This form is chosen since it makes it easy to visualize the operations in the DFT and its conversion to the FFT. To express the DFT in matrix form we define an input data vector x of dimension N the elements of which are the successive elements of the input sequence x[n]. Similarly we define a vector X of which the elements are the coefficients X[k] of the Discrete Fourier Transform. The DFT can thus be written in matrix form as X [k] = where F N is an N N matrix of which the elements are given by and The inverse relation is written x [n]e j2πnk/n (7.177) X = F N x (7.178) [F N] rs = w rs (7.179) w = e j 2π N. (7.180) x = 1 N F NX. (7.181) Note that pre-multiplication of a square matrix A by a diagonal Matrix D producing the matrix C = D A may be obtained by multiplying the successive elements of the diagonal matrix D by the successive rows of A. Conversely, post-multiplication of a square matrix A by a diagonal Matrix

26 Discrete-Time Fourier Transform 63 D producing the matrix C = A D may be obtained by multiplying the successive elements of the diagonal matrix D by the successive columns of A. The following example shows the factorization of the matrix F N, which leads to the Fast Fourier Transform. Example 7.30 FIGURE 7.46 Unit circle divided into N = 8 points. Let N = 8. The unit circle is divided as shown in Fig We have w 0 w 0 w 0 w 0 w 0 w 0 w 0 w 0 x 0 w 0 w 1 w 2 w 3 w 4 w 5 w 6 w 7 x 1 w 0 w 2 w 4 w 6 w 0 w 2 w 4 w 6 x 2 X = w 0 w 3 w 6 w 1 w 4 w 7 w 2 w 5 x 3 w 0 w 4 w 0 w 4 w 0 w 4 w 0 w 4 x 4 w 0 w 5 w 2 w 7 w 4 w 1 w 6 w 3 x 5 w 0 w 6 w 4 w 2 w 0 w 6 w 4 w 2 x 6 w 0 w 7 w 6 w 5 w 4 w 3 w 2 w 1 x 7 w 0 w 0 w 0 w 0 w 0 w 0 w 0 w 0 x 0 w 0 w 1 w 2 w 3 w 0 w 1 w 2 w 3 x 1 w 0 w 2 w 0 w 2 w 0 w 2 w 0 w 2 x 2 = w 0 w 3 w 2 w 1 w 0 w 3 w 2 w 1 x 3 w 0 w 0 w 0 w 0 w 0 w 0 w 0 w 0 x 4 w 0 w 1 w 2 w 3 w 0 w 1 w 2 w 3 x 5 w 0 w 2 w 0 w 2 w 0 w 2 w 0 w 2 x 6 w 0 w 3 w 2 w 1 w 0 w 3 w 2 w 1 x 7 since w 4 = w 0, w 5 = w 1, w 6 = w 2, and w 7 = w 3. We may re-write this matrix relation as the set of equations X 0 = x 0 + x x 7 X 1 = (x 0 x 4)w 0 + (x 1 x 5) w 1 + (x 2 x 6) w 2 + (x 3 x 7) w 3 X 2 = (x 0 + x 4)w 0 + (x 1 + x 5) w 2 (x 2 + x 6) w 0 (x 3 + x 7) w 2 X 3 = (x 0 x 4)w 0 + (x 1 x 5) w 3 (x 2 x 6) w 2 + (x 3 x 7) w 1 X 4 = (x 0 + x 4)w 0 (x 1 + x 5) w 0 + (x 2 + x 6) w 0 (x 3 + x 7) w 0 X 5 = (x 0 x 4)w 0 (x 1 x 5) w 1 + (x 2 x 6) w 2 (x 3 x 7) w 3 X 6 = (x 0 + x 4)w 0 (x 1 + x 5) w 2 (x 2 + x 6) w 0 + (x 3 + x 7) w 2 X 7 = (x 0 x 4)w 0 (x 1 x 5) w 3 (x 2 x 6) w 2 (x 3 x 7) w 1.

27 64 Signals, Systems, Transforms and Digital Signal Processing with MATLAB These operations can be expressed back in matrix form: x 0 + x 4 w w w w w 0 w 1 w 2 w 3 x 1 + x 5 w 0 w 2 w 0 w 2 x 2 + x 6 X = w 0 w 3 w 2 w 1 x 3 + x 7 w 0 w 0 w 0 w 0 x 0 x 4. w 0 w 1 w 2 w 3 x 1 x 5 w 0 w 2 w 0 w 2 x 2 x 6 w 0 w 3 w 2 w 1 x 3 x 7 }{{} g Calling the vector on the right g we can re-write this equation in the form: w 0 w 0 w 0 w 0 w 0 w 0 w 0 w 0 w 0 w 2 w 0 w 2 X = w 0 w 2 w 0 w 2 w 0 w 0 w 0 w 0 w 0 w 0 w 0 w 0 w 0 w 2 w 0 w 2 w 0 w 2 w 0 w 2 diag ( w 0, w 0, w 0, w 0, w 0, w 1, w 2, w 3) g. Let h = diag ( w 0, w 0, w 0, w 0, w 0, w 1, w 2, w 3) g. A graphical representation of this last equation is shown on the left side of Fig We can write X 0 = (h 0 + h 2)w 0 + (h 1 + h 3) w 0 X 1 = (h 4 + h 6)w 0 + (h 5 + h 7) w 0 X 2 = (h 0 h 2)w 0 + (h 1 h 3) w 2 X 3 = (h 4 h 6)w 0 + (h 5 h 7) w 2 X 4 = (h 0 + h 2)w 0 (h 1 + h 3) w 0 X 5 = (h 4 + h 6)w 0 (h 5 + h 7) w 0 X 6 = (h 0 h 2)w 0 (h 1 h 3) w 2 X 7 = (h 4 h 6)w 0 (h 5 h 7) w 2 which can be re-written in the form 0 0 h 0 + h 2 w w w 0 w 0 h 1 + h 3 w 0 w 2 h 0 h 2 X = w 0 w 2 h 1 h 3 w 0 w 0 h 4 + h 6. w 0 w 0 h 5 + h 7 w 0 w 2 h 4 h 6 w 0 w 2 h 5 h 7 }{{} l Denoting by l the vector on the right the relation between the vectors h and l can be represented graphically as shown in the figure. We can write

28 Discrete-Time Fourier Transform 65 FIGURE 7.47 Steps in factorization of the DFT. w 0 w 0 w 0 w 0 w 0 w 0 X = w 0 w 0 w 0 w 0 w 0 w 0 w 0 w 0 w 0 w 0 Let We have diag ( w 0, w 0, w 0, w 2, w 0, w 0, w 0, w 2) l. v = diag ( w 0, w 0, w 0, w 2, w 0, w 0, w 0, w 2) l. X 0 = (v 0 + v 1) X 1 = (v 4 + v 5) X 2 = (v 2 + v 3) X 3 = (v 6 + v 7) X 4 = (v 0 v 1) X 5 = (v 4 v 5) X 6 = (v 2 v 3) X 7 = (v 6 v 7). These relations are represented graphically in the figure. The overall factorization diagram is shown in Fig We note that the output of the diagram is not the vector X in normal order. The output vector is in fact a vector X which is the same as X but is in reverse bit order. As we shall see shortly. We now write this factorization more formally in order to obtain a factorization valid for an input sequence of a general length of N elements. Let T 2 = [ ] 1 1. (7.182) 1 1

29 66 Signals, Systems, Transforms and Digital Signal Processing with MATLAB FIGURE 7.48 An FFT factorization of the DFT. The Kronecker product A B of two matrices A and B result in a matrix having the elements b ij of B replaced by the product Ab ij. For example, let and The Kronecker product A B is given by : so that we may write for example [ ] a00 a 01 A = a 10 a 11 (7.183) [ ] b00 b 01 B =. (7.184) b 10 b 11 a 00b 00 a 01b 00 a 00b 01 a 01b 01 [ ] Ab00 Ab 01 a 10b 00 a 11b 00 a 10b 01 a 11b 01 A B = = Ab 10 Ab 11 a 00b 10 a 01b 10 a 00b 11 a 01b 11 I 4 T 2 = a 10b 10 a 11b 10 a 10b 11 a 11b (7.185). (7.186) Let D 2 = diag(w 0, w 0 ) = diag(1, 1) (7.187) D 4 = diag(w 0, w 0, w 0, w 2 ) (7.188) D 8 = diag(w 0, w 0, w 0, w 0, w 0, w 1, w 2, w 3 ). (7.189)

30 Discrete-Time Fourier Transform 67 Using these definitions we can write the matrix relations using the Kronecker product. We have g = (I 4 T 2)x (7.190) h = D 8 g (7.191) l = (I 2 T 2 I 2)h (7.192) v = (D 4 I 2)l (7.193) X = (T 2 I 4)v = col [X 0,, X 4,, X 2,, X 6,, X 1,, X 5,, X 3,, X 7] (7.194) where col denotes a column vector. The global factorization which produces the vector X is written : X = F 8x = (T 2 I 4)(D 4 I 2)(I 2 T 2 I 2)D 8(I 4 T 2)x. (7.195) Represented graphically, this factorization produces identically the same diagram as the one shown in Fig The factorization of the matrix F 8 is given by w F 8 = 1 1 w w w w (7.196) w w w and may be written in the closed form F 8 = 3 (D 2 i I 2 3 i) (I 2 i 1 T 2 I 2 3 i). (7.197) i=1 This form can be generalized. For N = 2 n, writing [1] and A matrix K 2 i = diag(w 0, w 2i, w 2 2i, w 3 2i,...) (7.198) D 2 n i = Quasidiag(I 2 n i 1, K 2 i). (7.199) X = Quasidiag(A, B, C,...) (7.200) is one which has the matrices A, B, C,... along its diagonal and zero elements elsewhere. We can write the factorization in the general form F N = n (D 2 i I 2 n i) (I 2 i 1 T 2 I 2 n i). (7.201) i=1 As noted earlier from factorization diagram, Fig. 7.48, the coefficients X i of the transform are in reverse bit order. For N = 8, the normal order (0, 1, 2, 3, 4, 5, 6, 7) in 3-bit binary is written: (000, 001, 010, 011, 100, 101, 110, 111). (7.202)

31 68 Signals, Systems, Transforms and Digital Signal Processing with MATLAB The bit reverse order is written: (000, 100, 010, 110, 001, 101, 011, 111) (7.203) which is in decimal: (0, 4, 2, 6, 1, 5, 3, 7). The DFT coefficients X[k] in the diagram, Fig can be seen to be in this reverse bit order. We note that the DFT coefficients X[k] are evaluated in log 2 8 = 3 iterations, each iteration involving 4 operations (multiplications). For a general value N = 2 n the FFT factorization includes log 2 N = n iterations, each containing N/2 operations for a total of (N/2) log 2 N operations. This factorization is a base-2 factorization applicable if N = 2 n, as mentioned above. If the number of points N of the finite duration input sequence satisfies that N = r n where r, called the radix or base, is an integer, then the FFT reduces the number of complex multiplications needed to evaluate the DFT from N 2 to (N/r) log r N. For N = 1024 and r = 2, the number of complex multiplications is reduced from about 10 6 to about = With r = 4 this is further reduced to = An Algorithm for a Wired-in Radix-2 Processor The following is a summary description of an algorithm and a wired-in processor for radix-2 FFT implementation (see CORINTHIOS A Time Series Analyzer ) The DFT F [k] of an N-point sequence f [k], namely, F [k] = Writing f n f[n], F k F [k] and constructing the vectors The TDF may be written in the matrix form where the elements of the matrix T N are given by f [n] e j2πnk/n. (7.204) f = col (f 0, f 1,..., f ) (7.205) F = col (F 0, F 1,..., F ). (7.206) F = T Nf. (7.207) (T N) nk = exp( 2πjnk/N). (7.208) Letting w = e j2π/n = cos (2π/N) j sin (2π/N) (7.209) (T N) nk = w nk (7.210) T N = w 0 w 0 w 0 w 0... w 0 w 0 w 1 w 2 w 3... w w 0 w 2 w 4 w 6... w 2()... w 0 w w 2() w 3()... w ()2.... (7.211) To reveal the symmetry in the matrix T N we re-arrange its rows by writing T N = P NP 1 N T N = P NT N (7.212)