![10.2 - Discrete Fourier Transform (DFT)
• A finite or periodic sequence has only N unique values,
f d h l l
x[n] for 0 ≤ n p N
• S t
Spectrum is completely defined by N distinct frequency
i l t l d fi d b di ti t f
samples
• Divide 0..2π into N equal steps {ω(k)} = 2πk/N
0 2π steps,
• Uniform sampling of DTFT spectrum:](https://image.slidesharecdn.com/dsp-u-lec10dftandfft-100105090404-phpapp01/85/Dsp-U-Lec10-DFT-And-FFT-3-320.jpg)
Dsp U Lec10 DFT And FFT
- 1. EC533: Digital Signal Processing 5 l l Lecture 10 DFT and FFT
- 2. 10.1 - Frequency Domain Vs. Time Domain CTFS DTFS CTFT DTFT
- 3. 10.2 - Discrete Fourier Transform (DFT) • A finite or periodic sequence has only N unique values, f d h l l x[n] for 0 ≤ n p N • S t Spectrum is completely defined by N distinct frequency i l t l d fi d b di ti t f samples • Divide 0..2π into N equal steps {ω(k)} = 2πk/N 0 2π steps, • Uniform sampling of DTFT spectrum:
- 4. 10.2 – DFT – contd. Hence, Where, i.e, 1/Nth of a revolution Twiddle Factor Properties of the Twiddle Factor: k+N k j W N = W N periodicity N k+ k W N 2 = − W N -1 1 2 W N = W N 2 -j N=4
- 5. DFT Example Find the DFT for the 4 points time sequence {1 0 0 1}, fs=8KHz 3 3 X (0) = ∑ x(nT ) e − j 0 = ∑ x(nT ) at k=0, n =0 n =0 = x(0) + x(T ) + x(2T ) + x(3T ) = 1+ 0 + 0 +1 = 2 3 at k 1, k=1, X (1) = ∑ x(nT ) e − jΩnT , Ω = 2π n =0 NT 3 − j 2πn X (1) = ∑ x(nT ) e N n =0 = 1 + 0 + 0 + 1e − j 2π 3 4 = 1 + e − j 3π 4 ⎛ 3π ⎞ ⎛ 3π ⎞ = 1 + cos⎜ ⎟ − j sin ⎜ ⎟ = 1+ j ⎝ 2 ⎠ ⎝ 2 ⎠
- 6. DFT Example – contd. 3 3 − j 2π 2 n at k=2, X (2) = ∑ x(nT ) e − j 2 ΩnT = ∑ x(nT ) e N n =0 n =0 3 − j 4πn = ∑ x(nT ) e N n =0 = 1 + 0 + 0 + 1e − j 4π 3 4 = 1 + e − j 3π = 1 − 1 = 0 3 − j 2π 3 n at k=3, X (3) = ∑ x(nT ) e N n =0 = 1 + 0 + 0 + 1e − j 9π 2 = 1 − j X (k ) = {2, 1 + j , 0, 1 − j}
- 7. DFT Example – contd. x(nT) |X(k)| 1 2 √2 x x x 0 125 250 375 t(µs) 0 12.57 25.14 37.71 50.28 kΩ(X 103 rad/s) φ(k)(⁰) x(n) ={ , 0, 0,1 1 } +45 Phase Angle indeterminate 37.71 0 x x 12.57 25.14 50.28 kΩ(X 103 rad/s) ‐45 X(k) ={2,1+ j, 0,1− j}
- 8. 10.3 - Inverse Discrete Fourier Transform ( (IDFT)) • • Ch k Check
- 9. 10.4 - DFT Computational Complexity The DFT has: • (N complex multiplies + N-1 complex adds per point) x N points (k = 0..N-1) N2 complex multiplies and N(N 1) complex additions N(N-1) where cpx mult: (a+jb)(c+jd) = ac - bd + j(ad+bc)= 4 real mults + 2 real adds cpx add = 2 real adds • Total: 4N2 real mults, 4N2-2N real adds •Looking at DFT Matrix, lots of repeated structure are found; means opportunities for efficient t iti f ffi i t algorithm to be used to reduce the DFT complexity
- 10. 10.5 – Fast Fourier Transform (FFT) • Reduce complexity of DFT from O(N2) to O(N log2N) O(N·log • Grows more slowly with larger N • Works by decomposing large DFT into several stages of smaller DFTs • Often provided as a highly optimized library
- 11. 10.6 - Decimation in Time (DIT) FFT • Can rearrange DFT formula in 2 halves:
- 12. 10.6 - Decimation in Time (DIT) FFT - contd. • We can evaluate an N-pt DFT as two N/2-pt DFTs (plus a few mults/adds) • But if DFTN{•} ~ O(N2) then DFTN/2{•} ~ O((N/2)2) = 1/4 O(N2) • Total computation ~ 2 * 1/4 O(N2) = 1/2 the computation (+ε) of direct DFT
- 13. 10.6.1 - One-Stage DIT Flowgraph
- 14. 10.6.2 - Multiple DIT Stages • If decomposing one DFTN into two smaller DFTN/2’s speeds things up... p • Why not further divide into DFTN/4’s ? • ie i.e. • so have • Similarly,
- 15. 10.6.2 - Multiple DIT Stages – contd. Two-Stage DIT Flowgraph
- 16. 10.6.2 - Multi-stage DIT FFT – contd. • Can keep doing this until we get down to 2-pt DFTs: x(0) x(1)
- 17. 10.7 - FFT Implementation Details • Basic Butterfly at any stage • Can be simplified to
- 18. 10. 8 - 8-pt DIT FFT Flowgraph 000 001 0 0 010 2 011 W 8 100 101 110 111 2 W 8 • -1’s absorbed into summation nodes 0 • WN disappears • Twiddle factor step=N/2 j (stage order), j≠1
- 19. 10.9 - FFT Complexity N • Total no. of butterflies = Total no. of butterflies log 2 N g 2 No of stages f t • As each butterfly gives one complex multiplication and 2 As each butterfly gives one complex multiplication and 2 complex addition • So, FFT has complex multiplications of So, FFT has complex multiplications of N Ο( log 2 N) instead of Ο(N 2 ) in case of DFT 2 and complex additions of Ο(Nl 2 N) i (Nlog instead of Ο(N(N - 1)) i case of DFT d f in f
- 20. 10.10 - Inverse FFT • • Thus Steps f calculating IFFT: St for l l ti IFFT 1. Take the conjugate of the freq. •Hence, Use FFT to calculate IFFT Samples 2. Calculate FFT of these samples 3. Divide the output by N 4. 4 Take the conjugate of the output