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EECS 451 DISCRETE FOURIER TRANSFORM (DFT)
DFT: Xk =
∑N − 1
n=0 x(n)e
−j 2 πnk/N (^) ; x(n) = 1 N
∑N − 1
k=0 Xke
j 2 πnk/N (^) ; length≤ N. DTFT: Xk = X(z)|z=ej 2 πk/N = X(ejω^ )|ω=2πk/N sampled on unit circle. DTFS: F{periodic extension of {x(n)}} =periodic extension of {Xk}. except: factor of 1/N moved. Note: k = 0, 1... N − 1 and n = 0, 1... N − 1. Text: Uses X(k) not Xk. I hate that–too easy to confuse with X(z)!
What: Use DFT to compute DTFT at ω = (^2) Nπk : equispaced samples on u.c. Why: Use for spectral analysis, and to recover x(n) from DTFT. BUT: Finite x(n) has finite length L →using N ≥ L →recover x(n) from Xk. length: But N < L →can only recover
k x(n^ +^ kN^ )^ →^ aliased x(n).
EX: x(n) = { 1 , 2 , 3 , 4 , 5 } → X(ejω^ ) = 1 + 2e−jω^ + 3e−^2 jω^ + 4e−j^3 ω^ + 5e−j^4 ω Sample DTFT on unit circle at ω = 0, π 2 , π, 32 π to get DFT: DFT: X 0 = 1 + 2 + 3 + 4 + 5 = 15; X 1 = 1 − 2 j − 3 + 4j + 5 = 3 + 2j; N=4: X 2 = 1 − 2 + 3 − 4 + 5 = 03; X 3 = 1 + 2j − 3 − 4 j + 5 = 3 − 2 j = X 1 ∗. IDFT: x(0) = 14 [15 + (3 + 2j) + 3 + (3 − 2 j)] = 6 incorrect (6=1+5 aliased). x(1) = 14 [15(1) + (3 + 2j)(j) + 3(−1) + (3 − 2 j)(−j)] = 2 correct. x(2) = 14 [15(1) + (3 + 2j)(−1) + 3(1) + (3 − 2 j)(−1)] = 3 correct. x(3) = 14 [15(1) + (3 + 2j)(−j) + 3(−1) + (3 − 2 j)(j)] = 4 correct. Why? Undersampled X(ejω^ ) on unit circle→aliasing (just like before).
Inter- Let x(n) have length L ≤ N. We’re given X(ej^2 πk/N^ ), k = 0, 1... N −1. polate Then interpolate X(ejω^ ) =
∑N − 1
k=0 X(e
j 2 πk/N (^) )S(ω − 2 πk N ) DTFT: S(ω) = (^) Nsin( sin(ωN/ω/2)2) e−jω(N^ −1)/^2 = DT F T { (^) N^1
∑N − 1
i=0 δ(n^ −^ i)}.
Zero: N > L → DF T {x(0)... x(L − 1), 0... 0(zero − pad)} →above. padding This smoothes DTFT (finer sampling in ω), BUT: no additional info.
EX: DF TN {
∑L− 1
i=0 δ(n^ −^ i)}^ =^
sin(πkL/N ) sin(πk/N ) e
−jπk(L−1)/N (^). Window blurs.
Cyclic y(n) = h(n) c©x(n) =
∑N − 1
i=0 h(i)x((n^ −^ i))N^ ⇔^ Yk^ =^ HkXk. convol: (x(n))N ⇔N-point periodic extension of x(n). ”Cyclic”=”circular.” Ex: { 1 , 2 , 3 } ©{c 4 , 5 , 6 } : Take one complete cycle of each: y(0) = 14 ,,^26 ,,^35 ,,^14 ,,^26 ,,^35 = 1 · 4 + 2 · 6 + 3 · 5 = 31. y(1) = 15 ,,^24 ,,^36 ,,^15 ,,^24 ,,^36 = 1 · 5 + 2 · 4 + 3 · 6 = 31. y(2) = 16 ,,^25 ,,^34 ,,^16 ,,^25 ,,^34 = 1 · 6 + 2 · 5 + 3 · 4 = 28. Check: H 0 X 0 = (1 + 2 + 3)(4 + 5 + 6) = 90 = (31 + 31 + 28) = Y 0 checks.
EECS 451 FAST FOURIER TRANSFORM (FFT)
Goal: Compute DFT Xk =
∑N − 1
n=0 x(n)W^
nk N where^ WN^ =^ e
−j 2 π/N (^). Names: FFT is an algorithm for computing the DFT, which is a transform. Why? Direct computation requires N 2 mults and N (N − 1) adds: too many! Cooley- 1965 at IBM. Serious DSP dates from this algorithm. Tukey: Divide up large DFT into smaller DFTs: N = N 1 N 2. coarse: n = n 1 + N 1 n 2 where n 1 = 0... N 1 − 1 and n 2 = 0... N 2 − 1. vernier: k = k 2 + N 2 k 1 where k 1 = 0... N 1 − 1 and k 2 = 0... N 2 − 1. indices nk = (n 1 + N 1 n 2 )(k 2 + N 2 k 1 ) = n 1 k 2 + N 1 n 2 k 2 + N 2 n 1 k 1 + N 1 N 2 n 2 k 1. exponent: W (^) Nnk = W (^) Nn^1 k^2 W (^) Nn^22 k 2 W (^) Nn^11 k^1 using W (^) NN 11 N 2 = WN 2 and W (^) NN 1 N^2 = 1.
DFT: Xk = Xk 2 +N 2 k 1 =
∑N 1 − 1
n 1 =
∑N 2 − 1
n 2 =0 x(n^1 +^ N^1 n^2 )W^
(n 1 +N 1 n 2 )(k 2 +N 2 k 1 ) N. rewrite: Xk 2 +N 2 k 1 =
∑N 1 − 1
n 1 =0 W^
n 1 k 1 N 1
[
W (^) Nn^1 k^2
∑N 2 − 1
n 2 =0 W^
n 2 k 2 N 2 x(n^1 +^ N^1 n^2 )
]
- Compute N 1 (for each n 1 ) N 2 -point DFTs of x(n 1 + N 1 n 2 ) (in n 2 ).
- Multiply result by twiddle factors W (^) Nn^1 k^2. These are twiddle mults.
- Compute N 2 (for each k 2 ) N 1 -point DFTs of the result. Now done!
- (N 1 N 2 )-point→ N 1 (N 2 -point)+N 2 (N 1 -point)+(N 1 −1)(N 2 −1) twiddle since n 1 = 0 or k 2 = 0 → W (^) Nn^1 k^2 = 1. This is important below. Visual: (N 1 × N 2 ) arrays: xn 1 ,n 2 = x(n 1 + N 1 n 2 ); Xk 1 ,k 2 = X(k 2 + N 2 k 1 ).
- Take N 2 -point DFT of each row (fixed n 1 ). Yields ˆxn 1 ,k 2.
- Multiply ˆxn 1 ,k 2 point-by-point by twiddle factor W (^) Nn^1 k^2.
- Take N 1 -point DFT of each column (fixed k 2 ). Yields Xk 1 ,k 2.
Radix-2 Cooley-Tukey Fast Fourier Transforms
N = 2 N 2 → Decimation-in-time FFT: N 1 = 2; N 2 = N/2; n 1 = 0, 1; k 1 = 0, 1.
Xk 2 = 1
∑N/ 2 − 1
n 2 =0 W^
n 2 k 2 N/ 2 x(2n^2 ) +^ W^
1 k 2 N
∑N/ 2 − 1
n 2 =0 W^
n 2 k 2 N/ 2 x(2n^2 + 1). Xk 2 +N/ 2 = 1
∑N/ 2 − 1
n 2 =0 W^
n 2 k 2 N/ 2 x(2n^2 )^ −^ W^
1 k 2 N
∑N/ 2 − 1
n 2 =0 W^
n 2 k 2 N/ 2 x(2n^2 + 1). N-point→ 2( N 2 -point)+ N 2 (2-point)+( N 2 − 1)twiddle mults. Total: N 2 log 2 N mults. Note: (N 1 − 1)(N 2 − 1) = (2 − 1)( N 2 − 1) = N 2 − 1 : 12 twiddle mults trivial!
N = N 2 2 → Decimation-in-freq. FFT: N 2 = 2; N 1 = N/2; n 2 = 0, 1; k 2 = 0, 1.
X 2 k 1 =
∑N/ 2 − 1
n 1 =0 W^
n 1 k 1 N/ 2 [x(n^1 ) +^ x(n^1 +^ N/2)]1. X 2 k 1 +1 =
∑N/ 2 − 1
n 1 =0 W^
n 1 k 1 N/ 2 [x(n^1 )^ −^ x(n^1 +^ N/2)]W^
1 n 1 N. N-point→ N 2 (2-point)+2( N 2 -point)+( N 2 − 1)twiddle mults. Total: N 2 log 2 N mults.
