The Fast Fourier Transform; Short-time Fourier transform

ELEN E4810: Digital Signal Processing
Topic 10:
The Fast Fourier Transform
1. Calculation of the DFT
2. The Fast Fourier Transform algorithm
3. Short-Time Fourier Transform
Dan Ellis 2012-11-28
1

1. Calculation of the DFT
Filter design so far has been oriented to
time-domain processing - cheaper!
But: frequency-domain processing
makes some problems very simple:
X[k] Fourier domain Y[k]
x[n] DFT IDFT y[n]
processing
use all of x[n], or use short-time windows
Need an efficient way to calculate DFT
Dan Ellis 2012-11-28
2

Computational Complexity
N1
n=0
X[k] = x[n]W N
N complex multiplies
+ N-1 complex adds per point (k)
× N points (k = 0.. N-1)
cpx mult: (a+jb)(c+jd) = ac - bd + j(ad + bc)
= 4 real mults + 2 real adds
cpx add = 2 real adds
N points: 4N 2 real mults, 4N 2-2N real adds
Dan Ellis 2012-11-28
4
kn

Goertzel’s Algorithm
x e [n]
x e [N] = 0
Now:
i.e.
where
+
W N -k
N1
=0
kN
X[ k]
= x[ ]WN
= W N
Dan Ellis 2012-11-28
5
k
x[ ]
WN X[ k]
= y [ k N]
y [ k n]
= x [ e n]
hk [n]
z -1
y k [n]
y k [-1] = 0
y k [N] = X[k]
( )
k N
x e [n] = {
looks like a
convolution
x[n] 0 ≤ n < N
0 n = N
h k [n] = { W N -kn n ≥ 0
0 n < 0

H z
( ) =
Goertzel’s Algorithm
Separate ‘filters’ for each X[k]
can calculate for just a few values of k
No large buffer, no coefficient table
Same complexity for full X[k]
(4N 2 mults, 4N 2 - 2N adds)
but: can halve multiplies by making the
denominator real:
1
1 W N
k z 1 =
1 W N
k z 1
1 2cos 2k
N z1 + z 2
evaluate only
for last step
Dan Ellis 2012-11-28
6
2 real mults
per step

2. Fast Fourier Transform FFT
Reduce complexity of DFT
from O(N 2) to O(N·logN)
grows more slowly with larger N
Works by decomposing large DFT into
several stages of smaller DFTs
Often provided as a highly optimized
library
Dan Ellis 2012-11-28
7

Decimation in Time (DIT) FFT
Can rearrange DFT formula in 2 halves:
N1
X[ k]
= x n
k = 0.. N-1
Arrange
terms
in pairs...
n=0
N
2 1
[ ] W N
= x 2m
m=0
N
2 1
nk
( [ ]
2mk
WN + x[ 2m +1]
( 2m+1)k
W ) N
N
21
Group terms
from each
pair
= x[ 2m]
mk k
WN + WN x[ 2m +1]
mk
WN
2
2
m=0
m=0
X0 [ N/2 ] X1 [ N/2 ]
N/2 pt DFT of x for even n N/2 pt DFT of x for odd n
Dan Ellis 2012-11-28
8

One-Stage DIT Flowgraph
X[ k]
= X [ 0 k ] N + WN Even
points
from
x[n]
Odd
points
from
x[n]
x[0]
x[2]
x[4]
x[6]
x[1]
x[3]
x[5]
x[7]
2
DFT N2
DFT N2
[ ]
k
X1 k N
2
X 0[0]
X 0[1]
X 0 [2]
X 0 [3]
X 1 [0]
X 1[1]
X 1[2]
X 1 [3]
W N 0
W N 1
W N 2
W N 3
W N 4
W N 5
W N 6
W N 7
Classic FFT structure
“twiddle factors”:
always apply to
odd-terms output
NOT mirror-image
X[0]
X[1]
X[2]
X[3]
X[4]
X[5]
X[6]
X[7]
Dan Ellis 2012-11-28
10
Same as
X[0..3]
except for
factors on
X 1 [•]
terms

Multiple DIT Stages
If decomposing one DFTN into two
smaller DFTN/2 ’s speeds things up ...
Why not further divide into DFTN/4 ’s ?
i.e. X[ k]
= X [ 0 k ] N + W k [ N X1 k ] N
make:
0 ≤ k < N
Similarly,
X [ 0 k]
= X00 k N
Dan Ellis 2012-11-28
11
2
[ ]
0 ≤ k < N/2
N/4-pt DFT of even points
in even subset of x[n]
X [ 1 k]
= X10 k N
4
2
[ ]
k
+ WN X01 k N
2
4
N/4-pt DFT of odd points
from even subset
[ ]
4
[ ]
k
+ WN X11 k N
2
4

Two-Stage DIT Flowgraph
x[0]
x[4]
x[2]
x[6]
x[1]
x[5]
x[3]
x[7]
different from before X same as before
0[0]
DFT W 0
N/2
N4 X00 X0 [1] W 0
N
X0[2] W 1
N
DFTN4 X01 X0[3] W 2
N
W 3
W 3
N/2
N
DFT N4
DFT N4
X 10
X 11
W 0
N/2
W 3
N/2
X 1[0]
X 1[1]
X 1 [2]
X 1 [3]
W N 4
W N 5
W N 6
W N 7
X[0]
X[1]
X[2]
X[3]
X[4]
X[5]
X[6]
X[7]
Dan Ellis 2012-11-28
12

Multi-stage DIT FFT
Can keep doing this until we get down
to 2-pt DFTs:
DFT 2
X[0] = x[0] + x[1]
X[1] = x[0] - x[1]
≡
“butterfly” element
1 = W 2 0
-1 = W 2 1
→ N = 2 M-pt DFT reduces to M stages of
twiddle factors & summation
(O(N 2) part vanishes)
→ real mults < M·4N , real adds < 2M·2N
→ complexity ~ O(N·M) = O(N·log 2 N)
Dan Ellis 2012-11-28
13

FFT Implementation Details
Basic butterfly (at any stage):
XX0 [r]
•
XX1 [r]
Can simplify:
X X0 [r]
XX1 [r]
W r
N -1
W N r
W N r+N/2
XX [r]
•
2 cpx mults
•
XX [r+N/2]
X X [r]
r+
WN N j
2 = e
2 ( r+N 2 )
N
2N /2
j2r j
= e N e N
r
= WN just one cpx mult!
XX [r+N/2]
i.e. SUB rather than ADD
Dan Ellis 2012-11-28
14

it-reversed indexing
8-pt DIT FFT Flowgraph
000
100
010
110
001
101
011
111
x[0]
x[4]
x[2]
x[6]
x[1]
x[5]
x[3]
x[7]
W 4
W 4
-1’s absorbed into summation nodes
W N 0 disappears
-
-
-
-
W8 W8 W8 ‘in-place’ algorithm: sequential stages
Dan Ellis 2012-11-28
15
-
-
-
-
2
3
-
-
-
-
X[0]
X[1]
X[2]
X[3]
X[4]
X[5]
X[6]
X[7]

FFT for Other Values of N
Having N = 2 M meant we could divide
each stage into 2 halves = “radix-2 FFT”
Same approach works for:
N = 3 M radix-3
N = 4 M radix-4 - more optimized radix-2
etc...
Composite N = a·b·c·d → mixed radix
(different N/r point FFTs at each stage)
.. or just zero-pad to make N = 2 M
Dan Ellis 2012-11-28
16
M

x n
[ ] = 1 N
Inverse FFT
Recall IDFT:
Thus:
N1
x[n] = 1
N
( ) *
N1
k=0
only differences
from forward DFT
nk
X[k]W N
Nx * [n] =
nk
X[k]W N = X * [k]W N
k=0
Forward DFT of x′[n] = X *[k]| k=n
i.e. time sequence made from spectrum
N1
k=0
Hence, use FFT to calculate IFFT:
N 1
k=0
X * [
nk
k]WN
*
Re{X[k]}
Im{X[k]}
pure real flowgraph
Re Re
1/N
DFT
Dan Ellis 2012-11-28
17
-1
Im
Im
-1/N
nk
Re{x[n]}
Im{x[n]}

3. Short-Time
Fourier Transform (STFT)
Fourier Transform (e.g. DTFT) gives
spectrum of an entire sequence:
How to see a time-varying spectrum?
e.g. slow AM of a sinusoid carrier:
x[ n]
= 1 cos 2n
cos 0n N
-2
0 200 400 600 800 1000
Dan Ellis 2012-11-28
19
2
1
0
-1
x[n]
n

Fourier Transform of AM Sine
Spectrum of
whole sequence
indicates
modulation
indirectly...
... as
cancellation
between
closelytuned
sines
600
400
200
Nsin2ºkn
N
-Nsin2º(k-1)n
2 N
-Nsin2º(k+1)n
2 N
X[k]
0
0 0.02 0.04 0.06 0.08
Dan Ellis 2012-11-28
20
1
0.5
0
-0.5
-1
1
0.5
0
-0.5
-1
1
0.5
0
-0.5
N
N/2
2cAcB
= cA+B
+cA-B
k/(N/2)
-1
0 128 256 384 512 640 768 896
M

Fourier Transform of AM Sine
Sometimes we’d rather separate
modulation and carrier:
x[n] = A[n]cos! 0 n
A[n] varies on a
different (slower) timescale
One approach:
chop x[n] into short sub-sequences ..
.. where slow modulator is ~ constant
DFT spectrum of pieces → show variation
Dan Ellis 2012-11-28
21
! 0
A[n]
!

FT of Short Segments
Break up x[n] into successive, shorter
chunks of length NFT , then DFT each:
2
1
0
-1
-2
0 128 256 384 512 640 768 896 1024 = N
100
50
x[n]
N FT
= N/8
x0 [n]
X 0 [k]
0
0 64
Shows amplitude modulation
of ! 0 energy
k
x 1 [n]
X 1 [k]
x 2 [n]
X 2 [k]
x 3 [n]
X 3 [k]
x 4 [n]
X 4 [k]
k 0 = 0 · N FT
2
x 5 [n]
X 5 [k]
x 6 [n]
X 6 [k]
x 7 [n]
X 7 [k]
Dan Ellis 2012-11-28
22
k
n

The Spectrogram
Plot successive DFTs in time-frequency:
X i[k]
X[k,n]
k
X 0 [k] X 1 [k] X 2 [k] X 3 [k] X 4 [k] X 5 [k] X 6 [k] X 7 [k]
15
10
5
0
0
k k
128 256 384 512 640 768 896 1024
time hopsize (between successive frames)
= 128 points
This image is called the Spectrogram
Dan Ellis 2012-11-28
23
n
120
100
80
60
40
20
0

Short-Time Fourier Transform
Spectrogram = STFT magnitude
plotted on time-frequency plane
STFT is (DFT form):
frequency
index
X[ k,n ] 0 = x[ n0 + n]
w[n] e
time
index
N FT 1
n=0
N FT points of x
starting at n
j 2kn
N FT
window DFT
kernel
intensity as a function of time & frequency
Dan Ellis 2012-11-28
24

DTFT
form of
STFT
STFT Window Shape
w[n] provides ‘time localization’ of STFT
e.g. rectangular
w[n]
selects x[n], n 0 ≤ n < n 0 +N W
But: resulting spectrum has same
problems as windowing for FIR design:
X e j ( ,n ) 0 = DTFT{ x[ n0 + n]
w[ n]
}
= e jn
0 j
X( j( )
e )W( e )d
spectrum of short-time window
is convolved with (twisted) parent spectrum
Dan Ellis 2012-11-28
25
n

STFT Window Shape
e.g. if x[n] is a pure sinusoid,
X(e
j) W(ej) blurring (mainlobe)
+ ghosting (sidelobes)
Hence, use tapered window for w[n]
e.g. Hamming
w[ n]
=
0.54 + 0.46 cos(2 n
2M +1 )
w[n] W(e j)
Dan Ellis 2012-11-28
26
sidelobes
< -40 dB
n
-10 -5 0 5 10

STFT Window Length
Can illustrate time-frequency tradeoff
on the time-frequency plane:
k
250
200
150
100
50
0
1
0.5
0
0 100 200 300
Alternate tilings
of time-freq:
n
disks show ‘blurring’
due to window length;
area of disk is constant
→ Uncertainty principle:
±f·±t ≥ k
half-length window → half as many DFT samples
Dan Ellis 2012-11-28
28

freq / Hz
freq / Hz
0.1
0
4000
3000
2000
1000
4000
3000
2000
1000
Spectrograms of Real Sounds
0
2.35 2.4 2.45 2.5 2.55 2.6
0
0 0.5 1 1.5 2 2.5
time / s
time / s
time-domain
successive
short
DFTs
individual t-f
cells merge
into continuous
image
Dan Ellis 2012-11-28
29
10
0
-10
-20
-30
-40
-50
intensity / dB

Window = 256 pt
Narrowband
Window = 48 pt
Wideband
Narrowband vs. Wideband
Effect of varying window length:
freq / Hz
freq / Hz
0.2
0
4000
3000
2000
1000
0
4000
3000
2000
1000
0
1.4 1.6 1.8 2 2.2 2.4 2.6
time / s
Dan Ellis 2012-11-28
30
10
0
-10
-20
-30
-40
-50
level
/ dB
M

Frequency
8000
6000
4000
2000
Spectrogram in Matlab
>> [d,sr]=wavread(’mpgr1_sx419.wav');
>> Nw=256;
>> specgram(d,Nw,sr)
>> caxis([-80 0])
>> colorbar
0
(hann) window length
actual sampling rate
(to label time axis)
0.5 1 1.5
Time
2 2.5 3
Dan Ellis 2012-11-28
31
0
-20
-40
-60
-80
dB

just one freq.
STFT as a Filterbank
Consider one ‘row’ of STFT:
Dan Ellis
X k n 0
where
N1
n=0
N1
[ ] = x[ n0 + n]
w[ n]
e
( )
= h [ k m]x
n0 m
m=0
2012-11-28
[ ]
h k n
[ ] = w n
[ ] e
j 2kn
N
j 2kn
N
Each STFT row is output of a filter
(subsampled by the STFT hop size)
Im{x[n]}
1
0
-1
-60
convolution
with
complex IR
-40
n
-20
32
0 -1
0
1
Re{x[n]}

STFT as a Filterbank
If
then
h [ k n]
= w[ ( )n]
e
Hk e j
( ) = W e
j 2kn
N
( ) j 2k
( ( N ) ) shift-in-!
Each STFT row is the same bandpass
response defined by W(ej!), frequency-shifted to a given DFT bin:
W(ej)
H1(ej) H2(ej) • • •
A bank of identical,
frequency-shifted
bandpass filters:
“filterbank”
Dan Ellis 2012-11-28
33

STFT Analysis-Synthesis
IDFT of STFT frames can reconstruct
(part of) original waveform
e.g. if
then
X[ k,n ] 0 = DFT x[ n0 + n]
w[ n]
{ } = x[ n0 + n]
w n
IDFT X[ k,n ] 0
{ }
Can shift by n0 , combine, to get x[n]: ^
^x[n]
[ ]
Could divide by w[n-n 0 ] to recover x[n]...
Dan Ellis 2012-11-28
34
n0
x[n]·w[n-n 0]
n

STFT Analysis-Synthesis
Dividing by small values of w[n] is bad
Prefer to
overlap windows:
i.e. sample X[k,n 0 ]
at n 0 = r·H where H = N/2 (for example)
Then
^x[n]
hopsize window length
x ˆ [ n]
= x[ n]w[
n rH]
r
= x[ n]
if w[ n rH]
=1
Dan Ellis 2012-11-28
35
r
x[n]·w[n-r·H]
n

STFT Analysis-Synthesis
Hann or Hamming windows
with 50% overlap
1
sum to constant
0.54 + 0.46cos(2 n N )
( )
N n 2
N )
( ) =1.08 0.4
0.2
0
0 20 40 60 80
+ 0.54 + 0.46cos(2
Can modify individual frames of X[k,n]
and then reconstruct
complex, time-varying modifications
tapered overlap makes things OK
Dan Ellis 2012-11-28
36
0.8
0.6
w[n] + w[n-N/2]
w[n] w[n-N/2]
n

STFT Analysis-Synthesis
e.g. Noise reduction:
STFT of
original speech
Speech corrupted
by white noise
Energy threshold
mask
k
100 200 300
Dan Ellis 2012-11-28
37
120
100
80
60
40
20
r
M