Topic 3: Fourier domain

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ELEN E4810: Digital Signal Processing
Topic 3: Fourier domain
1.
2.
3.
4.
The Fourier domain
Discrete-Time Fourier Transform (DTFT)
Discrete Fourier Transform (DFT)
Convolution with the DFT
Dan Ellis
2013-09-23
1
1. The Fourier Transform


Basic observation (continuous time):
A periodic signal can be decomposed
into sinusoids at integer multiples of the
fundamental frequency
i.e. if x˜ (t) = x˜ (t +T )
we can approach x˜ with
M
ak cos
x̃(t)
k=0
Dan Ellis
2 k
t+
T
2013-09-23
k
Harmonics
of the
fundamental
2
M
Fourier Series

ak cos
k=0
For a square wave,
= 0;
k
i.e.
ak =
x(t) = cos
2
t
T
( 1)
0
k
2
1
1
cos
3
1
k
2 k
t+
T
k
k = 1, 3, 5, . . .
otherwise
2
1
3t + cos
T
5
2
5t
T
...
1
0.5
0
0.5
1
1.5
Dan Ellis
1
0.5
0
2013-09-23
0.5
1
1.5
3
M
Fourier Domain

x is equivalently described
by its Fourier Series
parameters:
ak = ( 1)
k
2
1
1
k
k = 1, 3, 5, . . .
ak 1.0
¡k
k
1 2 3 4 5 6 7
k
π
Negative ak is
equivalent to phase of º
M

1 2 3 4 5 6 7
Complex form: x̃(t)
j 2T k t
ck e
k= M
Dan Ellis
2013-09-23
4
M
Fourier Analysis

x̃(t)
ck e
k= M
How to find {|ck |}, {arg{ck }} ?
Inner product with
(conjugate) complex sinusoids:
1
ck =
T
Dan Ellis
T /2
x(t)e
j 2T k t
dt
T /2
2013-09-23
j 2T k t
5
M
Fourier Analysis
1
ĉk =
T
T /2
x(t)e
x
1
=
2
e
cl ejl
dt ; call
jk
e
1
2
ej(l
k)
2
=
t
T
d
jk
l
cl
l
T
2
ck e
k= M
T /2
1
=
2
=
j 2T k t
x̃(t)
d
d
Integral of (l-k) complete
cycles of a complex sinusoid;
= 0 for l≠k ∵ real (cos) part is
complete cycles, imag (sin)
part is odd; = 1 for l=k ∵ ∫ 1dτ
= ck
Dan Ellis
j 2T k t
2013-09-23
6
Fourier Series Analysis

1
Thus, ck = T
T /2
x(t)e
j 2T k t
dt
T /2
j 2T k t
because complex sinusoids e
pick out the corresponding sinusoidal
components linearly combined in
M
j 2T k t
x(t) =
ck e
k= M
Dan Ellis
2013-09-23
7
Fourier Transform

Fourier series for periodic signals
extends naturally to Fourier Transform
for any (CT) signal (not just periodic):
dt
Fourier
Transform (FT)
j t
Inverse Fourier
Transform (IFT)
X(j ) =
x(t)e
1
x(t) =
2
X(j )e

j t
d
Discrete index k → continuous freq. Ω
Dan Ellis
2013-09-23
8
Fourier Transform
Mapping between two continuous
functions:
|X( )|

0
level
/ dB -20
0.02
x(t)
-40
-60
-80
0
0.01
4000
6000
8000
freq / Hz
arg{X( )}
0
-0.01
0
2000
0.002
0.004
0.006
0.008
time / sec
/2
0
/2
2π ambiguity
Dan Ellis
2013-09-23
0
2000
4000
6000
8000
freq / Hz
9
Fourier Transform of a sine

Z
Assume x(t) = e
1
Now, since x(t) =
j 0 t
•
X(W)e
2p •
...we know X(W) = 2pd(W
jWt
dW
W0)
...where ±(x) is the Dirac delta function
δ(x-x0)
(continuous time) i.e.
  ( x  x0 ) f ( x) dx = f ( x0 )

→ x(t) = Ae
Dan Ellis
j 0 t
x0
f(x)
x
X() = A (   0 )
2013-09-23
10
Fourier Transforms
Fourier Series
(FS)
Fourier
Transform (FT)
Discrete-Time
FT (DTFT)
Discrete FT
(DFT)
Dan Ellis
Time
Frequency
Continuous
periodic ~x(t)
Discrete
infinite ck
Continuous
Continuous
infinite x(t)
infinite X(Ω)
Discrete
Continuous
infinite x[n] periodic X(ejω)
Discrete
Discrete
finite/pdc ~x[n] finite/pdc X[k]
2013-09-23
11
2. Discrete Time FT (DTFT)

FT defined for discrete sequences:
j
X(e ) =

 x[n]e
 jn
DTFT
n=

Summation (not integral)
Discrete (normalized)
frequency variable !

Argument is ej!, not j!

Dan Ellis
2013-09-23
12
DTFT example
e.g. x[n] = Æn·μ[n], |Æ| < 1

X(e ) = n=  µ[n]e

j
= n=0 (e

Re
n
|X(ej )|
)
2
0
2
3
Dan Ellis
4
5
3
4
5
n
 jn

S =  c  cS =  c
n
n
n =1
 S  cS = c = 1
( |c | < 1 )
1
S =
1c
0
2
1

1 2 3 4 5 6 7
n =0
arg{X(ej )}
3
n
 j n
1
=
 j
1  e
1 - e-j
Im
-1
2013-09-23
13
Periodicity of X(ej!)
X(ej!) has periodicity 2º in ! :

X(e
j ( +2  )
) =  x[n]e
=  x[n]e
|X(ej )|
 j ( +2  )n
 jn
e
 j 2 n
j
= X(e )
arg{X(ej )}
3
2
2
1
2
0

3
4
3
4
5
5
Phase ambiguity of ej! makes it implicit
Dan Ellis
2013-09-23
14
Inverse DTFT (IDTFT)

Same basic “Fourier Synthesis” form:
1
x[n] =
2
  X(e

j
)e
jn
d
IDTFT

Note: continuous, periodic X(ej!)

discrete, infinite x[n] ...
IDTFT is actually Fourier Series analysis
(except for sign of !)
Dan Ellis
2013-09-23
15
IDTFT

Verify by substituting in DTFT:
1
x[n] =
2
1
=
2
  X(e

j
)e
jn
d
  (l x[l]e )e

 jl
jn
d
1  j (nl )
= l x[l]  e
d
2
= l x[l]sinc (n  l)= x[n]
Dan Ellis
2013-09-23
= 0 unless
n=l
i.e. = δ[n-l]
16
sinc
1
=
2

j (n
e
1
l)
dw =
2
Same as ∫cos
Dan Ellis
ej (n l)
j(n l)
1 ej (n l) e j (n
=
2
j(n l)
2j sin (n l)
= sinc (n
j(n l)
∴
1
2
l)
l)
imag jsin part cancels
2013-09-23
17
sinc

sin x
sincx
x
∆
=
1.0
4
3
sin(x)/x
2
2
3
4
x
sin(x)
y=x

= 1 when x = 0
= 0 when x = r·º, r ≠ 0, r = ±1, ±2, ±3,...
Dan Ellis
2013-09-23
18
DTFTs of simple sequences
j
X(e ) =
x[n] = ±[n]


 x[n]e
n=
 j 0
=e

i.e.
±[n]
Dan Ellis
1
X(ejω)
x[n]
-3 -2 -1
(for all !)
X(ej!)
x[n]
=1
 jn
1 2 3
n
-º
2013-09-23
º
!
19
DTFTs of simple sequences
IDTFT

1
j
jn
 x[n] = e
: x[n] =
X(e
)e
d



2

j
X(e
) = 2
  (
  0 ) over -º < ! < º
but X(ej!) must be periodic in !
j 0 n
e
k 2   (   0  2k)
j 0 n

If !0 = 0 then x[n] = 1 ∀ n
so
Dan Ellis
1
k 2   (  2k)
2013-09-23
X(ej!)
0 2º 4º
20
DTFTs of simple sequences

From before:
 µ[n]
n

1
 j
1  e
( |α | < 1)
μ[n] tricky - not finite
µ[n]
1
+ k  ( + 2k)
 j
1 e
DTFT of 1/2
Dan Ellis
2013-09-23
21
DTFT properties

Linear:
j
j
g[n] + h[n]  G(e ) + H (e )

Time shift:
g[n  n0 ]  e

Frequency shift:
e
Dan Ellis
j 0 n
 jn 0
g[n]  G (e
2013-09-23
j
G(e )
j (  0 )
)
‘delay’
in
frequency
22
DTFT example

x[n] = ±[n] + Æn μ[n-1]
Dan Ellis
?
= ±[n] + Æ(Æn-1 μ[n-1])
  j 1
1 
X(e ) = 1+   e


 j

1  e 
e j
1  e j + e j
= 1+
=
 j
1  e
1  e j
1
=
x[n] = Æn μ[n]
 j
1  e
j
2013-09-23
23
DTFT symmetry
If x[n] x[-n]
x*[n] 
Re{x[n]} j
X(e ) =

 x[n]e jn
n=
X(ej!) then...
from summation
X(e-j!)
X*(e-j!) (e-j!)* = ej!
1
j
!
XCS(e ) = [ X (e j ) + X * (e j )]
2
conjugate symmetry cancels Im parts on IDTFT
jIm{x[n]} xcs[n] xca[n] Dan Ellis
1
j
*
 j
= [ X (e )  X (e )]
2
Re{X(ej!)}
jIm{X(ej!)}
XCA(ej!)
2013-09-23
24
DTFT of real x[n]

When x[n] is pure real,
XCS
XR(ej!) = XR(e-j!)
XI(ej!) = -XI(e-j!)
Imag
xcs[n] ≡ xev[n] = xev[-n]
xca[n] ≡ xod[n] = -xod[-n]
X(ej!) = X*(e-j!)
x[n] real, even
X(ej!) even, real
xim[n]
Real
xre[n]
n
Dan Ellis
2013-09-23
25
DTFT and convolution

Convolution: x[n] = g[n] h[n]
X(e ) = n = ( g[n]  h[n]) e
j

 jn
= n ( k g[k]h[n  k])e
 jn
= k ( g[k]e  jk n h[n  k]e  j (n k ) )
j
j
= G(e )  H (e )
g[n] h[n]  G(e j )H (e j )
Dan Ellis
2013-09-23
Convolution
becomes
multiplication
26
Convolution with DTFT

j
j
Since g[n] h[n]  G(e )H (e )
we can calculate a convolution by:



g[n]
h[n]
Dan Ellis
finding DTFTs of g, h → G, H
multiply them: G·H
IDTFT of product is result, g[n] h[n]
DTFT
DTFT
G(e j )
Y (e j )
IDTFT
y[n]
H (e j )
2013-09-23
27
M
DTFT convolution example


1
x[n] =
X(e ) =
1  e j
h[n] = ±[n] - Ʊ[n-1]

j
Æn·μ[n]
H (e ) = 1   (e
j
 j 1
) 1
y[n] = x[n] ∗ h[n]
j
j
j
Y
(e
)
=
H
(e
)X(e
)
1
 j
=

1


e
=1
)
 j (
1  e
y[n] = ±[n] i.e. ...
Dan Ellis
2013-09-23
28
DTFT modulation

Modulation: x[n] = g[n]  h[n]
Could solve if g[n] was just sinusoids...
1 

j
j
jn
 jn
X(e ) = n 
G(e
)e
d


h[n]e




 2 write g[n] as IDTFT 
1 
j
 j (  )n
=
G(e ) n h[n]e
d


2
1 
j
j (  )
g[n]  h[n] 
G(e )H (e
)d


2
[
]
Dual of convolution in time
Dan Ellis
2013-09-23
29
Parseval’s relation

“Energy” in time and frequency domains
are equal:
1 
*
j
*
j
g[n]h
[n]
=
G(e
)H
(e
)d




2

n

If g = h, then g·g* = |g|2 = energy...
Dan Ellis
2013-09-23
30
Energy density spectrum

Energy of sequence  g =  g[n]
2
n
1
g =
2
  G(e


By Parseval

Define Energy Density Spectrum (EDS)
j
j 2
Sgg (e ) = G(e )
Dan Ellis
2013-09-23
j
2
) d
31
EDS and autocorrelation

Autocorrelation of g[n]:
rgg [] =

 g[n]g[n  ] = g[n] g[n]
n=
DTFT {rgg []} = G(e )G(e
j


 j
)
If g[n] is real, G(e-j!) = G*(ej!), so
j 2
j
DTFT {rgg []} = G(e ) = Sgg (e )
no phase
info.
Mag-sq of spectrum is DTFT of autoco
Dan Ellis
2013-09-23
32
3. Discrete FT (DFT)
Discrete FT
(DFT)



Discrete
Discrete
finite/pdc x[n] finite/pdc X[k]
A finite or periodic sequence has only
N unique values, x[n] for 0 ≤ n < N
Spectrum is completely defined by N
distinct frequency samples
Divide 0..2º into N equal steps,
2 k
{ k} =
N
Dan Ellis
2013-09-23
33
DFT and IDFT

Uniform sampling of DTFT spectrum:
N 1
X[k] = X(e j )  = 2 k =  x[n]e
N
j
2 k
n
N
n=0
N 1

2º/N
1
kn
X[k]
=
x[n]W
DFT:

N
WN
n=0
where WN = e
Dan Ellis
j
2
N
i.e. -1/Nth of a revolution
2013-09-23
34
IDFT
N 1


1
nk
Inverse DFT: IDFT x[n] =  X[k]WN
N k=0
Check:
1
kl
nk
x[n] = k l x[l]WN WN
N
Sum of complete set
N 1
N 1
1
k (ln ) of rotated vectors
=  x[l] WN
= 0 if l ≠ n; = N if l = n
N l=0
k=0
im
or finite
re
W
= x[n]
geometric series
(
)
N
WN2
0≤n<N
Dan Ellis
2013-09-23
= (1-WNlN)/(1-WNl)
35
DFT examples

Finite impulse x[n] =
N 1
X[k] =
1
0
n=0
n = 1...N
x[n]WNkn = WN0 = 1
1
k
n=0

Periodic sinusoid:
⇥
2 rn
N
x[n] = cos
X[k] =
(0 ≤ k < N)
Dan Ellis
1
2
=
(r
Z) = 1 W rn + W rn ⇥
N
N
N 1
rn
(W
N
n=0
2
+ WNrn )WNkn
k = r, k = N
0 otherwise
N
2
2013-09-23
r
36
DFT: Matrix form
X[k] = n=0 x[n]  W
N 1

X[0]
X[1]
X[2]
..
.
⇧
⇧
⇧
⇧
⇧
⇤
X[N
⇥
1
⌃ ⇧
1
⇧
⌃ ⇧
⌃ ⇧1
⌃=⇧
⌃ ⇧.
⌅ ⇤ ..
1]

1
WN1
WN2
..
.
(N 1)
1 WN
i.e.
Dan Ellis
kn
N
as a matrix multiply:
1
WN2
WN4
..
.
···
···
···
..
.
2(N 1)
···
WN
1
⇥
(N 1) ⌃
WN
⌃⇧
2(N 1) ⌃ ⇧
WN
⌃⇧
⌃⇧
..
.
(N
WN
⌃⇧
⌅⇤
1)2
x[N
X = DN  x
2013-09-23
x[0]
x[1]
x[2]
..
.
37
⇥
⌃
⌃
⌃
⌃
⌃
⌅
1]
Matrix IDFT


If
X = DN  x
1
then
x = DN  X
i.e. inverse DFT is also just a matrix,
DN1
1
⇧1
⇧
1 ⇧
⇧1
=
N⇧
⇧ ..
⇤.
1
WN 1
WN 2
..
.
1 WN
(N 1)
1
WN 2
WN 4
..
.
WN
2(N 1)
=1/NDN*
Dan Ellis
2013-09-23
···
···
···
..
.
···
1
⌃
⌃
1) ⌃
⌃
⌃
⌃
⌅
(N 1)
WN
2(N
WN
..
.
WN
(N 1)2
38
⇥
DFT and MATLAB

MATLAB is concerned with sequences
not continuous functions like X(ej!)

Instead, we use the DFT to sample
X(ej!) on an (arbitrarily-fine) grid:


Dan Ellis
X = freqz(x,1,w); samples the DTFT
of sequence x at angular frequencies in w
X = fft(x); calculates the N-point DFT
of an N-point sequence x
2013-09-23
39
M
DFT and DTFT
DTFT

 x[n]e
j
X(e ) =
 jn
• continuous freq !
• infinite x[n], -∞<n<∞
n=
N 1
X[k] =  x[n]W
• discrete freq k=N!/2º
• finite x[n], 0≤n<N
kn
N
DFT
n=0

DFT ‘samples’ DTFT at discrete freqs:
X[k]
j
X[k] = X(e )  = 2 k
N
Dan Ellis
k=1...
2013-09-23
X(ej!)
!
40
DTFT from DFT

N-point DFT completely specifies the
continuous DTFT of the finite sequence
N 1

  jn
1
j
kn
X(e ) =    X[k]WN e

n=0  N k=0
N 1
 j (  2 k ) n
1
N
“periodic
=  X[k] e
sinc”
N k=0
n=0
 k
N 1
sin N 2  j ( N21) k
1
=  X[k] 
e
 k
N k=0
sin 2
N 1
 k =   2Nk
interpolation
Dan Ellis
N 1
2013-09-23
41
Periodic sinc  jN
N 1
1e
 j n
=
e
 j
1e
k
k
k
n =0
jN k / 2
 jN k / 2
e
e
e
=  j k / 2  j k / 2
e
e
 e  j k / 2
 k
( N 1)
 j 2  k sin N 2
pure real
=e
 k
pure phase
sin 2
= N when Δ!k = 0; = (-)N when Δ!k/2 = º
= 0 when Δ!k/2 = r·º/N, r = ±1, ± 2, ...
other values in-between...
factor out
half the angle

 jN k / 2
Dan Ellis
2013-09-23
42
Periodic sinc
sin Nx
sin x
N
sinNx/sinx
sinNx
0
-N
X[k] = X(ej2
(N = 8)
k/N)
X(ej 0) =
X[k]·
1.5
DFT→ DTFT
= interpolation
by periodic
sinc
X[k]→X(ej!)
Dan Ellis
x
2
sinx
sin N
N sin
X[3]·
1
sin N
N sin
k/2
k/2
3/2
3/2
0.5
0
-0.5
-1
freq
0
k=1
= 2 /N
2013-09-23
0
k=3
= 6 /N
k=4
= 8 /N
43
DFT from overlength DTFT

If x[n] has more than N points, can still
j
form X[k] = X(e )  = 2 k
N

IDFT of X[k] will give N point x̃[n]

How does x̃[n] relate to x[n]?
Dan Ellis
2013-09-23
44
DFT from overlength DTFT
x[n]
DTFT
-A ≤ n < B
X(ejω)
1
x̃[n] =
N
=
sample
x[ ]WNk
k=0
=
=
0≤n<N
Dan Ellis
x[n
r=
x˜[n]
0≤n<N
N 1
x[ ]
x̃[n] =
X[k]
IDFT
1
N
N 1
k(
WN
WN nk
=1 for n-l = rN, r∈I
= 0 otherwise
n)
k=0
all values shifted by
rN ] exact multiples of N pts
to lie in 0 ≤ n < N
2013-09-23
45
DFT from DTFT example

If x[n] = { 8, 5, 4, 3, 2, 2, 1, 1} (8 point)
We form X[k] for k = 0, 1, 2, 3
by sampling X(ej!) at ! = 0, º/2, º, 3º/2

IDFT of X[k] gives 4 pt


x̃[n] =
r=
Overlap only for r = -1:
x̃[n] =
Dan Ellis
8
+
2
5 4
+ +
2 2
3
+
1
2013-09-23
= {10
rN ]
x[n
7
(N = 4)
5
4}
46
DFT from DTFT example

x[n]
-1


x[n+N]
(r = -1)
-5 -4 -3 -2 -1
1 2 3 4 5 6 7 8
n
1 2 3 4 5
x˜[n]
1 2 3

n
n
x˜[n] is the time aliased or ‘folded down’
version of x[n].
Dan Ellis
2013-09-23
47
Properties: Circular time shift



DFT properties mirror DTFT, with twists:
Time shift must stay within N-pt ‘window’
g[ n n0 N ]
WNkn0 G[k]
Modulo-N indexing keeps index between
0 and N-1:
g[ n
n0
N]
=
g[n n0 ]
g[N + n n0 ]
n n0
n < n0
0 ≤ n0 < N
Dan Ellis
2013-09-23
48
Circular time shift

Points shifted out to the right don’t
disappear – they come in from the left
g[n]
‘delay’ by 2
n
5-pt sequence
1 2 3 4

g[<n-2>5]
1 2 3 4
n
Like a ‘barrel shifter’:
origin pointer
Dan Ellis
2013-09-23
49
Circular time reversal

Time reversal is tricky in ‘modulo-N’
indexing - not reversing the sequence:
x˜ [n]
5-pt sequence
made periodic
-7 -6 -5 -4 -3 -2 -1
1 2 3 4 5 6 7 8 9 10 11
x˜ [ n
N
]
Time-reversed
periodic sequence
-7 -6 -5 -4 -3 -2 -1

1 2 3 4 5 6 7 8 9 10 11
Zero point stays fixed; remainder flips
Dan Ellis
2013-09-23
n
50
n
Duality

DFT and IDFT are very similar


both map an N-pt vector to an N-pt vector
Duality:
if
then

Circular
time reversal
g[n]  G [k ]
G [n]  N  g[ k
N
]
i.e. if you treat DFT sequence as a time
sequence, result is almost symmetric
Dan Ellis
2013-09-23
51
M
4. Convolution with the DFT



IDTFT of product of DTFTs of two N-pt
sequences is their 2N-1 pt convolution
IDFT of the product of two N-pt DFTs
can only give N points!
Equivalent of 2N-1 pt result time aliased:

(0 ≤ n < N)
 i.e. y [n ] = 
y
[n
+
rN
]
c
l
r=


must be, because G[k]H[k] are exact
samples of G(ej!)H(ej!)
This is known as circular convolution
Dan Ellis
2013-09-23
52
Circular convolution


Can also do entire convolution with
modulo-N indexing
Hence, Circular Convolution:
N 1
 g[m ]h[ n  m N ]
 G[k]H[k]
m=0

N h[n]
Written as g[n] 
Dan Ellis
2013-09-23
53
Circular convolution example

4 pt sequences:
g[n]={1 2 0 1} h[n]={2 2 1 0}
N 1
 g[m ]h[ n  m N ]
1 2 3
m=0
h[<n - 0>4]
h[<n - 1>4]
h[<n - 2>4]
h[<n - 3>4]
Dan Ellis
1
1 2 3
n
2
1 2 3
n
1 2 3
n
1 2 3
n
0
1
2013-09-23
n
1 2 3
n
g[n] 4 h[n]={4 7 5 4}
1 2 3
n
check: g[n] * h[n]
={2 6 5 4 2 1 0}
54
DFT properties summary




Circular convolution
N 1
m=0 g[m ]h[ n  m
Modulation
g[n]  h[n] 
Duality
Parseval
Dan Ellis
1
N
N
]
 G[k]H[k]
m=0 G[m]H[ k  m N ]
N 1
G [n]  N  g[ k
n=0 x[n]
N 1
2
=
1
N
N
]
k=0 X [k ]
2013-09-23
N 1
2
55
Linear convolution w/ the DFT





DFT → fast circular convolution
.. but we need linear convolution
Circular conv. is time-aliased linear
conv.; can aliasing be avoided?
e.g. convolving L-pt g[n] with M-pt h[n]:
y[n] = g[n] * h[n] has L+M-1 nonzero pts
Set DFT size N ≥ L+M-1 → no aliasing
Dan Ellis
2013-09-23
56
Linear convolution w/ the DFT

Procedure (N = L + M - 1):




g[n]
pad L-pt g[n] with (at least)
n
M-1 zeros
L
h[n]
→ N-pt DFT G[k], k = 0..N-1
pad M-pt h[n] with (at least)
n
M
L-1 zeros
yc[n]
→ N-pt DFT H[k], k = 0..N-1
Y[k] = G[k]·H[k], k = 0..N-1
n
N

IDFT{Y[k]} = 
yL [n + rN ] = yL [n] (0 ≤ n < N)
r=
Dan Ellis
2013-09-23
57
Overlap-Add convolution


Very long g[n] → break up into
segments, convolve piecewise, overlap
→ bound size of DFT, processing delay
Make gi [n] =
h[n]

i · N n < (i + 1) · N
otherwise
g[n]
0
g[n] =
g[n] =
i gi [n]
i
h[n]
gi [n]
Called Overlap-Add (OLA) convolution...
Dan Ellis
2013-09-23
58
Overlap-Add convolution
g[n]
h[n]
n
g0[n]
M
n
g0[n] * h[n]
n
g1[n]
n
g2[n]
L
2L
N=
L+M-1
g1[n] * h[n]
n
g2[n] * h[n]
n
3L
h[n] * g[n]
valid OLA sum
ML
Dan Ellis
2013-09-23
n
2L
3L
59
n
n
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