Discrete Fourier Transform (DFT)
Recall the DTFT:
X(ω) =
∞
X
x(n)e−jωn.
n=−∞
DTFT is not suitable for DSP applications because
• In DSP, we are able to compute the spectrum only at specific
discrete values of ω,
• Any signal in any DSP application can be measured only in
a finite number of points.
A finite signal measured at N points:

n < 0,
 0,
y(n), 0 ≤ n ≤ (N − 1),
x(n) =

0,
n ≥ N,
where y(n) are the measurements taken at N points.
EE 524, Fall 2004, # 5
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Sample the spectrum X(ω) in frequency so that
X(k) = X(k∆ω),
X(k) =
N
−1
X
∆ω =
−j2π kn
N
x(n)e
2π
N
=⇒
DFT.
n=0
The inverse DFT is given by:
N −1
kn
1 X
x(n) =
X(k)ej2π N .
N
k=0
1
x(n) =
N
=
N
−1
X
(N −1
X
k=0
m=0
N
−1
X
(
x(m)
m=0
|
EE 524, Fall 2004, # 5
)
−j2π km
N
x(m)e
1
N
N
−1
X
−j2π
e
j2π kn
N
e
k(m−n)
N
)
= x(n).
k=0
{z
δ(m−n)
}
2
The DFT pair:
X(k) =
N
−1
X
kn
x(n)e−j2π N
analysis
n=0
x(n) =
N −1
1 X
j2π kn
X(k)e N
N
synthesis.
k=0
Alternative formulation:
X(k) =
N
−1
X
x(n)W kn
2π
←− W = e−j N
n=0
x(n) =
N −1
1 X
X(k)W −kn.
N
k=0
EE 524, Fall 2004, # 5
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EE 524, Fall 2004, # 5
4
Periodicity of DFT Spectrum
X(k + N ) =
N
−1
X
−j2π
x(n)e
(k+N )n
N
n=0
N
−1
X
=
!
−j2π kn
N
x(n)e
e−j2πn
n=0
= X(k)e−j2πn = X(k) =⇒
the DFT spectrum is periodic with period N (which is expected,
since the DTFT spectrum is periodic as well, but with period
2π).
Example: DFT of a rectangular pulse:
x(n) =
X(k) =
1, 0 ≤ n ≤ (N − 1),
0, otherwise.
N
−1
X
kn
e−j2π N = N δ(k) =⇒
n=0
the rectangular pulse is “interpreted” by the DFT as a spectral
line at frequency ω = 0.
EE 524, Fall 2004, # 5
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DFT and DTFT of a rectangular pulse (N=5)
EE 524, Fall 2004, # 5
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Zero Padding
What happens with the DFT of this rectangular pulse if we
increase N by zero padding:
{y(n)} = {x(0), . . . , x(M − 1), 0,
. . . , 0} },
| 0,{z
N −M positions
where x(0) = · · · = x(M − 1) = 1. Hence, DFT is
Y (k) =
N
−1
X
−j2π kn
N
y(n)e
=
M
−1
X
n=0
=
EE 524, Fall 2004, # 5
sin(π kM
N )
sin(π Nk )
−j2π kn
N
y(n)e
n=0
−jπ
e
k(M −1)
N
.
7
DFT and DTFT of a Rectangular Pulse with
Zero Padding (N = 10, M = 5)
Remarks:
• Zero padding of analyzed sequence
“approximating” its DTFT better,
results
in
• Zero padding cannot improve the resolution of spectral
components, because the resolution is “proportional” to
1/M rather than 1/N ,
• Zero padding is very important for fast DFT implementation
(FFT).
EE 524, Fall 2004, # 5
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Matrix Formulation of DFT
Introduce the N × 1 vectors

x(0)

x(1)
x=
..

x(N − 1)


,



X(0)


X(1)

.
X=
..

X(N − 1)
and the N × N matrix




W=


0
0
0
W
W
W
W0
W1
W2
W0
W2
W4
..
..
..
W 0 W N −1 W 2(N −1)
0
···
W
· · · W N −1
· · · W 2(N −1)
..
..
(N −1)2
··· W




.


DFT in a matrix form:
X = Wx.
Result: Inverse DFT is given by
x=
EE 524, Fall 2004, # 5
1 H
W X,
N
9
which follows easily by checking W H W = WW H = N I, where
I denotes the identity matrix. Hermitian transpose:
xH = (xT )∗ = [x(1)∗, x(2)∗, . . . , x(N )∗].
Also, “∗” denotes complex conjugation.
Frequency Interval/Resolution: DFT’s frequency resolution
Fres ∼
1
NT
[Hz]
and covered frequency interval
∆F = N ∆Fres
1
= = Fs
T
[Hz].
Frequency resolution is determined only by the length of
the observation interval, whereas the frequency interval is
determined by the length of sampling interval. Thus
• Increase sampling rate =⇒ expand frequency interval,
• Increase observation time =⇒ improve frequency resolution.
Question: Does zero padding alter the frequency resolution?
EE 524, Fall 2004, # 5
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Answer: No, because resolution is determined by the length of
observation interval, and zero padding does not increase this
length.
Example (DFT Resolution): Two complex exponentials with
two close frequencies F1 = 10 Hz and F2 = 12 Hz sampled
with the sampling interval T = 0.02 seconds. Consider various
data lengths N = 10, 15, 30, 100 with zero padding to 512
points.
DFT with N = 10 and zero padding to 512 points.
Not resolved: F2 − F1 = 2 Hz < 1/(N T ) = 5 Hz.
EE 524, Fall 2004, # 5
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DFT with N = 15 and zero padding to 512 points.
Not resolved: F2 − F1 = 2 Hz < 1/(N T ) ≈
3.3 Hz.
DFT with N = 30 and zero padding to 512 points.
Resolved: F2 − F1 = 2 Hz > 1/(N T ) ≈ 1.7 Hz.
EE 524, Fall 2004, # 5
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DFT with N = 100 and zero padding to 512
points. Resolved: F2 − F1 = 2 Hz > 1/(N T ) =
0.5 Hz.
EE 524, Fall 2004, # 5
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DFT Interpretation Using
Discrete Fourier Series
Construct a periodic sequence by periodic repetition of x(n)
every N samples:
{e
x(n)} = {. . . , x(0), . . . , x(N − 1), x(0), . . . , x(N − 1), . . .}
|
{z
} |
{z
}
{x(n)}
{x(n)}
The discrete version of the Fourier Series can be written as
x
e(n) =
X
j2π kn
N
Xk e
k
1 X e
1 X e
j2π kn
=
X(k)e N =
X(k)W −kn,
N
N
k
k
e
where X(k)
= N Xk . Note that, for integer values of m, we
have
W
−kn
j2π kn
N
=e
j2π
=e
(k+mN )n
N
= W −(k+mN )n.
As a result, the summation in the Discrete Fourier Series (DFS)
should contain only N terms:
N −1
1 X e
j2π kn
x
e(n) =
X(k)e N
N
DFS.
k=0
EE 524, Fall 2004, # 5
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Inverse DFS
The DFS coefficients are given by
e
X(k)
=
N
−1
X
kn
x
e(n)e−j2π N
inverse DFS.
n=0
Proof.
N
−1
X
−j2π kn
N
x
e(n)e
n=0


N
−1 
N −1

X
1 X e
j2π pn
−j2π kn
N
N
X(p)e
=
e
N

n=0
p=0
N
−1
X
=
(
e
X(p)
p=0
1
N
N
−1
X
j2π
e
(p−k)n
N
)
e
= X(k).
n=0
|
{z
δ(p−k)
}
2
The DFS coefficients are given by
e
X(k)
=
N
−1
X
kn
x
e(n)e−j2π N
analysis,
n=0
x
e(n) =
N −1
1 X e
j2π kn
X(k)e N
N
synthesis.
k=0
EE 524, Fall 2004, # 5
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• DFS and DFT pairs are identical, except that
− DFT is applied to finite sequence x(n),
− DFS is applied to periodic sequence x
e(n).
• Conventional (continuous-time) FS vs. DFS
− CFS represents a continuous periodic signal using an
infinite number of complex exponentials,
whereas
− DFS represents a discrete periodic signal using a finite
number of complex exponentials.
EE 524, Fall 2004, # 5
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DFT: Properties
Linearity
Circular shift of a sequence: if X(k) = DFT {x(n)} then
km
X(k)e−j2π N = DFT {x((n − m) mod N )}
Also if x(n) = DFT −1{X(k)} then
km
x((n − m) mod N ) = DFT −1{X(k)e−j2π N }
where the operation mod N denotes the periodic extension
x
e(n) of the signal x(n):
x
e(n) = x(n mod N ).
EE 524, Fall 2004, # 5
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DFT: Circular Shift
N
−1
X
x((n − m)modN )W kn
n=0
= W km
N
−1
X
x((n − m)modN )W k(n−m)
n=0
EE 524, Fall 2004, # 5
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= W km
N
−1
X
x((n − m)modN )W k(n−m)modN
n=0
= W kmX(k),
where we use the facts that W k(lmodN ) = W kl and that the
order of summation in DFT does not change its result.
Similarly, if X(k) = DFT {x(n)}, then
j2π mn
N
X((k − m)modN ) = DFT {x(n)e
}.
DFT: Parseval’s Theorem
N
−1
X
x(n)y ∗(n) =
n=0
N −1
1 X
X(k)Y∗(k)
N
k=0
Using the matrix formulation of the DFT, we obtain
yH x =
=
EE 524, Fall 2004, # 5
H 1 H
1 H
W Y
W Y
N
N
1 H
1 H
H
Y W
W } X = Y X.
| {z
N2
N
NI
19
DFT: Circular Convolution
If X(k) = DFT {x(n)} and Y (k) = DFT {y(n)}, then
X(k)Y (k) = DFT {{x(n)} ~ {y(n)}}
Here, ~ stands for circular convolution defined by
{x(n)} ~ {y(n)} =
N
−1
X
x(m)y((n − m) mod N ).
m=0
DFT {{x(n)} ~ {y(n)}}
N
−1 h
i
X
PN −1
kn
=
x(m)y((n
−
m)
mod
N
)
W
m=0
|
{z
}
n=0
{x(n)}~{y(n)}
=
N
−1 h
X
PN −1
n=0
m=0 |
= Y (k)
N
−1
X
|m=0
EE 524, Fall 2004, # 5
i
y((n − m) mod N )W kn x(m)
{z
}
Y (k)W km
x(m)W km = X(k)Y (k).
{z
X(k)
}
20