Finite Length Discrete Fourier
Transform
Discrete Fourier Transform (DFT)
Usually, we do not have an infinite amount of data which is required by the
DTFT. Instead, we have 1 image, a segment of speech, etc. Also, most realworld data are not of the convenient form an u[n].
Finally, on a computer, we can not calculate an uncountably infinite
(continuum) of frequencies as required by the DTFT.
ACTUAL DATA ANALYSIS on a computer- Use a DFT to look at a
finite segment of data.
In our development in the previous section of x[n] periodic with x0 [n] the
part of the signal that was repeated, we could have assumed that our finite
segment of data came from “windowing” an infinite length sequence x[n]
x0 [n] = x[n]wR [n]
where wR [n] is a rectangular window:
1
wR [n] =
!
1, n = 0, 1, · · · , N − 1
0, otherwise
x0 [n] = x[n]wR [n] is just the samples of x[n] between n = 0 and n = N − 1.
x0 [n] is 0 everywhere else. Therefore, it is defined ∀n, and we can take its
DTFT:
X0 (Ω) = DTFT(x0 [n]) =
∞
"
−jΩn
x0 [n]e
=
n=−∞
∞
"
−jΩn
x[n]wR [n]e
n=−∞
=
N
−1
"
x[n]e−jΩn
n=0
So,
X0 (Ω) =
N
−1
"
x[n]e−jΩn =
n=0
N
−1
"
x0 [n]e−jΩn
n=0
as we saw before.
Let’s say now that we want to sample X0 (Ω) (which is continuous and
periodic with period 2π) so we store it on a computer.
Sample X0 (Ω):
Assume we want 8 points in frequency – then sample X0 (Ω) at 8 uniformly
spaced points on the unit circle:
Values of frequency are 0, π/4, π/2, · · · , 7π/4 or 2πk/8, k = 0, 1, · · · , 7.
2
If we let k = N, what happens? If k = N, we get repetition of the points
we sampled so only N samples are unique.
Define Discrete Fourier Transform (DFT) as
X[k] = X0 (
2πk
)
N
, k = 0, 1, . . . , N − 1, i.e. only look at the N distinct sampled
for Ω = 2πk
N
frequencies of X0 (Ω).
Significant Observations
1. The resolution of the samples of the frequency spectrum is 2π
since we
N
sample the spectrum at points that are spaced 2π
apart
in
frequency,
N
that is, ∆Ω = 2π
.
N
2. If we looked at the samples of X0 (Ω) for all k = −∞ to ∞ for frequencies 2πk
, we would get the closely related Discrete Fourier Series (DFS)
N
which is of course periodic with period N since X0 (Ω) is periodic.
3. If x is a periodic signal with period N, as we saw that DFS was sufficient to construct the DTFT of the signal. In other words, the DFT
procedure approximates a signal with a periodic approximation of itself.
Now
X[k] = X0 (Ω)|Ω =
=
=
N
−1
"
n=0
N
−1
"
2πk
, k = 0, 1, · · · , N − 1
N
x[n]e−jΩn |Ω= 2πk ,k=0,1,···,N −1
N
x[n]e−j
2πkn
N
n=0
, k = 0, 1, · · · , N − 1
Shorthand Notation for the DFT:
2π
Let WN = e−j N ⇒ N th root of unity (WNN = 1) since WNN = e−j2π = 1.
You may also write WN simply as W .
Then
X[k] =
N
−1
"
n=0
x[n]WNkn , k = 0, 1, · · · , N − 1
is the DFT of your windowed sequence x0 [n].
3
Ex.
Find
N
−1
"
(e
−j2πk
N
)n =
n=0
N
−1
"
W kn
n=0
This is just the DFT of x[n] = 1, n = 0, 1, · · · , 7.
4
Ex.
x[n] =
!
1,
n=0
0, n = 1, . . . , 7
Find X[k], k = 0, 1, . . . , 7.
5
Given y[n] = δ[n − 2] and N = 8, find Y [k].
6
Ex. x[n] = cWN−pn , n = 0, 1, . . . , N − 1,
2π
p is an integer with p ∈ [0, 1, . . . , N − 1] and WN = e−j N (as usual), find
its DFT.
7
Synthesis: INVERSE DFT
How can we recover x[n] from X[k]?
Synthesis formula is
x[n] =
−1
1 N"
X[k]WN−kn , n = 0, 1, . . . , N − 1
N k=0
Prove this gives back x[n]:
x[n] =
=
−1
−1 N
"
1 N"
x[l]WNkl WN−kn
N k=0 l=0
−1
N
−1
"
1 N"
k(l−n)
x[l]
WN
N l=0
k=0
Ex.
N
−1
"
k(l−n)
WN
k=0
8
=?
ORTHOGONALITY OF EXPONENTIALS AGAIN!
So,
N
−1
"
k(l−n)
WN
=
k=0
!
N, l = n
0, l =
̸ n
and
−1
N
−1
−1
"
1 N"
1 N"
k(l−n)
x[l]
WN
=
x[l]Nδ[n − l]
N l=0
N l=0
k=0
1
(Nx[n]) = x[n]
=
N
x[n] =
Ex. Find the IDFT of X[k] = 1, k = 0, 1, . . . , 7.
9
Ex. Given x[n] = δ[n] + 2δ[n − 1] + 3δ[n − 2] + δ[n − 3] and N = 4, find
X[k].
10
Ex. Given X[k] = 2δ[k] + 2δ[k − 2] and N = 4, find x[n].
11
Selected DFT Properties
1. Linearity.
ax1 [n] + bx2 [n] ↔ aX1 [k] + bX2 [k]
2. Time shift.
x[n − n0 ]mod
N
↔ WNkn0 X[k]
3. Frequency shift.
WN−nk0 x[n] ↔ X[k − k0 ]mod
N
4. Multiplication.
x1 [n]x2 [n] ↔
1
X1 [k] ⊗ X2 [k]
N
5. Circular convolution.
x1 [n] ⊗ x2 [n] ↔ X1 [k]X2 [k]
6. Real/Even.
x[n] = xe [n] + xo [n] ↔ X[k] = A[k] + jB[k]
xe [n] ↔ A[k]
xo [n] ↔ jB[k]
If x[n] is real, then
X[−k]mod
N
= X ∗ [k].
If x[n] is even, then X[k] is real. (For a finite length function to be
even, x[n] = x[−n]mod N )
12
Fast Fourier Transform
The work of Cooley and Tukey showed how to calculate the DFT with complexity N log N (called the Fast Fourier Transform) instead of complexity
N 2 using the direct algorithm. The fft command that you use in MATLAB
implements a Fast Fourier Transform.
Examine:
X[k] =
N
−1
"
x[n]W kn
n=0
There are approximately N 2 complex multiplications and additions required to implement this (N for each value of X[k]).
If N = 210 = 1024, then N 2 = 220 = 106 , a very large number!
However, the FFT would only require about 5000, a substantial savings
in complexity (the actual calculation is N2 log2 N).
There are a number of different FFT algorithms that exist including
decimation-in-time and decimation-in-frequency.
The primary idea is to split up the size-N DFT into
each.
N
2
DFTs of length 2
You split the sum into 2 subsequences of length N2 and continue all the
way down until you have N2 subsequences of length 2.
13
First break x[n] into even and odd subsequences:
"
X[k] =
x[n]W kn +
neven
"
x[n]W kn
nodd
Now let n = 2m for even numbers and n = 2m + 1 for odd numbers:
N
2
X[k] =
−1
"
x[2m]W
2mk
N
2
+
m=0
N
2
−1
"
−1
"
x[2m + 1]W k(2m+1) =
m=0
2 mk
x[2m](W )
+W
m=0
k
N
2
−1
"
x[2m + 1](W 2 )mk =
m=0
Xe [k] + W k Xo [k]
Xe [k] and Xo [k] are both the DFT of a
N
2
point sequence.
W k is often referred to as the “twiddle factor.”
Now break up the size
N
4
Xe [k] =
−1
"
N
2
subsequences in half by letting m = 2p:
4 kp
x[4p](W )
+W
p=0
2k
N
4
−1
"
x[4p + 2](W 4 )kp =
p=0
The first subsequence here is the terms x[0], x[4], . . . and the second is
x[2], x[6], . . . .
14
Also, we have that:
N
WN2 = −1
Y [k] =
1
"
y[n]W2kn = y[0] + W2k y[1]
n=0
W2 = e−
j2π
2
= e−jπ = −1
So we get,
Y [k] = y[0] + (−1)k y[1]
and:
Y [0] = y[0] + y[1]
Y [1] = y[0] − y[1]
15
Ex.
This was a problem I had on a DSP final exam in 1984:
Express the DFT of the 9-point sequence {x[0], x[1], . . . , x[8]} in terms of
the DFTs of 3-point sequences {x[0], x[3], x[6]}, {x[1], x[4], x[7]}, and {x[2], x[5], x[8]}
We start with:
X[k] =
2
"
m=0
x[3m]W93mk +
2
"
(3m+1)k
x[3m + 1]W9
m=0
+
2
"
m=0
16
(3m+2)k
x[3m + 2]W9
Applications of the DFT
Linear Convolution
You can use an FFT in MATLAB to compute linear convolution. If you
don’t use a sufficient number of points in the DFT, you will get overlap.
CIRCULAR CONVOLUTION
Since DFTs are a limited length sequence, convolution is done mod N ⇒
circular convolution.
x1 [n] ⊗ x2 [n] =
N
−1
"
p=0
x1 [p]x2 [n − p]mod
N
That is, when we flip and shift the sequence x2 [n], we do it mod N.
Also, note that:
x1 [n] ⊗ x2 [n] ↔ X1 [k]X2 [k]
17
Ex. Find x1 [n] ⊗ x2 [n] = z[n].
18
Ex. N = 8, find x1 [n] ⊗ x2 [n] = y[n].
19
Ex. Find y[n] = x[n] ⊗ x[n].
20
Overview of Chapters 3, 4, and 5
Fourier Analysis for Discrete-Time Signals
1. DTFT for infinite length sequences:
continuous frequency, periodic with period 2π,
X(Ω) =
"
x[n]e−jΩn
n
Important: DT convolution ↔ multiplication of DTFTs.
Inverse system:
h[n] ∗ hi [n] = δ[n] ↔ H(Ω)Hi (Ω) = 1
2. DTFT of a periodic DT sequence by allowing impulses in frequency:
x[n] =
"
k
ak ejΩk n ↔ 2π
where
ak = X0 (
"
k
ak δ(Ω − Ωk )
2πk
)
N
3. DFT for a finite length data
discrete frequency
Take DTFT of windowed infinite length sequence and then sample at
discrete frequencies,
Ω=
2πk
, k = 0, 1, . . . , N − 1
N
N discrete frequencies since exponential is periodic.
X[k] =
N
−1
"
x[n]e−j
n=0
2πkn
N
=
N
−1
"
n=0
Orthogonality of Exponentials:
N
−1
"
WNkn = Nδ[k]
n=0
21
x[n]WNkn