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Chapter 8
The Discrete Fourier Transform
2
Introduction
In Chapters 2 and 3 we discussed the representation of sequences and LTI systems in terms of Fourier and z-transforms.
For finite duration sequences, it is possible to develop an alternative Fourier representation, called the discrete Fourier transform (DFT).
The DFT is a sequence rather than a function of a continuous variable. It corresponds to samples, equally spaced in frequency, of the Fourier transform of a signal.
DFT plays a central role in the implementation of a variety of DSP algorithms, because efficient algorithms exist for the computation of DFT (chapter 9).
We will begin by considering the Fourier series representation of periodic sequences. Then we will consider the relationship between periodic sequences and finite-length sequences.
3
The Discrete Fourier Series (1)
We first review the Fourier series for periodic continuous-time signals.
For a continuous-time, T-periodic signal x(t), the Fourier series approximation can be written as
where 2/T is the fundamental frequency.
That is, a periodic signal can be represented by a Fourier series corresponding to a sum of harmonically related complex exponential sequences. Infinitely many harmonically related complex exponentials are required.
The frequency of the complex exponentials are integer multiples of the fundamental frequency (2/T).
ktTjk
k
eaT
tx )/2(1)(
4
The Discrete Fourier Series (2)
Consider a sequence that is periodic with period N, so that
for any integer value of n and r.
As with continuous-time periodic signals, such sequences can be represented by a Fourier series corresponding to a sum of harmonically related complex exponential sequences.
The frequency of the complex exponentials are integer multiples of the fundamental frequency (2/N); i.e., the Fourier series representation has the form
][~ nx
][~][~ rNnxnx
knNj
k
ekXN
nx )/2(][~1
][~
5
The Discrete Fourier Series (3)
The period is 2, and the fundamental frequency is 2/N.
Therefore, the discrete-time periodic signal case only requires N harmonically related complex exponentials.
Time Frequency
Continuous, aperiodic Aperiodic, continuous
Discrete, aperiodic Periodic, continuous
Continuous, Periodic Aperiodic, discrete
Discrete, Periodic Periodic, Discrete
6
The Discrete Fourier Series (4)
Denote . Then, for any integer l, we have
Consequently, the set of N periodic complex exponentials e0[n], e1[n], …, eN-1[n] defines all the distinct periodic complex exponentials.
Thus, the Fourier series representation of a periodic sequenceneed contain only N of these complex exponentials. Therefore, it has the form
What is ?
knNjk ene )/2(][
][][ )/2(2))(/2( neeeene kknNjnljnlNkNj
lNk
][~ nx
knNjN
k
ekXN
nx )/2(1
0
][~1
][~
][~
kX
7
The Discrete Fourier Series (5)
To obtain , we multiply both sides by , and summing from n = 0 to n = N –1, we obtain
Exploiting the orthogonality of the complex exponentials
Therefore, the Fourier series coefficients become
][~
kX rnNje )/2(
knNjN
k
enxkX )/2(1
0
][~][~
nrkNjN
k
N
n
rnNjN
n
ekXN
enx ))(/2(1
0
1
0
)/2(1
0
][~1
][~
otherwise,0
integeran ,,11 ))(/2(1
0
mmNrke
NnrkNj
N
n
][~1
][~
][~ ))(/2(1
0
1
0
)/2(1
0
rXeN
kXenx nrkNjN
n
N
k
rnNjN
n
8
The Discrete Fourier Series (6)
is periodic with period N. For integer l, we have
For convenience of notation, we can define . Then the analysis and synthesis pair of discrete Fourier series (DFS) is expressed as
analysis equation
synthesis equation
We can also use the notation
][~
kX
knN
N
k
WnxkX ][~][~ 1
0
][~
][~
][~][~
)/2(21
0
))(/2(1
0
kXeenx
enxlNkX
knNjnljN
k
nlNkNjN
k
)/2( NjN eW
knN
N
k
WkXN
nx
][
~1][~
1
0
][~
][~ kXnxDFS
9
Examples of DFS (1)
Example 1 - DFS of a Periodic Impulse Train
Consider the periodic impulse train
Since for 0 ≤ n ≤ N–1, the DFS coefficients are found as
If we synthesize the signal from the DFS coefficients, we have
1][][~ 0
1
0
N
knN
N
n
WWnkX
otherwise,0integerany ,,1
][][~ rrNnrNnnx
r
r
knN
N
k
knN
N
k
rNnWN
WkXN
nx ][1
][~1
][~1
0
1
0
][][~ nnx
10
Examples of DFS (2)
Example 1 - DFS of a Periodic Impulse Train
(N=8)
11
Examples of DFS (3)
Example 2 - Duality in the DFS
We see that the two equations of DFS pair are very similar, differing only in a constant multiplier and the sign of the exponents.
Consider the periodic impulse train
Then,
Compared with the previous example, we see that
][][~
rNkNkYr
1][][1
][~ 01
0
1
0
N
knN
N
k
knN
N
k
WWkWkNN
ny
][~][~
kxNkY ][~
][~ nXny
12
Examples of DFS (4)
Example 3 – DFS of a Periodic Rectangular Pulse Train
Consider the sequence shown in figure, whose period is N=10.
The DFS coefficients become
4
0
)10/4(
10
510
10 )10/sin(
)2/sin(
1
1][
~
n
kjk
kkn
k
ke
W
WWkX
13
Examples of DFS (5)
Example 3 – DFS of a Periodic Rectangular Pulse Train
14
Properties of DFS (1)
It is not surprising that many of the basic properties are analogous to properties of the Fourier and z-transforms.
Linearity
Consider two periodic sequences and , both with period N, such that
then
Shift of a Sequence
If , then
][~1 nx ][~
2 nx
][~
][~11 kXnx
DFS
][~
][~22 kXnx
DFS
][~
][~
][~][~2121 kXbkXanxbnxa
DFS
][~
][~ kXnxDFS
][~
][~ kXWmnx kmN
DFS
][~
][~ lkXnxWDFS
klN
15
Properties of DFS (2)
Symmetric Properties
If , then][~
][~ kXnxDFS
][~
][~ ** kXnxDFS
][~
][~ ** kXnxDFS
]}[~
][~
{][~
]}[~Re{ *21 kXkXkXnx e
DFS
]}[~
Re{}][~][~{][~ *21 kXnxnxnx
DFS
e
]}[~
][~
{][~
]}[~Im{ *21 kXkXkXnxj o
DFS
]}[~
Im{}][~][~{][~ *21 kXjnxnxnx
DFS
o
16
Properties of DFS (3)
Symmetric Properties (cont.)
When is real, then][~ nx
][~
][~ * kXkX
]}[~
Re{}][~][~{][~21 kXnxnxnx
DFS
e
]}[~
Im{}][~][~{][~21 kXjnxnxnx
DFS
o
]}[~
Re{]}[~
Re{ kXkX
]}[~
Im{]}[~
Im{ kXkX
|][~
||][~
| kXkX
][~
][~
kXkX
17
Properties of DFS (4)
Periodic Convolution
Consider two periodic sequences and , both with
period N, and their DFS coefficients are and ,
respectively. If we form the product , then
the periodic sequence with DFS coefficients is
A convolution in the above form is referred to as a periodic convolution.
The duality theorem suggests that, if , then
][~1 nx ][~
2 nx
][~
1 kX ][~
2 kX
][~
][~
][~
213 kXkXkX ][
~3 kX][~
3 nx
1
012
1
0213 ][~][~][~][~][~
N
m
N
m
mnxmxmnxmxnx
][~][~][~213 nxnxnx
][~
][~1
][~
][~1
][~
12
1
021
1
03 lkXlX
NlkXlX
NkX
N
l
N
l
18
Properties of DFS (5)
Periodic Convolution
- Example
19
Fourier Transform of Periodic Signals (1)
As discussed in Chapter 2, uniform convergence of the Fourier transform of a sequence requires that the sequence be absolutely summable, and mean-square convergence requires that the sequence be square summable. Periodic sequence satisfy neither condition.
However, in Chapter 2, we know that sequences that can be expressed as a sum of complex exponentials can be considered to have a Fourier transform representation in the form of a train of impulse.
Similarly, it is often useful to incorporate the DFS representation of periodic signals within the framework of the Fourier transform. This can be done by interpreting the Fourier transform of a periodic signal to be an impulse train in the frequency domain with the impulse values proportional to the DFS coefficients for the sequence.
20
Fourier Transform of a Constant
Consider the sequence x[n]=1for all n. This sequence is neither absolutely summable not square summable, and the Fourier transform does not converge in either the uniform or mean-square sense for this case. However, it is possible and useful to define the Fourier transform of the sequence x[n] to be the periodic impulse train
The impulses in this case are functions of a continuous variable and therefore are of “infinite height, zero width, and unit area,” consistent with the fact that Fourier transform does not converge. The above equation is justifies principally because this result leads to correct inverse Fourier transform.
r
j reX )2(2)(
21
Fourier Transform of Complex Exponential Sequences
Consider a sequence x[n] whose Fourier transform is the periodic impulse train
To obtain the inverse Fourier transform, we can assume that –<0< in this problem. Then, we need include only the r=0 term, and
For 0=0, this reduces to the sequence of a constant considered in the previous example.
njnj edenx 0)(22
1][ 0
r
j reX )2(2)( 0
22
Fourier Transform of Periodic Signals (2)
If is periodic with period N and its DFS coefficients are
, then the Fourier transform of is defined to be the
impulse train
To show that is a Fourier transform representation of
, we obtain the inverse Fourier transform as (0<<2/N)
][~
kX
N
kkX
NeX
k
j 2][
~2)(
~
][~ nx
][~][~12
][~1
2][
~2
2
1)(
~
2
1
)/2(2
0
2
0
2
0
2
0
2
0
nxekXN
deN
kkX
N
deN
kkX
NdeeX
knNj
k
nj
k
nj
k
njj
][~ nx
)(~ jeX
][~ nx
23
Fourier Transform of Periodic Signals (3)
Example – The Fourier Transform of a Periodic Impulse Train
Consider a periodic impulse train
we know that the DFS coefficients for this sequence are
Therefore, the Fourier transform of is
N
k
NeP
k
j 22)(
~
kkP allfor ,1][~
][][~ rNnnpr
][~ np
24
Fourier Transform of Periodic Signals (4)
Consider a finite-length signal x[n] such that x[n] = 0 except in the intervals 0 ≤ n ≤ N–1, and consider the convolution of x[n] with the periodic impulse train ; i.e.,
][][*][][~*][][~ rNnxrNnnxnpnxnxrr
][~ np
25
Fourier Transform of Periodic Signals (5)
Denote the Fourier transform of x[n] as X(ej), then the Fourier transform of becomes
Compare with the definition of , we conclude that
That is, DFS coefficients are the equally spaced samples of X(ej), the Fourier transform of
kN
jkNj eXeXkX)/2(
)/2( )()(][~
][~
kX
otherwise,0
10],[~][
Nnnxnx
][~ nx
N
keX
N
N
k
NeXePeXeX
k
kNj
k
jjjj
2)(
2
22)()(
~)()(
~
)/2(
)(~ jeX
26
Fourier Transform of Periodic Signals (6)
Example – Relationship between the DFS Coefficients and the Fourier Transform of One Period
Again consider
the corresponding one period sequence is
otherwise,0
40,1][
nnx
27
Fourier Transform of Periodic Signals (7)
Example – Relationship between the DFS Coefficients and the Fourier Transform of One Period (cont.)
The Fourier transform of x[n] is
and the DFS coefficients are
)2/sin(
)2/5sin(
)(
2
4
0
j
n
njj
e
eeX
)10/sin(
)2/sin(][
~ )10/4(
k
kekX kj
28
Sampling of the Fourier Transform (1)
We consider in more detail the relationship between X(ej) and . Consider an aperiodic sequence x[n] with Fourier
transform X(ej) , and assume that a sequence is obtained by sampling X(ej) at frequencies k=2k/N; i.e.,
Since the Fourier transform is equal to the z-transform evaluated on the unit circle, it follows that can also be obtained by sampling X(z) at N equally spaced points on the unit circle; i.e.,
It is clear that the same sequence repeats
as k varies outside the range 0 ≤ k ≤ N –1.
][~
kX][
~kX
)()(][~ )/2(
)/2(
kNj
kN
j eXeXkX
][~
kX
)()(][~ )/2(
)/2(
kNj
ezeXzXkX kNj
29
Sampling of the Fourier Transform (2)
The sequence of samples of X(z), being periodic with period N, could be the sequence of DFS coefficients of a sequence .
The sequence can be obtained as
][~
kX
][~ nx
][~][
1][
][1
)(1
][~1
][~
)(1
0
)/2(1
0
)/2(1
0
1
0
mnpmx
WN
mx
WemxN
WeXN
WkXN
nx
m
mnkN
N
km
knN
kmNj
m
N
k
knN
NkjN
k
knN
N
k
30
Sampling of the Fourier Transform (3)
Since
We have
][][~ rNmnmnpr
][][*][][~*][][~ rNnxrNnnxnpnxnxrr
31
Sampling of the Fourier Transform (4)
In this figure, the sequence x[n] is of length 9. Similar to the sampling problem of a continuous-time signal, when the value of N is larger than the length of x[n], the delayed replications of x[n] do not overlap, and one period of the sequence is recognizable as x[n].
On the other hand, when N is less than the length of x[n], the replicas of x[n] overlap and one period of is no longer identical to x[n]. The next figure shows the case with N = 7.
][~ nx
][~ nx
32
Sampling of the Fourier Transform (5)
To summarize, if x[n] has finite length [cf. continuous-time
signal sampling: X(j) is bandlimited] and we take sufficient number [a number greater than or equal to the length of x[n] ] of equally spaced samples of its Fourier transform [cf. continuous-time signal sampling: sampling rate should be larger than the Nyquist rate], then the Fourier transform is recoverable from these samples, and equivalently, x[n] is recoverable from the corresponding periodic sequence through the relation
To recover x[n], it is not necessary to know X(ej) at all frequencies if x[n] has finite length.
][~ nx
otherwise,0
10],[~][
Nnnxnx
33
Discrete Fourier Transform (1)
From the above discussion, we understand that, given a finite-length sequence x[n], we can form a periodic sequence , which can be represented by a DFS .
On the other hand, given the sequence of DFS coefficients , we can find and then obtain x[n].
When the DFS is used in this way to represent finite-length sequences, it is called the discrete Fourier transform (DFT).
It is always important to remember that the representation through samples of the Fourier transform is in effect a representation of the finite-duration sequences by a periodic sequence, one period which is finite-duration sequence that we wish to represent.
][~ nx][
~kX
][~
kX][~ nx
34
Discrete Fourier Transform (2)
We assume that x[n] = 0 outside the range 0 ≤ n ≤ N – 1. In many instances, we will assume that a sequence has length N even if its length is M ≤ N. In such cases, we simply recognize that the last (N – M ) samples are zero.
When there is no aliasing, the finite-length sequence x[n] and the periodic sequence are associated by
An alternative expression is
For convenience, we will denote it as
][~ nx
][][~ rNnxnxr
)] modulo [(][~ Nnxnx
]))[((][~Nnxnx
35
Discrete Fourier Transform (3)
The sequence of DFS coefficients are periodic with period N. To maintain a duality between the time and frequency domains, we will choose the Fourier coefficients that are associated with a finite-duration sequence to be a finite-duration sequence corresponding to the one period of . That is
and
][~
kX
][~
kX
otherwise,010],[
~][ NkkXkX
]))[((] modulo [][~
NkXNkXkX
36
Discrete Fourier Transform (4)
Recall that and are related as
The DFT pairs become
otherwise,0
10,][][~
][
1
0
NkWnxkXkX
N
n
knN
1
0
][~][~ N
n
knNWnxkX
][~
kX ][~ nx
1
0
][~1
][~N
n
knNWkX
Nnx
otherwise,0
10,][1
][~][
1
0
NnWkXN
nxnx
N
n
knN
37
Discrete Fourier Transform (5)
Generally, the DFT analysis and synthesis equations are written as follows
The relationship between x[n] and X[k] will sometimes be denoted as
It is noted that, if we evaluate x[n] outside 0 ≤ n ≤ N – 1, the result will not be zero, but rather a periodic extension of x[n]. Same can be said for X[k]. In defining the DFT representation, we are simply recognizing that we are only interested in values of x[n] for 0 ≤ n ≤ N – 1 and X[k] for 0 ≤ n ≤ N – 1.
1
0
][][N
n
knNWnxkX
1
0
][1
][N
k
knNWkX
Nnx
][][ kXnxDFT
38
Discrete Fourier Transform (6)
Example – the DFT of a Rectangular Pulse
(N=5)
39
Discrete Fourier Transform (7)
Example – the DFT of a Rectangular Pulse
(N=10)
40
Homework Assignments (10)
8.58.7
8.24
