2. Introduction
The discrete-time Fourier transform (DTFT) provided the frequency-
domain (ω) representation for absolutely summable sequences.
The z-transform provided a generalized frequency-domain (z)
representation for arbitrary sequences.
These transforms have two features in common.
First, the transforms are defined for infinite-length sequences.
Second, and the most important, they are functions of continuous variables
(ω or z).
In other words, the discrete-time Fourier transform and the z-transform
are not numerically computable transforms.
The Discrete Fourier Transform (DFT) avoids the two problems
mentioned and is a numerically computable transform that is suitable
for computer implementation.
2
4. Periodic Sequences
Let 𝑥(𝑛) is a periodic sequence with period N-Samples (the fundamental
period of the sequence).
Notation: a sequence with period N is satisfying the condition
where r is any integer.
4
),...1(),...,1(),0(),1(),...,1(),0(...,)(~ NxxxNxxxnx
x(n) x(n)
)(~)(~ rNnxnx
5. Periodic Sequences
From Fourier analysis we know that the periodic functions can be synthesized
as a linear combination of complex exponentials whose frequencies are
multiples (or harmonics) of the fundamental frequency (which in our case is
2π/N).
The discrete version of the Fourier Series can be written as
where { 𝑋(𝑘), 𝑘 = 0, ± 1, . . . , ∞} are called the discrete Fourier series
coefficients.
Note That, for integer values of m, we have
(it is called Twiddle Factor)
5
k
kn
N
k
N
kn
j
k
kn
N
j
k WkX
N
ekX
N
eXnx )(
~1
)(
~1
)(~ 2
2
nmNk
N
N
nmNk
j
N
kn
j
kn
N WeeW )(
)(
22
6. Periodic Sequences
As a result, the summation in the Discrete Fourier Series (DFS) should
contain only N terms:
The Harmonics are
6
1
0
)(
~1
)(~
N
k
kn
NWkX
N
nx
knNj
k ene )/2(
)(
,2,1,0 k
7. Inverse DFS
The DFS coefficients are given by
Proof
7
1
0
1
0
2
)(~)(~)(
~ N
k
kn
N
N
k
N
kn
j
WnxenxkX
DFSInverse
)(
~
)()(
~
)(~
1
)(
~
)(~
)(
~1
)(~
1
0
1
0
2
1
0
1
0
)(
21
0
2
1
0
21
0
21
0
2
kXkpPXenx
e
N
PXenx
eePX
N
enx
N
P
N
k
N
kn
j
N
P
N
k
N
nkP
jN
k
N
kn
j
N
k
N
kn
jN
P
N
Pn
jN
k
N
kn
j
8. Synthesis and Analysis (DFS-Pairs)
8
Notation
)/2( Nj
N eW
Synthesis
1
0
)(
~1
)(~
N
k
kn
NWkX
N
nx
Analysis
)(
~
)(~ kXnx DFS
Both have
Period N
1
0
)(~)(
~ N
k
kn
NWnxKX
12. DFS vs. FT
12
)(~ nx
0 N nN
0 n
)(nx
0
)()(
n
njj
enxeX
1
0
)(
N
n
nj
enx
1
0
)/2(
)(~)(
~ N
n
knNj
enxkX
Nk
j
eXkX
/2
)()(
~
13. Example
13
0 1 2 3 4 5 6 7 8 9 n
0 1 2 3 4 5 6 7 8 9 n
)(~ nx
)(nx
)2/sin(
)2/5sin(
1
1
)( 2
54
0
j
j
j
n
njj
e
e
e
eeX
)10/sin(
)2/sin(
)()(
~ )10/4(
10/2 k
k
eeXkX kj
k
j
14. Example
0 1 2 3 4 5 6 7 8 9 n
0 1 2 3 4 5 6 7 8 9 n
)(~ nx
)(nx
)2/sin(
)2/5sin(
1
1
)( 2
54
0
j
j
j
n
njj
e
e
e
eeX
14
)10/sin(
)2/sin(
)()(
~ )10/4(
10/2 k
k
eeXkX kj
k
j
15. 0 1 2 3 4 5 6 7 8 9 n
0 1 2 3 4 5 6 7 8 9 n
)(~ nx
)(nx
Example
15
)2/sin(
)2/5sin(
1
1
)( 2
54
0
j
j
j
n
njj
e
e
e
eeX
)10/sin(
)2/sin(
)()(
~ )10/4(
10/2 k
k
eeXkX kj
k
j
16. Relation To The Z-transform
Let 𝑥(𝑛) be a finite-duration sequence of duration 𝑁 such that
Then we can compute its z-transform:
Now we construct a periodic sequence 𝑥 𝑛 by periodically repeating
𝑥(𝑛) with period 𝑁, that is,
The DFS of 𝑥 𝑛 is given by
16
Elsewhere
NnNonzero
nx
,0
)1(0,
)(
1
0
)(z)(
N
n
n
znxX
Elsewhere
Nnnx
nx
,0
)1(0),(~
)(
1
0
2
)(k)(
~ N
n
n
k
N
j
enxX
k
N
j
ez
zXX 2
)(k)(
~
17. Relation To The Z-transform
which means that the DFS 𝑿(𝑘) represents 𝑁 evenly spaced samples of
the z-transform 𝑋(𝑧) around the unit circle.
17
Re(z)
Im(z)
z0
z1
z2
z3
z4
z5
z6
z7
1
N = 8
20. Recall of DTFT
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 measures at N-points:
Where y(n) are the measurements taken at N-points
20
Nn
Nnny
n
nx
,0
)1(0),(
00
)(
21. Computation of DFT
Recall of DTFT
The DFT can be computed by:
Truncate the summation so that it ranges over finite limits x[n] is a finite-length
sequence.
Discretize ω to ωk evaluate DTFT at a finite number of discrete frequencies.
For an N-point sequence, only N values of frequency samples of X(ejw) at N
distinct frequency points, are sufficient to determine x[n]
and X(ejw) uniquely.
So, by sampling the spectrum X(ω) in frequency domain
21
DFTenxX
N
X
N
n
N
kn
j
1
0
2
)(k)(
2
),X(kk)(
10, Nkk
22. Computation of DFT
The inverse DFT is given by:
Proof
22
IDFTekX
N
x
N
k
N
kn
j
1
0
2
)(
1
n)(
1
0
1
0
1
0
)(
2
1
0
21
0
2
)()()(n)(
1
)(n)(
)(
1
n)(
N
m
N
m
N
n
N
nmk
j
N
k
N
kn
jN
m
N
mk
j
nxnmmxx
e
N
mxx
eemx
N
x
24. Example
Find the discrete Fourier Transform of the following N-points discrete
time signal
Solution:
On the board
24
)1(,...),1(),0()( Nxxxnx
25. Nth Root of Unity
252
1
2
1
10)
19)
2
1
2
1
8)
7)
2
1
2
1
6)
15)
2
1
2
1
4)
3)
2
1
2
1
2)
11)
9
8
2
9
8
8
8
2
8
8
7
8
2
7
8
6
8
2
6
8
5
8
2
5
8
4
8
2
4
8
3
8
2
3
8
2
8
2
2
8
1
8
2
1
8
0
8
2
0
8
jeW
eW
jeW
jeW
jeW
eW
jeW
jeW
jeW
eW
j
j
j
j
j
j
j
j
j
j
1*
2/
2
)2/(
NN
k
N
k
N
k
N
Nk
N
k
N
Nk
N
WW
WW
WW
WW
N
j
N eW
2
26. Relation To The Z-transform
Let 𝑥(𝑛) be a finite-duration sequence of duration 𝑁 such that
Then we can compute its z-transform:
Now we construct a periodic sequence 𝑥 𝑛 by periodically repeating
𝑥(𝑛) with period 𝑁, that is,
The DFS of 𝑥 𝑛 is given by
26
Elsewhere
NnNonzero
nx
,0
)1(0,
)(
1
0
)(z)(
N
n
n
znxX
Elsewhere
Nnnx
nx
,0
)1(0),(~
)(
1
0
2
)(k)(
~ N
n
n
k
N
j
enxX
k
N
j
ez
zXX 2
)(k)(
~
27. Relation To The Z-transform
which means that the DFS 𝑿(𝑘) represents 𝑁 evenly spaced samples of
the z-transform 𝑋(𝑧) around the unit circle.
27
Re(z)
Im(z)
z0
z1
z2
z3
z4
z5
z6
z7
1
N = 8
29. DFT - Matrix Formulation
xWX
Nx
x
x
x
WWW
WWW
WWW
NX
X
X
X
NN
N
N
N
N
N
N
NNN
N
NNN
]1[
]2[
]1[
]0[
1
1
1
1111
)1(
)2(
)1(
)0(
)1)(1()1(21
)1(242
121
29
30. Computation Complexity
To calculate the DFT of N-Points discrete time signal, we need:
(N-1)2 Complex Multiplications
N(N-1) Complex Additions.
30
31. IDFT - Matrix Formulation
XW
N
x
XWx
NX
X
X
X
WWW
WWW
WWW
N
Nx
x
x
x
NN
N
N
N
N
N
N
NNN
N
NNN
*
1
)1)(1()1(2)1(
)1(242
)1(21
1
]1[
]2[
]1[
]0[
1
1
1
1111
1
)1(
)2(
)1(
)0(
31
32. Matrix Formulation
Result: Inverse DFT is given by
which follows easily by checking WHW = WWH = NI, where I denotes the
identity matrix. Hermitian transpose:
Also, “*” denotes complex conjugation.
32
33. DFT Interpretation
DFT sample X(k) specifies the magnitude and phase angle of the kth spectral
component of x[n].
The amount of power that x[n] contains at a normalized frequency, fk, can be
determined from the power density spectrum defined as
33
)(spectrumPhase
|)(|spectrumMagnitude
kX
kX
10,
|)(|
)(
2
Nk
N
kX
kSN
34. Periodicity of DFT Spectrum
The DFT spectrum is periodic with period N (which is expected, since the DTFT
spectrum is periodic as well, but with period 2π).
34
)()(N)k(
)(N)k(
)(N)k(
2
2
1
0
2
1
0
)(
2
kXekXX
eenxX
enxX
nj
nj
N
n
N
nk
j
N
n
N
nNk
j
35. Example
Example: DFT of a rectangular pulse:
the rectangular pulse is “interpreted” by the DFT as a spectral line at
frequency ω = 0. DFT and DTFT of a rectangular pulse (N=5)
35
36. Zero Padding
What happens with the DFT of this rectangular pulse if we increase N by
zero padding:
where x(0) = · · · = x(M − 1) = 1. Hence, DFT is
36
0,...,0,0,0),1(,...),1(),0()( Mxxxnx
(N-M) Positions
37. Zero Padding
DFT and DTFT of a Rectangular Pulse with Zero Padding (N = 10, M = 5)
37
38. Zero Padding
Zero padding of analyzed sequence results in “approximating” its DTFT
better,
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).
38
39. Frequency Interval/Resolution
Frequency Interval/Resolution: DFT’s frequency resolution
and covered frequency interval
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?
Answer: No, because resolution is determined by the length of observation
interval, and zero padding does not increase this length.
39
40. Example (DFT Resolution)
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/(NT) = 5 Hz.
40
41. Example (DFT Resolution)
DFT with N = 30 and zero padding to 512 points. Resolved: F2 − F1 = 2 Hz > 1/(NT)=1.7 Hz.
41
42. Example (DFT Resolution)
DFT with N=100 and zero padding to 512 points. Resolved: F2−F1 = 2 Hz > 1/(NT) = 0.5 Hz.
42
43. DSF VS DFT
DFS and DFT pairs are identical, except that
DFT is applied to finite sequence x(n),
DFS is applied to periodic sequence .
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.
43
)(~ nx