DSP_FOEHU - Lec 08 - The Discrete Fourier Transform

‫ر‬َ‫ـد‬ْ‫ق‬‫ِـ‬‫ن‬،،،‫لما‬‫اننا‬ ‫نصدق‬ْْ‫ق‬ِ‫ن‬‫ر‬َ‫د‬
LECTURE (08)
The Discrete Fourier Transform
Assist. Prof. Amr E. Mohamed

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

Representation of Periodic
Sequences --- DFS
3

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 

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






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

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
















 
 
























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

Example
9
k
k
n
kn
W
W
WkX
10
5
10
4
0
10
1
1
)(
~


 
0 1 2 3 4 5 6 7 8 9 n
)10/sin(
)2/sin()10/4(
k
k
e kj


 

Example
10
k
k
n
kn
W
W
WkX
10
5
10
4
0
10
1
1
)(
~


 
0 1 2 3 4 5 6 7 8 9 n
)10/sin(
)2/sin()10/4(
k
k
e kj


 

Example
11
k
k
n
kn
W
W
WkX
10
5
10
4
0
10
1
1
)(
~


 
0 1 2 3 4 5 6 7 8 9 n
)10/sin(
)2/sin()10/4(
k
k
e kj


 

DFS vs. FT
12
)(~ nx
0 N nN
0 n
)(nx





0
)()(
n
njj
enxeX





1
0
)(
N
n
nj
enx





1
0
)/2(
)(~)(
~ N
n
knNj
enxkX
Nk
j
eXkX
/2
)()(
~




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


 



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


 



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


 



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)(
~



 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



 
,)()( j
ez
j
zHeH
10,][
)()(
21
0
/2





 Nkenx
eXkX
N
nk
jN
n
Nk
j
k










N
kj
zk
2
exp
Re(z)
Im(z)
z0
z1
z2
z3
z4
z5
z6
z7
1
N = 8
18
Relation To The DTFT

Discrete Fourier Transform(DFT)
19

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
)(

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

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




The DFT Pair
23
N
j
N
N
k
kn
N
N
k
N
kn
j
N
n
kn
N
N
n
N
kn
j
eWwhere
NnWkX
N
x
ekX
N
xSynthesis
NkWnxX
enxXAnalysis



2
1
0
1
0
2
1
0
1
0
2
1,...,1,0)(
1
n)(
)(
1
n)(
1,...,1,0)(k)(
)(k)(





















Example
 Find the discrete Fourier Transform of the following N-points discrete
time signal
 Solution:
 On the board
24
   )1(,...),1(),0()(  Nxxxnx

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



 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)(
~



 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



 
,)()( j
ez
j
zHeH
10,][
)()(
21
0
/2





 Nkenx
eXkX
N
nk
jN
n
Nk
j
k










N
kj
zk
2
exp
Re(z)
Im(z)
z0
z1
z2
z3
z4
z5
z6
z7
1
N = 8
28
Relation To The DTFT

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

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

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

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

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

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























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

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

Zero Padding
 DFT and DTFT of a Rectangular Pulse with Zero Padding (N = 10, M = 5)
37

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

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

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

 DFT with N = 30 and zero padding to 512 points. Resolved: F2 − F1 = 2 Hz > 1/(NT)=1.7 Hz.
41

 DFT with N=100 and zero padding to 512 points. Resolved: F2−F1 = 2 Hz > 1/(NT) = 0.5 Hz.
42

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

Linear & Circular Convolution
44

Linear Convolution
 By Discrete Time Fourier Transform (DTFT)
45









-k
-k
x(k)k)-h(ny(n)
k)-x(nh(k)y(n)
x(n)*h(n)y(n)
)()()(  jjj
eXeHeY 
 )()( j
eYIDTFTny 

Circular Convolution
 Circular convolution of length N is
 By DFT
46
     
     
     








1
0k
C
1
0k
C
NC
kxk-nhny
k-nxkhny
nxnhny
N
N
N
N
)()()( kXkHkY 
 )()( kYIDFTnyC 

Convolution of two periodic sequences
47

How to Compute Circular Convolutions
 Method #1:
48

 Method #2:
 Compute the linear convolution and then alias it:
49

 Method #3:
 Compute 4-point DFTs, multiply, compute 4-point inverse DFT:
50

Using Cyclic Convs and DFTs to Compute Linear Convs:
51

DSP_FOEHU - Lec 08 - The Discrete Fourier Transform

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DSP_FOEHU - Lec 08 - The Discrete Fourier Transform