Digital Signal Processing Tutorial:Chapt 3 frequency analysis

10/7/2009
e-TECHNote from IRDC India
info@irdcindia.com 1
Chapt 03 & 04
Frequency Analysis of Signals
and System
& DFT
Digital Signal Processing
PrePrepared by
IRDC India

10/7/2009
Copyright© with Authors. All right reserved
For education purpose.
Commercialization of this material is
strictly not allowed without permission
from author.

10/7/2009
Chapt 3 : Frequency Analysis of Signals &
Systems
Contents:
• Frequency Analysis: CTS and DTS
• Properties of the Fourier Transform for DTS
• Frequency domain characteristics of LTI systems
• LTI system as a frequency selective filter
• Inverse systems and de-convolution

10/7/2009
Frequency Analysis: CTS and DTS
• Analysis tools
• Fourier Series
• Fourier Transform
• DTFS
• DTFT
• DFT
Jean Baptiste Joseph
Fourier (1768 - 1830).
Fourier was a French
mathematician, who was
taught by Lagrange and
Laplace

10/7/2009
e-TECHNote
This PPT is sponsored by
IRDC India
www.irdcindia.com

10/7/2009
Fourier Series
•Continuous time variable t
•Discrete frequency variable k
•Signal to be analyzed has to
be periodic
• Fourier series for continuous-time periodic signal
•
tTp
∑
∞
−∞=
=
k
tkFj
k eCtx 02
)( π
dtetx
T
C tkFj
Tp
k
p
02
)(
1 π−
∫=
Tp-> is the Fundamental Period of Signal
where

10/7/2009
•Continuous time variable t
•Continuous frequency variable F
•Signal to be analyzed is aperiodic
Fourier Transform
• Fourier Transform for continuous-time periodic
signal
t
∫
∞
∞−
= dFeFXtx Ftj π2
)()(
where
∫
∞
∞−
−
= dtetxFX Ftj π2
)()(

10/7/2009
• Fourier Series Periodic signal
• Fourier Transform Aperiodic signal
When Fourier series is applied to the aperiodic signal by assuming
its period = , then it is called as Fourier transform
Fourier Series & Fourier Transform
Fourier
Series
t
Tp F
Fourier
Series
Tp=∞
∆F=1/Tp
F
∆F=1/Tp=0
∞
)(lim)( txtx T
T ∞→
=

10/7/2009
•Discrete time variable t
•Discrete frequency variable k
•Signal to be analyzed has to
be periodic
DTFS
• Fourier series for discrete-time periodic signal
∑
−
=
=
1
0
2
)(
N
k
k
N
knj
eCnx
π
∑
−
=
−
=
1
0
2
)(
1 N
n
k
N
knj
enx
N
C
π
where

10/7/2009
•Discrete time variable t
•Continuous frequency variable F
•Signal to be analyzed is aperiodic
DTFT
• Fourier Transform for discrete-time periodic
signal
∫=
π
ω
ωω
π 2
)(
2
1
)( deXnx nj
Tp is the Fundamental Period of Signal
where
∑
∞
−∞=
−
=
n
jwn
enxX )()(ω

10/7/2009
Dirichlet Conditions
• To exist Fourier transform of the signal, it should satisfy
Dirichlet Conditions
1. The signal has finite number of discontinuities
2. It has finite number of maxima and minima
3. The signal should be absolutely integral
i.e.
∞<∫
∞
∞−
dttx )(
∑
∞
−∞=
∞<
n
dtnx )(
or

10/7/2009
Problem(1)
• Obtain exponential Fourier series for the waveform shown in fig
•
0 2Π 4Π ωt

10/7/2009
Solution (1)
dtetx
T
C tjk
T
k
0
0
)(
1
0
ω−
∫=
As per Fourier series
π20 =T ttx 0
2
5
)( ω
π
= for π20 ≤≤ wt
tdwet
T
C tjkw
k 0
2
0
02
5
0
0
1 −
∫=
π
π ω tdwteC tjkw
k 0
2
0
0
0
2
5
2
1
∫
−
=
π
ω
ππ
( )
π
π
2
0
022
1
)()2(
5 0






−−
−
=
−
tjkw
jk
e tjkw
k
jCk
π2
5
=
LLL ++++−−= −− twjtwjtjwtwj
ejejejejtx 0000 2
4
52
2
5
2
5
2
52
4
5
)( ππππ
0

10/7/2009
Problem (2)
Obtain Fourier transform for the gate function ( rectangular pulse)
as shown in fig
0-T/2 T/2 t
,1)( =tf 22
TT
t ≤≤−
=0 otherwise

10/7/2009
Solution (2)
We know
∫
∞
∞−
−
= dtetfwF jwt
)()(
∫−
−
=
2/
2/
.1
T
T
jwt
dte
2/
2/
T
T
jwt
jw
e
−
−
−
= [ ]2/2/1 jwTjwT
ee
jw
−
−
= −
jw
ee jwTjwT 2/2/
−
=
−
w
wT )2/sin(2
=
)2/(sin
2/
)2/sin(
wTcT
wt
wT
T ==
Amplitude
Phase

10/7/2009
Problem (3)
A finite duration sequence of Length L is given as
Determine the N-point DFT of this sequence for N≥L
,1)( =nx 10 −≤≤ Ln
=0 otherwise

10/7/2009
Solution (3)
The Fourier transform (DTFT) is given by
∑
∞
−∞=
−
=
n
jwn
enxX )()(ω
∑
−
=
−
=
1
0
.1
L
n
jwn
e ∑
−
=
−
=
1
0
)(
L
n
njw
e
We know summation formula
∑=
+
−
−
=
n
k
n
k
a
a
a
0
1
1
1
jw
jwL
e
e
X −
−
−
−
=
1
1
)(ω 2/2/2/2/
2/2/2/2/
jwjwjwjw
jwLjwLjwLjwL
eeee
eeee
−−
−−
−
−
=
)(
)(
2/2/2/
2/2/2/
jwjwjw
jwLjwLjwL
eee
eee
−−
−−
−
−
=
)2/sin(
)2/sin(2/)1(
w
wL
e Ljw −−
=

10/7/2009
Contd..
DFT of x(n) can be obtained by sampling X(w) at N
equally spaced frequencies
N
k
wk
π2
=∴ 1,.......1,0 −= Nk
Nkj
NkLj
e
e
kX /2
/2
1
1
)( π
π
−
−
−
−
= 1,.......1,0 −= Nk
,)( LkX = 0=KIf N=L,
0= 1.......2,1 −= LK
•If N>L, computational point of view sequence x(n) is extended by
appending N-L zeroes(zero padding).
• DFT would approach to DTFT as no of points in DFT tends to infinity

10/7/2009
Nyquist Criterion – Sampling Theorem
• Continuous time domain( analog ) signal is converted into discrete time
signal by sampling process. It should be sampled in such a way that the original
signal can be reconstructed from the samples
• Nyquist criterion or sampling theorem suggest the minimum sampling
frequency by which signal should be sampled in order to have proper reconstruction
if required.
Mathematically, sampled signal xs(t) is obtained by multiplying sampling
function gT(t) with original signal x(t)
)1.......().........()()( tgtxtx Ts =
Where gT(t) continuous train of pulse with period T (sampling period)
)(
1
)(
Triodsamplingpe
fequencysamplingfr s =

10/7/2009
Contd..
The Fourier series of periodic signal gT(t) can be given as
dtetg
T
C tnfj
T
T
n
sπ2
2/
2/
)(
1 −
−
∫=
)2........(..........)( 2
∑
∞
−∞=
=
n
tnfj
n
s
eCtg π
Where
gT(t)
x(t)
t
t
Thus, from eq(1) and eq(2), we get
∑
∞
−∞=
=
n
tnfj
ns
s
eCtxtx π2
)()(
)3.........(..........)( 2
∑
∞
−∞=
=
n
tnfj
n
s
etxC π

10/7/2009
Contd..
Taking Fourier transform of xs(t), we get
∫
∞
∞−
−
= dtetxfX ftj
ss
π2
)()(
∫ ∑
∞
∞−
−
∞
−∞=
= dteetxC ftj
n
tnfj
n
s ππ 22
)(
Interchanging the order of integration and summation , we get
∑ ∫
∞
−∞=
−−
∞
∞−
=
n
tnffj
n dtetxC s )(2
)( π
∑
∞
−∞=
−=
n
sns nffXCfX )()(
By definition

10/7/2009
Contd..
Thus, spectrum of sampled signal is the spectrum of x(t) plus the spectrum of
x(t) translated to each harmonic fo the sampling frequency as shown in figure
X(f)
f
-fh fh
-fh fh fs-fh fs+fhfs 2fs-fh 2fs+fh2fs
-2fs-fh -2fs+fh-2fs -fs-fh -fs
-fs+fh
There are three cases when sampling frequency fs is compared with
highest frequency present in original signal
Xs(f)

10/7/2009
Contd..
Case I: fs > 2fh
In this case all spectrums would be isolated from each other . This
leads to proper reconstruction of original signal
Case II: fs = 2fh
-fh fh fs-fh fs+fhfs 2fs-fh 2fs+fh2fs
-2fs-fh -2fs+fh-2fs -fs-fh -fs
-fs+fh
Xs(f)
Case III: fs < 2fh
-fh fh fs 2fs
-2fs -fs
Xs(f)
Overlapping of spectrum causes aliasing effect which distorts in
reconstruction of original signal

N-points in time
domain will give
N-points in DFT
and vice-versa
10/7/2009
DFT
DTFT has continuous frequency domain variable which makes
impossible to store X(w) with digital device. Thus continuous
frequency variable (f or w) from DTFT is sampled to get
discrete frequencies (wk).
DTFT calculated for discrete frequencies (wk) is called as
Discrete Fourier Transform
N
k
wk
π2
=
∑
−
=
−
=
1
0
/2
)()(
N
n
Nknj
enxkX π
1,.......1,0 −= Nk
∑
−
=
=
1
0
/2
)(
1
)(
N
k
Nknj
ekX
N
nx π
1,.......1,0 −= Nn
DFT
Inverse DFT ( IDFT)

10/7/2009
Physical Significance
• Signal to be analyzed is 64 points in length
x(n)
analog discrete
0 10 20 30 40 50 60 70
-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
0 10 20 30 40 50 60 70
-0 .5
-0 .4
-0 .3
-0 .2
-0 .1
0
0 .1
0 .2
0 .3
0 .4
0 .5

10/7/2009
k=0
)/2sin()/2cos(
2
NnkjNnke N
nk
j
ππ
π
+=
−
0 10 20 30 40 50 60 70
-1
-0.5
0
0.5
1
SINE SEQUENCE
Amplitude
n
0 10 20 30 40 50 60 70
0
0.5
1
COSINE SEQUENCE
Amplitude
n
0 10 20 30 40 50 60 70
-0. 5
-0. 4
-0. 3
-0. 2
-0. 1
0
0. 1
0. 2
0. 3
0. 4
0. 5
Im{X(0)}
Re{X(0)}

10/7/2009
k=1
)/2sin()/2cos(
2
NnkjNnke N
nk
j
ππ
π
+=
−
0 10 20 30 40 50 60 70
-1
-0.5
0
0.5
1
SINE SEQUENCE
Amplitude
n
0 10 20 30 40 50 60 70
-1
-0.5
0
0.5
1
COSINE SEQUENCE
Amplitude
n
0 10 20 30 40 50 60 70
-0. 5
-0. 4
-0. 3
-0. 2
-0. 1
0
0. 1
0. 2
0. 3
0. 4
0. 5
Im{X(1)}
Re{X(1)}

10/7/2009
k=2
)/2sin()/2cos(
2
NnkjNnke N
nk
j
ππ
π
+=
−
0 10 20 30 40 50 60 70
-1
-0.5
0
0.5
1
SINE SEQUENCE
Amplitude
n
0 10 20 30 40 50 60 70
-1
-0.5
0
0.5
1
COSINE SEQUENCE
Amplitude
n
0 10 20 30 40 50 60 70
-0. 5
-0. 4
-0. 3
-0. 2
-0. 1
0
0. 1
0. 2
0. 3
0. 4
0. 5
Im{X(2)}
Re{X(2)}

10/7/2009
k=3
)/2sin()/2cos(
2
NnkjNnke N
nk
j
ππ
π
+=
−
0 10 20 30 40 50 60 70
-1
-0.5
0
0.5
1
SINE SEQUENCE
Amplitude
n
0 10 20 30 40 50 60 70
-1
-0.5
0
0.5
1
COSINE SEQUENCE
Amplitude
n
0 10 20 30 40 50 60 70
-0. 5
-0. 4
-0. 3
-0. 2
-0. 1
0
0. 1
0. 2
0. 3
0. 4
0. 5
Im{X(3)}
Re{X(3)}

10/7/2009
k=31
)/2sin()/2cos(
2
NnkjNnke N
nk
j
ππ
π
+=
−
0 10 20 30 40 50 60 70
-1
-0.5
0
0.5
1
SINE SEQUENCE
Amplitude
n
0 10 20 30 40 50 60 70
-1
-0.5
0
0.5
1
COSINE SEQUENCE
Amplitude
n
0 10 20 30 40 50 60 70
-1
-0.5
0
0.5
1
SINE SEQUENCE
Amplitude
n
0 10 20 30 40 50 60 70
-1
-0.5
0
0.5
1
COSINE SEQUENCE
Amplitude
n

10/7/2009
Relation bet VA and DFT
Vector algebra -2/3D space DFT
No of analysis vectors 2/3 N
No of elements in each
vector
2/3 N
Projection measurement
method
Inner Product Inner Product

10/7/2009
Fourier’s Idea in 1807

10/7/2009
2D Fourier Transform
2D FT basis images for 8x8

10/7/2009
Problem(1)
Find the 6-point DFT of the sequence x(n)= cos(nπ/6)

10/7/2009
Solution(1)
N=6 )]6/5cos(),3/2cos(),2/cos(),3/cos(),6/cos(),0[cos()( πππππ=∴ nx
]86.05.005.086.01[ −−=
We know
∑
−
=
−
=
1
0
/2
)()(
N
n
Nknj
enxkX π
1,.......1,0 −= Nk
For k=0
∑∑ ==
==
5
0
5
0
0
)()()0(
nn
nxenxX
)86.0()5.0(05.086.01 −+−++++=
1=
For k=1
∑=
−
=
5
0
3/
)()1(
n
nj
enxX π
3/53/4
3/23/0
)5()4()3(
)2()1()0(
πππ
ππ
jjj
jj
exexex
exexex
−−−
−−
+++
++=

10/7/2009
Contd..
)86.05.0)(86.0()86.05.0)(5.0()1(
)86.05.0(5.0)86.05.0(86.01
jj
jj
−−+−−−+−+
+−+++=
34.21)1( jX +=
Similarly, calculate X(2)---X(5)

10/7/2009
Problem(2)
Calculate DFT of x(n)= {1 2 2 1}. From calculated DFT of x(n),
determine x(n) again using IDFT. Verify your answer.

10/7/2009
Solution(2)
}2221{)( =nx
We know
∑
−
=
−
=
1
0
/2
)()(
N
n
Nknj
enxkX π
1,.......1,0 −= Nk
For k=0
∑∑ ==
==
3
0
3
0
0
)()()0(
nn
nxenxX
61221 =+++=
For k=1
∑=
−
=
3
0
4/
)()1(
n
nj
enxX π
4/3
2/4/0
)3(
)2()1()0(
π
ππ
j
jj
ex
exexex
−
−−
+
++=
N=4

10/7/2009
Contd..
2/3
2/0
)3(
)2()1()0(
π
ππ
j
jj
ex
exexex
−
−−
+
++=
jj .1)1.(2).(21 +−+−+=
jX −−= 1)1(
Similarly, calculate X(2) & X(3)
Then use IDFT to calculate x(n)

10/7/2009
Twiddle Factors
Twiddle Factor N
j
N eW
π2
=
( ) 1
00 2
==∴ N
j
N eW
π
( ) NN
jj
N eeW
ππ 22 11
==∴
( ) NN
jj
N eeW
ππ 42 22
==∴
So on…….
For N=4
For N=8
W4
0
W4
3
W4
2
W4
1
W8
0
W8
7
W8
6
W8
5
W8
4
W8
3
W8
2
W8
1
( ) N
l
N
jljl
N eeW
ππ 22
==

10/7/2009
Contd..
1
4
)5(4mod
4
5
4 WWW ==
Proof:
( ) 4
10
4
2 55
4
ππ jj
eeW ==∴
πππ 2
1
2
1
.2)2( jjj
eee ==
+
4
1.2
4
2
.1
ππ jj
ee ==
1
4W= N
ljl
N eW
π2
=Q

10/7/2009
Contd..
3
4
6
8 WW =
Proof:
( ) 8
12
8
2 66
8
ππ jj
eeW ==∴ 4
3.2
4
6 ππ jj
ee ==
3
4W= N
ljl
N eW
π2
=Q
2
8
6
8 WW −=
Proof:
( ) 8
12
8
2 66
8
ππ jj
eeW ==∴ 8
4
8
4 )1( π
ππ jjj
eee ==
+
2
8)1( W−= N
ljl
N eW
π2
=Q
2
8W−=

10/7/2009
Matrix Method for DFT calculation
∑
−
=
−
=
1
0
/2
)()(
N
n
Nknj
enxkX π
1,.......1,0 −= Nk
∑
−
=
=∴
1
0
)()(
N
n
kn
NWnxkX 1,.......1,0 −= Nk
We know
N
j
N eW
π2−
=and
In matrix form ]][[][ xWX kn
N=


























=












)3(
)2(
)1(
)0(
)3(
)2(
)1(
)0(
9
4
6
4
3
4
0
4
6
4
4
4
2
4
0
4
3
4
2
4
1
4
0
4
0
4
0
4
0
4
0
4
x
x
x
x
WWWW
WWWW
WWWW
WWWW
X
X
X
X4-point DFT

10/7/2009
Contd..


























=












)3(
)2(
)1(
)0(
)3(
)2(
)1(
)0(
1
4
2
4
3
4
0
4
2
4
0
4
2
4
0
4
3
4
2
4
1
4
0
4
0
4
0
4
0
4
0
4
x
x
x
x
WWWW
WWWW
WWWW
WWWW
X
X
X
X


























−−
−−
−−
=












)3(
)2(
)1(
)0(
)3(
)2(
)1(
)0(
1
4
0
4
1
4
0
4
0
4
0
4
0
4
0
4
1
4
0
4
1
4
0
4
0
4
0
4
0
4
0
4
x
x
x
x
WWWW
WWWW
WWWW
WWWW
X
X
X
X

10/7/2009
Problems
Calculate DFT of x(n)= {1 2 2 1} & x(n)={ 0 1 2 3 } by using matrix
method.

10/7/2009
Solution(1)
x(n)= {1 2 2 1}


























=












)3(
)2(
)1(
)0(
)3(
)2(
)1(
)0(
1
4
2
4
3
4
0
4
2
4
0
4
2
4
0
4
3
4
2
4
1
4
0
4
0
4
0
4
0
4
0
4
x
x
x
x
WWWW
WWWW
WWWW
WWWW
X
X
X
X
























−−
−−
−−
=












1
2
2
1
11
1111
11
1111
)3(
)2(
)1(
)0(
jj
jj
X
X
X
X
10
4 =W12
4 −=W
jW −=1
4
jW =3
4
jXXjXX +−==−−== 1)3(,0)2(,1)1(,6)0(

10/7/2009
Solution(2)
x(n)={ 0 1 2 3 }


























=












)3(
)2(
)1(
)0(
)3(
)2(
)1(
)0(
1
4
2
4
3
4
0
4
2
4
0
4
2
4
0
4
3
4
2
4
1
4
0
4
0
4
0
4
0
4
0
4
x
x
x
x
WWWW
WWWW
WWWW
WWWW
X
X
X
X
























−−
−−
−−
=












3
2
1
0
11
1111
11
1111
)3(
)2(
)1(
)0(
jj
jj
X
X
X
X
10
4 =W12
4 −=W
jW −=1
4
jW =3
4
jXXjXX 22)3(,2)2(,22)1(,6)0( −−=−=+−==

10/7/2009
DFT using FFT flow diagram
x(0)
x(1)
0
2W
1
2W
X(0)
X(1)
2-point DFT
x(0)
x(1)
1
1−
X(0)
X(1)
0
2W1
2W
Looks
like
Butterfly
?
x(0)
x(1)
1−
X(0)
X(1)

2
4W
10/7/2009
Contd..
x(0)
x(2)
0
4W
X(0)
X(1)
4-point DFT
0
4W2
4W
1
4W
3
4W
x(1)
x(3)
X(2)
X(3)
0
4W
1
4W
2
4W
3
4W
0
4W
2
4W

10/7/2009
Contd..
x(0)
x(2)
X(0)
X(1)
4-point DFT
10
4 =W12
4 −=W
jW −=1
4
jW =3
4
x(1)
x(3)
X(2)
X(3)
1
j−
1−
j1−
1−

10/7/2009
Problems
Calculate DFT of x(n)= {1 2 2 1} & x(n)={ 0 1 2 3 } by FFT flow diagram

10/7/2009
Solution (1)
x(0)=1
x(2)=2
X(0)=6
X(1)=-1+j
x(1)=2
x(3)=1
X(2)=0
X(3)=-1+j
1
j−
1−
j1−
1−
x(n)= {1 2 2 1}
1+2=3
1-2=-1
2+1=3
2-1=1
3+3=6
-1-j1=-1-j
3-3=0
-1+j1=-1+j
jXXjXX +−==−−== 1)3(,0)2(,1)1(,6)0(

10/7/2009
Solution(2)
x(0)=0
x(2)=2
X(0)=6
X(1)=-2+j2
x(1)=1
x(3)=3
X(2)=-2
X(3)=-2-j2
1
j−
1−
j1−
1−
x(n)= {0 1 2 3 }
0+2=2
0-2=-2
1+3=4
1-3=-2
2+4=6
-2+j2
2-4=-2
-2-j2
jXXjXX 22)3(,2)2(,22)1(,6)0( −−=−=+−==

10/7/2009
Shuffled sequence
The input sequence placement at FFT flow graph is not in order but in
shuffled order.
Perfect shuffling can be obtained by using bit reversal algorithm.
( to be used in FFT implementation)
2 point FFT
0
1
4 point FFT
0
1
2
3
0 2
1 3
0 2 1 3
8 point FFT
0
1
2
3
4
5
6
7
0 4
1 5
2 6
3 7
0 4 2 6
1 5 3 7
0 1 0 4 2 6 1 5 3 7
16-point shuffled sequence ?????

10/7/2009
Bit Reversal Algorithm
2 point FFT 4 point FFT 8 point FFT
0
1
0
1
2
3
00
01
10
11
00
10
01
11
0
2
1
3
0
1
2
3
4
5
6
7
000
001
010
011
100
101
110
111
000
100
010
110
001
101
011
111
0
4
2
6
1
5
3
7
Bit Reversal
Algorithm
Bit Reversal
Algorithm

10/7/2009
8-point FFT flow diagram
4
8W
x(0)
x(4)
0
8W
X(0)
X(1)
x(2)
x(6)
X(2)
X(3)
0
8W
2
8W
4
8W
6
8W
0
8W
4
8W
4
8W
x(1)
x(5)
0
8W
X(4)
X(5)
x(3)
x(7)
X(6)
X(7)
0
8W
2
8W
4
8W
6
8W
0
8W
4
8W
0
8W
1
8W
2
8W
3
8W
4
8W
5
8W
6
8W
7
8W

10/7/2009
Contd..
For N=8
W8
7
W8
6
W8
5
W8
4
W8
3
W8
2
W8
1
W8
0
1=
707.0707.0 j−==−− 707.0707.0 j
j−=
j=
=+− 707.0707.0 j 707.0707.0 j+=
=−1

10/7/2009
Problem
Calculate DFT of x(n)= {1 2 2 1 0 1 2 3 } by FFT flow diagram

10/7/2009
x(n)= {1 2 2 1 0 1 2 3 }
1−
x(0)=1
x(4)=0
x(2)=2
x(6)=2
x(1)=2
x(5)=1
x(3)=1
x(7)=3
1−
1−
1−
1
j−
j
1−
1
j−
j
1−
1
707.0707.0 j−
707.0707.0 j−−
j−
j
707.0707.0 j+−
707.0707.0 j+
1−
1+0=1
1-0=1
2+2=4
2-2=0
2+1=3
2-1=1
1+3=4
1-3=-2
1+4=5
1-0=1
1-4=-3
1+0=1
3+4=7
1+j2
3-4=-1
1-j2
X(0)=5+7=12
X(2)=-3+j
X(4)=5-7=-2
X(6)=-3-j
707.0121.1707.0121.21
)21)(707.0707.0(1)3(
jj
jjX
+−=+−=
−−−+=
707.0121.1707.0121.21
)21)(707.0707.0(1)5(
jj
jjX
−−=−−=
++−+=
707.0121.3707.0121.21
)21)(707.0707.0(1)7(
jj
jjX
−=−+=
−++=
707.0121.3707.0121.21
)21)(707.0707.0(1)1(
jj
jjX
+=++=
+−+=

10/7/2009
Properties of DFT
1) Periodicity
and x[n] is periodic such that x[n+N]=x[n] for all n
If x[n] X(k) ,
Then, X[k+N] = X(k) for all k
i.e. DFT of periodic sequence is also periodic with same period

10/7/2009
2) Linearity
][][
][][
22
11
kXnx
kXnx
DFT
DFT
 →←
 →←If
and
then for any real-valued or complex valued constants a1 and a2 ,
][][][][ 22112211 kXakXanxanxa DFT
+ →←+

10/7/2009
Problem
a) Find the DFT of x1(n)={ 0 1 2 3} and x2(n)= { 1 2 2 1}.
b) Calculate the DFT of x3(n)={ 2 5 6 5} using results obtained
in a) otherwise not.
}1,0,1,6{)(2 jjkX +−−−=
}22,2,22,6{)(1 jjkX −−−+−=
Since x1(n) + 2x2(n)= x3(n) , X3(k)= X1(k) + 2X2(k)
}1,0,1,6{2}22,2,22,6{)(3 jjjjkX +−−−+−−−+−=
}4,2,4,18{)(3 −−−=kX

10/7/2009
3) Circular time shift, x((n-k))N
x(n)={1 2 3 4 }
xp(n)
xp(n)
Circular delay by
1,x((n-1))
Circular advance
by 1 ,x((n+1))
x(0)=1x(2)=3
x(1)=2
x(3)=4
x(1)=2x(3)=4
x(2)=3
x(0)=1
x(n)
x((n+1))4
x(3)=4x(1)=2
x(0)=3
x(2)=3
x((n-1))4

10/7/2009
Contd..
x(2)=3
x(0)=1
x(3)=4
x(1)=2
x((n-2))4
x(2)=3
x(0)=1
x(3)=4
X(1)=2
x((n+2))4 x(3)=4
x(1)=2
x(0)=3
x(2)=1
x((n+3))4
x(1)=2
x(3)=4
x(2)=3
x(0)=1
x((n-3))4

10/7/2009
Contd..
Circular shift property
N
kljDFT
N
DFT
ekXlnx
kXnx
π2
].[))((
][][
−
 →←−
 →←If
then

10/7/2009
Problem
If x(n)= { 1 2 3 4} ,find X(k). Also using this result find the DFT of
h(n)={ 3 4 1 2}.

10/7/2009
Solution
N
kljDFT
N ekXlnx
π2
].[))((
−
 →←−
x(n)= { 1 2 3 4} }22,2,22,10{)( jjkX −−−+−=
Circular shift DFT property
l = 2 4
22
].[))2(()( 4
kjDFT
ekXnxnh
π−
 →←−=
x(n) = h((n-2))4h(n)={ 3 4 1 2}
}1111{)( 4
22
−−==
− kj
ekc
π
}22,2,22,10{)( jjkH +−−=

10/7/2009
4) Time Reversal
)())(()())((
][][
kNXkXnNxnx
kXnx
N
DFT
N
DFT
−=− →←−=−
 →←If
then
i.e. reversing the N-point sequence in time domain is
equivalent to reversing the DFT sequence

10/7/2009
Problem
h(n)={ 1 4 3 2}. Verify your answer with DFT calculation

10/7/2009
5) Circular frequency shift
N
DFTj
DFT
lkXenx
kXnx
N
nl
))(()(
][][
2
− →←
 →←
π
If
then

10/7/2009
Problem
h(n)={ 1 -2 3 -4}. Verify your answer with DFT calculation

10/7/2009
6) Symmetry properties
)()()()()( njxnjxnxnxnx o
I
e
I
o
R
e
R +++=
)()()()()( kjXkjXkXkXkX o
I
e
I
o
R
e
R +++=
EvenalEvenal DFT
,Re,Re  →←
EvenaginaryEvenaginary DFT
,Im,Im  →←
OddaginaryOddal DFT
,Im,Re  →←
OddalOddaginary DFT
,Re,Im  →←

10/7/2009
Problems
}131312{)() =nxa
Find DFT for followings
}12313212{)() =nxb
}12303210{)() −−−=nxc
}234432{)() jjjjjjjjnxd =
}234432{)() jjjjjjjjnxe −−−=
}1151111{)() −−−−=kXa
}8284.118284.318284.318284.115{)() −−−−−=kXb
}65.9465.1065.1465.90{)() jjjjjjkXc −−−=
}828.33828.13828.1382.321{)() jjjjjjjjkXd −−−−−=

10/7/2009
7)Parseval’s Relation
∑∑
−
=
−
=
=
1
0
2
1
0
2
|)(|
1
|)(|
N
k
N
n
kX
N
nx

10/7/2009
Problem
2) Determine the missing value from following sequence
x(n)= { 1 3 _ 2 } if its DFT is X(k)={ 8 -1-j -2 -1+j}
1) x(n)= {1 2 3 1}. Prove parseval’s relation for this
sequence and its DFT .

10/7/2009
8) Circular Convolution
][][
][][
22
11
kXnx
kXnx
DFT
DFT
 →←
 →←If
and
then
][].[][][ 2121 kXkXnxnx DFT
 →←⊗
where
∑
−
=
−==⊗
1
0
21321 ))(()(][][][
N
k
Nknxkxnxnxnx

10/7/2009
Circular convolution of two sequences
Circular convolution of sequences x1(n)={2 1 2 1} and x2(n)={1 2 3 4 }
x(0)=2x(2)=2
x(1)=1
x(3)=1
x1(n)
x(0)=1x(2)=3
x(1)=2
x(3)=4
x2(n)
x2(0)=1x2(2)=3
x2(3)=4
x2(1)=2
x2((-n)) 26
4
2
x1(n)x2((-n))
x3(0)=2+4+6+2
= 14

10/7/2009
Contd..
x2(1)=2x2(3)=4
x2(0)=1
x2(2)=3
x2((1-n))
48
1
3
x1(n)x2((1-n))
x3(1)=4+1+8+3
= 16
62
2
4
x1(n)x2((2-n))
x3(2)=6+2+2+4
= 14
84
3
1
x1(n)x2((3-n))
x3(3)=8+3+4+1
= 16
x2(2)=3x2(0)=1
x2(1)=2
x2(3)=4
x2((2-n))
x2(3)=4x2(1)=2
x2(2)=3
x2(0)=1
x2((3-n))
16)3(,14)2(,16)1(,14)0( 3333 ==== xxxx

10/7/2009
Problem
Find the circular convolution of sequences x1(n)={2 1 2 1} and
x2(n)={1 2 3 4 } using DFT .
Steps:
1. Find DFT of both sequences
2. Multiply both DFT’s
3. Take inverse DFT of product
( DFT product IDFT )

10/7/2009
Contd..
x1(0)=2
x1(2)=2
X(0)=6
X(1)=0
x1(1)=1
x1(3)=1
X(2)=2
X(3)=0
1
j−
1−
j1−
1−
x1(n)= {2 1 2 1}
2+2=4
2-2=0
1+1=2
1-1=0
4+2=6
0-0=0
4-2=2
0+0=0
0)3(,2)2(,0)1(,6)0( 1111 ==== XXXX

10/7/2009
Contd..
x2(0)=1
x2(2)=3
X(0)=10
X(1)=-2+j2
x2(1)=2
x2(3)=4
X(2)=-2
X(3)=-2-j2
1
j−
1−
j1−
1−
x2(n)= {1 2 3 4}
1+3=4
1-3=-2
2+4=6
2-4=-2
4+6=10
-2+j2
4-6=-2
-2-j2
22)3(,2)2(,22)1(,10)0( 2222 jXXjXX −−=−=+−==

10/7/2009
Contd..
X3(0)=60
X3(2)=-4
x3(0)=14
x3(1)=16
X3(1)=0
X3(3)=0
x3(2)=14
x3(3)=16
1
j
1−
j−1−
1−
X3(k)= X1(k).X2(k)= {60 0 -4 0}=
60-4=56
60+4=64
0+0=0
0-0=0
56+0=56
64-0=64
56-0=56
64+0=64
16)3(,14)2(,16)1(,14)0( 3333 ==== xxxx
IDFT( Twiddle factors in reverse
order and divide by N at the end)
1
--
4

10/7/2009
Linear convolution using Circular convolution
Steps:
1. Append zeros to both sequence such that both sequence
will have same length of N1+N2-1
2. Number of zeros to be appended in each sequence is
addition of length of both sequences minus (one plus
length of respective sequence)
3. As both sequences are of same length equal to the length
of linear convoluted signal, apply circular convolution to
both modified signals using DFT ( DFT product IDFT )

10/7/2009
Problem
Calculate the linear convolution of following signals by using DFT
method only. x(n)={ 1 2 3 2 -2} and h(n)={ 1 2 2 1}
Verify your answer with tabulation method.

10/7/2009
Linear Filtering
If the system has frequency response H(w) and input
signal spectrum is X(w)
H(w)
X(w) Y(w)
Then , out spectrum of the system given by
Y(w)=H(w).X(w)
•In application, convolution would be used to calculate the
output of the system or DFT would be used to calculate the
spectrum of output.
•Even, DFT can be used for convolution

10/7/2009
Contd..
•Instead of taking DFT/convolution for whole input sequence,
DFT/convolution can be applied to smaller blocks of input
sequence. This would yields two advantages
•DFT/convolution size would be smaller and hence
computational complexity
• In online filtering delay can be kept small as only small
number of points will required to store in buffer for
DFT/convolution calculations
•If the input length is very large as compared to impulse
response of the system, then computational complexity of
the DFT/convolution would be more.
•There are two methods to do linear filtering by braking up input
sequence into smaller blocks
• Overlap add method
• Overlap save method

10/7/2009
Overlap-add Method
•In the overlap-add method the input x(n) is broken up into consecutive
non-overlapping blocks xi(n)
•The output yi(n) for each input xi(n) is computed separately by convolving
(non-cyclic/linear) xi(n) with h(n).
•The output blocks yi(n) would be lager than corresponding input blocks
xi(n)
•Hence, the each output block yi(n) will be overlapped with next and
previous blocks to get y(n)

10/7/2009
Contd..
h(n)xi(n)
yi(n)
x(n)
y(n)
Adding overlapped points
Overlapping length=M-1
Length of each block= N
Length of impulse
response = M
Length of
each block=
N+M-1

10/7/2009
Problem
An FIR digital filter has the unit impulse response sequence h(n)={ 2 2 1}.
Determine the output sequence in response to the input sequence x(n)= {3 0
-2 0 2 1 0 -2 -1 0}

10/7/2009
Solution
x(n)= {3 0 -2 0 2 1 0 -2 -1 0} h(n)={ 2 2 1}
x1(n)= {3 0}
x2(n)= {-2 0}
x3(n)= {2 1}
x4(n)= {0 -2}
x5(n)= {-1 0}
x’1(n)= {3 0 0 0}
x’2(n)= {-2 0 0 0}
x’3(n)= {2 1 0 0}
x’4(n)= {0 -2 0 0}
x'5(n)= {-1 0 0 0}
Length of convolution between each block and
impulse response would be 4.
h’(n)={ 2 2 1 0}
X’1(k)= {3 3 3 3}
X’2(k)= {-2 -2 -2 -2}
X’3(k)= {3 2-j 1 2+j}
X’4(k)= {-2 j2 2 -j2}
X'5(k)= {-1 -1 -1 -1}
H’(k)={ 5 1-j2 1 1+j2}h(n)={ 2 2 1}

10/7/2009
Contd..
Y1(k)=X’1(k)H’(k) = {15 3-j6 3 3+j6}
Y2(k)=X’2(k)H’(k) = {-10 -2+j4 -2 -2-j4}
Y3(k)=X’3(k)H’(k) = { 15 -j5 1 j5}
Y4(k)=X’4(k)H’(k) = {-10 4+j2 2 4-j2}
Y5(k)=X’5(k)H’(k) = {-5 -1+j2 -1 -1-j2}
y1(n)= { 6 6 3 0}
y2(n)= {-4 -4 -2 0}
y3(n)= { 4 6 4 1}
y4(n)= {0 -4 -4 -2}
y5(n)= { -2 -2 -1 0}
Overlapping
length= M-1
=3-1=2
y(n)= {6 6 -1 -4 2 6 4 -3 -6 -4 -1 0}
y1(n)= { 6 6 3 0}
y2(n)= {-4 -4 -2 0}
y3(n)= { 4 6 4 1}
y4(n)= {0 -4 -4 -2}
y5(n)= { -2 -2 -1 0}

10/7/2009
Problem
Solve previous problem with 8-point DFT

10/7/2009
Overlap-save Method
h(n)xi(n)
yi(n)
x(n)
y(n)
Discard overlapped points
Length of each block= N
Length of impulse
response = M
Length of
each block=
N+M-1
M-1 zeros

10/7/2009
Problem
An FIR digital filter has the unit impulse response sequence h(n)={ 2 2 1}.
Determine the output sequence in response to the input sequence x(n)= {3 0
-2 0 2 1 0 -2 -1 0}

10/7/2009
Solution
x(n)= {3 0 -2 0 2 1 0 -2 -1 0} h(n)={ 2 2 1}
x1(n)= {3 0 -2 0}
x2(n)= {2 1 0 -2}
x3(n)= { -1 0 0 0}
x’1(n)= {0 0 3 0 -2 0}
x’2(n)= {-2 0 2 1 0 -2}
x’3(n)= {0 -2 -1 0 0 0}
Length of convolution between modified each block
and impulse response would be 8.
h(n)={ 2 2 1}
y’1(n)= {0 0 6 6 -1 -4 -2 0}
y’2(n)= {-4 -4 2 6 4 -3 -4 -2}
y’3(n)= {0 -4 -6 -4 -1 0 0 0}
y(n) = { 6 6 -1 -4 2 6 4 -3 -6 -4 -1 0 }

10/7/2009
End of Chapter 03
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