UNIT - 2: Properties of Discrete Fourier Transforms(DFT)[?, ?, ?, ?]
Dr. Manjunatha. [email protected]
ProfessorDept. of ECE
J.N.N. College of Engineering, Shimoga
September 14, 2014
DSP Syllabus Introduction
Digital Signal Processing: Introduction [?, ?, ?, ?]
Slides are prepared to use in class room purpose, may be used as areference material
All the slides are prepared based on the reference material
Most of the figures/content used in this material are redrawn, someof the figures/pictures are downloaded from the Internet.
This material is not for commercial purpose.
This material is prepared based on Digital Signal Processing forECE/TCE course as per Visvesvaraya Technological University (VTU)syllabus (Karnataka State, India).
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 2 / 49
DSP Syllabus
DSP Syllabus
PART - A
UNIT - 2: Properties of Discrete Fourier Transforms (DFT)
Properties of DFT.
Multiplication of two DFTs- the circular convolution.
Additional DFT properties
Use of DFT in linear filtering,
Overlap-save and overlap-add method. 7 Hours
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 3 / 49
Properties of Discrete Fourier Transform (DFT)
Properties of Discrete Fourier Transform (DFT)
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 4 / 49
Properties of Discrete Fourier Transform (DFT)
Periodicity
x(n)DFT↔ X (k)
ifx(n + N) = x(n)
ThenX (k + N) = X (k)
X (k) =
N−1∑n=0
x(n)e−j 2πN
kn
X (k + N) =
N−1∑n=0
x(n)e−j 2πN
(k+N)n =
N−1∑n=0
x(n)e−j2πne−j 2πN
kn
X (k + N) =
N−1∑n=0
x(n)e−j 2πN
kn = X (k) since e−j2πn = 1
Hence we note that X(k+N) = X(k)
This tells us that DFT is periodic with period N. This is known as the cyclical property ofDFT.
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 5 / 49
Properties of Discrete Fourier Transform (DFT)
Linearity
x1(n)DFT↔ X1(k)
x2(n)DFT↔ X2(k)
ax1(n) + bx2(n)DFT↔ aX1(k) + bX2(k)
Proof
X (k) =
N−1∑n=0
ax1(n) + bx2(n)W knN
=
N−1∑n=0
ax1(n)W knN +
N−1∑n=0
bx2(n)W knN
= aX1(k) + bX2(k)
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 6 / 49
Properties of Discrete Fourier Transform (DFT)
Circular Symmetries Of a SequenceConsider sequence x(n) and its DFT is X(K). When IDFT taken it get periodic sequence xp(n)
xp(n) =∞∑
l=−∞x(n − lN)
x(n) and xp(n) are related by
x(n) =
{xp(n) for 0 ≤ n ≤ N − 10 otherwise
let xp(n) shifted by k units to the right then x′p(n)
x′p(n) = xp(n − k)
=∞∑
l=−∞x(n − k − lN)
x′(n) =
{x′p(n) for 0 ≤ n ≤ N − 1
0 otherwise
x′(n) = x(n − k − lN)
= x(n − k, modulo N)
= x((n − k))N
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 7 / 49
Properties of Discrete Fourier Transform (DFT)
n
Am
plitu
de
0 1 2 3 4
The sequencex(n) circularlyshifted by two
samples
n
Am
plitu
de
0 1 2 3
The sequencex(n)
5
23
4
5 4
23
n
Am
plitu
de
-4 -3 -2 -1 0 1 2 3 4 5 6 7
The sequencex(n)
5
23
45
23
4 5
23
4
Am
plitu
de-2 -1 0 1 2 3 4 5 6 7
The sequencex(n)
5
23
45
23
4 5 4
( )px n
( )x n0 3n≤ ≤
' ( ) ( 2)p px n x n= −
Figure 1: Circular shift of a sequence
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 8 / 49
Properties of Discrete Fourier Transform (DFT)
x′(n) = x((n − k))N
Consider x′(n) with k=2 and N=4 then
x′(n) = x((n − 2))4
x′(0) = x(−2)4 = x(2) x
′(1) = x(−1)4 = x(3)
x′(2) = x(0)4 = x(0) x
′(3) = x(1)4 = x(1)
These shifting operations are as shown in Figures
x(0)=5
x(1)=4
x(2)=3
x(3)=2
x(0)=2
x(1)=5
x(2)=4
x(3)=3
x(0)=4
x(1)=3
x(2)=2
x(3)=5
x(0)=3
x(1)=2
x(2)=5
x(3)=4
x(n) x(n-1)
x(n-2) x(n-3)
Figure 2: Circular shift of a sequence
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 9 / 49
Properties of Discrete Fourier Transform (DFT)
The circularly shifting in clockwise is represented by x((n + 1))4 and is as shown in Figure
x(0)=5
x(1)=4
x(2)=3
x(3)=2
X((n+1))4
Figure 3: Circular shift of a sequence
The circularly folded sequence is represented by x((−n))4 and is as shown in Figure
x(0)=5
x(1)=4
x(2)=3
x(3)=2 x(n)
x(0)=5
x(1)=4
x(2)=3
x(3)=2
x((-n))4
Folded sequence( x(n) is plotted in clockwise)
Figure 4: Folded sequence
x((−n))N = x(N − n) 0 ≤ n ≤ N − 1
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 10 / 49
Properties of Discrete Fourier Transform (DFT) Symmetry Property
Symmetry Property
Let the sequence x(n) be of complex valued and is expressed as
x(n) = xR(n) + jxI (n)
Its DFT isX (k) = XR(k) + jXI (k)
X (k) =
N−1∑n=0
x(n)e−j 2πN
kn
=
N−1∑n=0
[xR(n) + jxI (n)]e−j 2πN
kn
=
N−1∑n=0
[xR(n) + jxI (n)]
[cos
(2π
Nkn
)− jsin
(2π
Nkn
)]
=
N−1∑n=0
[xR(n)cos
(2π
Nkn
)+ jxI (n)sin
(2π
Nkn
)]
−jN−1∑n=0
[xR(n)sin
(2π
Nkn
)− xI (n)cos
(2π
Nkn
)]
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 11 / 49
Properties of Discrete Fourier Transform (DFT) Symmetry Property
X (k) = XR(k) + jXI (k) real and imaginary parts of X (k) is
XR(k) =
N−1∑n=0
[xR(n)cos
(2π
Nkn
)+ xI (n)sin
(2π
Nkn
)]
XI (k) = −N−1∑n=0
[xR(n)sin
(2π
Nkn
)− xI (n)cos
(2π
Nkn
)]
x(n) = xR(n) + jxI (n) real and imaginary parts of sequence x(n) is
xR(n) =1
N
N−1∑k=0
[XR(k)cos
(2π
Nkn
)− XI (k)sin
(2π
Nkn
)]
xI (n) =1
N
N−1∑k=0
[XR(k)sin
(2π
Nkn
)+ XI (k)cos
(2π
Nkn
)]
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 12 / 49
Properties of Discrete Fourier Transform (DFT) Symmetry Property
Real and even sequenceIf x(n) = x(N − n) then its DFT becomes
X (k) =
N−1∑n=0
x(n)cos
(2π
Nkn
)
Real and even sequenceIf x(n) = −x(N − n) then its DFT becomes
X (k) = −jN−1∑n=0
x(n)sin
(2π
Nkn
)
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 13 / 49
Properties of Discrete Fourier Transform (DFT) Symmetry Property
Symmetry property of real value of x(n)
X (k) =
N−1∑n=0
x(n)e−j2πkn/N
Let k = N − k :
X (N − k) =
N−1∑n=0
x(n)e−j2π(N−k)n/N =
N−1∑n=0
x(n)e j2πkn/Ne−j2πn
But e−j2πn = 1 for n = 0, 1, 2, ...
Therefore
X (N − k) =
N−1∑n=0
x(n)e j2πkn/N = X∗(k) complex conjugate of X (k)
X (N − k) = X∗(k)
When N is even, |Xk| is symmetric about N/2.
The phase, Xk, has odd symmetry about N/2.
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 14 / 49
Properties of Discrete Fourier Transform (DFT) Symmetry Property
The first five points of the eight point DFT of a real valued sequence are {0.25, 0.125 - j0.3018,0, 0.125 - j0.0518, 0} Determine the remaining three points
X(0)=0.25 X(1)=0.125 - j0.3018, X(2)=0, X(3)=0.125 - j0.0518, X(4)=0}The remaining three points X(5), X(6) and X(7) are determined using symmetry property
X (N − k) = X∗(k)X (8− k) = X∗(k)
By taking complex conjugate on both sidesX∗(8− k) = X (k)X (k) = X∗(8− k)
For k=5X (5) = X∗(8− 5) = X∗(3)But X (3) = 0.125− j0.0518 and X∗(3) = 0.125 + j0.0518 = X (5)
For k=6X (6) = X∗(8− 6) = X∗(2)But X (2) = 0 and X∗(2) = 0 = X (6)
For k=7X (7) = X∗(8− 7) = X∗(1)But X (1) = 0.125− j0.3018 and X∗(1) = 0.125 + j0.3018 = X (7)
Hence The remaining the DFTs are {0.125 + j0.0518, 0, 0.125 + j0.3018}
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 15 / 49
Properties of Discrete Fourier Transform (DFT) Symmetry Property
The first five points of the eight point DFT of a real valued sequence are {0.25, -j0.3018, 0, 0,0.125-j0.0518} Determine the remaining three points
X(0)=0.25 X(1)=-j0.3018, X(2)=0, X(3)=0, X(4)=0.125-j0.0518}The remaining three points X(5), X(6) and X(7) are determined using symmetry propertyX (N − k) = X∗(k)X (8− k) = X∗(k)
By taking complex conjugate on both sidesX∗(8− k) = X (k)X (k) = X∗(8− k)
For k=5X (5) = X∗(8− 5) = X∗(3)X (3) = 0 X∗(3) = 0
For k=6X (6) = X∗(8− 6) = X∗(2)X (2) = 0 X∗(2) = 0
For k=7X (7) = X∗(8− 7) = X∗(1)X (1) = −j0.3018 X∗(1) = +j0.3018
Hence the remaining the DFTs are {0.25,−j0.3018, 0, 0.125− j0.0518, 0, 0, j0.3018}
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 16 / 49
Properties of Discrete Fourier Transform (DFT) Symmetry Property
Circular ShiftCircular shift of x(n) can be defined:
xm(n) = x((n + m))NRN(n)
Xm(k) = DFT [xm(n)] = DFT [x((n + m))NRN(n)] = W−mkN X (k)
DFT [x((n + m))NRN(n)] = DFT [x̃(n + m)RN(n)]
= DFS[x̃(n + m)]RN(k)
= W−mkN X̃ (k)RN(k)
= W−mkN X (k)
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 17 / 49
Properties of Discrete Fourier Transform (DFT) Symmetry Property
Shift of a sequence
x̃ [n]DFS↔ X̃ [k]
x̃ [n −m]DFS↔ W km
N X̃ [k]
WN = e−j(2π/N)
W−nlN x̃ [n]
DFS↔ X̃ [k − l ]
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 18 / 49
Properties of Discrete Fourier Transform (DFT) Symmetry Property
Circular Frequency Shift
x(n)DFT↔ X (k)
then
x(n)e j2πn/NDFT↔ X ((k − l))N
Shifting the frequency components of DFT circularly is equivalent to multiplying the timedomain sequence by e j2πn/N
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 19 / 49
Properties of Discrete Fourier Transform (DFT) Circular Correlation
Circular Correlation
x(n)DFT←−−→
NX (k)
y(n)DFT←−−→
NY (k)
r̃xy (l)DFT←−−→
N= R̃xy (k) = X (k)Y ∗(k)
where rxy (l) is the circular cross correlation which is given as
r̃xy (l) =
N−1∑n=0
x(n)y∗((n − l))N
Multiplication of DFT one sequence and conjugate DFT of another sequence is equivalentto circular cross correlation of these two sequences in time domain
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 20 / 49
Properties of Discrete Fourier Transform (DFT) Circular Correlation
Proof:
r̃xy (l) =
N−1∑n=0
x(n)y∗((n − l))N
y∗((n − l))N can be written as y∗((−[l − n]))N
r̃xy (l) =
N−1∑n=0
x(n)y∗((−[l − n]))N
Based on circular convolution the above equation can be written as
r̃xy = x(l) N y∗(−l)
DFT{rxy (l)} = DFT{x(l)}DFT{y∗(−l)}
R̃xy (k) = X (k)DFT{y∗(−l)}
DFT{y∗(−l)} =
N−1∑l=0
y∗(−l)e−j 2πN
kl
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 21 / 49
Properties of Discrete Fourier Transform (DFT) Circular Correlation
let n = −lwhen l = 0 n = 0 and
when l = N − 1 n = −(N − 1)
DFT{y∗(−l)} =
−(N−1)∑n=0
y∗(n)e j2πN
kN
DFT{y∗(−l)} =
N−1∑l=0
y∗(n)e j2πN
kN
=
[N−1∑n=0
y(n)e−j 2πN
nk
]∗= [Y (k)]∗ = [Y ∗(k)]
R̃xy = X (k)Y ∗(k)
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 22 / 49
Properties of Discrete Fourier Transform (DFT) Complex Conjugate Properties
Complex Conjugate Properties
x(n)DFT↔ X (k)
then
x∗(n)DFT↔ X∗((−k))N = X∗(N − k)
and
x∗((−n))N = x∗(N − k)DFT↔ X∗(k)
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 23 / 49
Properties of Discrete Fourier Transform (DFT) Complex Conjugate Properties
DFT{x∗(n)} =
N−1∑n=0
x∗(n)e−j 2πN
kn
e j2πnNN = 1
DFT{x∗(n)} =
N−1∑n=0
x∗(n)e−j 2πN
kne j2πnNN
DFT{x∗(n)} =
N−1∑n=0
x∗(n)e−j 2πN
kne j2πnNN
=
N−1∑n=0
x∗(n)e j2πnN
(N−k)
=
[N−1∑n=0
x(n)e−j 2πnN
(N−k)
]∗= [X (N − k)]∗ = [X∗(N − k)]
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 24 / 49
Properties of Discrete Fourier Transform (DFT) Time Reversal of a sequence
Time Reversal of a sequence
x(n)DFT⇔ X (k)
x((−n))N = x(N − n) =DFT↔ X ((−k))N = X (N − k)
Proof
If the sequence is circularly folded its DFT is also circularly folded.
x((−n))N = x(N − n)
DFT{x(N − n)} =
N−1∑n=0
x(N − n)e−j 2πN
kn
Let m=N-n then the summation limits are
n=0 m=N
n=N-1 m=N-N+1=1
DFT{x(N − n)} =1∑
m=N
x(m)e−j 2πN
k(N−m)
x(N − n) is circular and DFT is periodic. The summation is performed from 0 to N-1 i.efor N samples. If the index is changed from (0+N) to (N-1+N). The limits are same
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 25 / 49
Properties of Discrete Fourier Transform (DFT) Time Reversal of a sequence
DFT{x(N − n)} =
N−1∑m=0
x(m)e−j 2πN
k(N−m) =
N−1∑m=0
x(m)e−j2πke−j2πkm/N
=
N−1∑m=0
x(m)e j2πkm/N
Multiply RHS by e−j2πm ∵ e−j2πm = 1
DFT{x(N − n)} =
N−1∑m=0
x(m)e−j2πkm/Ne−j2πm
=
N−1∑m=0
x(m)e−j2πm(N−k)/N
DFT{x(N − n)} = X (N − k)
DFT is periodic over period N
DFT{x(N − n)} = X (N − k)
= X ((−k))N
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 26 / 49
Properties of Discrete Fourier Transform (DFT) Circular Correlation
Multiplication Of Two Sequences
x(n)DFT←−−→
NX (k)
y(n)DFT←−−→
NY (k)
Then
y(n)y(n)DFT←−−→
N
1
NX (k) N Y (k)
Multiplication of two sequences in time domain is equivalent to circular convolution of their
DFTs in frequency domain
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 27 / 49
Circular Convolution Circular Convolution
Circular Convolution
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 28 / 49
Circular Convolution Circular Convolution
For two finite-duration sequences x1(n) and (x2(n) both of length N, with DFTs X1(k)and X2(k)
x1(n)DFT↔ X1(k) and x2(n)
DFT↔ X2(k)
Then
X1(k)X2(k) = x1(n) N x2(n)
Consider X3(k)X3(k) = X1(k)X2(k)
Now consider x3(m) is an IDFT of X3(k) and is represented as
x3(m) =1
N
N−1∑k=0
X3(k)e j2πN
km
x3(m) =1
N
N−1∑k=0
X1(k)X2(k)e j2πN
km
x3(m) =1
N
N−1∑k=0
[N−1∑n=0
x1(n)e−j 2πN
knN−1∑l=0
x2(l)e−j 2πN
kl
]e j
2πN
km
x3(m) =
N−1∑n=0
x1(n)
N−1∑l=0
x2(l)
[N−1∑k=0
e j2πN
k(m−n−l)
]
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 29 / 49
Circular Convolution Circular Convolution
x3(m) =
N−1∑n=0
x1(n)
N−1∑l=0
x2(l)
[N−1∑k=0
e j2πN
k(m−n−l)
]N−1∑k=0
ak =
{N for a = 11−aN
1−afor a 6= 1
when (m-n-l)=N,2N,3N,.... multiple of N then a=1because
e j2πN
kN = e j2πN
2kN .... = 1
N−1∑K=0
e j2πN
k(m−n−l) = N when (m − n − l) is multiple ofN
N−1∑K=0
e j2πN
k(m−n−l) =1− e j2πk(m−n−l)
1− e j2πN
k(m−n−l)when (m − n − l) is not multiple of N
e j2πk(m−n−l) = 1N−1∑K=0
e j2πN
k(m−n−l) =1− 1
1− e j2πN
k(m−n−l)= 0 when (m − n − l) is not multiple of N
N−1∑k=0
e j2πN
k(m−n−l) =
{N when (m − n − l) is not multiple of N0 otherwise
x3(m) =1
N
N−1∑n=0
x1(n)
N−1∑l=0
x2(l) when (m − n − l) is not multiple of N
=
N−1∑n=0
x1(n)
N−1∑l=0
x2(l)
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 30 / 49
Circular Convolution Circular Convolution
x3(m) =1
N
N−1∑n=0
x1(n)
N−1∑l=0
x2(l)N when (m − n − l) is not multiple of N
=
N−1∑n=0
x1(n)
N−1∑l=0
x2(l)
(m-n-l) is multiple of N i.e., (m-n-l)=pN where p is some integer it may positive ornegative
m − n − l = −pN Then l = m − n + pN
x3(m) =
N−1∑n=0
x1(n)x2(m − n + pN)
x2(m − n + pN) represents x2 shifted circularly by n samples
x2(m − n + pN) = x2(m − n, mpdulo N)
x2(m − n + pN) = x2((m − n))N
x3(m) =
N−1∑n=0
x1(n)x2((m − n))N m = 0, 1, . . .N − 1
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 31 / 49
Circular Convolution Circular Convolution
Circular convolution of x1(n) x2(n)is represented by x1(n) N x2(n) and is given by
x3(m) = x1(n) N x2(n) =
N−1∑n=0
x1(n)x2((m − n))N m = 0, 1, . . .N − 1
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 32 / 49
Circular Convolution Circular Convolution
x3(m) = x1(n) N x2(n) =
N−1∑n=0
x1(n)x2((m − n))N m = 0, 1, . . .N − 1
x1=[4 3 5 2] and x2=[3 4 1 2]x1(n) = 4 3 5 2x2(−n) = 3 2 1 4
x3(0) =3∑
x=0x1(n)x2(−n)4 = 12 +6 +5 +8 =31
x1(n) = 4 3 5 2x2(1− n) = 4 3 2 1
x3(1) =3∑
x=0x1(n)x2(1− n)4 = 16 +9 +10 +2 =37
x1(n) = 4 3 5 2x2(2− n) = 1 4 3 2
x3(2) =3∑
x=0x1(n)x2(2− n)4 = 4 +12 +15 +4 =35
x1(n) = 4 3 5 2x2(−n) = 2 1 4 3
x3(3) =3∑
x=0x1(n)x2(3− n)4 = 8 +3 +20 +6 =37
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 33 / 49
Circular Convolution Circular Convolution
x1=[4 3 5 2] and x2=[3 4 1 2]
x1(0)=4
x1(1)=3
x1(2)=5
x1(3)=2
x2(0)=3
x2(1)=4
x2(2)=1
x2(3)=2
x2((-n))4x1(n)
x1(0)=4
x1(1)=3
x1(2)=5
x1(3)=2
x2(0)=3
x2(1)=4
x2(2)=1
x2(3)=2
x2((3-n))4x1(n)
x1(0)=4
x1(1)=3
x1(2)=5
x1(3)=2
x2(0)=3
x2(1)=4
x2(2)=1
x2(3)=2
x2((2-n))4x1(n)
x1(0)=4
x1(1)=3
x1(2)=5
x1(3)=2
x2(0)=3
x2(1)=4
x2(2)=1
x2(3)=2
x2((1-n))4x1(n)
2x4=8
3x3=9
5x1=5
3x2=6
4x3=12
5x4=205x3=15
3x4=12
3x1=3
4x4=16
2x1=2
5x2=10
4x2=8
2x3=6
4x1=4
2x2=4
Figure 5: Circular shift of a sequence
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 34 / 49
Circular Convolution Circular Convolution
Using matrix approach for circular convolution
x3(m) =
N−1∑n=0
x1(m)x2(m − n), 0 ≤ m ≤ N − 1
y(n) = h(n) N x(n), 0 ≤ m ≤ N − 1
y(m) =
N−1∑n=0
x(n)h(m − n), 0 ≤ m ≤ N − 1
y(0) = x(0)h(0) + x(1)h(−1) + x(2)h(−2) + . . . + x(N − 2)h(−(N − 2)) + x(N − 1)h(−(N − 1))
= x(0)h(0) + x(1)h(N − 1) + x(2)h(N − 2) + . . . x(N − 2)h(2) + x(N − 1)h(1)
y(1) = x(0)h(1) + x(1)h(0) + x(2)h(N − 2) + . . . + x(N − 2)h(3) + x(N − 1)h(2)
= x(0)h(0) + x(1)h(N − 1) + x(2)h(N − 2) + . . . x(N − 2)h(2) + x(N − 1)h(1)
y(N − 1) = x(0)h(N − 1) + x(1)h(N − 2) + x(2)h(N − 3) + . . . + x(N − 2)h(1) + x(N − 1)h(0)
y(0)y(0)y(0)
.
.
.y(N − 2)y(N − 2)
=
h(0) h(N − 1) h(N − 2) · · · h(2) h(1)h(1) h(0) h(N − 1) h(3) h(2)h(2) h(1) h(0) h(4) h(3)
h(N − 2) h(N − 3) h(N − 4) h(0) h(N − 1)h(N − 1) h(N − 2) h(N − 3) h(1) h(0)
=
x(0)x(1)x(2)
.
.
.x(N − 2)x(N − 1)
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 35 / 49
Circular Convolution Circular Convolution
x=[4 3 5 2] and h=[3 4 1 2]
The first row of h matrix is obtained by folding the h(n) i.e, [2, 1, 4, 3]and shift circularlyright once it becomes [3, 2, 1, 4]
Similarly shift circularly right once to get second row of h matrix and continue for theremaining rows of h matrix.
y(0)y(1)y(2)y(3)
=
3 2 1 44 3 2 11 4 3 22 1 4 3
4352
=
12 + 6 + 5 + 8 = 31
16 + 9 + 10 + 2 = 374 + 12 + 15 + 4 = 358 + 3 + 20 + 6 = 37
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 36 / 49
Circular Convolution Circular Convolution
Using DFT and IDFT method
First determine the DFT of the given sequences and multiply both the DFTs i,e.,
Y (k) = X (k).H(k)
Use IDFT to find the y(n) i.e. convolved sequence
y(n) = IDFT (X (k).H(k))
X (0)X (1)X (2)X (3)
=
1 1 1 11 −j −1 j1 −1 1 −11 j −1 −j
4352
=
14−1− 1j4−1 + 1j
X (0)X (1)X (2)X (3)
=
1 1 1 11 −j −1 j1 −1 1 −11 j −1 −j
3412
=
102− 2j−2(2 + 2j)
Y (k) = X (k).H(k)
Y (0)Y (1)Y (2)Y (3)
=
14× 10(−1− 1j)× (2− 2j)4× (−2)(−1 + 1j)(2 + 2j)
=
140(−1− 1j)× (2− 2j)−8(−1 + 1j)(2 + 2j)
=
140−4−8−4
y(n) = IDFT (X (k).H(k))
y(0)y(1)y(2)y(3)
=1
4
1 1 1 11 j −1 −j1 −1 1 −11 −j −1 j
140−4−8−4
=
31373537
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 37 / 49
Circular Convolution Circular Convolution
%--------------------------------------------------------------------------
% GENERALAZED CIRCULAR CONVOLUTION COMPUTING CODE IN MATLAB WITHOUT
% USING MATLAB BUILTIN FUNCTION [cconv(a,b,n)]
%--------------------------------------------------------------------------
clc; clear all; close all;
x=[4 3 5 2]
h=[3 4 1 2]
N1=length(x);%To find the length of the sequence
N2=length(h);%To find the length of the sequence
N=max(N1,N2);
a=[x, zeros(1,N2)]%Padding zeros to make sequence length
b=[h, zeros(1,N1)]%Padding zeros to make sequence length
for i=1:N
y(i)=0;
for j=1:N
k=i-j+1;
if(k<=0)
k=k+N;
end
y(i)=y(i)+a(j)*b(k);
end
end
y
y1=cconv(x,h, N)% This is to verify the result
subplot(3,1,1); stem(x);
xlabel(’------------->n’);ylabel(’Sequence x[n]’);
subplot(3,1,2);stem(h);
xlabel(’------------->n’);ylabel(’Sequence h[n]’);
subplot(3,1,3); stem(y);
xlabel(’------------->n’);ylabel(’Y[n]’);title(’Convolution without using "conv" function’);
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 38 / 49
Circular Convolution Circular Convolution
% -----------------------------------------------------------------
% Program for Circular Convolution using DFT & IDFT of equal length
%------------------------------------------------------------------
% 1. Define the sequence x1 and x2
% 2. Take DFT of x1 and x2 i.e., X1=DFT(x1) & X2=DFT(x2)
% 3. Multiply X1 & X2
% 4. Take IDFT for the resultant.
%------------------------------------------------------------------
clc; clear all; close all;
% Let us define the input sequence x1 & x2
x1 = input(’Enter the first sequence:’);
x2 = input(’Enter the second sequence:’);
%Now let us take DFT of x1 and x2 i.e., X1=DFT(x1) & X2=DFT(x2)
X1 = fft(x1)
X2 = fft(x2)
% Now let us multiply X1 & X2
Y = X1.*X2
% Now let us take IDFT for the resultant
y = ifft(Y)
% Now let us plot this result
subplot(5,2,1); stem(x1); title(’First Sequence’);
subplot(5,2,2); stem(x2); title(’Second Sequence’);
subplot(5,2,3); stem(abs(X1)); title(’Magnitude of DFT of Sequence x1’);
subplot(5,2,4); stem(abs(X2)); title(’Magnitude of DFT of Sequence x2’);
subplot(5,2,5); stem(angle(X1)); title(’Angle of DFT of Sequence x2’);
subplot(5,2,6); stem(angle(X2)*pi/180); title(’Angle of DFT of Sequence x2’);
subplot(5,2,7); stem(abs(Y)); title(’Magnitude of multiplied o/p of X1 & X2 = Y’);
subplot(5,2,8); stem(angle(Y)); title(’Angle of multiplied o/p of X1 & X2 = Y’);
subplot(5,2,9); stem(abs(y)); title(’Magnitude of IDFT of Sequence Y’);
subplot(5,2,10); stem(angle(y)); title(’Angle of IDFT of Sequence Y’);
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 39 / 49
Circular Convolution Circular Convolution
Let x(n) be the sequence x(n) = δ(n) + 2δ(n − 2) + δ(n − 3)
(i) Find the 4 point DFT of x(n)
(ii) If y(n) is the 4 point circular convolution with itself find y(n) and the four point DFTY(K)
Solution:x(n) = δ(n) + 2δ(n − 2) + δ(n − 3)x(0) = δ(0) + 2δ(0− 2) + δ(0− 3) = 1 + 0 + 0 = 1x(1) = δ(1) + 2δ(1− 2) + δ(1− 3) = 0 + 0 + 0 = 0x(2) = δ(2) + 2δ(2− 2) + δ(2− 3) = 0 + 2 + 0 = 2x(3) = δ(3) + 2δ(3− 2) + δ(3− 3) = 0 + 0 + 1 = 1
The 4 point DFT x(n) can be obtained by matrix method
X [N] = [WN ]x(n)
X (0)X (1)X (2)X (3)
=
1 1 1 11 −j −1 j1 −1 1 −11 j −1 −j
1021
=
4−1 + j2−1− j
The DFT of the sequence x(n) = [1 0 2 1] is [4, − 1 + j , 2, = 1− j]
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 40 / 49
Circular Convolution Circular Convolution
The circular convolution of x(n) with itself Y (n)
Y (n) = x(n) ~ x(n)DFT←−−− ⇐⇒ X (k).X (k) = Y (k)
4−1 + j2−1− j
.
4−1 + j2−1− j
=
16−j24j2
Determine the y(n) by taking IDFT of Y (k)
x(0)x(1)x(2)x(3)
=1
4
1 1 1 11 j −1 −j1 −1 1 −11 −j −1 j
16−j24j2
=
1021
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 41 / 49
Circular Convolution Circular Convolution
Let x(n) be the sequence x(n) = 2δ(n) + δ(n − 1) + δ(n − 3) Find the sequence
y(n) = x(n) 5 x(n) i.e. 5 point circular convolution of x(n) with itself
Solution:x(n) = 2δ(n) + δ(n − 1) + δ(n − 3)x(0) = 2δ(0) + δ(0− 1) + δ(0− 3) = 2 + 0 + 0 = 2x(1) = 2δ(1) + δ(1− 1) + δ(1− 3) = 0 + 1 + 0 = 1x(2) = 2δ(2) + δ(2− 1) + δ(2− 3) = 0 + 0 + 0 = 0x(3) = 2δ(3) + δ(3− 1) + δ(3− 3) = 0 + 0 + 1 = 1x(n) = [2, 1, 0, 1]
The 5 point circular convolution is achieved by appending zero at the end of the sequencex(n) i.e., x(n) = [2 1 0 1 0]Using Matrix approach
x=[2 1 0 1 0] folded sequence is =[0 1 0 1 2]
Then shift circularly right once it becomes [2 0 1 0 1]y(0)y(1)y(2)y(3)y(4)
=
2 0 1 0 11 2 0 1 00 1 2 0 11 0 1 2 00 1 0 1 2
21010
=
4 + 0 + 0 + 0 + 0 = 42 + 2 + 0 + 1 + 0 = 50 + 1 + 0 + 0 + 0 = 12 + 0 + 0 + 2 + 0 = 40 + 1 + 0 + 1 + 0 = 2
y(n) = x(n) 5 x(n) is [4 5 1 4 2]
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 42 / 49
Circular Convolution Circular Convolution
k=0 k=1 k=2 k=3 k=4n=0 W 0
5 W 05 W 0
5 W 05 W 0
5n=1 W 0
5 W 15 W 2
5 W 35 W 4
5n=2 W 0
5 W 15 W 4
5 W 45 W 8
5n=3 W 0
5 W 35 W 6
5 W 95 W 12
5n=4 W 0
5 W 45 W 8
5 W 125 W 16
5
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 43 / 49
Circular Convolution Circular Convolution
Compute the circular convolution between the following sequences using DFT and IDFT methodx(n) = {1
↑, 2, 3, 4} y(n) = {−1
↑,−2,−3,−4} x(n) and y(n) are periodic sequences with period
N=4.
The 4 point DFT x(n) can be obtained by matrix method
X [N] = [WN ]x(n)
X (0)X (1)X (2)X (3)
=
1 1 1 11 −j −1 j1 −1 1 −11 j −1 −j
1234
=
10−2 + j2−2−2− j2
The DFT of the sequence x(n) = [1 2 3 4] is [10, − 2 + j2, − 2, − 2− j2]
Y (0)Y (1)Y (2)Y (3)
=
1 1 1 11 −j −1 j1 −1 1 −11 j −1 −j
−1−2−3−4
=
−102− j222 + j2
The DFT of the sequence y(n) = [1 0 2 1] is [−10, 2− j2, 2, 2 + j2]
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 44 / 49
Circular Convolution Circular Convolution
The circular convolution of x(n) with itself Y (n)
Z(n) = x(n) ~ x(n)DFT←−−− ⇐⇒ X (k).X (k) = Z(k)
10−2 + j2−2−2− j2
.
102− j222 + j2
=
−100j8−4−j8
Determine the y(n) by taking IDFT of Y (k)
Z(0)Z(1)Z(2)Z(3)
=1
4
1 1 1 11 j −1 −j1 −1 1 −11 −j −1 j
−100j8−4−j8
=
−26−28−26−20
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 45 / 49
Circular Convolution Circular Convolution
Evaluate circular convolution y(n) = x(n) N x(n) where x(n) = u(n)− u(n − 4) and
h(n) = u(n)− u(n − 3) assuming N=8 show your calculations by int0 approximate usage ofequations and relevant sketches. Plot y(n). (ii) Verify the result using DFT and IDFT method
u(n)
0 1 2 3 4 5 6 7 8 9 10 11 nu(n-4)
0 1 2 3 4 5 6 7 8 9 10 11 n
0 1 2 3 4 5 6 7 8 9 10 11 n
x(n)=u(n-u(n-4)
Unit step sequence u(n)
Unit step sequence u(n) delayed by 4 samples
x(n) is generated by subtracting u(n) by u(n-4)
Figure 6: Sequence x(n)=u(n)+u(n-4)
u(n)
0 1 2 3 4 5 6 7 8 9 10 11 n
Unit step sequence u(n)
u(n-3)
0 1 2 3 4 5 6 7 8 9 10 11 n
h(n)=u(n-u(n-3)
0 1 2 3 4 5 6 7 8 9 10 11 n
Unit step sequence u(n) delayed by 3 samples
h(n) is generated by subtracting u(n) by u(n-3)
Figure 7: Sequence h(n)=u(n)+u(n-3)
y(n)
0 1 2 3 4 5 6 n
Sketch of convolved sequence h(n)
01
23
4
Figure 8: Convolved Sequence y(n)Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 46 / 49
Circular Convolution Circular Convolution
Using Matrix approach
y(n) = x(n) 8 h(n)
x(n) = [1, 1, 1, 1] and h(n) = [1, 1, 1]
The 8 point circular convolution is achieved by appending zero at the end of the sequencex(n) i.e., x(n) = [1, 1, 1, 1, 0, 0, 0, 0] and h(n) = [1, 1, 1, 0, 0, 0, 0, 0]
The first row of h matrix is obtained by folding the h(n) i.e, [0, 0, 0, 0, 0, 0, 1, 1, 1]andshift circularly right once it becomes [1, 0, 0, 0, 0, 0, 1, 1]
Similarly shift circularly right once to get second row of h matrix and continue for theremaining rows of h matrix.
y(0)y(1)y(2)y(3)y(4)y(5)y(6)y(7)
=
1 0 0 0 0 0 1 11 1 0 0 0 0 0 11 1 1 0 0 0 0 00 1 1 1 0 0 0 00 0 1 1 1 0 0 00 0 0 1 1 1 0 00 0 0 0 1 1 1 00 0 0 0 0 1 1 1
11110000
=
12332100
The convolved sequence is y(n) = [1, 2, 3, 3, 2, 1, 0, 0]
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 47 / 49
Circular Convolution Circular Convolution
If X(k)is the DFT the sequence x(n)determine the N point DFT of the sequence
xc (n) = x(n)cos(
2πlnN
)and xs(n) = x(n)sin
(2πlnN
)for 0 ≤ n ≤ N − 1) in terms of X(k)
xc (n) = x(n)cos
(2πln
N
)= x(n)
1
2
(e
j2πlnN + e−
j2πlnN
)
Xc (k) =
N−1∑n=0
xc (n)e−j2πkn
N
Xc (k) =1
2
N−1∑n=0
x(n)(e
j2πlnN + e−
j2πlnN
)e−
j2πknN
=1
2
N−1∑n=0
x(n)ej2πlnN e−
j2πknN +
1
2
N−1∑n=0
x(n)e−j2πlnN e−
j2πknN
x(n)e j2πn/NDFT↔ X ((k − l))N = x(n)e−j2πn/N DFT↔ X ((k + l))N
Xc (k) =1
2X ((k − l))N +
1
2X ((k + l))N
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 48 / 49
Circular Convolution Circular Convolution
xs(n) = x(n)sin
(2πln
N
)= x(n)
1
2j
(e
j2πlnN − e−
j2πlnN
)
Xs(k) =
N−1∑n=0
xs(n)e−j2πkn
N
Xs(k) =1
2j
N−1∑n=0
x(n)(e
j2πlnN − e−
j2πlnN
)e−
j2πknN
=1
2j
N−1∑n=0
x(n)ej2πlnN e−
j2πknN −
1
2j
N−1∑n=0
x(n)e−j2πlnN e−
j2πknN
x(n)e j2πn/NDFT↔ X ((k − l))N = x(n)e−j2πn/N DFT↔ X ((k + l))N
Xc (k) =1
2jX ((k − l))N −
1
2jX ((k + l))N
Dr. Manjunatha. P (JNNCE) UNIT - 2: Properties of Discrete Fourier Transforms (DFT)[?, ?, ?, ?]September 14, 2014 49 / 49
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