UNIT - 2: Properties of Discrete Fourier...

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


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.



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)



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



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



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)




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


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


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

)]



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

)]



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

)



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.



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}



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}



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)



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 ]



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


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



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



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)


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)


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)]


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


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



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


Circular Convolution 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)

]



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)



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



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



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



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




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)



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



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



%--------------------------------------------------------------------------

% 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’);



% -----------------------------------------------------------------

% 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’);



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]



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



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]



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



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]



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



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


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]



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



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


UNIT - 2: Properties of Discrete Fourier...

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