Discrete fourier transform

Discrete Fourier Transform
Unit-4
1
Mohammad Akram,Assistant Professor,ECE
Department,JIT

Introduction
• As the name implies, the Discrete Fourier Transform is purely
discrete: discrete time data sets are converted into a discrete
frequency representation
• Mathematically, The DFT of discrete time sequence x(n) is denoted
by X(k).It is given by,
Here k=0,1,2,….N-1
Since this summation is taken for ‘N’ points, it is called ‘N’ point
DFT.





1
0
2
)()(
N
n
N
knj
enxkX

2
Department,JIT

We can obtain discrete sequence x(n) from its DFT. It is called
as
Inverse Discrete Fourier Transform (IDFT).It is given by,
Here n=0,1,2,…N-1
•This is called as N-point IDFT.
•Now we will define the new term ‘W’ as
WN=e-j2π/N
•This is called as “twiddle factor”.
•Twiddle factor makes the computation of DFT a bit easy and fast.
•Using twiddle factor we can write equation of DFT and IDFT as
follows:
Here k=0,1,2,….N-1




1
0
2
)(
1
)(
N
k
N
knj
ekX
N
nx





1
0
)()(
N
n
kn
NWnxkX
3
Department,JIT






1
0
)(
1
)(
N
k
kn
NWkX
N
nx





1
0
2
)()(
N
n
N
knj
enxkX

And
Here n=0,1,2,…N-1
Numericals using defintion of DFT:
1.Obtain DFT of unit impulse δ(n).
Solution:
•Here x(n)= δ(n) ……….(1)
•According to the definition of DFT ,we have
……..(2)
•But δ(n)=1 only at n=0.Thus equation (2) becomes,
X(k)= δ(0)e0
= 1
δ(n) 1
•This is the standard DFT pair.
DFT
4
Department,JIT

2.Obtain DFT of delayed unit impulse δ(n-n0)
Solution: We know that δ(n-n0) indicates unit impulse delayed
by ‘n0’samples.
Here x(n)= δ(n-n0) ……..(1)
Now we have ……..(2)
But δ(n-n0) =1 only at n=n0.Thus eq.(2) becomes
Similarly, DFT of





1
0
2
)()(
N
n
N
knj
enxkX

Nknj
ekX /2 0
.1)( 

Nknj
enn /2
0
0
)( 
 

Nknj
enn /2
0
0
)( 
 
5
Department,JIT

3.Find the 4 point DFT of following window function
w(n)=u(n)-u(n-N).
Solution:
According to the definition of DFT,
……..(1)
•The given equation is x(n)=w(n)=1 for . Here
N=4,so we will get 4 point DFT.
……..(2)
•The range of ‘k’ is from ‘0’ to ‘N-1’.So in this case ‘k’ will vary
from 0 to 3.
For k=0,





1
0
2
)()(
N
n
N
knj
enxkX

10  Nn



3
0
4
2
.1)(
n
knj
ekX



3
0
0
.1)0(
n
eX 411111
3
0
 n
6
Department,JIT

For k=1,
For k=2,



3
0
4
2
)1(
n
nj
eX

4644420
)1(  jjj
eeeeX 




















4
6
sin
4
6
cos
4
4
sin
4
4
cos
4
2
sin
4
2
cos1)1(

jjjX
011)1(  jjX
 




3
0
3
0
4
22
)2(
n
nj
n
nj
eeX 

 320
)2( jjj
eeeeX 

      3sin3cos2sin2cossincos1)2( jjjX 
      011110101011)2( X
7
Department,JIT

For k=3,  




3
0
4
63
0
4
32
)3(
n
nj
n
nj
eeX

2
9
34
6
0
)3(


 j
j
j
eeeeX




  












2
9
sin
2
9
cos3sin3cos
4
6
sin
4
6
cos1)3(



jjjX
      000101)3(  jjX
 0,0,0,4)(  kX
8
Department,JIT

Department,JIT
9
Cyclic Property of Twiddle factor
(DFT as linear transformation matrix)
•The twiddle factor is denoted by WN and is given by,
WN=e-j2π/N ……….(1)
•Now the discrete time sequence x(n) can be denoted by x N .Here
‘N’ stands for ‘N’ point DFT.
•Range of ‘n’ is from 0 to ‘N-1’.
•x N can be represented in the matrix form as follows:
•This is a “N×1” matrix and ‘n’ varies from ‘0’ to ‘N-1’
1
)1(
.
.
)2(
)1(
)0(























N
N
Nx
x
x
x
x

Department,JIT
10
Now the DFT of x(n) is denoted by X(k).In the matrix form
X(k) can be represented as follows:
•This is also a “N×1” matrix and ‘k’ varies from ‘0’ to ‘N-1’
•From the definition of DFT,
1
)1(
.
.
)2(
)1(
)0(























N
N
NX
X
X
X
X




1
0
)()(
N
n
kn
NWnxkX

Department,JIT
11
•We can also represent in the matrix form. Remember that
‘k’ varies from 0 to N-1 and ‘n’ also varies from 0 to N-1.
•Note that each value is obtained by taking multiplication of k
and n.
•For example, if k=2, n=2 then we get =
W
kn
N
.
.
.
. . .
W
kn
N
 
   



























WWWW
WWWW
WWWW
WWWW
N
N
N
N
N
NN
N
NNNN
N
NNNN
NNNN
2
11210
12420
1210
0000
...
....
....
....
...
...
...
k=0
k=1
k=2
k=N-1
n=2n=1n=0 n=N-1
W
kn
N WN
4

Department,JIT
12
Thus, DFT can be represented in the matrix form as
•Similarly IDFT can be represented in the matrix form as
•Here WN
* is the complex conjugate of WN .
Periodicity property of Twiddle factor WN
•WN possesses the periodicity property.
•After N-points, it repeats its value
 xX NNN
W
 Xx NNN
W
N
*1


Department,JIT
13
Problem-1:Calculate the four point DFT of four point sequence
x(n)=(0,1,2,3).
Solution:The four point DFT in matrix form is given by
{6, 2j-2, -2, -2j-2}
 xX W 444
















jj
jj
11
1111
11
1111












3
2
1
0
X 4
X 4
































22
2
22
6
320
3210
320
3210
j
j
jj
jj
X 4

Department,JIT
14
Properties of DFT
DFT
N
1-Linearity: If x1(n) X1(k) and x2(n) X2(k) then,
a1x1(n) + a2x2(n) a1X1(k) + a2X2(k)
Here a1and a2 are constants.
2-Periodicity:
If x(n) X(k) then
x(n+N) = x(n) for all n
X(k+N) = X(k) for all k
DFT
N
DFT
N
DFT
N

15
Mohammad Akram,Assistant
Professor,JIT,Barabanki

Discrete fourier transform

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