DFT - Discrete Fourier Transform and its Properties

1. By
Ms.J.Shiny Christobel, AP/ECE
Sri Ramakrishna Institute of Technology, Coimbatore

2.  1.Linearity
    
   
       
k
bX
k
aX
n
bx
n
ax
k
X
n
x
k
X
n
x
2
1
DFT
2
1
2
DFT
2
1
DFT
1



 




 



 

2.Duality
   
   
 
 
N
DFT
DFT
k
Nx
n
X
k
X
n
x



 



 

3.Circular Shift of a Sequence
   
 
 
     m
N
/
k
2
j
DFT
N
DFT
e
k
X
1
-
N
n
0
m
n
x
k
X
n
x




 






 


3.  Periodicity
• X(K) is N-point DFT of a finite sequence x(n)
 then x(n+N)=x(n) for all n
 DFT [x(k+N)]=x(k)
 Time reversal of the sequence
• The time reversal of an N-point sequence x(n) is
attained by wrapping the sequence x(n) around the
circle in clockwise direction
 Then X((-n))N=x(N-n)
 DFT [X(N-m)]=X(N-k)

4.  Circular frequency shift
• If DFT[x(n)]=X(k)
 then DFT[ x(n)𝒆
𝒋𝟐𝝅𝒍𝒏
𝑵 ]=X((k-l))N
 Complex conjugate
• If DFT[x(n)]=X(k)
 then DFT[ x*(n)]=X*(N-k)=X*((-k))N
 Circular convolution
• x1(n)&x2(n) are finite duration sequence both of
length N
 The DFTS X1(k),X2(k)
• Circular convolution of x1(n)&x2(n) represented as
 x3(n)=x1(n) x2(n)
 Then DFT[x1(n) x2(n) ]=X1(k)X2(k)
N
N
N

6.  Relation ship between DTFT& DFT
• DTFT is a continuous periodic function of𝜔.
• DFT is obtained by sampling DTFT at a finite
number of equally spaced point over one period

7.  Example1
 1.Find the DFT of sequence 𝑥 𝑛 = 1,1,0,0
• Solution
• Let us assume N=L=4
• We have to find
• formula 1.
 2. 𝑒−𝑗𝜃
= 𝑐𝑜𝑠𝜃 − 𝑗 𝑠𝑖𝑛𝜃
• Step1.Find x(0),x(1),x(2),x(3). Where k=0,1,2,..N-1
 Find x(0)
 x(0)= x(n)𝑒−𝑗2𝜋𝑛.𝑘/𝑛
3
𝑛=0 = x(0)𝑒−𝑗2𝜋0.0/4
+ x(1)𝑒−𝑗2𝜋1.0/4
+
x(2)𝑒−𝑗2𝜋2.0/4
+ x(3)𝑒−𝑗2𝜋3.0/4
.
 = x(0)𝑒0
+ x(1)𝑒0
+ x(2)𝑒0
+ x(3)𝑒0
.
 X(0)=1+1+0+0=2.
  ]
n
[
x
k
X DFT


 

   






1
N
0
n
kn
N
/
2
j
e
]
n
[
x
k
X

8.  Find x(1)
 x(1)= 𝒙(𝒏)
𝟑
𝒏=𝟎 𝒆−𝒋𝟐𝝅𝒏.𝒌/𝟒
= x(0)𝒆−𝒋𝟐𝝅𝟎.𝟏/𝟒
+ x(1)𝒆−𝒋𝟐𝝅𝟏.𝟏/𝟒
+
x(2)𝒆−𝒋𝟐𝝅𝟐.𝟏/𝟒+ x(3)𝒆−𝒋𝟐𝝅𝟑.𝟏/𝟒.
 =x(0).1+ x(1).𝒄𝒐𝒔 𝝅/𝟐-j𝒔𝒊𝒏 𝝅/𝟐+ x(2).𝒄𝒐𝒔 𝝅-
j𝒔𝒊𝒏 𝝅+ x(3).𝒄𝒐𝒔𝟑 𝝅/𝟐-j𝒔𝒊𝒏 𝟑𝝅/𝟐.
 We know 𝒙 𝒏 = 𝟏, 𝟏, 𝟎, 𝟎
 Therefore x(1) =1.1+1.(0-j)+0(𝒄𝒐𝒔 𝝅-j𝒔𝒊𝒏 𝝅)+0.(
𝒄𝒐𝒔𝟑 𝝅/𝟐-j𝒔𝒊𝒏 𝟑𝝅/𝟐).
 x(1) =1-j.
 same way find x(2) =0, x(3) = 1+j
 final answer assign X(K)= 𝒙 𝟎 , 𝒙 𝟏 , 𝒙 𝟐 , 𝒙(𝟑)
 X(K)= 𝟐, 𝟏 − 𝒋, 𝟎, 𝟏 + 𝒋

9.  Example2
 1.Find the IDFT of sequence 𝑋 𝐾 = 2,0,2,0 .
 Solution
 Let us assume N= 4
 We have to find x(n)
 Formula
 1.X(n)=𝑰/𝑵 𝐱(𝐊)
𝑵−𝟏
𝑲=𝟎 𝒆𝒋𝟐𝝅𝒏.𝒌/𝑵
𝟐. 𝒆−𝒋𝜽
= 𝒄𝒐𝒔𝜽 − 𝒋 𝒔𝒊𝒏𝜽
 Step1.Find x(0),x(1),x(2),x(3). Where N=0,1,2,..N-1
 Find X(0) N=4
 X(n)= 𝑰/𝟒 𝐱(𝐊)𝒆𝒋𝟐𝝅𝒏.𝒌/𝟒
𝟑
𝑲=𝟎
 =𝑰/𝟒 𝐱(𝐊)𝒆𝒋𝝅𝒏.𝒌/𝟐
𝟑
𝑲=𝟎
 for n=0
 X(0)=
𝑰
𝟒
[ 𝐱 𝟎 𝒆𝒋𝝅
𝟎.𝟎
𝟐 + 𝐱 𝟏 𝒆𝒋𝝅
𝟎.𝟏
𝟐 + 𝐱 𝟐 𝒆𝒋𝝅
𝟎.𝟐
𝟐 + 𝐱 𝟑 𝒆𝒋𝝅
𝟎.𝟑
𝟐 ]
 =
𝑰
𝟒
[ 𝟐. 𝒆𝟎 + 𝟎. 𝒆𝟎 + 𝟐. 𝒆𝟎+ 𝟎. 𝒆𝟎]=
𝑰
𝟒
[ 𝟐. 𝟏 + 𝟎 + 𝟐. 𝟏+ 𝟎]
 =4/4=1
  ]
n
[
x
k
X DFT


 


10.  for n=1
 X(1)=
𝑰
𝟒
[ 𝐱 𝟎 𝒆𝒋𝝅
𝟏.𝟎
𝟐 + 𝐱 𝟏 𝒆𝒋𝝅
𝟏.𝟏
𝟐 + 𝐱 𝟐 𝒆𝒋𝝅
𝟏.𝟐
𝟐 + 𝐱 𝟑 𝒆𝒋𝝅
𝟏.𝟑
𝟐 ] 𝐜𝐨𝐬 𝝅/𝟐-
j𝐬𝐢𝐧 𝝅/𝟐

=
𝑰
𝟒
[ 𝟐. 𝟏 + 𝟎. [𝐜𝐨𝐬 𝝅/𝟐+j𝐬𝐢𝐧 𝝅/𝟐] +
𝟐. [𝐜𝐨𝐬 𝝅+j𝐬𝐢𝐧 𝝅]+ 𝟎. [𝐜𝐨𝐬 𝟑𝝅/𝟐+j𝐬𝐢𝐧 𝟑𝝅/𝟐]]
 =
𝑰
𝟒
[ 𝟐 + 𝟎 + 𝟐 −𝟏 𝟎]
 =0
 same way find x(2) =1, x(3) = 0
 final answer assign x(n)= 𝒙 𝟎 , 𝒙 𝟏 , 𝒙 𝟐 , 𝒙 𝟑
 x(n)= 𝟏, 𝟎, 𝟏, 𝟎